Autonomous heating for a private house is available, comfortable and varied. You can install a gas boiler and not depend on the vagaries of nature or system failures district heating... The main thing is to choose the right equipment and calculate the heating capacity of the boiler. If the capacity exceeds the heat needs of the premises, then the money for the installation of the unit will be wasted. In order for the heat supply system to be comfortable and financially profitable, at the design stage it is necessary to calculate the power gas boiler heating.

Basic values ​​for calculating heating power

The easiest way to get data on the heating performance of the boiler by the area of ​​the house: taken 1 kW of power for every 10 sq. m... However, this formula has serious errors, because modern construction technologies, type of terrain, climatic changes in temperature, the level of thermal insulation, the use of double-glazed windows, and the like.

To make a more accurate calculation of the boiler heating power, you need to take into account whole line important factors affecting the final result:

• dimensions of the dwelling;
• the degree of home insulation;
• the presence of double-glazed windows;
• thermal insulation of walls;
• type of building;
• air temperature outside the window during the coldest season;
• type of heating circuit wiring;
• the ratio of the area of ​​load-bearing structures and openings;
• heat loss of the building.

In houses with forced ventilation the calculation of the heating capacity of the boiler must take into account the amount of energy required to heat the air. Experts advise making a gap of 20% when using the obtained result of the boiler's thermal power in case of unforeseen situations, severe cold snap or a decrease in gas pressure in the system.

With an unreasonable increase in thermal power, it is possible to reduce the efficiency of the heating unit, increase the cost of purchasing system elements, and lead to rapid wear of components. That is why it is so important to correctly calculate the power of the heating boiler and apply it to the indicated dwelling. The data can be obtained using a simple formula W = S * W beats, where S is the area of ​​the house, W is the factory power of the boiler, W beats is the specific power for calculations in a certain climatic zone, it can be adjusted according to the characteristics of the user's region. The result should be rounded up to a larger value in case of heat leakage in the house.

For those who do not want to waste time on mathematical calculations, you can use the online gas boiler power calculator. Just keep individual data on the characteristics of the room and get a ready-made answer.

The formula for obtaining the power of the heating system

The online heating boiler power calculator makes it possible to get the necessary result in a matter of seconds, taking into account all the above characteristics, which affect the final result of the data obtained. In order to use such a program correctly, it is necessary to enter the prepared data into the table: the type of window glazing, the level of thermal insulation of the walls, the ratio of the areas of the floor and the window opening, the average temperature outside the house, the number of side walls, the type and area of ​​the room. And then press the "Calculate" button and get the result on the heat loss and heat output of the boiler.

Thanks to such a formula, each consumer will be able to obtain the necessary indicators in a short time and apply them in the design work. heating system.

The heating boiler is the basis of the heating system, it is the main device, the performance of which will determine the ability of the communication network to provide the house with the amount of heat that is needed. And if you calculate the power of the heating boiler correctly and correctly, then this will exclude the occurrence of unnecessary costs associated with the purchase of devices and their operation. The boiler, selected according to preliminary calculations, will operate with such heat transfer, which is included in it by the manufacturer - this will contribute to the preservation of its technical parameters.

What is the calculation based on?

The calculation of the power of the heating boiler is important point... Power, as a rule, can be compared with the entire heat transfer of the heating system, which will provide a house with a certain size, with a given number of floors, and heat engineering properties.

To equip a one-story suburban or private house, you do not need a very powerful heating boiler.

So, in calculating the boiler performance for autonomous home area is the main parameter if we consider the heating technology of a building in accordance with the climate of the region. So, the area of ​​the house is the most important parameter in order to calculate the boiler for heating.

Characteristics that will affect the calculation

Those who want to calculate the boiler for heating a house with maximum accuracy can use the methodology provided by SNiP II-3-79. In this case, professional calculations will take into account the following factors:

• Average temperature of the region during the coldest time.
• The insulating properties of the materials used to build the building envelope.
• Type of heating circuit wiring.
• The ratio of the area of ​​load-bearing structures and openings.
• Separate information about each room.

How to calculate the power of a heating boiler? To execute accurate calculations, even such information as data on units of household and digital appliances is used - after all, all this also somehow releases heat into the premises.

However, we note that not every owner of a heating system requires professional calculations - it is usually customary to purchase autonomous heating circuits with devices with a power reserve.

So, the efficiency of heating boilers can be higher than the calculated values, especially since they, as a rule, are rounded.

What is taken into account without fail?

How to calculate the power of a heating boiler, what data must be present without fail? One rule should be remembered: every 10 square meters of a cottage with insulating characteristics, the standard ceiling height limit (up to 3 m) will require approximately 1 kW for heating. To the power of the boiler, which is intended for working together in heating and hot water supply, you will need to add at least 20%.

An autonomous heating circuit, which has an unstable pressure in the heating boiler, will need to be equipped with a device so that its power reserve is at least 15 percent higher than the calculated value. To the power of the boiler, which provides heating and hot water supply, it is required to add 15%.

We take into account heat loss

Note that regardless of whether the power is calculated electric boiler, a gas, diesel or wood-fired boiler - in any case, the operation of the heating system will be accompanied by heat losses:

• It is necessary to ventilate the premises, but if the windows are open all the time, the house will lose about 15% of its energy.
• If the walls are poorly insulated, then 35% of the heat will go away.
• 10% of the heat will go through the window openings, and even more if the frames are of the old model.
• If the floor is not insulated, then 15% of the heat will be given to the basement or ground.
• 25% of the heat will go through the roof.

The simplest formula

Heat engineering calculations in any case will have to be rounded and also increased in order to provide a power reserve. That is why, in order to determine the power of a heating boiler, it will be possible to use a very simple formula:

W = S * Wud.

Here S is the total area of ​​the heated building, which takes into account living and utility rooms in sq. M.

W is the power of the heating boiler, kW.

Wood. Is the average statistical power density, this parameter is used for calculations taking into account a certain climatic zone, kW / sq.m. And it's worth noting that given characteristic based on many years of experience operation of various heating systems in the regions. And when we multiply the area by this indicator, we get the average power value. It will need to be adjusted based on the features listed above.

Calculation example

Consider an example using a heating boiler power calculator. Natural gas is the most affordable fuel used in Russia. For this reason, it is so widespread and in demand. Therefore, we will calculate the power of the gas boiler. And as an example, let's take a private house with an area of ​​140 sq. M. Territory - Krasnodar Territory. Also, in the example, we take into account that our boiler will provide not only heating the house, but also plumbing fixtures with water. We will do the calculations for a system with natural circulation, the pressure here will not be maintained by the circulation pump.

Specific power - 0.85 kW / sq.m.

So, 140 sq.m / 10 sq.m = 14 is an intermediate calculation factor. It will provide for the condition that for every 10 square meters of heated premises, 1 kW of heat will be required, which will be provided by the boiler.

14 * 0.85 = 11.9 kW.

We get thermal energy, which will be needed by the house, which has standard thermal properties. To provide hot water for showers and sinks, we will add another 20%.

11.9 + 11.9 * 0.2 = 14.28 kW.

We do not use a circulating pump, so we must remember that the pressure here can be unstable. Therefore, we must add another 15% to ensure the supply of thermal energy.

14.28 + 11.9 * 0.15 = 16.07 kW.

Also be aware that there will be some heat leaks. That is why we must round our result to more meaning... Thus, we need a heating boiler with a minimum power of 17 kW.

As a rule, the calculation of the heating boiler power is carried out at the stage of building design. Indeed, in order for the heating system to work effectively, specific conditions are required - the arrangement of the combustion room, the supply of the premises with a chimney and ventilation.

In any heating system that uses a liquid heat carrier, its "heart" is the boiler. It is here that the energy potential of fuel (solid, gaseous, liquid) or electricity is converted into heat, which is transferred to the coolant, and is already carried by it to all heated rooms of a house or apartment. Naturally, the capabilities of any boiler are not unlimited, that is, they are limited by its technical and operational characteristics indicated in the product passport.

One of the key characteristics is the heat output of the unit. Simply put, it must have the ability to generate such an amount of heat per unit of time, which would be sufficient to fully heat all the premises of a house or apartment. The selection of a suitable model "by eye" or according to some overly generalized concepts can lead to an error in one direction or another. Therefore, in this publication we will try to offer the reader, albeit not professional, but nevertheless with a sufficiently high degree of accuracy, an algorithm on how to calculate the power of a boiler for heating a house.

A trivial question - why know the required boiler power

Despite the fact that the question does seem to be rhetorical, it still seems necessary to provide a couple of explanations. The fact is that some owners of houses or apartments still manage to make mistakes, going to one extreme or another. That is, when purchasing equipment, either knowingly insufficient thermal performance, hoping to save money, or greatly overestimated, so that, in their opinion, it is guaranteed, with a large margin, to provide themselves with heat in any situation.

Both are completely wrong, and negatively affects both the provision of comfortable conditions residence, and on the durability of the equipment itself.

• Well, with the lack of calorific value, everything is more or less clear. With the onset of winter cold weather, the boiler will work at its full capacity, and it is not a fact that there will be a comfortable microclimate in the premises. This means that you will have to "catch up with heat" with the help of an electric heating appliances, which will entail unnecessary considerable expenses. And the boiler itself, operating at the limit of its capabilities, is unlikely to last long. In any case, after a year or two, homeowners will unequivocally realize the need to replace the unit with a more powerful one. One way or another, the cost of a mistake is quite impressive.

• Well, why not buy a boiler with a large margin, how can this hinder? Yes, of course, high-quality heating of the premises will be provided. But now let's list the "cons" of this approach:

Firstly, a boiler of higher power by itself can cost significantly more, and it is difficult to call such a purchase rational.

Secondly, with an increase in power, the dimensions and weight of the unit almost always increase. These are unnecessary difficulties during installation, "stolen" space, which is especially important if the boiler is planned to be placed, for example, in the kitchen or in another room of the living area of ​​the house.

Thirdly, you can face the uneconomic operation of the heating system - part of the energy resources expended will be spent, in fact, in vain.

Fourthly, excess capacity means regular long-term shutdowns of the boiler, which, in addition, are accompanied by cooling of the chimney and, accordingly, abundant formation of condensate.

Fifth, if powerful equipment is never properly loaded, it does not benefit it. Such a statement may seem paradoxical, but it is so - wear becomes higher, the duration of accident-free operation is significantly reduced.

Prices for popular heating boilers

Excess boiler power will be appropriate only if it is planned to connect a water heating system to it for household needs- boiler indirect heating... Well, or when it is planned to expand the heating system in the future. For example, the owners are planning to build a residential extension to the house.

Methods for calculating the required boiler power

In truth, it is always better to trust specialists to carry out heat engineering calculations - there are too many nuances to be taken into account. But, it is clear that such services are not provided free of charge, so many owners prefer to take responsibility for choosing the parameters of boiler equipment.

Let's see what methods of calculating thermal power are most often offered on the Internet. But first, let's clarify the question of what exactly should influence this parameter. This will make it easier to understand the advantages and disadvantages of each of the proposed calculation methods.

What principles are key in making calculations

So, the heating system has two main tasks. Let us clarify right away that there is no clear separation between them - on the contrary, there is a very close relationship.

• The first is to create and maintain a comfortable temperature for living in the premises. Moreover, this level of heating should apply to the entire volume of the room. Of course, due to physical laws, temperature gradation in height is still inevitable, but it should not affect the feeling of the comfort of being in the room. It turns out that it should be able to warm up a certain volume of air.

The degree of temperature comfort is undoubtedly a subjective value, that is different people it can be assessed in its own way. Nevertheless, it is generally accepted that this indicator is in the range of +20 ÷ 22 ° С. Usually, it is precisely this temperature that is operated when carrying out heat engineering calculations.

This is also indicated by the standards established by the current GOST, SNiP and SanPiN. For example, the table below shows the requirements of GOST 30494-96:

Room type	Air temperature level, ° С
	optimal	permissible
Living spaces	20 ÷ 22	18 ÷ 24
Living quarters for regions with minimum winter temperatures of -31 ° C and below	21 ÷ 23	20 ÷ 24
Kitchen	19 ÷ 21	18 ÷ 26
Toilet	19 ÷ 21	18 ÷ 26
Bathroom, combined bathroom	24 ÷ 26	18 ÷ 26
Office, rooms for rest and study sessions	20 ÷ 22	18 ÷ 24
The corridor	18 ÷ 20	16 ÷ 22
Lobby, staircase	16-18	14 ÷ 20
Pantries	16-18	12 ÷ 22
Living quarters (the rest are not standardized)	22 ÷ 25	20 ÷ 28

• The second task is to constantly compensate for possible heat losses. To create an “ideal” house, in which there would be no heat leaks at all, is a problem that is practically insoluble. You can only reduce them to the ultimate minimum. And practically all elements of the building structure become leakage paths to one degree or another.

Building structure element	Approximate share of total heat losses
Foundation, plinth, floors of the first floor (on the ground or over an unheated felling)	from 5 to 10%
Joints building structures	from 5 to 10%
Sections of the passage of engineering communications through construction structures (sewer pipes, water supply, gas supply, electrical or communication cables, etc.)	up to 5%
External walls, depending on the level of thermal insulation	from 20 to 30%
Windows and doors to the street	about 20 ÷ 25%, of which about half - due to insufficient sealing of boxes, poor fit of frames or canvases
Roof	up to 20%
Chimney and ventilation	up to 25 ÷ 30%

Why were all these rather lengthy explanations given? And only in order for the reader to have complete clarity that when calculating, willy-nilly, it is necessary to take into account both directions. That is, both the "geometry" of the heated premises of the house, and the approximate level of heat losses from them. And the amount of these heat leaks, in turn, depends on a number of factors. This is the difference in temperatures outside and in the house, and the quality of thermal insulation, and the features of the whole house as a whole and the location of each of its premises, and other evaluation criteria.

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Now, armed with this preliminary knowledge, let's move on to considering different methods calculating the required thermal power.

Calculation of power by the area of ​​heated premises

It is proposed to proceed from their conditional ratio, that for high-quality heating of one square meter of the area of ​​the room, it is necessary to consume 100 W of thermal energy. Thus, it will help to calculate which:

Q =Stot / 10

Q- the required heat output of the heating system, expressed in kilowatts.

Stot- the total area of ​​the heated premises of the house, square meters.

However, reservations are made:

• The first is that the ceiling height of the room should be 2.7 meters on average, a range of 2.5 to 3 meters is allowed.
• The second - you can make an amendment for the region of residence, that is, accept not a rigid rate of 100 W / m², but a "floating" one:

That is, the formula will take on a slightly different form:

Q =Stot ×Qsp / 1000

Qud - the value of the specific heat output taken from the table above square meter area.

• Third, the calculation is valid for houses or apartments with an average degree of insulation of the enclosing structures.

Nevertheless, despite the above-mentioned reservations, such a calculation is by no means accurate. Agree that it is largely based on the "geometry" of the house and its premises. But heat loss is practically not taken into account, except for the rather "blurred" ranges of specific thermal power by region (which also have very vague boundaries), and remarks that the walls should have an average degree of insulation.

But be that as it may, this method is still popular, precisely because of its simplicity.

It is clear that the operational reserve of the boiler power must be added to the calculated value obtained. It should not be overestimated too much - experts advise to stop at the range from 10 to 20%. This, by the way, applies to all methods of calculating power. heating equipment, which will be discussed below.

Calculation of the required thermal power by the volume of premises

By and large, this method of calculation is largely the same as the previous one. True, the initial value here is not the area, but the volume - in fact, the same area, but multiplied by the height of the ceilings.

And the norms of specific thermal power are taken here as follows:

• for brick houses- 34 W / m³;
• for panel houses- 41 W / m³.

Even based on the proposed values ​​(from their formulation), it becomes clear that these norms were established for apartment buildings, and are mainly used to calculate the heat demand for premises connected to central system branches or to an autonomous boiler station.

It is quite obvious that "geometry" is again put at the forefront. And the whole system of accounting for heat losses is reduced only to differences in the thermal conductivity of brick and panel walls.

In a word, this approach to calculating thermal power does not differ in accuracy either.

Calculation algorithm taking into account the characteristics of the house and its individual rooms

Description of the calculation method

So, the methods proposed above give only a general idea of ​​the required amount of thermal energy for heating a house or apartment. They have a common vulnerability - almost complete ignorance of possible heat losses, which are recommended to be considered "average".

But it is quite possible to carry out more accurate calculations. This will help the proposed calculation algorithm, which is embodied, in addition, in the form of an online calculator, which will be offered below. Just before starting the calculations, it makes sense to step by step consider the very principle of their implementation.

First of all, an important note. The proposed method involves the assessment not of the entire house or apartment in terms of the total area or volume, but of each heated room separately. Agree that rooms of equal area, but differing, say, in the number of external walls, will require different amounts of heat. You cannot put an equal sign between rooms that have a significant difference in the number and area of ​​windows. And there are many such criteria for evaluating each of the rooms.

So it will be more correct to calculate required power for each of the premises separately. Well, then a simple summation of the obtained values ​​will lead us to the desired indicator of the total heat power for the entire heating system. That is, in fact, for her "heart" - the cauldron.

One more note. The proposed algorithm does not claim to be "scientific", that is, it is not directly based on any specific formulas established by SNiP or other guiding documents. However, it has been proven in practice and shows results with a high degree of accuracy. Differences with the results of professionally performed heat engineering calculations are minimal, and do not in any way affect the right choice equipment at its rated thermal power.

The "architecture" of the calculation is as follows - the base is taken, where the aforementioned value of the specific thermal power, equal to 100 W / m2, is taken, and then a whole series of correction factors are introduced, to one degree or another reflecting the amount of heat loss in a particular room.

If you express this with a mathematical formula, it will turn out something like this:

Qk= 0.1 × Sk× k1 × k2 × k3 × k4 × k5 × k6 × k7 × k8 × k9 × k10 × k11

Qk- the required thermal power required for full heating of a particular room

0.1 - conversion of 100 W to 0.1 kW, just for the convenience of obtaining the result in kilowatts.

Sк- the area of ​​the room.

k1 ÷k11- correction factors for adjusting the result, taking into account the characteristics of the room.

Presumably, there should be no problems with determining the area of ​​the premises. So let's move on to a detailed examination of the correction factors.

• k1 is a coefficient that takes into account the height of the ceilings in the room.

It is clear that the height of the ceilings directly affects the volume of air that the heating system must warm up. For the calculation, it is proposed to take the following values ​​of the correction factor:

• k2 is a coefficient that takes into account the number of walls in the room in contact with the street.

The larger the contact area with external environment, the higher the level of heat loss. Everyone knows that it is always much cooler in a corner room than in one with only one outer wall. And some premises of a house or apartment may even be internal, having no contact with the street.

According to the mind, of course, one should take not only the number of external walls, but also their area. But our calculation is still simplified, so we will limit ourselves only to the introduction of a correction factor.

The coefficients for different cases are shown in the table below:

We do not consider the case when all four walls are external. This is no longer a residential building, but just some kind of barn.

• k3 is a coefficient that takes into account the position of the outer walls relative to the cardinal points.

Even in winter, don't discount the potential impact of energy. sun rays... On a clear day, they penetrate through the windows into the premises, thereby being included in the general supply of heat. In addition, the walls receive a charge solar energy, which leads to a decrease in the total amount of heat loss through them. But all this is true only for those walls that "see" the Sun. On the north and north-east side of the house, there is no such influence, for which a certain correction can also be made.

The values ​​of the correction factor for the cardinal points are in the table below:

• k4 - coefficient taking into account the direction of winter winds.

Perhaps this amendment is not mandatory, but for houses located in open areas, it makes sense to take it into account.

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Almost in any locality there is a predominance of winter winds - this is also called the "wind rose". Local meteorologists have such a scheme without fail - it is drawn up based on the results of many years of weather observations. Quite often, the locals themselves are well aware of which winds most often disturb them in winter.

And if the wall of the room is located on the windward side, and is not protected by some natural or artificial barriers from the wind, then it will be cooled much more strongly. That is, and heat losses the premises are growing. To a lesser extent, this will be expressed at the wall located parallel to the direction of the wind, in the minimum - located on the leeward side.

If there is no desire to "bother" with this factor, or there is no reliable information about the winter wind rose, then you can leave the coefficient equal to one. Or, conversely, take it as maximum, just in case, that is, for the most unfavorable conditions.

The values ​​of this correction factor are in the table:

• k5 is a coefficient that takes into account the level of winter temperatures in the region of residence.

If heat engineering calculations are carried out according to all the rules, then the assessment of heat losses is carried out taking into account the temperature difference in the room and outside. It is clear that the colder in terms of climatic conditions the region, the more heat is required to be supplied to the heating system.

In our algorithm, this will also be taken into account to a certain extent, but with an acceptable simplification. Depending on the level of minimum winter temperatures falling on the coldest decade, a correction factor k5 is selected .

Here it is pertinent to make one remark. The calculation will be correct if the temperatures that are considered normal for the given region are taken into account. There is no need to recall the abnormal frosts that happened, say, several years ago (and that is why, by the way, they are remembered). That is, the lowest, but normal temperature for a given area should be chosen.

• k6 is a coefficient that takes into account the quality of the thermal insulation of the walls.

It is quite clear that what more efficient system wall insulation, the lower the level of heat loss will be. Ideally, to which one should strive, thermal insulation should generally be complete, carried out on the basis of the performed thermal engineering calculations, taking into account climatic conditions region and design features of the house.

When calculating the required heat output of the heating system, the existing thermal insulation of the walls should also be taken into account. The following gradation of correction factors is proposed:

An insufficient degree of thermal insulation or its complete absence at all, in theory, should not be observed at all in a residential building. V otherwise the heating system will be very costly, and even without the guarantee of creating truly comfortable living conditions.

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If the reader wishes to independently assess the level of thermal insulation of his home, he can use the information and calculator, which are placed in the last section of this publication.

• k7 andk8 are coefficients that take into account heat loss through the floor and ceiling.

The following two coefficients are similar - their introduction into the calculation takes into account the approximate level of heat losses through the floors and ceilings of the premises. There is no need to describe in detail here - both the possible options and the corresponding values ​​of these coefficients are shown in the tables:

To begin with, the coefficient k7, which corrects the result depending on the characteristics of the floor:

Now is the coefficient k8, correcting for the neighborhood from above:

• k9 is a coefficient that takes into account the quality of the windows in the room.

Here, too, everything is simple - the higher the quality of the windows, the less heat loss through them. Old wooden frames, as a rule, do not have good thermal insulation characteristics. The situation is better with modern window systems equipped with double-glazed windows. But they can also have a certain gradation - according to the number of chambers in the glass unit and according to other design features.

For our simplified calculation, the following values ​​of the coefficient k9 can be applied:

• k10 is a factor correcting the area of ​​the room's glazing.

The quality of windows does not yet fully reveal all the volumes of possible heat loss through them. The glazing area is very important. Agree, it is difficult to compare a small window and a huge panoramic window that is almost the entire wall.

To make an adjustment for this parameter, you first need to calculate the so-called room glazing coefficient. It is not difficult - it is just that the ratio of the glazing area to the total area of ​​the room is found.

kw =sw /S

kw- coefficient of glazing of the room;

sw- total area of ​​glazed surfaces, m²;

S- area of ​​the room, m².

Everyone can measure and sum up the area of ​​windows. And then it is easy to find the required glazing coefficient by simple division. And he, in turn, makes it possible to enter the table and determine the value of the correction factor k10 :

Glazing coefficient value kw	The value of the coefficient k10
- up to 0.1	0.8
- from 0.11 to 0.2	0.9
- from 0.21 to 0.3	1.0
- from 0.31 to 0.4	1.1
- from 0.41 to 0.5	1.2
- over 0.51	1.3

• k11 - coefficient taking into account the presence of doors to the street.

The last of the considered coefficients. The room may have a door leading directly to the street, on cold balcony, into an unheated corridor or entrance, etc. Not only is the door itself often a very serious "cold bridge" - with its regular opening, a fair amount of cold air will penetrate into the room every time. Therefore, a correction should be made for this factor: such heat losses, of course, require additional compensation.

The values ​​of the coefficient k11 are given in the table:

This coefficient should be taken into account if the doors in winter time use regularly.

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* * * * * * *

So, all the correction factors have been considered. As you can see, there is nothing super complicated here, and you can safely proceed to the calculations.

One more tip before starting calculations. Everything will be much easier if you first draw up a table, in the first column of which you sequentially indicate all the rooms of the house or apartment to be sealed off. Further, according to the columns, place the data required for calculations. For example, in the second column - the area of ​​the room, in the third - the height of the ceilings, in the fourth - orientation to the cardinal points - and so on. It is not difficult to draw up such a tablet, having in front of you a plan of your residential estates. It is clear that the calculated values ​​of the required heat output for each room will be entered in the last column.

The table can be drawn up in an office application, or even simply drawn on a piece of paper. And do not rush to part with it after the calculations - the obtained heat power indicators will still come in handy, for example, when purchasing heating radiators or electric heating devices used as backup source heat.

To make it as easy as possible for the reader to carry out such calculations, a special online calculator is placed below. With it, with the initial data previously collected in a table, the calculation will take literally a few minutes.

Calculator for calculating the required thermal power for the premises of a house or apartment.

The calculation is carried out for each room separately.
Sequentially enter the requested values ​​or mark the necessary options in the proposed lists.
Click on "CALCULATE THE REQUIRED THERMAL OUTPUT"

Room area, m2

100 W per sq. m

Indoor ceiling height

Number of external walls

The outer walls face:

The position of the outer wall relative to the winter "wind rose"

Level negative temperatures air in the region during the coldest week of the year

After carrying out calculations for each of the heated premises, all indicators are summed up. This will be the value of the total thermal power required to fully heat a house or apartment.

As already mentioned, a margin of 10 ÷ 20 percent should be added to the resulting final value. For example, the calculated power is 9.6 kW. If you add 10%, then this will be 10.56 kW. With the addition of 20% - 11.52 kW. Ideally, the nominal thermal power of the purchased boiler should just be in the range from 10.56 to 11.52 kW. If there is no such model, then the closest one in terms of power is purchased in the direction of its increase. For example, for this particular example, 11.6 kW are perfect - they are presented in several lines of models from different manufacturers.

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What is the best way to assess the degree of thermal insulation of the walls of a room?

As promised above, this section of the article will help the reader to assess the level of thermal insulation of the walls of his residential estates. To do this, you will also have to carry out one simplified heat engineering calculation.

Calculation principle

According to the requirements of SNiP, the resistance to heat transfer (which is also called thermal resistance) of building structures of residential buildings should not be lower than the standard indicator. And these normalized indicators are established for the regions of the country, in accordance with the peculiarities of their climatic conditions.

Where can you find these values? Firstly, they are in special tables-appendices to SNiP. Secondly, information about them can be obtained from any local construction or architectural design company. But it is quite possible to use the proposed schematic map covering the entire territory of the Russian Federation.

In this case, we are interested in walls, therefore we take from the diagram the value of thermal resistance precisely "for walls" - they are indicated by purple numbers.

Now let's take a look at what this thermal resistance is composed of, and what it is equal to from the point of view of physics.

So, the resistance to heat transfer of some abstract homogeneous layer NS equals:

Rх = hх / λх

Rx- resistance to heat transfer, measured in m² × ° K / W;

hx- layer thickness, expressed in meters;

λx- coefficient of thermal conductivity of the material from which this layer is made, W / m × ° K. This is a tabular value, and for any of the building or thermal insulation materials it is easy to find it on the Internet reference resources.

Conventional Construction Materials used for the construction of walls, most often, even with their large (within reason, of course) thickness does not reach normative indicators resistance to heat transfer. In other words, the wall cannot be called fully thermally insulated. This is exactly what the insulation is used for - an additional layer is created that "makes up for the deficit" necessary to achieve the normalized indicators. And due to the fact that the thermal conductivity coefficients of high-quality insulation materials are low, you can avoid the need to erect very thick structures.

You may be interested in information about what is

Let's take a look at a simplified diagram of an insulated wall:

1 - in fact, the wall itself, which has a certain thickness and was erected from one or another material. In most cases, "by default" it itself is not able to provide the normalized thermal resistance.

2 - a layer of insulation material, the coefficient of thermal conductivity and the thickness of which should provide a "shortage coverage" up to the normalized indicator R. Let's make a reservation right away - the location of the thermal insulation is shown outside, but it can be placed with inside walls, and even be located between two layers supporting structure(for example, laid out of bricks according to the principle of "well masonry").

3 - external facade decoration.

4 - interior decoration.

Finish layers often do not have any significant effect on the overall thermal resistance. Although, when performing professional calculations, they are also taken into account. In addition, the finish can be different - for example, warm plaster or cork slabs are very capable of strengthening the overall thermal insulation of the walls. So for the "purity of the experiment" it is quite possible to take into account both of these layers.

But there is also an important note - the layer is never taken into account. facade decoration, if a ventilated gap is located between it and the wall or insulation. And this is often practiced in ventilated facade systems. In this design exterior decoration will have no effect on the overall level of thermal insulation.

So, if we know the material and thickness of the main wall itself, the material and thickness of the layers of insulation and decoration, then using the above formula it is easy to calculate their total thermal resistance and compare it with the normalized indicator. If it is not less - no question, the wall has full thermal insulation. If it is not enough, you can calculate which layer and which insulation material is able to fill this shortage.

You might be interested in information about how the

And to make the task even easier - below is an online calculator that will perform this calculation quickly and accurately.

Several explanations for working with him at once:

• To begin with, the normalized value of the resistance to heat transfer is found according to the map of the scheme. In this case, as already mentioned, we are interested in walls.

(However, the calculator has versatility. And, it allows you to evaluate the thermal insulation of both floors, and roofing... So, if necessary, you can use - add the page to your bookmarks).

• The next group of fields specifies the thickness and material of the main supporting structure - the wall. The thickness of the wall, if it is equipped according to the principle of "well masonry" with insulation inside, the total is indicated.
• If the wall has a thermal insulation layer (regardless of its location), then the type of insulation material and thickness are indicated. If there is no insulation, then the default thickness is left equal to "0" - go to the next group of fields.
• And the next group is "dedicated" exterior decoration walls - the material and thickness of the layer are also indicated. If there is no finishing, or there is no need to take it into account, everything is left by default and move on.
• Do the same with interior decoration walls.
• Finally, all that remains is to choose an insulation material that is planned to be used for additional thermal insulation. Possible options are indicated in the drop-down list.

A zero or negative value immediately indicates that the thermal insulation of the walls complies with the standards, and additional insulation is simply not required.

A positive value close to zero, say, up to 10 ÷ 15 mm, also does not give any particular reason to worry, and the degree of thermal insulation can be considered high.

Insufficiency up to 70 ÷ 80 mm should already make the owners think. Although such insulation can be attributed to average efficiency, and it should be taken into account when calculating the thermal power of the boiler, it is still better to plan the work to strengthen the thermal insulation. The thickness of the additional layer is already shown. And the implementation of these works will immediately give a tangible effect - both by increasing the comfort of the microclimate in the premises, and by lowering the consumption of energy resources.

Well, if the calculation shows a shortage above 80 ÷ 100 mm, there is practically no insulation or it is extremely ineffective. There can be no two opinions here - the prospect of holding insulation works comes to the fore. And it will be much more profitable than purchasing a boiler of increased power, some of which will simply be spent literally on "warming up the street." Naturally, accompanied by ruinous energy bills.

Compared to heating electrical appliances, own heating system is more profitable in terms of savings, and in maximum convenience when heating rooms.

The efficiency and profitability of a heating system in a home depends on correct calculations, adherence to precise rules and instructions.

Calculation of heating by the area of ​​the house is a laborious and complex process. Don't skimp on materials. High quality equipment and installing it affects the financial budget, but then serves the home well and comfortably.

When equipping the house with a heating system, construction works and the installation of heating must go strictly according to the project and taking into account all safety rules for use.

Consider the following points:

• house building material,
• footage of window openings;
• climatic features of the area where the house is located;
• location window frames by compass;
• what is the structure of the "warm floor" system.

Subject to all of the above rules and calculations for the implementation of heating, some knowledge in the field of engineering is required. But there is also a simplified system - the calculation of heating by area, which can be done independently, again, adhering to the rules and compliance with all norms.

Choosing a boiler requires an individual approach.

If the house has gas, then the most the best way- this is gas boiler... In the absence of a centralized gas pipeline, we choose an electric boiler, a heat generator for solid or liquid fuel. Taking into account regional peculiarities, access to the supply of materials, it is possible to install a combi boiler. Combined generator heat will always allow you to maintain a comfortable temperature, in any emergency and force majeure situations. Here you need to build on the simple type of operation, the heat transfer coefficient.

After determining the type of boiler, it is necessary to calculate the heating by the area of ​​the room. The formula is simple, but it takes into account the temperature of the cold period, the coefficient of heat loss for large windows and their location, wall thickness and ceiling height.

Each boiler has a certain power. If you make the wrong choice, the room will be either cold or excessively hot. Thus, if the specific power of the boiler is 10 cubic meters. taking into account the area of ​​the heated room of 100 sq.m., you can choose the most optimal heat generator.

From the formula that engineers use - Wcat = (SxWud) / 10, kW... - it follows that a boiler with a capacity of 10 kW heats a room of 100 sq.m.

The required number of heating radiator sections.

To make it clearer, let's solve the problem using specific numbers as an example. Assuming that room area 14 sq. m.... and ceiling height 3 meters, the volume is determined by multiplication.

14 x 3 = 42 cubic meters.

V middle lane Russia, Ukraine, Belarus thermal power per cubic meter corresponds to 41 W... Determine: 41x 42 = 1722 W. Found out that for a room of 14 sq.m. need a 1700 W radiator... Each individual section (rib) has a power of 150 watts. By dividing the results obtained, we get the number of sections required for the acquisition. Heating calculation by area is not the same everywhere. For premises over 100 sq.m. required installation of a circulation pump, which serves as a "force" for the movement of the coolant through the pipes. Its installation takes place in the opposite direction from the heating devices to the heat generator. The circulation pump increases the life of the heating system by reducing the contact of hot heat carriers with the devices.

When installing the heating system " warm floor»The heating coefficient of the house is growing at times. Connect the system underfloor heating can already be used for existing types of heating. A pipe is removed from the heating radiators and the floor heating wiring is supplied. This is the most convenient and profitable option, taking into account the savings in time and money.

How to calculate the power of a gas boiler for the given parameters of the heated room? I know at least three different methods that give different levels of reliability of the results, and today we will get to know each of them.

general information

Why do we calculate the parameters specifically for gas heating?

The fact is that gas is the most economical (and, accordingly, the most popular) heat source. A kilowatt-hour of thermal energy obtained during its combustion costs the consumer 50-70 kopecks.

For comparison - the price of a kilowatt-hour of heat for other energy sources:

• Solid fuel- 1.1-1.6 rubles per kilowatt-hour;
• Diesel fuel- 3.5 rubles / kWh;
• Electricity- 5 rubles / kWh.

In addition to being economical, gas equipment attracts with ease of use. The boiler requires maintenance no more than once a year, does not need kindling, cleaning the ash pan and replenishing the fuel supply. Devices with electronic ignition work with external thermostats and are able to automatically maintain a constant temperature in the house, regardless of the weather.

Does the calculation of a gas boiler for a house differ from the calculation of a solid fuel, liquid fuel or electric boiler?

In general, no. Any heat source must compensate for heat loss through the floor, walls, windows and ceiling of the building. Its thermal power has nothing to do with the energy carrier used.

In the case of a double-circuit boiler supplying the house hot water for economic purposes, we need a reserve of power to heat it up. The excess capacity will ensure the simultaneous flow of water in DHW system and heating the coolant for heating.

Calculation methods

Scheme 1: by area

We will be helped in this normative documents half a century ago. According to the Soviet SNiP, heating should be designed at the rate of 100 watts of heat per square meter of the heated room.

Let's, for example, calculate the power for a house measuring 6x8 meters:

1. The area of ​​the house is equal to the product of its overall dimensions. 6x8x48 m2;
2. With a specific power of 100 W / m2, the total boiler power should be 48x100 = 4800 watts, or 4.8 kW.

The choice of the boiler power according to the area of ​​the heated room is simple, understandable and ... in most cases it gives the wrong result.

Because he neglects a number of important factors that affect real heat loss:

• The number of windows and doors... More heat is lost through glazing and doorways than through a main wall;
• Ceiling height... In Soviet-built apartment buildings, it was standard - 2.5 meters with a minimum error. But in modern cottages you can find ceilings with a height of 3, 4 or more meters. The higher the ceiling, the larger the heated volume;

• Climatic zone... With the same quality of thermal insulation, heat loss is directly proportional to the difference between indoor and outdoor temperatures.

V apartment building heat loss is affected by the location of the dwelling relative to the outer walls: end and corner rooms lose more heat. However, in a typical cottage, all rooms have common walls with the street, therefore the corresponding correction factor is included in the base value of the heat output.

Scheme 2: by volume, taking into account additional factors

How to calculate with your own hands a gas boiler for heating a private house, taking into account all the factors I mentioned?

First and foremost: in the calculation, we take into account not the area of ​​the house, but its volume, that is, the product of the area by the height of the ceilings.

• Base value boiler power per one cubic meter of heated volume - 60 watts;
• Window increases heat loss by 100 watts;
• Door adds 200 watts;
• Heat loss is multiplied by the regional coefficient... It is determined by the average temperature of the coldest month:

Image	Coefficient and climatic zone
	0,6-0,9 - for regions with an average January temperature of about 0 ° C (Krasnodar Territory, Crimea).
	1,2-1,3 - for the average temperature of the coldest month in -15-20 ° С (Moscow and Leningrad regions).
	1,5-1,6 - for regions with an average January temperature of -25-30 ° C (Novosibirsk region, Khabarovsk Territory).
	2 - for -40 and below (Chukotka, Yakutia).

Let's calculate the boiler power again for our 6x8 meter house, specifying a few additional parameters:

• House location- the city of Sevastopol (average January temperature - +3 degrees Celsius);
• Number of windows- 5. One door leads to the street;
• Ceiling height- 3.2 meters.

1. House volume(with outside walls) is equal to the product of its three dimensions: 6x8x3.2 = 153.6 cubic meters;

1. Base power for this volume - 153.6x60 = 9216 W;
2. Taking into account windows and doors it will increase by 5x100 + 200 = 700 watts. 9216 + 700 = 9916;
3. Regional coefficient for the warm climate of Crimea, we take it equal to 0.6.

9916 * 0.6 = 6000 (rounded) watts.

As you can see, the complicated calculation scheme gave a result that was noticeably different from the previous one. How accurate is it?

The calculation will give a reliable result for a house, the insulation quality of which roughly corresponds to the quality of insulation of Soviet-built houses. The scheme is based on the same 100 watts per square of area, recalculated taking into account standard height ceilings of 2.5 meters at 40 W / m3 and multiplied by a factor of 1.5 to compensate for the heat loss of a private house through the roof and floor.

How to determine the need for heat at home with non-standard insulation?

Scheme 3: by volume, taking into account the quality of insulation

The most universal formula for calculating the heat output of a boiler is Q = V * Dt * k / 860.

In this formula:

• Q is the heat loss of the house in kilowatts;
• V is the volume to be heated by the boiler, in cubic meters;
• Dt is the calculated delta of the temperature between the heated room and the air outside the outer walls;
• k is the dissipation coefficient, which is determined by the quality of the house insulation.

How to choose the coefficient k?

Select its value for your conditions, referring to the following table:

Image	The value of the coefficient and the description of the building
	3-4 - building without insulation (warehouse made of profiled sheet, panel house with walls made of boards in one layer)
	2.0-2.9 - walls made of timber 10 cm thick or bricks 25 cm thick, wooden frames, single glazing
	1,0-1,9 - brick walls 50 cm thick, double glazed windows in the windows
	0.6-0.9 - facade insulated with foam or mineral wool, plastic windows with triple or energy-saving double-glazed windows

How to choose the calculated value outside temperature? In the calculations, it is customary to use the temperature of the coldest five-day winter for a given region. Rare extreme frosts are not taken into account: if the thermometer falls below the usual marks, auxiliary heat sources (heaters, fan heaters, etc.) can be used.

Where can I get the relevant information? The instruction is quite predictable: the necessary data can be found in SNiP 23-01-99, normative document dedicated to building climatology.

For the convenience of readers, I will give here a short excerpt from the text of SNiP.

Town	Temperature of the coldest 5 days of winter, ° С
Maykop	-22
Barnaul	-42
Blagoveshchensk	-37
Tynda	-46
Shimanovsk	-41
Arkhangelsk	-37
Astrakhan	-26
Ufa	-39
Belgorod	-28
Bryansk	-30
Ulan-Ude	-40
Vladimir	-34
Vologda	-37
Voronezh	-31
Makhachkala	-19
Irkutsk	-38
Kaliningrad	-24
Petropavlovsk-Kamchatsky	-22
Pechora	-48
Kostroma	-35
Agatha	-55
Turukhansk	-56
St. Petersburg	-30
Susuman	-57
Moscow	-32
Novosibirsk	-42
Vladivostok	-26
Komsomolsk-on-Amur	-37
Yalta	-8
Sevastopol	-11

Let's go back to our example with a house in Sevastopol, once again clarifying a few details:

• Glazing of windows- single, in large-slotted wooden frames;
• Wall material- quarrystone, about half a meter thick.

Let's get down to the calculations.

1. For the calculated internal temperature, we will take the corresponding sanitary standards+ 20 ° C. Taking into account the data from the above table, the Dt parameter will be equal to 20 - -11 = 31 degrees;
2. The dissipation coefficient is assumed to be 2.0: the thermal conductivity of rubble walls is much higher than that of brick ones;

1. We calculated the volume of the house earlier. It is equal to 153.6 cubic meters;
2. Let's substitute the values ​​of the variables into our formula. Q = 153.6x31 * 2/860 = 11 kW.

As you can see, the correction for significant heat loss has almost doubled the calculated power of the gas boiler.

Two circuits

It's very simple: a 20% stock is put into the project for the operation of the second flow line. In our case, the required power will be 11x1.2 = 13.2 kW.