How to calculate a heating boiler. How to choose a boiler for heating a private house in terms of power. What power reserve should a gas boiler have

In any heating system using a liquid heat carrier, its “heart” is the boiler. It is here that the energy potential of the fuel (solid, gaseous, liquid) or electricity is converted into heat, which is transferred to the coolant, and is already distributed to all heated rooms of the house or apartment. Naturally, the possibilities 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 thermal power of the unit. Simply put, it must be able to produce in a unit of time such an amount of heat that 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, although not professional, but still with a sufficiently high degree of accuracy, an algorithm on how to calculate the boiler power for heating a house.

A banal question - why know the required boiler power

Despite the fact that the question does seem rhetorical, it still seems necessary to give a couple of explanations. The fact is that some owners of houses or apartments still manage to make mistakes, falling into one or another extreme. That is, purchasing equipment of either obviously insufficient thermal performance, in the hope of saving 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 of these are completely wrong, and negatively affect both the provision of comfortable living conditions and 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 operate at its full capacity, and it is not a fact that there will be a comfortable microclimate in the rooms. This means that you will have to “catch up with heat” with the help of electric heaters, which will entail considerable extra costs. And the boiler itself, functioning at the limit of its capabilities, is unlikely to last long. In any case, after a year or two, homeowners clearly 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, what can prevent it? Yes, of course, high-quality space heating will be provided. But now we list the "cons" of this approach:

Firstly, a boiler of greater power can cost much more in itself, and it is difficult to call such a purchase rational.

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

Thirdly, you may encounter uneconomical operation of the heating system - part of the energy spent will be spent, in fact, wasted.

Fourthly, excess power is regular long 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 him. Such a statement may seem paradoxical, but it is true - wear becomes higher, the duration of trouble-free operation is significantly reduced.

Prices for popular heating boilers

An excess of boiler power will be appropriate only if it is planned to connect a water heating system for household needs to it - an indirect heating boiler. Well, or when it is planned to expand the heating system in the future. For example, in the plans of the owners - the construction of a residential extension to the house.

Methods for calculating the required boiler power

In truth, it is always better to entrust the conduct of heat engineering calculations to specialists - there are too many nuances to take 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 affect this parameter. So it will be 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 faces two main tasks. Let us immediately clarify that there is no clear division between them - on the contrary, there is a very close relationship.

• The first is the creation and maintenance of 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 comfort 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, of course, a subjective value, that is, different people can evaluate it in their own way. But still, it is generally accepted that this indicator is in the region of +20 ÷ 22 ° С. Usually, it is precisely this temperature that is used during thermal 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	admissible
Living spaces	20÷22	18:24
Residential premises for regions with minimum winter temperatures from -31 °С and below	21÷23	20÷24
Kitchen	19:21	18:26
Toilet	19:21	18:26
Bathroom, shared bathroom	24÷26	18:26
Office, recreation and study rooms	20÷22	18:24
Corridor	18:20	16:22
lobby, stairwell	16÷18	14:20
Storerooms	16÷18	12÷22
Residential premises (the rest are not standardized)	22÷25	20÷28

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

Building element	Approximate share of total heat loss
Foundation, basement, floors of the first floor (on the ground or over an unheated basement)	from 5 to 10%
Joints of building structures	from 5 to 10%
Sections of the passage of engineering communications through building structures (sewerage, water supply, gas supply pipes, 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 in the calculations, willy-nilly, it is necessary to take into account both directions. That is, the "geometry" of the heated premises of the house, and the approximate level of heat loss from them. And the amount of these heat leaks, in turn, depends on a number of factors. This is the temperature difference in the street 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.

You might be interested in information on which are suitable

Now, armed with this preliminary knowledge, we turn to the consideration of various methods for 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 spend 100 W of thermal energy. Thus, it will help to calculate which:

Q=Stotal / 10

Q- the required thermal power of the heating system, expressed in kilowatts.

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

However, there are caveats:

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

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

Q=Stot ×Qud / 1000

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

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

However, despite the above reservations, such a calculation cannot be called accurate. Agree that it is largely based on the "geometry" of the house and its premises. But heat losses are practically not taken into account, except for the rather “blurred” ranges of specific thermal power by region (which are also with 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 for its simplicity.

It is clear that it is necessary to add the operating power reserve of the boiler to the calculated value obtained. It should not be excessively overestimated - experts advise stopping at a range of 10 to 20%. This, by the way, applies to all methods for calculating the power of heating equipment, which will be discussed below.

Calculation of the required heat output by the volume of the premises

By and large, this method of calculation largely repeats the previous one. True, the initial value here is no longer 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 here are accepted as follows:

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

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

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

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

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

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 - the almost complete disregard for possible heat losses, which are recommended to be considered "average".

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

First of all, an important note. The proposed methodology involves the assessment not of the entire house or apartment in terms of 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 a different amount of heat. It is impossible to 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 would be more correct to calculate the required power for each of the premises separately. Well, then a simple summation of the obtained values ​​\u200b\u200bwill lead us to the desired indicator of the total heat output for the entire heating system. That is, in fact, for its "heart" - the boiler.

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 governing documents. However, it has been field tested and shows results with a high degree of accuracy. Differences with the results of professionally carried out heat engineering calculations are minimal, and do not affect the correct choice of equipment in terms of its rated thermal power.

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

If this is expressed by a mathematical formula, then it will turn out something like this:

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

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

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

Sk- area of ​​the room.

k1 hk11- correction factors for adjusting the result, taking into account the characteristics of the room.

With the determination of the area of ​​\u200b\u200bthe room, presumably, there should be no problems. So let's move on to a detailed discussion 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 amount of air that the heating system must warm up. For the calculation, it is proposed to accept the following values ​​of the correction factor:

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

The larger the area of ​​contact with the external environment, the higher the level of heat loss. Everyone knows that it is always much cooler in a corner room than in a room with only one outer wall. And some rooms of a house or apartment may even be internal, not having 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 restrict ourselves only to the introduction of a correction factor.

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

The case when all four walls are external is not considered. 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, you should not discount the possible impact of the energy of the sun's rays. On a clear day, they penetrate through the windows into the premises, thereby being included in the overall heat supply. In addition, the walls receive a charge of 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. There is no such influence on the north and northeast side of the house, which can also be corrected.

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

• k4 is a coefficient that takes 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.

You may be interested in information about what they are

In almost any area there is a predominance of winter winds - this is also called the "wind rose". Local meteorologists must have such a scheme - it is compiled 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 any natural or artificial barriers from the wind, then it will cool much more. That is, the heat loss of the room increases. To a lesser extent, this will be expressed near the wall located parallel to the direction of the wind, and to a 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, on the contrary, take it to the 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 in accordance with all the rules, then the assessment of heat losses is carried out taking into account the temperature difference in the room and on the street. It is clear that the colder the climatic conditions of 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 would be appropriate to make one remark. The calculation will be correct if temperatures are taken into account, which are considered normal for a given region. There is no need to recall the anomalous frosts that happened, say, a few years ago (and that's why, by the way, they are remembered). That is, the lowest, but normal temperature for the area should be selected.

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

It is quite clear that the more efficient the wall insulation system, the lower the level of heat loss. Ideally, to which one should strive, thermal insulation in general should be complete, carried out on the basis of heat engineering calculations performed, taking into account the climatic conditions of the region and the 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, in theory, should not be observed at all in a residential building. Otherwise, the heating system will be very expensive, and even without a guarantee of creating really comfortable living conditions.

You might be interested in information about the heating system

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

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

The next two coefficients are similar - their introduction into the calculation takes into account the approximate level of heat loss 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 - the coefficient k8, which corrects 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 better the windows, the less heat loss through them. Old wooden frames usually do not have good thermal insulation properties. This is better with modern window systems equipped with double-glazed windows. But they can also have a certain gradation - according to the number of cameras in a double-glazed window and according to other design features.

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

• k10 is a coefficient that corrects for the room's glazing area.

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

To make an adjustment for this parameter, first you need to calculate the so-called room glazing coefficient. It's easy - just find the ratio of the glazing area to the total area of ​​​​the room.

kw =sw/S

kw- coefficient of glazing of the room;

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

S- room area, m².

Anyone can measure and sum the area of ​​windows. And then it is easy to find the desired 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 :

Value of glazing factor 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, to a cold balcony, to an unheated corridor or entrance, etc. Not only is the door itself often a very serious “cold bridge” - if it is opened regularly, a fair amount of cold air will enter the room each time. Therefore, this factor should also be corrected: 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 are regularly used in winter.

You may be interested in information about what is

* * * * * * *

So, all correction factors are 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 soldered. Next, in columns, place the data that is required for calculations. For example, in the second column - the area of ​​\u200b\u200bthe 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 make such a plate, having in front of you a plan of your residential properties. 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 compiled in an office application, or even simply drawn on a piece of paper. And do not rush to part with it after making the calculations - the obtained indicators of thermal power will still be useful, for example, when purchasing heating radiators or electric heaters used as a backup heat source.

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 heat output 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 required options in the proposed lists.
Click "CALCULATE THE REQUIRED THERMAL OUTPUT"

Room area, m²

100 watts per sq. m

Ceiling height in the room

Number of external walls

External walls look at:

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

The level of negative air temperatures in the region in the coldest week of the year

Assessment of the degree of thermal insulation of walls

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 you get 10.56 kW. With the addition of 20% - 11.52 kW. Ideally, the nominal thermal power of the purchased boiler should 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 in the direction of its increase is purchased. For example, specifically for this example, they are perfect with a power of 11.6 kW - they are presented in several lines of models from various manufacturers.

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How to correctly assess the degree of thermal insulation of the walls of the room?

As promised above, this section of the article will help the reader with an assessment of the level of thermal insulation of the walls of his residential properties. To do this, you will also have to carry out one simplified thermal calculation.

The principle of the calculation

According to the requirements of SNiP, the resistance to heat transfer (which is also called thermal resistance) of building structures of residential buildings must not be lower than the standard indicator. And these normalized indicators are set 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-applications 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 map-scheme, covering the entire territory of the Russian Federation.

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

Now let's take a look at what this thermal resistance consists 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 X equals:

Rх = hх / λх

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

hx- layer thickness, expressed in meters;

λх- 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 building materials used for the construction of walls, most often, even with their large (within reasonable, of course) thickness, do not reach the standard indicators of heat transfer resistance. In other words, the wall cannot be called fully thermally insulated. This is exactly what insulation is used for - an additional layer is created that “fills in the deficit” necessary to achieve normalized performance. And due to the fact that the coefficients of thermal conductivity of high-quality insulation materials are low, it is possible to avoid the need to build very thick structures.

You might be interested in knowing what is

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

1 - in fact, the wall itself, having a certain thickness and erected from one or another material. In most cases, “by default”, she herself is not able to provide normalized thermal resistance.

2 - a layer of insulating material, the coefficient of thermal conductivity and thickness of which should provide "shortage coverage" up to the normalized indicator R. Let's make a reservation right away - the location of the thermal insulation is shown on the outside, but it can also be placed on the inside of the wall, and even located between two layers of the supporting structure (for example , laid out of brick 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 boards are very capable of enhancing 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 an important note - the layer of facade decoration is never taken into account if there is a ventilated gap between it and the wall or insulation. And this is often practiced in ventilated facade systems. In this design, the exterior finish will not have any 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 insulation and finishing layers, 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 questions, the wall has full thermal insulation. If not enough, you can calculate which layer and which insulating material can fill this shortage.

You may be interested in information on how

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

Just a few explanations on how to work with it:

• To begin with, a normalized value of heat transfer resistance is found from the scheme map. 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 it - add the page to your bookmarks).

• The next group of fields specifies the thickness and material of the main supporting structure - walls. The thickness of the wall, if it is equipped according to the principle of "well masonry" with insulation inside, is indicated as a total.
• 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” to the exterior decoration of the wall - the material and thickness of the layer are also indicated. If there is no finish, or there is no need to take it into account, everything is left by default and move on.
• Do the same with the interior decoration of the wall.
• Finally, it remains only to choose the insulation material that is planned to be used for additional thermal insulation. The available options are listed in the dropdown 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 much 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 taken into account when calculating the thermal power of the boiler, it is still better to plan work to strengthen thermal insulation. What thickness of the additional layer is needed has already been 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 reducing 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 inefficient. There can be no two opinions here - the prospect of carrying out insulation work comes to the fore. And it will be much more profitable than purchasing a high-capacity boiler, some of which will simply be spent literally on “heating the street”. Naturally, accompanied by ruinous bills for wasted energy.

Reading time: 3 min

For heating residential and office premises, equipment with an electric water heater is used. To ensure the balance of temperature and energy consumption, an electric boiler is calculated. When determining the operating parameters, not only the area of ​​​​the rooms is taken into account, but also the physical properties of the materials of the walls, floor and ceiling of the room.

What is the power of an electric boiler

An electric boiler is a reservoir with a heat exchanger through which tap water or a special heat carrier with increased thermal characteristics is pumped.

The boiler is connected to a household alternating current network, it heats water with heating elements or electrodes isolated from water. The design of the equipment includes a temperature controller.

The power consumption depends on the degree of cooling of the coolant during circulation through the heating radiators in the building. Part of the energy is spent on heat losses in the boiler design (heating of the walls or protective casings of the heating elements). An information plate is installed on the outside of the equipment, which indicates the operating parameters of the product and the power consumption.

Methods for determining the power of an electric boiler

The calculation of the operating power of the heating boiler is carried out to ensure a balanced heating system that is able to maintain a comfortable temperature in the room under various external conditions.

The equipment must ensure uniform heating of the rooms, a change in the direction of the wind must not adversely affect the conditions in the premises. Before choosing equipment, the owner of the house needs to know how to calculate the power of the electric boiler, taking into account the characteristics of the room.

For the calculation, 2 main methods are used:

• by the area of ​​​​the house or rooms connected to the heating circuit and the boiler;
• by the volume of the premises.

An auxiliary method for determining the power along the hot water circuit is designed to calculate additional productivity. The resulting parameter is added to the pre-calculated value of energy consumption for home heating.

Then the ability of the electrical wiring connected to the building to withstand the maximum load during the operation of the heating elements of the boiler is checked.

Calculation of the boiler according to the area of ​​\u200b\u200bthe house

The basic technique is to determine the power of an electric heating boiler by the area of ​​\u200b\u200bthe premises. To determine the value, the base value of the power required to heat a room of 10 m² is used.

The coefficient does not depend on the climatic zone, it is roughly considered that it is necessary to spend 1 kW of power to warm up 10 m². The coefficient does not take into account the thermal conductivity of wall materials and the height of the room, therefore, additional correction factors determined empirically are used to refine the calculation.

For example, if the ceiling height is more than 2.7 m, an additional correction parameter is introduced, equal to the ratio of the actual height to the value of 2.7 m. The climate coefficient depends on the location of the house, the value is in the range from 0.7 for the southern regions to 2.0 - northern regions. If the heating unit will also be used for hot water supply, then a power reserve of 25-30% is added to the obtained indicator.

There is another way to calculate, based on the formula S*K*100, where the parameter S is the area of ​​the premises, and K is the heat loss coefficient, which varies depending on the minimum air temperature threshold. The base value is 0.7, used in areas with a minimum temperature of -10°C. With a decrease in the climatic norm for every 5 ° C, the coefficient increases by 0.2.

The method is not applicable when calculating the boiler for rooms with the following design features:

1. The presence of plastic or wooden windows with duplicated double-glazed windows.
2. Use of an additional heat-insulating layer with a thickness of 150 mm or more, located inside or outside the brick wall (thickness 2 brick sizes).
3. Preservation of an unheated attic space and the absence of heat-insulating material on the roof finish.
4. Increasing the height of living rooms to 2.7 m or more.

Calculation of boiler power by volume

The calculation of the power of an electric heating boiler by the volume of residential premises is based on the heat loss coefficient, which is:

1. From 0.6 to 0.9 - for brick buildings with improved thermal insulation. The house uses plastic 2-chamber windows, a roof made of heat-insulating material can be used.
2. From 1 to 1.9 - for buildings built of brick (double masonry), with a standard roof and wooden windows.
3. From 2 to 2.9 - for rooms with poor thermal insulation (for example, with walls 1 brick thick).
4. From 3 to 4 - for buildings built of wood or made of corrugated metal sheet with a layer of heat-insulating material.

When calculating, a formula of the form is used V*K*T/860, which takes into account the volume of the house V, the correction factor K and the difference in temperature inside the house and outside the room. For the calculation, the minimum air temperature characteristic of the location of the house is taken.

The resulting value is redundant, but in the event of prolonged frosts, it will be possible to maintain the temperature in the house within the specified parameters. The above method for calculating the power of an electric boiler for heating a house does not take into account the supply of additional warm liquid for washing dishes or a shower cabin.

For residential premises in panel or brick houses, the calculation is carried out according to SNiP standards. The rules lay down the necessary power to heat 1 m³ of air within 41 and 34 W (for a house made of panels and sand-lime bricks, respectively).

Then the owner of the premises measures the height and area, and a safety margin of 10% is added to the obtained value (in case the air temperature drops in winter). When installing energy-saving windows, it is allowed to install a boiler with a capacity less than the calculated one.

For corner rooms, the number of walls in contact with the street is taken into account. If only 1 wall goes to the outside of the house, then a coefficient of 1.1 is required. Each additional wall increases the value of the adjustment parameter by 0.1. To reduce heat losses, it is recommended to analyze the room with a special device, and then mount an insulator layer.

Calculation for DHW

The calculation of an electric boiler for heating a private house, simultaneously used for hot water supply, takes into account the following factors:

1. The amount and temperature of warm water needed to ensure the life of people living in the room.
2. Based on the first parameter, the volume of +90°C hot water is determined, which is then diluted with a cold liquid flow to obtain warm water.
3. Based on the value obtained, the electric boiler is calculated. When determining the parameters, the decrease in the temperature of tap water in winter is not taken into account.

For example, a residential building daily consumes 200 liters of warm water (Vg) heated to +40°C (Tg). It is supposed to obtain the required temperature by mixing hot and cold water. The owner plans to purchase a boiler that heats the liquid up to +95°C (Tc), the cold water supply line is supplied with water with a temperature of +10°C (Tx).

The volume of hot water is determined by the formula Vg*(Tg-Tx)/(Tk-Tx)=200*(40-10)/(95-10). The calculation shows that to ensure the supply of hot water per day, it is required to warm up 71 liters of liquid to a temperature of +95°C.

Further calculation is carried out on the basis of the coefficient of specific heat capacity of water (4.218 kJ for each kg when heated by 1 ° C), the weight of the liquid and the temperature difference. The resulting value is then converted into kilowatts according to the tables, it is recommended to round the parameter upwards.

For the situation described above, an additional power of about 5 kW is required. The obtained value implies heating the water in 1 hour, if the liquid is used evenly throughout the day, then it is allowed to reduce additional energy costs by 2 times.

One of the first parameters that people pay attention to when choosing heating equipment is performance. The calculation of the power of a gas heating boiler is performed in several ways. Comfort during operation depends on accurate calculations.

How to choose the power of a gas boiler

The calculation of the power of a gas heating boiler from the area is carried out in three different ways:

European manufacturers often calculate the performance of boiler equipment from the volume of the room. Therefore, in the technical documentation, the possibility of heating in m³ is indicated. This factor is taken into account when choosing a unit manufactured in the EU countries.

Most consultants who sell heating equipment independently calculate the required performance using the formula 1 kW = 10 m². Additional calculations are carried out according to the amount of coolant in the heating system.

Calculation of a single-circuit heating boiler

As noted above, independent calculations of the operating parameters of heating equipment are performed according to the formula 1 kW \u003d 10 m². To the result obtained, 15-20% of the reserve is added, due to which the heat generator, even in severe frosts, does not work at full load, which prolongs its service life.

• For 60 m² - the unit will be able to satisfy the need for heat on 6 kW + 20% = 7.5 kilowatts. If there is no model with a suitable performance size, preference is given to heating equipment with a large power value.
• In a similar way, calculations are made for 100 m² - the required power of boiler equipment, 12 kW.
• For heating 150 m², a gas boiler is needed, with a capacity of 15 kW + 20% (3 kilowatts) = 18 kW. Accordingly, for 200 m², a 22 kW boiler is required.

These calculations are only suitable for single-circuit models that are not connected to an indirect heating boiler.

How to calculate the power of a double-circuit boiler

The formula for calculating the required power of a double-circuit gas boiler in terms of the heating area and hot water draw-off points is as follows, 10 m² = 1 kW + 20% (power reserve) + 20% (for water heating). It turns out that 40% is added immediately to the calculated performance.

The power of a double-circuit gas boiler for heating and hot water heating for 250 m² will be 25 kW + 40% (10 kilowatts) = 35 kW. Calculations are suitable for two-circuit equipment. To calculate the performance of a single-circuit unit connected to an indirect heating boiler, a different formula is used.

Calculation of the power of an indirect heating boiler and a single-circuit boiler

To calculate the required power of a single-circuit gas boiler with an indirect heating boiler, you must perform the following steps:

• Determine what volume of the boiler will be sufficient to meet the needs of the residents of the house.
• In the technical documentation for the storage tank, the required capacity of the boiler equipment is indicated in order to maintain the heating of hot water, without taking into account the necessary heat for heating. A 200 liter boiler will require an average of about 30 kW.
• The performance of the boiler equipment required for heating the house is calculated.

The resulting numbers are added up. The amount equal to 20% is subtracted from the result. This must be done for the reason that the heating will not simultaneously work for heating and DHW. The calculation of the thermal power of a single-circuit heating boiler, taking into account an external water heater for hot water supply, is done taking into account this feature.

What power reserve should a gas boiler have

The performance margin is calculated depending on the configuration of the heating equipment:

• For single-circuit models, the margin is about 20%.
• For two-circuit units, 20% + 20%.
• Boilers with connection to an indirect heating boiler - in the storage tank configuration, the required additional performance margin is indicated.

The specified power reserve is valid for rooms up to 300 m². Houses with a larger area require competent heat engineering calculations.

Calculation of gas demand based on boiler power

The formula for calculating gas consumption, depending on the power of the boiler used, takes into account the efficiency of the heating equipment. For standard models of classical heating heat generators, the efficiency will be 92%, for condensing ones up to 108%.

In practice, this means that 1 m³ of gas is equal to 10 kW of thermal energy, assuming 100% heat transfer. Accordingly, with an efficiency of 92%, fuel costs will be 1.12 m³, and at 108% no more than 0.92 m³.

The method for calculating the volume of consumed gas takes into account the performance of the unit. So, a 10 kW heating device, within an hour, will burn 1.12 m³ of fuel, a 40 kW unit, 4.48 m³. This dependence of gas consumption on the power of boiler equipment is taken into account in complex heat engineering calculations.

The ratio is also built into the online heating costs. Manufacturers often indicate the average gas consumption for each model produced.

In order to fully calculate the approximate material costs of heating, it will be necessary to calculate the electricity consumption in volatile heating boilers. At the moment, boiler equipment operating on main gas is the most economical way of heating.

For heated buildings of a large area, calculations are carried out only after an audit of the heat loss of the building. In other cases, when calculating, they use special formulas or online services.

These are mobile boiler plants designed to provide both residential and industrial facilities with heat and hot water. All equipment is placed in one or more blocks, which are then joined together, resistant to fires and temperature changes. Before dwelling on this type of energy supply, it is necessary to correctly calculate the power of the boiler house.

Block-modular boiler houses are divided according to the type of fuel used and can be solid fuel, gas, liquid fuel and combined.

For a comfortable stay at home, in the office or at work during the cold season, you need to take care of a good and reliable heating system for the building or room. For the correct calculation of the thermal power of the boiler house, you need to pay attention to several factors and building parameters.

Buildings are designed in such a way as to minimize heat loss. But taking into account timely wear or technological violations during the construction process, the building may have vulnerabilities through which heat will escape. To take this parameter into account in the general calculation of the power of a modular boiler house, you must either get rid of heat losses or include them in the calculation.

To eliminate heat losses, it is necessary to conduct a special study, for example, using a thermal imager. It will show all the places through which heat flows, and in need of insulation or sealing. If it was decided not to eliminate heat losses, then when calculating the power of a modular-type boiler house, it is necessary to add 10 percent to the resulting power to cover heat losses. Also, when calculating, it is necessary to take into account the degree of insulation of the building and the number and size of windows and large gates. If there are large gates for the arrival of trucks, for example, about 30% of the power is added to cover heat losses.

Calculation by area

The easiest way to find out the required heat consumption is to calculate the power of the boiler house according to the area of ​​\u200b\u200bthe building. Over the years, specialists have already calculated standard constants for some indoor heat exchange parameters. So, on average, for heating 10 square meters, you need to spend 1 kW of thermal energy. These figures will be relevant for buildings built in compliance with heat loss technologies and with a ceiling height of no more than 2.7 m. Now, based on the total area of ​​the building, you can get the required capacity of the boiler house.

Volume calculation

More accurate than the previous method of calculating power is the calculation of the power of the boiler house by the volume of the building. Here you can immediately take into account the height of the ceilings. According to SNiPs, an average of 34 watts has to be spent on heating 1 cubic meter in a brick building. In our company, we use various formulas to calculate the required heat output, taking into account the degree of insulation of the building and its location, as well as the required temperature inside the building.

What else needs to be taken into account when calculating?

For a complete calculation of the power of a block model boiler house, it will be necessary to take into account several more important factors. One of them is hot water supply. To calculate it, it is necessary to take into account how much water will be consumed daily by all family members or production. Thus, knowing the amount of water consumed, the required temperature and taking into account the time of year, it is possible to calculate the correct power of the boiler house. It is generally customary to add about 20% to the resulting figure for heating water.

A very important parameter is the placement of the heated object. To use geographic data in the calculation, you need to refer to SNiPs, in which you can find a map of average temperatures for the summer and winter periods. Depending on the placement, you need to apply the appropriate coefficient. For example, for central Russia, the number 1 is relevant. But the northern part of the country already has a coefficient of 1.5-2. So, having received a certain figure during past studies, it is necessary to multiply the received power by a coefficient, as a result, the final power for the current region will become known.

Now, before calculating the power of the boiler house for a particular house, you need to collect as much data as possible. There is a house in the Syktyvkar region, built of brick, according to technology and all measures to avoid heat loss, with an area of ​​100 sq. m. and a ceiling height of 3 m. Thus, the total volume of the building will be 300 meters cubed. Since the house is brick, you need to multiply this figure by 34 watts. It turns out 10.2 kW.

Taking into account the northern region, frequent winds and a short summer, the received power must be multiplied by 2. Now it turns out that 20.4 kW must be spent for a comfortable stay or work. At the same time, it should be taken into account that some part of the power will be used to heat water, and this is at least 20%. But for a reserve, it is better to take 25% and multiply by the current required power. The result is a figure of 25.5. But for reliable and stable operation of the boiler plant, you still need to take a margin of 10 percent so that it does not have to work for wear and tear in a constant mode. The total is 28 kW.

In such a not cunning way, the power necessary for heating and heating water turned out, and now you can safely choose block-modular boiler houses, the power of which corresponds to the figure obtained in the calculations.