Calculation of thermal energy for heating. Independent calculation of the thermal load for heating: hourly and annual indicators Methodology for thermal calculation of a building

The question of calculating the amount of payment for heating is very important, since consumers often receive quite impressive amounts for this utility service, at the same time having no idea how the calculation was made.

Since 2012, when the Decree of the Government of the Russian Federation of May 6, 2011 No. 354 “On the provision of utility services to owners and users of premises in apartment buildings and residential buildings” came into force, the procedure for calculating the amount of heating fees has undergone a number of changes.

Calculation methods changed several times, heating provided for general house needs appeared, which was calculated separately from heating provided in residential premises (apartments) and non-residential premises, but then, in 2013, heating again began to be calculated as a single utility service without splitting the fee.

The calculation of the heating fee has changed since 2017, and in 2019 the calculation procedure changed again; new formulas for calculating the heating fee have appeared, which are not so easy for an ordinary consumer to understand.

So, let's sort it out in order.

In order to calculate the heating fee for your apartment and choose the necessary calculation formula, you must first know:

1. Does your house have a centralized heating system?

This means whether thermal energy for heating needs in your apartment building already in finished form using centralized systems or the thermal energy for your home is produced independently using the equipment included in the common property owners of premises in apartment building.

2. Is your apartment building equipped with a common building (collective) metering device and are there any individual devices metering thermal energy in residential and non-residential premises of your home?

The presence or absence of a common house (collective) metering device in the house and individual metering devices in the premises of your home significantly affects the method of calculating the amount of heating fees.

3. How are you charged for heating – during the heating period or evenly throughout the calendar year?

The payment method for heating utilities is accepted by the authorities state power subjects Russian Federation. That is, in different regions of our country, heating fees may be charged differently - throughout the year or only during the heating season, when the service is actually provided.

4. Are there any rooms in your house that do not have heating devices (radiators, radiators), or that have their own sources of thermal energy?

It was from 2019, in connection with court decisions, the trials of which took place in 2018, that the calculation began to include premises in which there are no heating devices (radiators, radiators), which is provided for technical documentation per home, or residential and non-residential premises, the reconstruction of which, providing for the installation of individual sources of thermal energy, was carried out in accordance with the requirements for reconstruction established by the legislation of the Russian Federation in force at the time of such reconstruction. Let us remind you that previously the methods for calculating the amount of heating fees did not provide for a separate calculation for such premises, so the charges were calculated on a general basis.

In order to make the information on calculating the heating fee more understandable, we will consider each method of charging separately, using one or another calculation formula using a specific example.

When choosing a calculation option, you must pay attention to all the components that determine the calculation methodology.

Below are various calculation options, taking into account individual factors that determine the choice of calculating the heating fee:

Calculation No. 1: Amount of heating fee in residential/non-residential premises during the heating season.

Calculation No. 2: Amount of heating fee in residential/non-residential premises, there is no administrative budget for an apartment building, the amount of the fee is calculated during the calendar year(12 months).
Read about the procedure and example of calculation →

Calculation No. 3: Amount of heating fee in residential/non-residential premises, an ODPU is installed on an apartment building, There are no individual metering devices in all residential/non-residential premises.

Method thermal calculation is the determination of the surface area of ​​each individual heating device, which releases heat into the room. The calculation of thermal energy for heating in this case takes into account the maximum temperature level of the coolant, which is intended for those heating elements for which the thermotechnical calculation of the heating system is carried out. That is, if the coolant is water, then its average temperature in the heating system is taken. In this case, the coolant consumption is taken into account. Likewise, if the coolant is steam, then the calculation of heat for heating uses the value of the highest temperature of the steam at a certain pressure level in the heating device.

Calculation method

To calculate heat energy for heating, it is necessary to take the heat demand indicators of a separate room. In this case, the heat transfer of the heat pipe located in this room should be subtracted from the data.

The surface area that gives off heat will depend on several factors - first of all, on the type of device used, on the principle of connecting it to the pipes and on how exactly it is located in the room. It should be noted that all these parameters also affect the heat flux density emanating from the device.

Calculation of heating devices of the heating system - the heat transfer of the heating device Q can be determined using the following formula:

Q pr = q pr* A p .

However, it can be used only if the indicator of the surface density of the thermal device q pr (W/m 2) is known.

From here we can also calculate calculated area A r. It is important to understand that the calculated area of ​​any heating device does not depend on the type of coolant.

A p = Q np /q np ,

in which Q np is the level of heat transfer of the device required for a certain room.

Thermal calculation of heating takes into account that to determine the heat transfer of the device for a specific room, the formula is used:

Q pp = Q p - µ tr *Q tr

in this case, the indicator Q p is the heat demand of the room, Q tr is the total heat transfer of all elements heating system located in the room. Calculation of the heat load for heating implies that this includes not only the radiator, but also the pipes that are connected to it, and the transit heat pipeline (if any). In this formula, µtr is the correction factor, which provides for partial heat transfer from the system, designed to maintain a constant temperature in the room. In this case, the size of the correction may vary depending on how exactly the pipes of the heating system were laid in the room. In particular - when open method– 0.9; in the groove of the wall - 0.5; embedded in concrete wall – 1,8.

Calculation of the required heating power, that is, the total heat transfer (Q tr - W) of all elements of the heating system is determined using the following formula:

Q tr = µk tr *µ*d n *l*(t g - t c)

In it, k tr is an indicator of the heat transfer coefficient of a certain section of a pipeline located indoors, d n is the outer diameter of the pipe, l is the length of the section. Indicators tg and tv show the temperature of the coolant and air in the room.

Formula Q tr = q in *l in + q g *l g used to determine the level of heat transfer of the heat pipe present in the room. To determine indicators, you should refer to special reference literature. In it you can find a definition of the thermal power of a heating system - a definition of heat transfer vertically (q in) and horizontally (q g) of a heat pipeline laid in the room. The data found shows the heat transfer of 1 m of pipe.

Before calculating Gcal for heating, for many years, calculations made using the formula A p = Q np /q np and measurements of heat-transferring surfaces of the heating system were carried out using a conventional unit - equivalent square meters. At the same time, the ECM was conditional equal to the surface heating device with a heat output of 435 kcal/h (506 W). The calculation of Gcal for heating assumes that the temperature difference between the coolant and air (t g - t in) in the room was 64.5 ° C, and the relative water flow in the system was equal to Grel = l.0.

Calculation of thermal loads for heating implies that smooth-tube and panel heating devices, which had greater heat output than standard radiators from the times of the USSR, had an ecm area that differed significantly from their physical area. Accordingly, the ecm area of ​​less efficient heating devices was significantly lower than their physical area.

However, such dual measurement of the area of ​​heating devices was simplified in 1984, and the ECM was abolished. Thus, from that moment on, the area of ​​the heating device was measured only in m2.

After the area of ​​the heating device required for the room has been calculated and the thermal power of the heating system has been calculated, you can begin to select the required radiator from the catalog of heating elements.

It turns out that most often the area of ​​the purchased element is slightly larger than that which was obtained by calculation. This is quite easy to explain - after all, such a correction is taken into account in advance by introducing a multiplying factor µ 1 into the formulas.

Very common today sectional radiators. Their length directly depends on the number of sections used. In order to calculate the amount of heat for heating - that is, calculate optimal quantity sections for a specific room, the formula is used:

N = (A p /a 1)(µ 4 / µ 3)

In it, a 1 is the area of ​​one section of the radiator selected for installation indoors. Measured in m2. µ 4 – correction factor that is applied to the method of installing the heating radiator. µ 3 – correction factor, which indicates the actual number of sections in the radiator (µ 3 - 1.0, provided that A p = 2.0 m 2). For standard radiators of the M-140 type, this parameter is determined by the formula:

µ 3 =0.97+0.06/A p

During thermal tests, standard radiators are used, consisting on average of 7-8 sections. That is, the calculation of heat consumption for heating determined by us - that is, the heat transfer coefficient - is realistic only for radiators of this particular size.

It should be noted that when using radiators with fewer sections, there is a slight increase in the level of heat transfer.

This is due to the fact that in the outer sections the heat flow is somewhat more active. In addition, the open ends of the radiator contribute to greater heat transfer into the room air. If the number of sections is greater, a weakening of the current is observed in the outer sections. Accordingly, to achieve the required level of heat transfer, the most rational option is to slightly increase the length of the radiator by adding sections, which will not affect the power of the heating system.

For those radiators whose area of ​​one section is 0.25 m 2, there is a formula for determining the coefficient µ 3:

µ 3 = 0.92 + 0.16 /A p

But it should be borne in mind that it is extremely rare when using this formula that an integer number of sections is obtained. Most often, the required quantity turns out to be fractional. Calculation heating devices heating system suggests that in order to obtain a more accurate result, a slight (no more than 5%) reduction in the coefficient A p is permissible. This action leads to limiting the level of deviation temperature indicator in room. When the heat for heating the room is calculated, after receiving the result, a radiator is installed with the number of sections as close as possible to the obtained value.

Calculation of heating power by area assumes that the architecture of the house also imposes certain conditions on the installation of radiators.

In particular, if there is an external niche under the window, then the length of the radiator should be less than the length of the niche - no less than 0.4 m. This condition is valid only when the pipe is connected directly to the radiator. If a duck liner is used, the difference in the length of the niche and the radiator should be at least 0.6 m. In this case, the extra sections should be separated as a separate radiator.

For certain models of radiators, the formula for calculating heat for heating - that is, determining the length - is not applied, since this parameter is pre-determined by the manufacturer. This fully applies to radiators such as RSV or RSG. However, there are often cases when, to increase the area of ​​a heating device of this type, simply parallel installation of two panels next to each other is used.

If a panel radiator is determined to be the only one acceptable for a given room, then to determine the number of radiators required, use:

N = A p / a 1 .

In this case, the radiator area is a known parameter. If two parallel radiator blocks are installed, the A p indicator is increased, determining the reduced heat transfer coefficient.

In the case of using convectors with a casing, the calculation of heating power takes into account that their length is also determined exclusively by the existing model range. In particular, the floor convector “Rhythm” is presented in two models with casing lengths of 1 m and 1.5 m. Wall convectors may also differ slightly from each other.

In the case of using a convector without a casing, there is a formula that helps determine the number of elements of the device, after which you can calculate the power of the heating system:

N = A p / (n*a 1)

Here n is the number of rows and tiers of elements, which make up the area of ​​the convector. In this case, a 1 is the area of ​​one pipe or element. In this case, when determining the estimated area of ​​the convector, it is necessary to take into account not only the number of its elements, but also the method of their connection.

If a smooth-tube device is used in a heating system, the duration of its heating pipe is calculated as follows:

l = А р *µ 4 / (n*a 1)

µ 4 is the correction factor that is introduced if there is decorative shelter pipes; n – number of rows or tiers of heating pipes; and 1 is a parameter characterizing the area of ​​one meter of horizontal pipe with a predetermined diameter.

To obtain a more accurate (and not fractional) number, a slight (no more than 0.1 m2 or 5%) reduction in indicator A is allowed.

Example No. 1

It is necessary to determine the correct number of sections for the M140-A radiator, which will be installed in a room located on top floor. In this case, the wall is external, there is no niche under the window sill. And the distance from it to the radiator is only 4 cm. The height of the room is 2.7 m. Q n = 1410 W, and t = 18 ° C. Conditions for connecting the radiator: connection to a single-pipe riser of a flow-regulated type (D y 20, KRT tap with 0.4 m inlet); The heating system is routed from the top, t = 105°C, and the coolant flow through the riser is G st = 300 kg/h. The temperature difference between the coolant in the supply riser and the one in question is 2°C.

We define average radiator temperature:

t av = (105 - 2) - 0.5x1410x1.06x1.02x3.6 / (4.187x300) = 100.8 °C.

Based on the data obtained, we calculate the heat flux density:

t av = 100.8 - 18 = 82.8 °C

It should be noted that what happened minor change water consumption level (360 to 300 kg/h). This parameter has virtually no effect on q np.

Q pr =650(82.8/70)1+0.3=809W/m2.

Next, we determine the level of heat transfer horizontally (1g = 0.8 m) and vertically (1v = 2.7 - 0.5 = 2.2 m) located pipes. To do this you should use the formula Q tr =q in xl in + q g xl g.

We get:

Q tr = 93x2.2 + 115x0.8 = 296 W.

We calculate the area of ​​the required radiator using the formula A p = Q np /q np and Q pp = Q p - µ tr xQ tr:

A p = (1410-0.9x296)/809 = 1.41 m 2.

We calculate the required number of sections of the M140-A radiator, taking into account that the area of ​​one section is 0.254 m2:

m 2 (µ4 = 1.05, µ 3 = 0.97 + 0.06 / 1.41 = 1.01, we use the formula µ 3 = 0.97 + 0.06 / A r and determine:

N=(1.41/0.254)x(1.05/1.01)=5.8.
That is, the calculation of heat consumption for heating showed that in order to achieve the maximum comfortable temperature a radiator consisting of 6 sections should be installed.

Example No. 2

It is necessary to determine the brand of an open wall convector with a casing KN-20k “Universal-20”, which is installed on a single-pipe flow-through riser. There is no tap near the installed device.

Determines the average water temperature in the convector:

tcp = (105 - 2) - 0.5x1410x1.04x1.02x3.6 / (4.187x300) = 100.9 °C.

In the Universal-20 convectors, the heat flux density is 357 W/m2. Available data: µt cp ​​= 100.9-18 = 82.9 ° C, Gnp = 300 kg/h. Using the formula q pr =q nom (µ t av /70) 1+n (G pr /360) p we recalculate the data:

q np = 357(82.9 / 70)1+0.3(300 / 360)0.07 = 439 W/m2.

We determine the level of heat transfer of horizontal (1 g - = 0.8 m) and vertical (l in = 2.7 m) pipes (taking into account D y 20) using the formula Q tr = q in xl in +q g xl g. We obtain:

Q tr = 93x2.7 + 115x0.8 = 343 W.

Using the formula A p = Q np /q np and Q pp = Q p - µ tr xQ tr, we determine the estimated area of ​​the convector:

A p = (1410 - 0.9x343) / 439 = 2.51 m 2.

That is, the “Universal-20” convector, the casing length of which is 0.845 m, was accepted for installation (model KN 230-0.918, the area of ​​which is 2.57 m2).

Example No. 3

For a steam heating system, it is necessary to determine the number and length of cast iron finned pipes, provided that the installation is open and made in two tiers. In this case, the excess steam pressure is 0.02 MPa.

Additional characteristics: t on = 104.25 °C, t on = 15 °C, Q p = 6500 W, Q tr = 350 W.

Using the formula µ t n = t us - t v, we determine the temperature difference:

µ t n = 104.25-15 = 89.25 °C.

We determine the heat flux density using the known transmission coefficient of this type of pipe in the case when they are installed in parallel one above the other - k = 5.8 W/(m2-°C). We get:

q np = k np x µ t n = 5.8-89.25 = 518 W/m2.

The formula A p = Q np /q np helps determine required area device:

A p = (6500 - 0.9x350) / 518 = 11.9 m 2.

To determine the quantity necessary pipes, N = A p / (nхa 1). In this case, you should use the following data: the length of one tube is 1.5 m, the heating surface area is 3 m 2.

We calculate: N= 11.9/(2x3.0) = 2 pcs.

That is, in each tier it is necessary to install two pipes, each 1.5 m long. In this case, we calculate the total area of ​​​​this heating device: A = 3.0x*2x2 = 12.0 m 2.

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

But that's not all. It is not enough to correctly purchase the necessary heating boiler - it is very important to optimally select and correctly arrange heat exchange devices in the premises - radiators, convectors or “warm floors”. And again, relying only on your intuition or the “good advice” of your neighbors is not the most reasonable option. In a word, it’s impossible to do without certain calculations.

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

The simplest calculation methods

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

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

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

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

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

The most important “enemy” of the heating system is heat loss through building structures

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

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

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

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

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

Q = S× 100

Q– required heating power for the room;

S– room area (m²);

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

For example, a room 3.2 × 5.5 m

S= 3.2 × 5.5 = 17.6 m²

Q= 17.6 × 100 = 1760 W ≈ 1.8 kW

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

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

Q = S × h× 41 (or 34)

h– ceiling height (m);

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

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

Q= 17.6 × 3.2 × 41 = 2309 W ≈ 2.3 kW

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

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

You may be interested in information about what they are

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

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

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

General principles and calculation formula

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

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

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

• “a” is a coefficient that takes into account the number of external walls in a particular room.

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

The coefficient is taken equal to:

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

- external wall one: a = 1.0;

— external walls two: a = 1.2;

— external walls three: a = 1.4.

• “b” is a coefficient that takes into account the location of the external walls of the room relative to the cardinal directions.

You might be interested in information about what types of

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

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

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

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

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

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

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

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

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

- windward side of the house: c = 1.2;

- leeward walls of the house: c = 1.0;

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

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

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

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

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

— from – 35 °C and below: d = 1.5;

— from – 30 °С to – 34 °С: d = 1.3;

— from – 25 °С to – 29 °С: d = 1.2;

— from – 20 °С to – 24 °С: d = 1.1;

— from – 15 °С to – 19 °С: d = 1.0;

— from – 10 °С to – 14 °С: d = 0.9;

- no colder - 10 °C: d = 0.7.

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

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

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

— external walls do not have insulation: e = 1.27;

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

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

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

• coefficient "f" - correction for ceiling heights

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

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

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

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

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

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

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

• « g" is a coefficient that takes into account the type of floor or room located under the ceiling.

As shown above, the floor is one of the significant sources of heat loss. This means that it is necessary to make some adjustments to account for this feature of a particular room. The correction factor “g” can be taken equal to:

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

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

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

• « h" is a coefficient that takes into account the type of room located above.

The air heated by the heating system always rises, and if the ceiling in the room is cold, then increased heat loss is inevitable, which will require an increase in the required thermal power. Let us introduce the coefficient “h”, which takes into account this feature of the calculated room:

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

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

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

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

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

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

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

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

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

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

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

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

Whatever quality windows No matter how they were, it will still not be possible to completely avoid heat loss through them. But it’s quite clear that you can’t compare a small window with panoramic glazing covering almost the entire wall.

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

x = ∑SOK /SP

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

SP– area of ​​the room.

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

— x = 0 ÷ 0.1 →j = 0,8 ;

— x = 0.11 ÷ 0.2 →j = 0,9 ;

— x = 0.21 ÷ 0.3 →j = 1,0 ;

— x = 0.31 ÷ 0.4 →j = 1,1 ;

— x = 0.41 ÷ 0.5 →j = 1,2 ;

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

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

Door to the street or open balcony is capable of making adjustments to the thermal balance of the room - each opening of it is accompanied by the penetration of a considerable volume of cold air into the room. Therefore, it makes sense to take into account its presence - for this we introduce the coefficient “k”, which we take equal to:

- no door: k = 1,0 ;

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

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

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

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

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

• « m" - correction factor for the peculiarities of the installation location of heating radiators

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

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

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

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

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

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

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

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

Let's create a table something like this:

Then, using the calculator below, we make calculations for each room (already taking into account the 10% reserve). It won't take much time using the recommended app. After this, all that remains is to sum up the obtained values ​​for each room - this will be the required total power of the heating system.

The result for each room, by the way, will help you choose the right number of heating radiators - all that remains is to divide by the specific thermal power of one section and round up.

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

Why is thermal calculation necessary?

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

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

Initial data for thermal calculation of the heating system

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

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

Calculation of heating system power based on housing area

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

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

Step 2. The resulting figure is multiplied by 100-150 - that’s exactly how many watts from total power A heating system is needed for every m2 of housing.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Thermal load refers to the amount of thermal energy required to maintain a comfortable temperature in a house, apartment or separate room. The maximum hourly heating load refers to the amount of heat required to maintain normal values ​​for an hour under the most unfavorable conditions.

Factors affecting thermal load

• Wall material and thickness. For example, a 25-centimeter brick wall and a 15-centimeter aerated concrete wall can transmit different amounts of heat.
• Roof material and structure. For example, heat loss flat roof from reinforced concrete slabs differ significantly from the heat loss of an insulated attic.
• Ventilation. The loss of thermal energy with exhaust air depends on the performance of the ventilation system and the presence or absence of a heat recovery system.
• Glazing area. Windows lose more thermal energy compared to solid walls.
• Insolation level in different regions. Determined by the degree of absorption solar heat external coverings and orientation of building planes in relation to the cardinal directions.
• Temperature difference between the street and the room. It is determined by the heat flow through the enclosing structures under the condition of constant resistance to heat transfer.

Heat load distribution

For water heating, the maximum thermal power of the boiler should be equal to the sum of the thermal power of all heating devices in the house. For distribution of heating devices the following factors influence:

• Living rooms in the middle of the house - 20 degrees;
• Corner and end living rooms - 22 degrees. Moreover, due to more high temperature the walls do not freeze;
• Kitchen - 18 degrees, since it has its own heat sources - gas or electric stoves etc.
• Bathroom - 25 degrees.

At air heating the heat flow that enters a separate room depends on bandwidth air sleeve. Often the simplest way to adjust it is to adjust the position of the ventilation grilles with temperature control manually.

In a heating system that uses a distribution heat source (convectors, heated floors, electric heaters, etc.), the required temperature mode is set on the thermostat.

Calculation methods

To determine the thermal load, there are several methods that have of varying complexity calculations and reliability of the results obtained. The following are the three simplest methods for calculating the thermal load.

Method No. 1

According to current SNiP, there is a simple method for calculating the thermal load. On 10 square meters take 1 kilowatt of thermal power. Then the obtained data is multiplied by the regional coefficient:

• Southern regions have a coefficient of 0.7-0.9;
• For moderately cold climates (Moscow and Leningrad region) coefficient is 1.2-1.3;
• Far East and regions of the Far North: for Novosibirsk from 1.5; for Oymyakon up to 2.0.

Example calculation:

1. The area of ​​the building (10*10) is 100 square meters.
2. The basic thermal load indicator is 100/10=10 kilowatts.
3. This value is multiplied by a regional coefficient of 1.3, resulting in 13 kW of thermal power, which is required to maintain a comfortable temperature in the house.

Note! If you use this technique to determine the thermal load, you must also take into account a power reserve of 20 percent to compensate for errors and extreme cold.

Method No. 2

The first method of determining the thermal load has many errors:

• Different buildings have different heights ceilings. Considering that it is not the area that is heated, but the volume, this parameter is very important.
• More heat passes through doors and windows than through walls.
• Can't compare city ​​apartment with a private house, where below, above and outside the walls there are not apartments, but the street.

Method adjustment:

• The basic heat load indicator is 40 watts per 1 cubic meter of room volume.
• Each door leading to the street adds 200 watts to the base thermal load, each window 100 watts.
• Corner and end apartments of an apartment building have a coefficient of 1.2-1.3, which is affected by the thickness and material of the walls. A private house has a coefficient of 1.5.
• Regional coefficients are equal: for the Central regions and the European part of Russia - 0.1-0.15; for the Northern regions – 0.15-0.2; For Southern regions– 0.07-0.09 kW/sq.m.

Example calculation:

Method No. 3

Do not delude yourself - the second method of calculating the heat load is also very imperfect. It very roughly takes into account the thermal resistance of the ceiling and walls; temperature difference between outside air and indoor air.

It is worth noting that in order to maintain a constant temperature inside the house, an amount of thermal energy is required that will be equal to all losses through ventilation system and fencing devices. However, in this method, the calculations are simplified, since it is impossible to systematize and measure all factors.

On heat loss wall material influences– 20-30 percent heat loss. 30-40 percent goes through ventilation, through the roof - 10-25 percent, through windows - 15-25 percent, through the floor on the ground - 3-6 percent.

To simplify heat load calculations, heat loss through the enclosure is calculated and then this value is simply multiplied by 1.4. Temperature delta is easy to measure, but take data about thermal resistance only possible in reference books. Below are some popular ones thermal resistance values:

• The thermal resistance of a wall of three bricks is 0.592 m2*C/W.
• A wall of 2.5 bricks is 0.502.
• A wall of 2 bricks is equal to 0.405.
• A wall of one brick (25 cm thick) is equal to 0.187.
• Log house, where the diameter of the log is 25 cm - 0.550.
• A log house where the diameter of the log is 20 centimeters is 0.440.
• A log house where the thickness of the log house is 20 cm is 0.806.
• A log house where the thickness is 10 cm is 0.353.
• A frame wall, 20 cm thick, insulated with mineral wool - 0.703.
• Walls made of aerated concrete, the thickness of which is 20 cm - 0.476.
• Walls made of aerated concrete, the thickness of which is 30 cm - 0.709.
• Plasters with a thickness of 3 cm - 0.035.
• Ceiling or attic floor – 1,43.
• Wooden floor - 1.85.
• Double wooden door – 0.21.

Calculation according to the example:

Conclusion

As can be seen from the calculations, methods for determining the thermal load have significant errors. Fortunately, excess indicator will not harm the boiler power:

• Job gas boiler at reduced power it is carried out without a drop in the coefficient useful action, what about work condensing devices at partial load it is carried out in economy mode.
• The same applies to solar boilers.
• The efficiency of electric heating equipment is 100 percent.

Note! Operating solid fuel boilers at a power less than the rated power value is contraindicated.

Calculation of the heat load for heating is an important factor, the calculations of which must be performed before starting to create a heating system. If you approach the process wisely and carry out all work competently, trouble-free operation of the heating is guaranteed, and you also save significant money on extra costs.