Calculate hot (cold) water in a special order.
Recalculate the amount of payment for utility services in proportion to the number of days of temporary absence of the consumer. In this case, take into account the number of days of absence from the place of residence, excluding the day of departure from the place of residence and the day of arrival.
Recalculate only if there is consumer statement about this. A person can submit an application:
- before the start of the period of temporary absence;
- within 30 days after the end of the period of temporary absence.
He must attach documents to the application confirming the fact of absence from his place of residence. These could be, for example:
- a copy of a travel certificate or a business trip certificate certified at the place of work;
- certificate of being treated in a hospital facility;
- travel tickets issued in the name of the consumer (copies thereof);
- invoices for accommodation in a hotel, hostel or other place of temporary stay or copies thereof;
- certificate of registration at the place of residence;
- a certificate from the organization providing security for the residential premises in which the consumer was temporarily absent;
- a certificate from a consular or diplomatic office confirming the citizen’s temporary stay outside Russia;
- a copy of the passport with border crossing marks;
- a certificate from a dacha, gardening, vegetable gardening partnership, confirming the period of temporary stay of a citizen at the location of the dacha, gardening, vegetable gardening partnership;
- other documents confirming the consumer’s temporary absence.
Utility payments for general house needs do not need to be recalculated.
This procedure is established in paragraphs 86-93 of the Rules, approved by Decree of the Government of the Russian Federation of May 6, 2011 No. 354.
How to calculate the volume of transferred thermal energy
There are three options for calculating the volume of transferred thermal energy:
Calculation according to the new rules
Calculation of the volume of transferred thermal energy according to the new rules (option 1) assumes that payment for heating apartments in an apartment building is made only during the heating season. Accordingly, the volume calculation itself must also be made only during the heating season. The procedure for calculating these indicators differs depending on whether an individual (apartment) metering device is installed in the premises (and in the house - a collective common house metering device) or not (clauses 41-44 of the Rules approved by Decree of the Government of the Russian Federation of May 6, 2011 No. 354).
Calculation according to the new rules in the presence of meters
If there are meters, use following rules calculating the volume of thermal energy.
Determine the volume of transferred thermal energy for individual needs based on individual or general readings apartment meters.
Take meter readings at least once every six months. At the same time, residents can take monthly meter readings themselves and transfer them to the management company (HOA, TSN). Check information from residents at least once every six months. Otherwise, the procedure and conditions for receiving meter readings must be fixed in the management agreement for an apartment building.
This is stated in subparagraph “h” of paragraph 19, subparagraphs “g” and “e(1)” of paragraph 31 and subparagraph “k(1)” of paragraph 33 of the Rules, approved by Decree of the Government of the Russian Federation of May 6, 2011 No. 354.
If the tenant does not submit meter readings, the volume of thermal energy for the month will be:
- average monthly consumption - the first six months of non-submission of data;
- consumption by consumption standards - further (the seventh and subsequent months of non-submission of data).
This is stated in paragraph 59, paragraph 2 of paragraph 60 of the Rules, approved by Decree of the Government of the Russian Federation of May 6, 2011 No. 354.
If a resident's individual meter fails, determine the amount of thermal energy consumed as:
Calculate the average monthly consumption based on the readings of a specific meter for the heating period. And if the meter has been in operation for less than six months - for the actual period of its operation, but not less than three months heating season. Number of months heating season per year is determined by regional regulations.
If the tenant does not allow you to check the condition and readings of the meters more than twice, draw up an act of refusal of admission and calculate the consumption according toconsumption standards taking into account increasing factors.
Increasing coefficients for heating consumption standards in residential premises are:
Increasing coefficients are not applied if the tenant does not have technical feasibility meter installation. The lack of technical ability to install meters is confirmed by an act in the form approved by order of the Ministry of Regional Development of Russia dated December 29, 2011 No. 627.
This procedure is provided for in paragraphs 59, 60, 60.2 and 81 of the Rules, approved by Decree of the Government of the Russian Federation of May 6, 2011 No. 354, paragraph 3.1 of the appendix to the Rules, approved by Decree of the Government of the Russian Federation of May 23, 2006 No. 306.
Determine the volume of thermal energy transferred for common house needs using data from collective (common house) meters. Take readings from the collective meter from the 23rd to the 25th of the current month. Enter the obtained data into a special journal. This is stated in subparagraph “e” of paragraph 31 of the Rules, approved by Decree of the Government of the Russian Federation of May 6, 2011 No. 354.
The volume of thermal energy transferred for general house needs consists of several components. Including from the volume caused by overexpenditure (or shortfall) inside apartments that do not have individual metering devices, the calculation of which occurs according to standards, and not according to individual devices accounting. Due to the presence of this component, the volume of thermal energy transferred for general house needs can be not only positive, but also negative (if in apartments not equipped with metering devices, the actual consumption is less than the standards).
If a common house meter fails, determine the volume of thermal energy as:
- average monthly consumption - the first three months of meter failure;
- consumption by consumption standards taking into account increasing factors - further (the fourth and subsequent months of meter breakdown).
Increasing coefficients for heating consumption standards for general house needs are:
Increasing factors are not applied if it is not technically possible to install a counter. The lack of technical ability to install meters is confirmed by an act in the form approved by order of the Ministry of Regional Development of Russia dated December 29, 2011 No. 627.
This procedure follows from clauses 44, 59.1, 60.1 and 81 of the Rules, approved by Decree of the Government of the Russian Federation of May 6, 2011 No. 354, clause 3.1 of the appendix to the Rules, approved by Decree of the Government of the Russian Federation of May 23, 2006 No. 306.
If there is a positive difference between the readings of the common house and individual meters, in order to determine the amount of payment for utilities, it is necessary to calculate the amount of thermal energy transferred for common house needs and pertaining to a specific premises. In this case, you can distribute between all premises amounts not exceeding standard indicators. Excess amounts can be distributed among consumers only if a decision has been made to do so general meeting owners. Otherwise the specified difference management company(HOA, TSN) must be covered from its own funds (clause 44 of the Rules, approved by Decree of the Government of the Russian Federation of May 6, 2011 No. 354).
If the volume of thermal energy according to the general building meter turned out to be less than what the residents consumed according to the readings of individual meters and consumption according to standards, the distribution must be carried out in proportion to the size of the total area of each residential premises (apartment). That is, it only needs to be distributed between residential premises.
If the amount for reduction obtained as a result of the calculation is more than what a particular subscriber consumed, then reduce it only to 0, without transferring the balance to past or future periods.
This conclusion follows from paragraph 47 of the Rules, approved by Decree of the Government of the Russian Federation of May 6, 2011 No. 354.
An example of calculating the volume of transferred thermal energy. An apartment building has a common building meter; there are no individual (apartment) meters
The Alpha company manages apartment building. The house has a common house heat meter. The total area of all premises in the house (including those related to common property) is 4900.6 sq. m. m. The total area of all residential and non-residential premises in the house - 2710.8 sq. m.
In February, the consumption of 25 Gcal was recorded according to the general house meter.
The volume of transferred thermal energy in relation to a 1-room apartment, not equipped with a meter, with an area of 42 sq. m is:
25 Gcal × 42 sq. m: 2710.8 sq. m = 0.38733 Gcal.
Calculation according to new rules in the absence of meters
If individual and communal meters are not installed, calculate the volume of transferred thermal energy according to the standards (clause 42(1) of the Rules approved by Decree of the Government of the Russian Federation of May 6, 2011 No. 354). The standards are set by regional authorities (clause 5 of the Decree of the Government of the Russian Federation of May 6, 2011 No. 354).
For more details on the calculation procedure according to the standards, seetable .
Calculation using the payment frequency coefficient
Calculation using the payment frequency coefficient (option 2) can be used only if there are no communal and individual (apartment) metering devices.
Calculate the volume of transferred thermal energy according to the standards, and charge the fee monthly.
Calculate the payment frequency coefficient using the formula:
This is stated in subparagraph “a” of paragraph 1 of Decree of the Government of the Russian Federation of August 27, 2012 No. 857 and paragraphs 1 and 2 of the Rules approved by Decree of the Government of the Russian Federation of August 27, 2012 No. 857.
The number of months of the heating season per year is determined by regional regulations.
For more details on calculating the volume of transferred thermal energy using the payment frequency coefficient, seetable .
Calculation according to the old rules
Calculation according to the old rules (option 3) involves charging for heating in all months of the year (subparagraph “b”, paragraph 1 of the Decree of the Government of the Russian Federation of August 27, 2012 No. 857). It can be applied if there is a decision of regional authorities on this (see, for example, order of the Ministry of Housing and Communal Services of the Moscow Region dated September 13, 2012 No. 33).
This calculation option can only be used until it is canceled by the regional authorities, but it definitely loses force on July 1, 2016 (clause 6 of the Decree of the Government of the Russian Federation of May 6, 2011 No. 354, subparagraph “b”, clause 2 of the Decree of the Government of the Russian Federation dated December 17, 2014 No. 1380).
For more information on calculating the volume of transferred thermal energy according to the old rules, seetable .
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 unnecessarily 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 an optimal level of air temperature throughout the entire volume of the heated room. Of course, the temperature level may vary somewhat with altitude, but this difference should not be significant. An average of +20 °C is considered quite comfortable conditions - this is the temperature that is usually taken as the initial one in thermal calculations.
In other words, the heating system must be able to warm up a certain volume of air.
If we approach it with complete accuracy, then for individual rooms in residential buildings standards for the required microclimate have been established - they are defined by GOST 30494-96. An excerpt from this document is in the table below:
| Purpose of the room | Air temperature, °C | Relative humidity, % | Air speed, m/s | |||
|---|---|---|---|---|---|---|
| optimal | acceptable | optimal | permissible, max | optimal, max | permissible, max | |
| For the cold season | ||||||
| Living room | 20÷22 | 18÷24 (20÷24) | 45÷30 | 60 | 0.15 | 0.2 |
| The same, but for living rooms in regions with minimum temperatures of - 31 °C and below | 21÷23 | 20÷24 (22÷24) | 45÷30 | 60 | 0.15 | 0.2 |
| Kitchen | 19÷21 | 18÷26 | N/N | N/N | 0.15 | 0.2 |
| Toilet | 19÷21 | 18÷26 | N/N | N/N | 0.15 | 0.2 |
| Bathroom, combined toilet | 24÷26 | 18÷26 | N/N | N/N | 0.15 | 0.2 |
| Facilities for recreation and study sessions | 20÷22 | 18÷24 | 45÷30 | 60 | 0.15 | 0.2 |
| Inter-apartment corridor | 18÷20 | 16÷22 | 45÷30 | 60 | N/N | N/N |
| Lobby, staircase | 16÷18 | 14÷20 | N/N | N/N | N/N | N/N |
| Storerooms | 16÷18 | 12÷22 | N/N | N/N | N/N | N/N |
| For the warm season (Standard only for residential premises. For others - not standardized) | ||||||
| Living room | 22÷25 | 20÷28 | 60÷30 | 65 | 0.2 | 0.3 |
- 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:
| Building design element | Approximate value of heat loss |
|---|---|
| Foundation, floors on the ground or above unheated basement (basement) rooms | from 5 to 10% |
| “Cold bridges” through poorly insulated joints building structures | from 5 to 10% |
| Input locations engineering communications(sewage, water supply, gas pipes, electrical cables, etc.) | up to 5% |
| External walls, depending on the degree of insulation | from 20 to 30% |
| Poor quality windows and external doors | about 20÷25%, of which about 10% - through unsealed joints between the boxes and the wall, and due to ventilation |
| Roof | up to 20% |
| Ventilation and chimney | up to 25 ÷30% |
Naturally, in order to cope with such tasks, the heating system must have a certain thermal power, and this potential must not only meet the general needs of the building (apartment), but also be correctly distributed among the rooms, in accordance with their area and a number of other important factors.
Usually the calculation is carried out in the direction “from small to large”. Simply put, the required amount of thermal energy for each heated room is calculated, the obtained values are summed up, approximately 10% of the reserve is added (so that the equipment does not work at the limit of its capabilities) - and the result will show how much power the heating boiler is needed. And the values for each room will become starting point to calculate the required number of radiators.
The most simplified and most frequently used method in a non-professional environment is to adopt a norm of 100 W of thermal energy per square meter of area:
The most primitive way of calculating is the ratio of 100 W/m²
Q = S× 100
Q– required heating power for the room;
S– room area (m²);
100 — power density per unit area (W/m²).
For example, a room 3.2 × 5.5 m
S= 3.2 × 5.5 = 17.6 m²
Q= 17.6 × 100 = 1760 W ≈ 1.8 kW
The method is obviously very simple, but very imperfect. It is worth mentioning right away that it is conditionally applicable only when standard height ceilings - approximately 2.7 m (acceptable - in the range from 2.5 to 3.0 m). From this point of view, the calculation will be more accurate not from the area, but from the volume of the room.
It is clear that in this case the power density is calculated at cubic meter. It is taken equal to 41 W/m³ for reinforced concrete panel house, or 34 W/m³ - in brick or made of other materials.
Q = S × h× 41 (or 34)
h– ceiling height (m);
41 or 34 – specific power per unit volume (W/m³).
For example, the same room in 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 that are 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 have no relation to any quantities standardly accepted in physics. The meaning of each coefficient will be discussed separately.
- “a” is a coefficient that takes into account the number of external walls in a particular room.
Obviously, the more external walls there are in a room, the larger area, through which it occurs heat losses. In addition, the presence of two or more external walls also means corners - extremely vulnerable places from the point of view of the formation of “cold bridges”. Coefficient “a” will correct for this specific feature rooms.
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 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 qualitatively, based on thermal 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, may have different heights. Therefore, the thermal power to warm up a particular room of the same area will also differ in this parameter.
It would not be a big mistake to accept the following values for the correction factor “f”:
— ceiling heights up to 2.7 m: f = 1.0;
— flow height from 2.8 to 3.0 m: f = 1.05;
- ceiling heights from 3.1 to 3.5 m: f = 1.1;
— ceiling heights from 3.6 to 4.0 m: f = 1.15;
- ceiling height more than 4.1 m: f = 1.2.
- « g" is a coefficient that takes into account the type of floor or room located under the ceiling.
As shown above, the floor is one of the significant sources of heat loss. This means that it is necessary to make some adjustments to account for this feature of a particular room. The correction factor “g” can be taken equal to:
- cold floor on the ground or above unheated room(for example, basement or basement): g= 1,4 ;
- insulated floor on the ground or above an unheated room: g= 1,2 ;
— the heated room is located below: g= 1,0 .
- « h" is a coefficient that takes into account the type of room located above.
The air heated by the heating system always rises, and if the ceiling in the room is cold, then increased heat loss is inevitable, which will require an increase in the required thermal power. Let us introduce the coefficient “h”, which takes into account this feature of the calculated room:
— the “cold” attic is located on top: h = 1,0 ;
— there is an insulated attic or other insulated room on top: h = 0,9 ;
— any heated room is located on top: h = 0,8 .
- « i" - coefficient taking into account the design features of windows
Windows are one of the “main routes” for heat flow. Naturally, much in this matter depends on the quality of the window design. Old wooden frames, which were previously universally installed in all houses, are significantly inferior in terms of their thermal insulation to modern multi-chamber systems with double-glazed windows.
Without words it is clear that the thermal insulation qualities of these windows differ significantly
But there is no complete uniformity between PVH windows. For example, 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 glass: i = 1,0 ;
— modern window systems with two-chamber or three-chamber double-glazed windows, including those with argon filling: i = 0,85 .
- « j" - correction factor for 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 is quite clear that one cannot compare a small window with panoramic glazing almost the entire wall.
First you need to find the ratio of the areas of all the windows in the room and the room itself:
x = ∑SOK /Sn
∑ SOK– total area of windows in the room;
Sn– area of the room.
Depending on the obtained value, the correction factor “j” is determined:
— x = 0 ÷ 0.1 →j = 0,8 ;
— x = 0.11 ÷ 0.2 →j = 0,9 ;
— x = 0.21 ÷ 0.3 →j = 1,0 ;
— x = 0.31 ÷ 0.4 →j = 1,1 ;
— x = 0.41 ÷ 0.5 →j = 1,2 ;
- « k" - coefficient that corrects for the presence of an entrance door
A door to the street or to an unheated balcony is always an additional “loophole” for the cold
Door to the street or open balcony is capable of making adjustments to the thermal balance of the room - each opening of it is accompanied by the penetration of a considerable volume of cold air into the room. Therefore, it makes sense to take into account its presence - for this we introduce the coefficient “k”, which we take equal to:
- no door: k = 1,0 ;
- one door to the street or to the balcony: k = 1,3 ;
- two doors to the street or balcony: k = 1,7 .
- « l" - possible amendments to the heating radiator connection diagram
Perhaps this may seem like an insignificant detail to some, but still, why not immediately take into account the planned connection diagram for the heating radiators. The fact is that their heat transfer, and therefore their participation in maintaining a certain temperature balance in the room, changes quite noticeably when different types insertion of supply and return pipes.
| Illustration | Radiator insert type | The value of the coefficient "l" |
|---|---|---|
 | Diagonal connection: supply from above, return from below | l = 1.0 |
 | Connection on one side: supply from above, return from below | l = 1.03 |
 | Two-way connection: both supply and return from below | l = 1.13 |
 | Diagonal connection: supply from below, return from above | l = 1.25 |
 | Connection on one side: supply from below, return from above | l = 1.28 |
 | One-way connection, both supply and return from below | l = 1.28 |
- « m" - correction factor for the peculiarities of the installation location of heating radiators
And finally, the last coefficient, which is also related to the peculiarities of connecting heating radiators. It is probably clear that if the battery is installed openly and is not blocked by anything from above or from the front, then it will give maximum heat transfer. However, such an installation is not always possible - more often the radiators are partially hidden by window sills. Other options are also possible. In addition, some owners, trying to fit heating elements into the created interior ensemble, hide them completely or partially decorative screens– this also significantly affects the thermal output.
If there are certain “outlines” of how and where radiators will be mounted, this can also be taken into account when making calculations by introducing a special coefficient “m”:
| Illustration | Features of installing radiators | The value of the coefficient "m" |
|---|---|---|
| The radiator is located openly on the wall or is not covered by a window sill | m = 0.9 | |
| The radiator is covered from above with a window sill or shelf | m = 1.0 | |
| The radiator is covered from above by a protruding wall niche | m = 1.07 | |
| The radiator is covered from above by a window sill (niche), and from the front part - by a decorative screen | m = 1.12 | |
| The radiator is completely enclosed in a decorative casing | m = 1.2 |
So, the calculation formula is clear. Surely, some of the readers will immediately grab their head - they say, it’s too complicated and cumbersome. However, if you approach the matter systematically and in an orderly manner, then there is no trace of complexity.
Any good homeowner must have a detailed graphic plan of his “possessions” with dimensions indicated, and usually oriented to the cardinal points. 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 arbitrarily).
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:
| The room, its area, ceiling height. Floor insulation and “neighborhood” above and below | The number of external walls and their main location relative to the cardinal points and the “wind rose”. Degree of wall insulation | Number, type and size of windows | Availability of entrance doors (to the street or to the balcony) | Required thermal power (including 10% reserve) |
|---|---|---|---|---|
| Area 78.5 m² | 10.87 kW ≈ 11 kW | |||
| 1. Hallway. 3.18 m². Ceiling 2.8 m. Floor laid on the ground. Above is an insulated attic. | One, South, average degree of insulation. Leeward side | No | One | 0.52 kW |
| 2. Hall. 6.2 m². Ceiling 2.9 m. Insulated floor on the ground. Above - insulated attic | No | No | No | 0.62 kW |
| 3. Kitchen-dining room. 14.9 m². Ceiling 2.9 m. Well-insulated floor on the ground. Upstairs - insulated attic | Two. South-West. Average degree of insulation. Leeward side | Two, single-chamber double-glazed window, 1200 × 900 mm | No | 2.22 kW |
| 4. Children's room. 18.3 m². Ceiling 2.8 m. Well-insulated floor on the ground. Above - insulated attic | Two, North - West. High degree insulation. Windward | Two, double-glazed windows, 1400 × 1000 mm | No | 2.6 kW |
| 5. Bedroom. 13.8 m². Ceiling 2.8 m. Well-insulated floor on the ground. Above - insulated attic | Two, North, East. High degree of insulation. Windward side | Single, double-glazed window, 1400 × 1000 mm | No | 1.73 kW |
| 6. Living room. 18.0 m². Ceiling 2.8 m. Well-insulated floor. Above is an insulated attic | Two, East, South. High degree of insulation. Parallel to the wind direction | Four, double-glazed window, 1500 × 1200 mm | No | 2.59 kW |
| 7. Combined bathroom. 4.12 m². Ceiling 2.8 m. Well-insulated floor. Above is an insulated attic. | One, North. High degree of insulation. Windward side | One. Wooden frame with double glazing. 400 × 500 mm | No | 0.59 kW |
| TOTAL: |
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 necessary total power heating systems.
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 one section and round up.
Any owner of a city apartment has been surprised at least once by the numbers on the heating receipt. It is often unclear on what basis heating fees are calculated for us and why often the residents of the neighboring house pay much less. However, the numbers do not come out of nowhere: there is a standard for the consumption of thermal energy for heating, and it is on its basis that the final amounts are formed, taking into account the approved tariffs. How to understand this complex system?
Where do the standards come from?
Standards for heating residential premises, as well as standards for the consumption of any public services, be it heating, water supply, etc., is a relatively constant value. They are adopted by the local authorized body with the participation of resource supply organizations and remain unchanged for three years.
To put it more simply, the company supplying heat to a given region submits documents to local authorities justifying the new standards. During the discussion, they are accepted or rejected at city council meetings. After this, the consumed heat is recalculated, and the tariffs that consumers will pay are approved.
Thermal energy consumption standards for heating are calculated based on the climatic conditions of the region, type of house, wall and roof material, wear utility networks and other indicators. The result is the amount of energy that has to be spent on heating 1 square of living space in a given building. This is the standard.
The generally accepted unit of measurement is Gcal/sq. m – gigacalorie per square meter. The main parameter is the average ambient temperature during the cold period. Theoretically, this means that if the winter was warm, you will have to pay less for heating. However, in practice this usually does not work out.
What should be the normal temperature in the apartment?
Apartment heating standards are calculated taking into account the fact that a comfortable temperature must be maintained in the living space. Its approximate values:
- In the living room optimal temperature amounts to from 20 to 22 degrees;
- Kitchen - temperature from 19 to 21 degrees;
- Bathroom - from 24 to 26 degrees;
- Toilet - temperature from 19 to 21 degrees;
- Corridor – from 18 to 20 degrees.
If in winter time in your apartment the temperature is below the specified values, which means your home receives less heat than heating standards require. As a rule, in such situations, worn-out city heating networks are to blame, when precious energy is wasted into the air. However, the heating standards in the apartment are not met, and you have the right to complain and demand a recalculation.
Method thermal calculation is the determination of the surface area of each individual heating device, which releases heat into the room. Calculation of thermal energy for heating in in this case takes into account the maximum coolant temperature level, 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 highest temperature 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 only be used if the indicator is known surface density heat device q pr (W/m 2).
From here you can calculate the 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 of the system, designed to maintain constant temperature indoors. 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 required power heating, that is, the total heat transfer (Qtr - 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 pipeline located indoors, d n - O.D. pipes, l – length of the segment. 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 pipe 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 the heat-transfer 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 transfer 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 several more 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.
Today, sectional radiators are very common. Their length directly depends on the number of sections used. In order to calculate the amount of heat for heating - that is, to calculate the optimal number of sections for a certain 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 installation method 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 type M-140, 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 real 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 temperature deviation in the 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 certain conditions The installation of radiators is also affected by the architecture of the house.
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, in order to increase the area of the heating device of this type Simply parallel installation of two panels side by side 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, floor convector“Rhythm” is presented in two models with casing lengths of 1 m and 1.5 m. Wall-mounted 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 a decorative pipe cover; n – number of rows or tiers of heating pipes; and 1 is a parameter characterizing the area of one meter horizontal pipe at 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
Need to determine correct amount 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 density heat flow:
t av = 100.8 - 18 = 82.8 °C
It should be noted that there was a slight change in the level of water consumption (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 are counting required quantity radiator sections M140-A, 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 open wall convector with casing KN-20k “Universal-20”, which is installed on a single-pipe riser flow type. There is no tap near the installed device.
Defines average temperature water in the convector:
tcp = (105 - 2) - 0.5x1410x1.04x1.02x3.6 / (4.187x300) = 100.9 °C.
In 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 open type and is produced 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 the required area of the 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.