This article is the seventh publication of the "Myths of Housing and Public Utilities" cycle dedicated to debunking. Myths and false theories, widespread in the housing and communal services of Russia, contribute to the growth of social tension, the development of "" between consumers and public utilities, which leads to extremely negative consequences in the housing industry. Articles of the cycle are recommended, first of all, for consumers of housing and communal services (HCS), however, HCS specialists may find something useful in them. In addition, the dissemination of publications of the “Housing and Utilities Myths” cycle among consumers of housing and communal services can contribute to a deeper understanding of the housing and communal services sector by residents of apartment buildings, which leads to the development of constructive interaction between consumers and utility service providers. A complete list of articles in the Myths of Housing and Public Utilities series is available
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This article discusses a somewhat unusual question, which, nevertheless, as practice shows, worries a rather significant part of utility consumers, namely: why is the unit for measuring the consumption standard for heating utility services “Gcal/sq.m”? Misunderstanding this issue led to the advancement of an unfounded hypothesis that the alleged unit of measurement of the norm of heat energy consumption for heating was chosen incorrectly. The assumption under consideration leads to the emergence of some myths and false theories of the housing sector, which are refuted in this publication. Additionally, the article provides explanations of what is a public heating service and how this service is technically provided.
The essence of false theory
It should be noted right away that the incorrect assumptions analyzed in the publication are relevant for cases where there are no heating meters - that is, for those situations when it is used in the calculations.
It is difficult to clearly formulate the false theories that follow from the hypothesis of the wrong choice of the unit of measurement for the heating consumption standard. The consequences of such a hypothesis are, for example, the statements:
⁃ « The volume of the heat carrier is measured in cubic meters, heat energy in gigacalories, which means that the standard for heating consumption should be in Gcal / cubic meter!»;
⁃ « The heating utility is consumed to heat the space of the apartment, and that space is measured in cubic meters, not square meters! The use of area in calculations is illegal, volume must be used!»;
⁃ « Fuel for the preparation of hot water used for heating can be measured either in units of volume (cubic meters) or in units of weight (kg), but not in units of area (square meters). Norms are calculated illegally, incorrectly!»;
⁃ « It is absolutely incomprehensible, in relation to what area the standard is calculated - to the area of the battery, to the cross-sectional area of \u200b\u200bthe supply pipeline, to the area of \u200b\u200bthe land on which the house stands, to the area of \u200b\u200bthe walls of this house or, perhaps, to the area of its roof. It is only clear that it is impossible to use the area of the premises in the calculations, since in a multi-storey building the premises are located one above the other, and in fact their area is used in the calculations many times - approximately as many times as there are floors in the house».
Various conclusions can follow from the above statements, some of which boil down to the phrase “ Everything is wrong, I will not pay”, and the part, in addition to the same phrase, also contains some logical arguments, among which the following can be distinguished:
1) since the denominator of the unit of measure of the standard indicates a lower degree of magnitude (square) than it should be (cube), that is, the applied denominator is less than the one to be applied, then the value of the standard, according to the rules of mathematics, is overestimated (the smaller the denominator of the fraction, the greater the value the fraction itself);
2) an incorrectly chosen unit of measurement of the standard involves additional mathematical operations before being substituted into formulas 2, 2(1), 2(2), 2(3) of Appendix 2 of the Rules for the provision of utility services to owners and users of premises in apartment buildings and residential houses approved by the Government of the Russian Federation of 05/06/2011 N354 (hereinafter referred to as Rules 354) of the values \u200b\u200bof NT (standard consumption of utility services for heating) and TT (tariff for thermal energy).
As such preliminary transformations, actions that do not stand up to criticism are proposed, for example *
:
⁃ The value of NT is equal to the square of the standard approved by the subject of the Russian Federation, since the denominator of the unit of measurement indicates " square meter";
⁃ The value of TT is equal to the product of the tariff by the standard, that is, TT is not a tariff for heat energy, but a certain unit cost of heat energy spent on heating one square meter;
⁃ Other transformations, the logic of which could not be comprehended at all, even when trying to apply the most incredible and fantastic schemes, calculations, theories.
Since an apartment building consists of a combination of residential and non-residential premises and common areas (common property), while the common property belongs to the owners of individual premises of the house on the right of common ownership, the entire volume of heat energy entering the house is consumed by the owners of the premises of such a house. Consequently, the payment for heat energy consumed for heating should be made by the owners of the MKD premises. And here the question arises - how to distribute the cost of the entire volume of heat energy consumed by an apartment building among the owners of the premises of this MKD?
Guided by quite logical conclusions that the consumption of heat energy in each specific room depends on the size of such a room, the Government of the Russian Federation has established the procedure for distributing the volume of heat energy consumed by the entire house among the premises of such a house in proportion to the area of these premises. This is provided for by both Rules 354 (the distribution of readings from a common house heating meter in proportion to the share of the area of \u200b\u200bthe premises of specific owners in the total area of \u200b\u200ball the premises of the house in the property), and Rule 306 when setting the standard for heating consumption.
Paragraph 18 of Annex 1 to Rule 306 states:
« 18. Standard for the consumption of utility services for heating in residential and non-residential premises (Gcal per 1 sq.m of the total area of all residential and non-residential premises in apartment building or a residential building per month) is determined by the following formula (formula 18):
where:
- the amount of heat energy consumed in one heating period by apartment buildings that are not equipped with collective (common house) heat energy meters, or residential buildings that are not equipped with individual heat energy meters (Gcal), determined by formula 19;
- the total area of all residential and non-residential premises in apartment buildings or the total area of \u200b\u200bresidential buildings (sq.m);
- a period equal to the duration of the heating period (the number of calendar months, including incomplete ones, in the heating period)».
Thus, it is precisely the above formula that determines that the standard for the consumption of utility services for heating is measured precisely in Gcal / sq. Meter, which, among other things, is directly established by subparagraph “e” of paragraph 7 of Rule 306:
« 7. When choosing a unit of measure for utility consumption standards, the following indicators are used:
e) with regard to heating:
in living quarters - Gcal per 1 sq. meter the total area of all rooms in an apartment building or residential building».
Based on the foregoing, the standard for consumption of utility services for heating is equal to the amount of heat energy consumed in an apartment building per 1 square meter of premises in the property in a month of the heating period (when choosing a payment method, it is applied evenly throughout the year).
Calculation examples
As indicated, we will give an example of calculation by the correct method and by the methods offered by false theorists. To calculate the cost of heating, we will accept the following conditions:
Let the heating consumption standard be approved in the amount of 0.022 Gcal/sq.m., the tariff for heat energy be approved in the amount of 2500 rubles/Gcal., let's take the area of the i-th room equal to 50 sq.m. To simplify the calculation, we will accept the conditions that payment for heating is carried out, and there is no technical possibility in the house to install a common house heat energy meter for heating.
In this case, the amount of payment for the utility service for heating in the i-th residential building not equipped with an individual heat energy meter and the amount of payment for the utility service for heating in the i-th residential or non-residential premises in an apartment building that is not equipped with a collective (common house) heat energy meter, when paying during the heating period, it is determined by formula 2:
Pi = Si× NT× tt,
where:
Si is the total area of the i-th premise (residential or non-residential) in an apartment building or the total area of a residential building;
NT is the standard for the consumption of utility services for heating;
TT is the tariff for thermal energy, established in accordance with the legislation of the Russian Federation.
The following calculation is correct (and universally applicable) for the example under consideration:
Si = 50 square meters
NT = 0.022 Gcal/sq.m
TT = 2500 RUB/Gcal
Pi = Si × NT × TT = 50 × 0.022 × 2500 = 2750 rubles
Let's check the calculation by dimensions:
"square meter"× "Gcal/sq.meter"× × "RUB/Gcal" = ("Gcal" in the first multiplier and "Gcal" in the denominator of the second multiplier are reduced) = "RUB."
The dimensions are the same, the cost of the Pi heating service is measured in rubles. The result of the calculation: 2750 rubles.
Now let's calculate according to the methods proposed by false theorists:
1) The value of NT is equal to the square of the standard approved by the subject of the Russian Federation:
Si = 50 square meters
NT \u003d 0.022 Gcal / square meter × 0.022 Gcal / square meter \u003d 0.000484 (Gcal / square meter)²
TT = 2500 RUB/Gcal
Pi = Si x NT x TT = 50 x 0.000484 x 2500 = 60.5
As can be seen from the presented calculation, the cost of heating turned out to be equal to 60 rubles 50 kopecks. The attractiveness of this method lies precisely in the fact that the cost of heating is not 2750 rubles, but only 60 rubles 50 kopecks. How correct is this method and how accurate is the calculation result obtained from its application? To answer this question, it is necessary to carry out some transformations acceptable by mathematics, namely: we will carry out the calculation not in gigacalories, but in megacalories, respectively converting all the quantities used in the calculations:
Si = 50 square meters
NT \u003d 22 Mcal / square meter × 22 Mcal / square meter \u003d 484 (Mcal / square meter)²
TT \u003d 2.5 rubles / Mcal
Pi = Si x NT x TT = 50 x 484 x 2.500 = 60500
And what will we get as a result? The cost of heating is already 60,500 rubles! We note right away that in the case of applying the correct method, mathematical transformations should not affect the result in any way:
(Si = 50 square meters
NT \u003d 0.022 Gcal / square meter \u003d 22 Mcal / square meter
TT = 2500 RUB/Gcal = 2.5 RUB/Mcal
Pi = Si× NT× TT=50× 22 × 2.5 = 2750 rubles)
And if, in the method proposed by false theorists, the calculation is carried out not even in megacalories, but in calories, then:
Si = 50 square meters
NT = 22,000,000 cal/m2 × 22,000,000 cal/m2 = 484,000,000,000,000 (cal/m2)²
TT = 0.0000025 RUB/cal
Pi = Si × NT × TT = 50 × 484,000,000,000,000 × 0.0000025 = 60,500,000,000
That is, heating a room with an area of 50 square meters costs 60.5 billion rubles a month!
In fact, of course, the considered method is incorrect, the results of its application do not correspond to reality. Additionally, we will check the calculation by dimensions:
"square meter"× "Gcal/sq.meter"× "Gcal/sq.meter"× “ruble/Gcal” = (“sq.m.” in the first multiplier and “sq.m.” in the denominator of the second multiplier are reduced) = “Gcal”× "Gcal/sq.meter"× "Rub/Gcal" = ("Gcal" in the first multiplier and "Gcal" in the denominator of the third multiplier are reduced) = "Gcal/sq.meter"× "rub."
As you can see, the dimension "rub." as a result, it does not work, which confirms the incorrectness of the proposed calculation.
2) The value of TT is equal to the product of the tariff approved by the subject of the Russian Federation and the consumption standard:
Si = 50 square meters
NT = 0.022 Gcal/sq.m
TT = 2500 rubles / Gcal × 0.022 Gcal / sq. meter = 550 rubles / sq. meter
Pi = Si x NT x TT = 50 x 0.022 x 550 = 60.5
The calculation by this method gives exactly the same result as the first considered incorrect method. You can refute the second method applied in the same way as the first one: convert gigacalories to mega- (or kilo-) calories and check the calculation by dimensions.
conclusions
The myth of the wrong choice Gcal/sq.meter» has been refuted as a unit of measure for the consumption standard for heating utility services. Moreover, the logic and validity of the use of just such a unit of measurement has been proved. The incorrectness of the methods proposed by the false theorists has been proved, their calculations have been refuted by the elementary rules of mathematics.
It should be noted that the vast majority of false theories and myths of the housing sector aim to prove that the amount of fees charged to owners for payment is overstated - it is this circumstance that contributes to the “survivability” of such theories, their spread and the growth of their supporters. It is quite reasonable that consumers of any services want to minimize their costs, however, attempts to use false theories and myths do not lead to any savings, but are only aimed at introducing into the minds of consumers the idea that they are being deceived, unreasonably charging them money. funds. It is clear that the courts supervisory authorities authorized to deal with conflict situations between performers and consumers of public services will not be guided by false theories and myths, therefore, there can be no savings and no other positive consequences for either the consumers themselves or for other participants in housing relations.
Any owner of a city apartment at least once was surprised at the figures on the receipt for heating. It is often not clear on what basis we are charged for heating and why often the residents of a neighboring house pay much less. However, the figures are not taken from nowhere: there is a norm 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 deal with this complex system?
Where do regulations come from?
The norms for heating residential premises, as well as the norms for the consumption of any utility service, whether it be heating, water supply, etc., are a relatively constant value. They are accepted by the local authorized body starring resource supplying organizations and remain unchanged for three years.
More simply, the company supplying heat this region, submits to local authorities authorities documents substantiating the new regulations. During the discussion, they are accepted or rejected at meetings of the city council. After that, the consumed heat is recalculated, and the tariffs for which consumers will pay are approved.
The norms for the consumption of thermal energy for heating are calculated based on the climatic conditions of the region, the type of house, the material of the walls and roof, the deterioration of 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 this building. This is the norm.
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, then 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?
The standards for heating an apartment are calculated taking into account the fact that a comfortable temperature should be maintained in the living room. Its approximate values are:
- In a living room, the optimum temperature is 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;
- The corridor - from 18 to 20 degrees.
If in winter the temperature in your apartment is below the indicated values, it means that your house receives less heat than the norms for heating prescribe. As a rule, in such situations, worn-out city heating systems are to blame, when precious energy is wasted into the air. However, the heating norm in the apartment is not met, and you have the right to complain and demand recalculation.
When planning overhaul in your house or apartment, as well as when planning the construction of a new house, it is necessary to calculate the power of heating radiators. This will allow you to determine the number of radiators that can provide heat to your home in the most severe frosts. To carry out calculations, it is necessary to find out the necessary parameters, such as the size of the premises and the power of the radiator, declared by the manufacturer in the attached technical documentation. The shape of the radiator, the material from which it is made, and the level of heat transfer in these calculations are not taken into account. Often the number of radiators is equal to the number of window openings in the room, therefore, the calculated power is divided by the total number of window openings, so you can determine the size of one radiator.
It should be remembered that you do not need to make a calculation for the entire apartment, because each room has its own heating system and requires an individual approach. So if you have a corner room, then about twenty percent must be added to the resulting power value. The same amount should be added if your heating system is intermittent or has other efficiency deficiencies.
The calculation of the power of heating radiators can be carried out in three ways:
Standard calculation of heating radiators
According to building codes and other rules, you need to spend 100W of your radiator power per 1 square meter of living space. In this case necessary calculations produced using the formula:
C * 100 / P \u003d K, where
K is the power of one section of your radiator battery, according to its characteristics;
C is the area of the room. It is equal to the product of the length of the room and its width.
For example, a room is 4 meters long and 3.5 wide. In this case, its area is: 4 * 3.5 = 14 square meters.
The power of one section of the battery you have chosen is declared by the manufacturer at 160 watts. We get:
14*100/160=8.75. the resulting figure must be rounded up and it turns out that such a room will require 9 sections of a heating radiator. If this is a corner room, then 9*1.2=10.8, rounded up to 11. And if your heating system is not efficient enough, then add 20 percent of the original number again: 9*20/100=1.8 rounds up to 2.
Total: 11+2=13. For a corner room with an area of 14 square meters, if the heating system works with short-term interruptions, you will need to purchase 13 battery sections.
Approximate calculation - how many battery sections per square meter
It is based on the fact that heating radiators in mass production have certain dimensions. If the room has a ceiling height of 2.5 meters, then only one section of the radiator is required for an area of 1.8 square meters.
The calculation of the number of radiator sections for a room with an area of 14 square meters is equal to:
14 / 1.8 = 7.8, rounded up to 8. So for a room with a ceiling height of 2.5m, eight sections of the radiator will be needed. It should be borne in mind that this method is not suitable if the heater has a low power (less than 60W) due to a large error.
Volumetric or for non-standard rooms
This calculation is used for rooms with high or very low ceilings. Here, the calculation is based on the data that heating one meter of a cubic room requires a power of 41W. For this, the formula is applied:
K=O*41, where:
K - the required number of radiator sections,
O is the volume of the room, it is equal to the product of the height times the width times the length of the room.
If the room has a height of 3.0m; length - 4.0m and width - 3.5m, then the volume of the room is:
3.0*4.0*3.5=42 cubic meters.
Calculate the total heat demand for this room:
42*41=1722W, given that the power of one section is 160W, you can calculate the required number by dividing the total power requirement by the power of one section: 1722/160=10.8, rounded up to 11 sections.
If radiators are selected that are not divided into sections, the total number must be divided by the power of one radiator.
It is better to round the received data up, as manufacturers sometimes overestimate the declared power.
aquagroup.ru
Calculation of the number of sections of heating radiators - why you need to know this
At first glance, it is easy to calculate how many radiator sections to install in a particular room. The larger the room, the more sections the radiator should consist of. But in practice, how warm it will be in a particular room depends on more than a dozen factors. Given them, it is possible to calculate the required amount of heat from radiators much more accurately.
General information
The heat transfer of one section of the radiator is indicated in the technical characteristics of products from any manufacturer. The number of radiators in a room usually corresponds to the number of windows. Radiators are most often located under the windows. Their dimensions depend on the area of the free wall between the window and the floor. It should be borne in mind that the radiator must be lowered from the window sill by at least 10 cm. And between the floor and the bottom line of the radiator, the distance must be at least 6 cm. These parameters determine the height of the device.The heat output of one section of a cast-iron radiator is 140 watts, more modern metal ones - from 170 and above.
You can calculate the number of sections of heating radiators, leaving the area of \u200b\u200bthe room or its volume.
According to the norms, it is considered that 100 watts of thermal energy is needed to heat one square meter of a room. If we proceed from the volume, then the amount of heat per 1 cubic meter will be at least 41 watts.
But none of these methods will be accurate if you do not take into account the characteristics of a particular room, the number and size of windows, the material of the walls, and much more. Therefore, when calculating the radiator sections according to the standard formula, we will add the coefficients created by one or another condition.
Room area - calculation of the number of sections of heating radiators
Such a calculation is usually applied to premises located in standard panel residential buildings with a ceiling height of up to 2.6 meters.
The area of the room is multiplied by 100 (the amount of heat for 1m2) and divided by the heat output of one section of the radiator indicated by the manufacturer. For example: the area of the room is 22 m2, the heat transfer of one section of the radiator is 170 watts.
22X100/170=12.9
This room needs 13 radiator sections.
If one section of the radiator has 190 watts of heat transfer, then we get 22X100 / 180 \u003d 11.57, that is, we can limit ourselves to 12 sections.
You need to add 20% to the calculations if the room has a balcony or is located at the end of the house. A battery installed in a niche will reduce heat transfer by another 15%. But in the kitchen it will be 10-15% warmer.
We make calculations according to the volume of the room
For a panel house with a standard ceiling height, as already mentioned above, the heat calculation is based on the need for 41 watts per 1m3. But if the house is new, brick, double-glazed windows are installed in it, and the outer walls are insulated, then 34 watts per 1 m3 are already needed.
The formula for calculating the number of radiator sections looks like this: the volume (area multiplied by the height of the ceiling) is multiplied by 41 or 34 (depending on the type of house) and divided by the heat transfer of one section of the radiator indicated in the manufacturer's passport.
For example:
The area of the room is 18 m2, the ceiling height is 2.6 m. The house is a typical panel building. The heat output of one section of the radiator is 170 watts.
18X2.6X41 / 170 \u003d 11.2. So, we need 11 radiator sections. This is provided that the room is not corner and it does not have a balcony, otherwise it is better to install 12 sections.
Calculate as accurately as possible
And here is the formula by which you can calculate the number of radiator sections as accurately as possible:
The area of the room multiplied by 100 watts and by the coefficients q1, q2, q3, q4, q5, q6, q7 and divided by the heat transfer of one section of the radiator.
More about these ratios:
q1 - type of glazing: with triple glazing, the coefficient will be 0.85, with double glazing - 1 and with ordinary glazing - 1.27.
q2 - thermal insulation of walls:
- modern thermal insulation - 0.85;
- laying in 2 bricks with insulation - 1;
- non-insulated walls - 1.27.
q3 - the ratio of the areas of windows and floor:
- 10% - 0,8;
- 30% - 1;
- 50% - 1,2.
q4 - minimum outside temperature:
- -10 degrees - 0.7;
- -20 degrees - 1.1;
- -35 degrees - 1.5.
q5 - the number of external walls:
q6 - type of room, which is located above the calculated one:
- heated - 0.8;
- attic heated - 0.9;
- attic unheated - 1.
q7 - ceiling height:
- 2,5 – 1;
- 3 – 1,05;
- 3,5 – 1,1.
If all of the above coefficients are taken into account, it will be possible to calculate the number of radiator sections in the room as accurately as possible.
semidelov.ru
Calculation of the standard for heat consumption
Dear Igor Viktorovich!
I asked your specialists for data on the determination of standards for heat consumption. The answer has been received. But he also contacted MPEI, where they also gave a link to the calculations. I bring it:
Borisov Konstantin Borisovich.
Moscow Power Engineering Institute (Technical University)
To calculate the norm of heat consumption for heating, you must use the following document:
Decree No. 306 "Rules for establishing and determining standards for the consumption of utilities" (formula 6 - "Formula for calculating the heating standard"; table 7 - "The value of the normalized specific consumption of thermal energy for heating apartment building or residential building).
To determine the payment for heating for a dwelling (apartment), you must use the following document:
Decree No. 307 "Rules for the provision of public services to citizens" (Appendix No. 2 - "Calculation of the amount of payment for utilities", Formula 1).
In principle, the very calculation of the norm of heat consumption for heating an apartment and determining the payment for heating is not complicated.
If you want, let's try to roughly (roughly) estimate the main numbers:
1) The maximum hourly heating load of your apartment is determined:
Qmax \u003d Qsp * Skv \u003d 74 * 74 \u003d 5476 kcal / h
Qsp \u003d 74 kcal / h - normalized specific consumption of thermal energy for heating 1 sq. m of an apartment building.
The value of Qsp is taken according to table 1 for buildings built before 1999, with a height (number of storeys) of 5-9 floors at an outdoor temperature Tnro = -32 C (for the city of K).
Sq= 74 sq. m - the total area of the premises of the apartment.
2) The amount of thermal energy required to heat your apartment during the year is calculated:
Qav = Qmax×[(Tv-Tav.o)/(Tv-Tnro)]×Nо×24 = 5476×[(20-(-5.2))/(20-(-32))]×215* 24 \u003d 13,693,369 kcal \u003d 13.693 Gcal
TV = 20 C - normative value indoor air temperature in residential premises (apartments) of the building;
Tsr.o = -5.2 C - outdoor air temperature, average for the heating period (for the city of K);
No = 215 days - the duration of the heating period (for the city of K).
3) The standard for heating 1 sq. meters:
Heating_standard \u003d Qav / (12 × Skv) \u003d 13.693 / (12 × 74) \u003d 0.0154 Gcal / sq.m
4) The payment for heating the apartment is determined according to the standard:
Po \u003d Skv × Standard_heating × Tariff_heat \u003d 74 × 0.0154 × 1223.31 \u003d 1394 rubles
The data are taken from Kazan.
Following this calculation and in relation specifically to house No. 55 in the village of Vaskovo, with the introduction of the parameters of this structure, we obtain:
Arkhangelsk
177 - 8 253 -4.4 273 -3.4
12124.2 × (20-(-8) / 20-(-45) × 273 × 24 = 14.622…./ (12= 72.6)=0.0168
0.0168 is exactly such a standard that we obtain in the calculation, and it is precisely the most severe climatic conditions that are taken into account: the temperature is -45, the length of the heating period is 273 days.
I perfectly understand that deputies who are not specialists in the field of heat supply can be asked to introduce a standard of 0.0263.
But calculations are given, which indicate that the standard of 0.0387 is the only correct one, and this raises very big doubts.
Therefore, I kindly ask you to recalculate the standards for heat supply to residential buildings Nos. 54 and 55 in the village of Vaskovo to the corresponding values of 0.0168, since in the near future it is not planned to install heat meters in these residential buildings, but to pay 5300 rubles for heat supply very difficult.
Sincerely, Alexey Veniaminovich Popov.
www.orlov29.ru
How to calculate the heating system at home?
In the process of developing a heating system project, one of the key points is the thermal power of the batteries. This is necessary in order to ensure that the temperature inside the dwelling required by the sanitary standards of the Russian Federation is from +22 ° С. But the devices differ from each other not only in the material of manufacture, dimensions, but also in the amount of thermal energy released per 1 sq. m. Therefore, before the acquisition, the calculation of radiators is carried out.
Where to start
The optimal microclimate in the living room is ensured by properly selected radiators. For each product, the manufacturer encloses a passport with technical specifications. It indicates the power of a radiator of any kind, based on the size of one section or block. This information is important for calculating the dimensions of the unit, their number, taking into account some other factors.
From SNiP 41-01-2003 it is known that the heat flux entering the rooms and kitchens should be taken at least 10 W per 1 m2 of floor, that is, the calculation of the heating system of a private house is simple - you need to take the rated power of the battery, estimate the area of \u200b\u200bthe apartment and calculate the number of radiators. But everything is much more complicated: it is selected not by square meters, but by such a parameter as thermal loss. The reasons:
1. The task of the heating structure is to compensate for the heat losses of housing and raise the temperature inside to a comfortable one. Most actively, heat escapes through window openings and cold walls. At the same time, a house insulated according to the rules without drafts requires much less power of radiators.
2. The calculation includes:
- ceiling height;
- region of residence: the average street temperature in Yakutia is -40 °С, in Moscow - -6 °С. Accordingly, the dimensions and power of the radiators must be different;
- ventilation system;
- composition and thickness of enclosing structures.
Having received a given value, they begin to calculate the key parameters.
How to correctly calculate the power and number of sections
Sellers of heating equipment prefer to focus on the average indicators indicated in the instructions for the device. That is, if it is indicated that 1 segment of an aluminum battery can warm up to 2 square meters. m of the room, then additional calculations are not required, but this is not the case. During the tests, conditions are taken that are close to ideal: the inlet temperature is not less than +70 or +90 °С, the return temperature is +55 or +70 °С, the internal temperature is +20 °С, the insulation of the enclosing structures complies with SNiPs. In reality, the situation is very different.
- Rare CHP plants maintain a constant temperature corresponding to 90/70 or 70/55.
- Boilers used for heating a private house do not produce more than +85 ° C, therefore, until the coolant reaches the radiator, the temperature drops by a few more degrees.
- Aluminum batteries have the highest power - up to 200 watts. But they cannot be used in a centralized system. Bimetallic - an average of about 150 W, cast iron - up to 120.
1. Calculation by area.
AT different sources you can find both a greatly simplified calculation of the power of a heating battery per square meter, and a very complex one with the inclusion of logarithmic functions. The first is based on the axiom: 100 W of heat is needed for 1 m2 of floor. The standard must be multiplied by the area of \u200b\u200bthe room, and the required intensity of the radiator is obtained. The value is divided by the power of 1 section - the required number of segments is found.
There is a 4 x 5 room, Global bimetallic radiators with a 150 W segment. Power \u003d 20 x 100 \u003d 2,000 watts. Number of sections = 2,000 / 150 = 13.3.
The calculation of the number of sections of bimetallic radiators shows that 14 nodes are needed for this example. An impressive accordion will be placed under the window. Obviously, this approach is very conditional. Firstly, the volume of the room, thermal losses through the outer walls and window openings are not taken into account. Secondly, the “100 to 1” standard is the result of a complex but outdated engineering heat engineering calculation for a certain type of structure with rigid parameters (dimensions, thickness and material of partitions, insulation, roofing, etc.). For most dwellings, the rule is not suitable, and the result of its application will be insufficient or excessive heating (depending on the degree of insulation of the house). To check the correctness of the calculations, we take complex calculation methods.
2. Calculation of heat losses.
The calculation formula includes average correction factors and is expressed as follows:
Q = (22 + 0.54Dt)(Sp + Sns + 2So), where:
- Q is the required heat transfer of radiators, W;
- Dt is the difference between the air temperature in the room and the calculated outdoor temperature, deg;
- Sp - floor area, m2;
- Sns is the area of the walls outside, m2;
- So is the area of window openings, m2.
Number of sections:
- X=Q/N
- where Q is the heat loss of the room;
- N is the power of 1 segment.
There is a room 4 x 5 x 2.5 m, a window opening 1.2 x 1, one outer wall, Global bimetallic radiators with a section power of 150 watts. Thermal conductivity coefficient according to SNiP - 2.5. Air temperature - -10 ° С; inside - +20 °С.
- Q \u003d (22 + 0.54 x 30) x (20 + 10 + 2.4) \u003d 1237.68 watts.
- Number of sections = 1237.68 / 150 = 8.25.
Rounded up to the nearest integer, we get 9 sections. You can check another calculation option with climatic coefficients.
3. Calculation of room heat loss in accordance with SNiP "Construction Climatology" 23-01-99.
First you need to calculate the level of thermal losses of the room through the outer and inner walls. The same indicator is calculated separately for window openings and doors.
Q \u003d F x kthermal conductivity x (tin-tout), where:
- F is the area of external fences minus window openings, m2;
- k - taken according to SNiP "Construction climatology" 23-01-99, W/m2K;
- tvn - indoor temperature, on average, the value is taken from +18 to +22 ° С;
- tnar - outdoor temperature, the value is taken from the same SNiP or on the website of the city's meteorological service.
The results obtained for walls and openings are added up, and the total amount of heat loss comes out.
To determine the estimated heat consumption for heating a building, you can use the formula
Q \u003d q from * V zd (t ext - t n) * 10 -3, kW,
where q from is the specific thermal characteristic of the building, W / m 3 o C
V zd - the total external volume of the building, m 3.
The specific thermal characteristic of the building is found by the formula
q from \u003d P / S 1 / Rst + ρ (1 / Rok - 1 / Rst)] + 1 / h (0.9 * 1 / Rpl + 0.6 * 1 / Rpt),
where P, S, h - perimeter, area, height of the building, m
ρ - the degree of glazing of the building, equal to the ratio of the total area of light openings to the area of \u200b\u200bvertical fences of the building, ρ \u003d F rest / Fvert.en.
Rst, Rok, Rpl, Rpt - heat transfer resistance of walls, windows, floors, ceilings.
The value of the specific thermal characteristic determines the average heat loss of 1 m 3 of the building, referred to the calculated temperature difference equal to 1 o C.
It is convenient to use the characteristic q from for the thermotechnical assessment of possible design and planning solutions for the building.
According to the calculated heat consumption, a heating system boiler is selected (Appendix 1) and it is installed in the boiler room, taking into account design standards (Appendix 2).
3. Heat balance of premises
In buildings and premises with a constant thermal regime, heat losses and heat gains are compared in the design mode. For residential and public buildings, it is assumed that there are no heat sources in the premises, and the heat output of the heating system must compensate for heat losses through external fences.
Heat losses through the enclosing structures of the premises are the sum of heat losses through individual enclosures Q, determined with rounding up to 10 W according to the formula:
Q \u003d F * 1 / R * (t int - tn) * (1 + β) * n W, where
F - estimated area of the fence, m 2 (for the rules for measuring fences, see Appendix 3)
R - resistance to heat transfer of the building envelope, m 2 o C / W
t ext - room temperature, 0 C
t n V - estimated outside temperature of the coldest five-day period, 0 С
β - additional heat losses in shares of the main losses,
n - coefficient taken depending on the position of the outer surface of the enclosing structures to the outside air
Heat loss calculations are summarized in a table (see Appendix 4)
Additional heat loss β
1. Additive for orientation - for all vertical railings
C, NE, B, NW - 0.1
2. The addition in the corner rooms of public and industrial buildings (having two or more external walls) is accepted for all vertical fences in the amount of β = 0.15.
3. Addition for cold air intake through the entrances to the building (permanently operated) is accepted
for double doors with a vestibule between them 0.27 N
the same without tambour 0.34 N
for single doors 0.22 N
where H is the height of the building in m.
Coefficient n
|
Walling | |
|
Exterior walls | |
|
Ceilings over cold cellars communicating with outside air, attic ceilings | |
|
Ceilings over unheated basements with skylights in the walls | |
|
Ceilings over unheated basements without skylights in the walls | |
|
Walls separating from unheated rooms communicating with outside air | |
|
Walls that separate from unheated rooms that do not communicate with the outside air |
Creating a heating system in your own home or even in a city apartment is an extremely responsible task. At the same time, it would be completely unreasonable to purchase boiler equipment, as they say, “by eye”, that is, without taking into account all the features of housing. In this, it is quite possible to fall into two extremes: either the power of the boiler will not be enough - the equipment will work “to its fullest”, without pauses, but will not give the expected result, or, conversely, an overly expensive device will be purchased, the capabilities of which will remain completely unclaimed.
But that's not all. It is not enough to purchase the necessary heating boiler correctly - it is very important to optimally select and correctly place 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, certain calculations are indispensable.
Of course, ideally, such heat engineering calculations should be carried out by appropriate specialists, but this often costs a lot of money. Isn't it interesting to try to do it yourself? This publication will show in detail how heating is calculated by the area of \u200b\u200bthe room, taking into account many important nuances. By analogy, it will be possible to perform, built into this page, will help you perform the necessary calculations. The technique cannot be called completely “sinless”, however, it still allows you to get a result with a completely acceptable degree of accuracy.
The simplest methods of calculation
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, and their separation is very conditional.
- The first is maintaining an optimal level of air temperature in the entire volume of the heated room. Of course, the temperature level may vary slightly with altitude, but this difference should not be significant. Quite comfortable conditions are considered to be an average of +20 ° C - it is this temperature that, as a rule, is taken as the initial temperature in thermal calculations.
In other words, the heating system must be able to heat a certain volume of air.
If we approach with complete accuracy, then for individual rooms in residential buildings the standards for the necessary microclimate are established - they are defined by GOST 30494-96. An excerpt from this document is in the table below:
| Purpose of the room | Air temperature, °С | Relative humidity, % | Air speed, m/s | |||
|---|---|---|---|---|---|---|
| optimal | admissible | optimal | admissible, max | optimal, max | admissible, 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 from -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 bathroom | 24÷26 | 18:26 | N/N | N/N | 0.15 | 0.2 |
| Premises for rest and study | 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, stairwell | 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 (The standard is only for residential premises. For the rest - it is not standardized) | ||||||
| Living room | 22÷25 | 20÷28 | 60÷30 | 65 | 0.2 | 0.3 |
- The second is the compensation of heat losses through the structural elements of the building.
The main "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 go in all directions - their approximate distribution is shown in the table:
| Building element | Approximate value of heat loss |
|---|---|
| Foundation, floors on the ground or over unheated basement (basement) premises | from 5 to 10% |
| "Cold bridges" through poorly insulated joints of building structures | from 5 to 10% |
| Engineering communications entry points (sewerage, 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 non-sealed 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 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 needs. And the values for each room will be the starting point for calculating the required number of radiators.
The most simplified and most commonly used method in a non-professional environment is to accept the norm of 100 W of thermal energy per square meter of area:
The most primitive way of counting is the ratio of 100 W / m²
Q = S× 100
Q- the required thermal power for the room;
S– area of the room (m²);
100 — specific power per unit area (W/m²).
For example, 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 with a standard ceiling height - approximately 2.7 m (permissible - 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 value of specific power is calculated per cubic meter. It is taken equal to 41 W / m³ for a reinforced concrete panel house, or 34 W / m³ - in brick or made of other materials.
Q = S × h× 41 (or 34)
h- ceiling height (m);
41 or 34 - specific power per unit volume (W / m³).
For example, the same room, in a panel house, with a ceiling height of 3.2 m:
Q= 17.6 × 3.2 × 41 = 2309 W ≈ 2.3 kW
The result is more accurate, since it already takes into account not only all the linear dimensions of the room, but even, to a certain extent, the features of the walls.
But still, it is still far from real accuracy - many nuances are “outside the brackets”. How to perform closer to real conditions calculations are 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 are useful for the initial “estimate”, but you should still rely on them completely with very great care. Even to a person who does not understand anything in building heat engineering, the indicated average values \u200b\u200bmay certainly seem doubtful - they cannot be equal, say, for the Krasnodar Territory and for the Arkhangelsk Region. In addition, the room - the room is different: one is located on the corner of the house, that is, it has two external walls, and the other is protected from heat loss by other rooms on three sides. In addition, the room may have one or more windows, both small and very large, sometimes even panoramic. And the windows themselves may differ in the material of manufacture and other design features. And this is not a complete list - just such features are visible even to the "naked eye".
In a word, there are a lot of nuances that affect the heat loss of each particular room, and it is better not to be too lazy, but to carry out a more thorough calculation. Believe me, according to the method proposed in the article, this will not be so difficult to do.
General principles and calculation formula
The calculations will be based on the same ratio: 100 W per 1 square meter. But that's just the formula itself "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 quite 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" - a coefficient that takes into account the number of external walls in a particular room.
Obviously, the more external walls in the room, the larger the area through which heat loss occurs. In addition, the presence of two or more external walls also means corners - extremely vulnerable places in terms of the formation of "cold bridges". The coefficient "a" will correct for this specific feature of the room.
The coefficient is taken equal to:
- external walls No(indoor): a = 0.8;
- outer wall one: a = 1.0;
- external walls two: a = 1.2;
- external walls three: a = 1.4.
- "b" - coefficient taking into account the location of the external walls of the room relative to the cardinal points.
You may be interested in information about what are
Even on the coldest winter days solar energy still affects the temperature balance in the building. It is quite natural that the side of the house that faces south receives a certain amount of heat from the sun's rays, and heat loss through it is lower.
But the walls and windows facing north never “see” the Sun. The eastern part of the house, although it "grabs" the morning sun's rays, still does not receive any effective heating from them.
Based on this, we introduce the coefficient "b":
- the outer walls of the room look at North or East: b = 1.1;
- the outer walls of the room are oriented towards South or West: b = 1.0.
- "c" - coefficient taking into account the location of the room relative to the winter "wind rose"
Perhaps this amendment is not so necessary for houses located in areas protected from the winds. But sometimes the prevailing winter winds can make their own “hard adjustments” to the thermal balance of the building. Naturally, the windward side, that is, "substituted" to the wind, will lose much more body, compared to the leeward, opposite.
Based on the results of long-term meteorological observations in any region, the so-called “wind rose” is compiled - a graphical diagram showing the prevailing wind directions in winter and summer time of the year. This information can be obtained from the local hydrometeorological service. However, many residents themselves, without meteorologists, know very well where the winds mainly blow from in winter, and from which side of the house the deepest snowdrifts usually sweep.
If there is a desire to carry out calculations with higher accuracy, then the correction factor “c” can also be included in the formula, taking it equal to:
- windward side of the house: c = 1.2;
- leeward walls of the house: c = 1.0;
- wall located parallel to the direction of the wind: c = 1.1.
- "d" - a correction factor that takes into account the peculiarities of the climatic conditions of the region where the house was built
Naturally, the amount of heat loss through all the building structures of the building will greatly depend on the level of winter temperatures. It is quite clear that during the winter the thermometer indicators “dance” in a certain range, but for each region there is an average indicator of the lowest temperatures characteristic of the coldest five-day period of the year (usually this is characteristic of January). For example, below is a map-scheme of the territory of Russia, on which approximate values are shown in colors.
Usually this value is easy to check with the regional meteorological service, but you can, in principle, rely on your own observations.
So, the coefficient "d", taking into account the peculiarities of the climate of the region, for our calculations in we take equal to:
— from – 35 °С 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;
- not colder - 10 ° С: d=0.7.
- "e" - coefficient taking into account the degree of insulation of external walls.
The total value of the heat loss of the building is directly related to the degree of insulation of all building structures. One of the "leaders" in terms of heat loss are walls. Therefore, the value of the thermal power required to maintain comfortable living conditions in the room depends on the quality of their thermal insulation.
The value of the coefficient for our calculations can be taken as follows:
- external walls are not insulated: e = 1.27;
- medium degree of insulation - walls in two bricks or their surface thermal insulation with other heaters is provided: e = 1.0;
– insulation was carried out qualitatively, on the basis of heat engineering calculations: e = 0.85.
Later 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 height
Ceilings, especially in private homes, can have different heights. Therefore, the thermal power for heating one or another room of the same area will also differ in this parameter.
It will not be a big mistake to accept the following values of the correction factor "f":
– ceiling height up to 2.7 m: f = 1.0;
— flow height from 2.8 to 3.0 m: f = 1.05;
– ceiling height from 3.1 to 3.5 m: f = 1.1;
– ceiling height from 3.6 to 4.0 m: f = 1.15;
– ceiling height over 4.1 m: f = 1.2.
- « g "- coefficient taking 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. So, it is necessary to make some adjustments in the calculation of this feature of a particular room. The correction factor "g" can be taken equal to:
- cold floor on the ground or over an unheated room (for example, basement or basement): g= 1,4 ;
- insulated floor on the ground or over an unheated room: g= 1,2 ;
- a heated room is located below: g= 1,0 .
- « h "- coefficient taking 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 losses are inevitable, which will require an increase in the required heat output. We introduce the coefficient "h", which takes into account this feature of the calculated room:
- a "cold" attic is located on top: h = 1,0 ;
- an insulated attic or other insulated room is located on top: h = 0,9 ;
- any heated room is located above: h = 0,8 .
- « i "- coefficient taking into account the design features of windows
Windows are one of the "main routes" of heat leaks. Naturally, much in this matter depends on the quality of the window structure itself. Old wooden frames, which were previously installed everywhere in all houses, are significantly inferior to modern multi-chamber systems with double-glazed windows in terms of their thermal insulation.
Without words, it is clear that the thermal insulation qualities of these windows are significantly different.
But even between PVC-windows there is no complete uniformity. 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
No matter how high-quality the windows are, it will still not be possible to completely avoid heat loss through them. But it is quite clear that it is impossible to compare a small window with panoramic glazing almost on 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- the total area of windows in the room;
SP- area of the room.
Depending on the value obtained and the correction factor "j" is determined:
- x \u003d 0 ÷ 0.1 →j = 0,8 ;
- x \u003d 0.11 ÷ 0.2 →j = 0,9 ;
- x \u003d 0.21 ÷ 0.3 →j = 1,0 ;
- x \u003d 0.31 ÷ 0.4 →j = 1,1 ;
- x \u003d 0.41 ÷ 0.5 →j = 1,2 ;
- « k" - coefficient that corrects for the presence of an entrance door
The door to the street or to an unheated balcony is always an additional "loophole" for the cold
The door to the street or to an open balcony is able to make its own adjustments to the heat balance of the room - each of its opening is accompanied by the penetration of a considerable amount 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 balcony: k = 1,3 ;
- two doors to the street or to the balcony: k = 1,7 .
- « l "- possible amendments to the connection diagram of heating radiators
Perhaps this will seem like an insignificant trifle to some, but still - why not immediately take into account the planned scheme for connecting heating radiators. The fact is that their heat transfer, and hence their participation in maintaining a certain temperature balance in the room, changes quite noticeably with different types tie-in 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 the bottom | 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 features of the installation site of heating radiators
And finally, the last coefficient, which is also associated with the features of connecting heating radiators. It is probably clear that if the battery is installed openly, is not obstructed by anything from above and from the front, then it will give maximum heat transfer. However, such an installation is far from always possible - more often, radiators are partially hidden by window sills. Other options are also possible. In addition, some owners, trying to fit heating priors into the created interior ensemble, hide them completely or partially with decorative screens - this also significantly affects the heat output.
If there are certain “baskets” on how and where the radiators will be mounted, this can also be taken into account when making calculations by entering a special coefficient “m”:
| Illustration | Features of installing radiators | The value of the coefficient "m" |
|---|---|---|
| The radiator is located on the wall openly or is not covered from above by a window sill | m = 0.9 | |
| The radiator is covered from above by a window sill or a shelf | m = 1.0 | |
| The radiator is blocked from above by a protruding wall niche | m = 1.07 | |
| The radiator is covered from above with a window sill (niche), and from the front - with a decorative screen | m = 1.12 | |
| The radiator is completely enclosed in a decorative casing | m = 1.2 |
So, there is clarity with the calculation formula. Surely, some of the readers will immediately take up their heads - they say, it's too complicated and cumbersome. However, if the matter is approached systematically, in an orderly manner, then there is no difficulty at all.
Any good homeowner must have a detailed graphical plan of their "possessions" with affixed dimensions, and usually oriented to the cardinal points. It is not difficult to specify the climatic features of the region. It remains only to walk through all the rooms with a tape measure, to clarify some of the nuances for each room. Features of housing - "vertical neighborhood" from above and below, the location of the entrance doors, the proposed or existing scheme for installing heating radiators - no one except the owners knows better.
It is recommended to immediately draw up a worksheet, where you enter all the necessary data for each room. The result of the calculations will also be entered into it. Well, the calculations themselves will help to carry out the built-in calculator, in which all the coefficients and ratios mentioned above are already “laid”.
If some data could not be obtained, then, of course, they can not be taken into account, but in this case, the “default” calculator will calculate the result, taking into account the least favorable conditions.
It can be seen with an example. We have a house plan (taken completely arbitrary).
The region with the level of minimum temperatures in the range of -20 ÷ 25 °С. Predominance of winter winds = northeasterly. The house is one-story, with an insulated attic. Insulated floors on the ground. The optimal diagonal connection of radiators, which will be installed under the window sills, has been selected.
Let's create a table like this:
| The room, its area, ceiling height. Floor insulation and "neighborhood" from 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 | Existence of entrance doors (to the street or to the balcony) | Required heat output (including 10% reserve) |
|---|---|---|---|---|
| Area 78.5 m² | 10.87 kW ≈ 11 kW | |||
| 1. Hallway. 3.18 m². Ceiling 2.8 m. Warmed floor on the ground. Above is an insulated attic. | One, South, the average degree of insulation. Leeward side | Not | One | 0.52 kW |
| 2. Hall. 6.2 m². Ceiling 2.9 m. Insulated floor on the ground. Above - insulated attic | Not | Not | Not | 0.62 kW |
| 3. Kitchen-dining room. 14.9 m². Ceiling 2.9 m. Well insulated floor on the ground. Svehu - insulated attic | Two. South, west. Average degree of insulation. Leeward side | Two, single-chamber double-glazed window, 1200 × 900 mm | Not | 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 glazing, 1400 × 1000 mm | Not | 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 | One, double-glazed window, 1400 × 1000 mm | Not | 1.73 kW |
| 6. Living room. 18.0 m². Ceiling 2.8 m. Well insulated floor. Top - insulated attic | Two, East, South. High degree of insulation. Parallel to wind direction | Four, double glazing, 1500 × 1200 mm | Not | 2.59 kW |
| 7. Bathroom combined. 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 | Not | 0.59 kW |
| TOTAL: |
Then, using the calculator below, we make a calculation for each room (already taking into account a 10% reserve). With the recommended app, it won't take long. After that, it remains to sum the obtained values \u200b\u200bfor each room - this will be the required total power of the heating system.
