Program for calculating thermal power by volume. Calculation of boiler power for heating a house
In any heating system using coolant liquid, its “heart” is the boiler. It is here that the energy potential of fuel (solid, gaseous, liquid) or electricity is converted into heat, which is transferred to the coolant, and is already distributed throughout all the heated rooms of the house or apartment. Naturally, the capabilities of any boiler are not unlimited, that is, they are limited by its technical and operational characteristics specified in the product data sheet.
One of the key characteristics is the thermal power of the unit. Simply put, it must be able to generate in a unit of time such an amount of heat that would be sufficient to fully heat all the rooms of a house or apartment. Selection suitable model“by eye” or according to some overly generalized concepts can lead to an error in one direction or another. Therefore, in this publication we will try to offer the reader, although not professional, but still with a fairly high degree of accuracy, an algorithm on how to calculate the power of a boiler for heating a house.
A trivial question - why know the required boiler power?
Despite the fact that the question really seems rhetorical, there is still a need to give a couple of explanations. The fact is that some owners of houses or apartments still manage to make mistakes, going to one extreme or another. That is, purchasing equipment either of obviously insufficient thermal performance, in the hope of saving money, or greatly overestimated, so that, in their opinion, they are guaranteed to provide themselves with heat in any situation with a large margin.
Both of these are completely wrong and have a negative impact on both the provision of comfortable living conditions and the durability of the equipment itself.
- Well, with insufficiency calorific value everything is more or less clear. When winter cold sets in, the boiler will begin to operate at full capacity, and it is not a fact that there will be a comfortable microclimate in the rooms. This means that you will have to “bring up the heat” with the help of electric heating devices, which will entail significant extra costs. And the boiler itself, operating at the limit of its capabilities, is unlikely to last long. In any case, after a year or two, homeowners will clearly realize the need to replace the unit with a more powerful one. One way or another, the cost of an error is quite impressive.
- Well, why not buy a boiler with a large reserve, what can this hinder? Yes, of course, high-quality heating of the premises will be provided. But now let’s list the “cons” of this approach:
Firstly, a higher-power boiler itself can cost significantly more, and it’s difficult to call such a purchase rational.
Secondly, with increasing power, the dimensions and weight of the unit almost always increase. These are unnecessary difficulties during installation, “stolen” space, which is especially important if the boiler is planned to be placed, for example, in the kitchen or in another room in the living area of the house.
Thirdly, you may encounter uneconomical operation of the heating system - part of the expended energy resources will be spent, in fact, in vain.
Fourthly, excess power means regular long shutdowns of the boiler, which, in addition, are accompanied by cooling of the chimney and, accordingly, abundant formation of condensate.
Fifth, if powerful equipment is never properly loaded, it does not benefit it. Such a statement may seem paradoxical, but it is so - wear becomes higher, the duration of trouble-free operation is significantly reduced.
Prices for popular heating boilers
Excess boiler power will be appropriate only if it is planned to connect a water heating system for household needs - an indirect heating boiler. Well, or when it is planned to expand the heating system in the future. For example, the owners plan to build a residential extension to the house.
Methods for calculating the required boiler power
In truth, it is always better to trust specialists to carry out thermal engineering calculations - there are too many nuances to take into account. But, it is clear that such services are not provided free of charge, so many owners prefer to take responsibility for choosing the parameters of boiler equipment.
Let's see what methods of calculating thermal power are most often offered on the Internet. But first, let’s clarify the question of what exactly should influence this parameter. This will make it easier to understand the advantages and disadvantages of each of the proposed calculation methods.
What principles are key when carrying out calculations?
So, the heating system faces two main tasks. Let us immediately clarify that there is no clear division between them - on the contrary, there is a very close relationship.
- The first is creating and maintaining a comfortable temperature in the premises. Moreover, this level of heating should extend to the entire volume of the room. Of course, due to physical laws, temperature gradation in height is still inevitable, but it should not affect the feeling of comfort in the room. It turns out that it should be able to warm up a certain volume of air.
The degree of temperature comfort is, of course, a subjective value, that is different people they can evaluate it in their own way. But it is still generally accepted that this indicator is in the range of +20 ÷ 22 °C. Typically, this is the temperature that is used when carrying out thermal calculations.
This is also evidenced by the standards established by the current GOST, SNiP and SanPiN. For example, the table below shows the requirements of GOST 30494-96:
| Room type | Air temperature level, °C | |
|---|---|---|
| optimal | acceptable | |
| Living spaces | 20÷22 | 18÷24 |
| Residential premises for regions with minimum winter temperatures of - 31 °C and below | 21÷23 | 20÷24 |
| Kitchen | 19÷21 | 18÷26 |
| Toilet | 19÷21 | 18÷26 |
| Bathroom, combined toilet | 24÷26 | 18÷26 |
| Office, recreation and study areas | 20÷22 | 18÷24 |
| Corridor | 18÷20 | 16÷22 |
| Lobby, staircase | 16÷18 | 14÷20 |
| Storerooms | 16÷18 | 12÷22 |
| Residential premises (the rest are not standardized) | 22÷25 | 20÷28 |
- The second task is the constant compensation of possible heat losses. Creating an “ideal” house in which there would be no heat leaks is a problem that is practically unsolvable. You can only reduce them to the bare minimum. And almost all elements of a building’s structure become leakage paths to one degree or another.
| Building design element | Approximate share of total heat losses |
|---|---|
| Foundation, plinth, floors of the first stage (on the ground or above an unheated basement) | from 5 to 10% |
| Joints building structures | from 5 to 10% |
| Areas where utilities pass through building structures (sewage pipes, water supply pipes, gas supply pipes, electrical or communication cables, etc.) | up to 5% |
| External walls, depending on the level of thermal insulation | from 20 to 30% |
| Windows and doors to the street | about 20÷25%, of which about half is due to insufficient sealing of boxes, poor fit of frames or canvases |
| Roof | up to 20% |
| Chimney and ventilation | up to 25÷30% |
Why were all these rather lengthy explanations given? But only so that the reader has complete clarity that when making calculations, willy-nilly, it is necessary to take into account both directions. That is, the “geometry” of the heated rooms of the house, and the approximate level of heat loss from them. And the amount of these heat leaks, in turn, depends on a number of factors. This is the difference in temperatures outside and in the house, and the quality of thermal insulation, and the features of the entire house as a whole and the location of each of its rooms, and other evaluation criteria.
You might be interested in information about which ones are suitable
Now, armed with this preliminary knowledge, let's move on to consider various methods calculating the required thermal power.
Calculation of power based on the area of heated premises
It is proposed to proceed from their conditional relationship that for high-quality heating of one square meter of room area it is necessary to consume 100 W of thermal energy. Thus, it will help to calculate which one:
Q=Stotal / 10
Q- the required thermal power of the heating system, expressed in kilowatts.
Stotal- the total area of the heated premises of the house, square meters.
However, reservations are made:
- First, the ceiling height of the room should be on average 2.7 meters, a range from 2.5 to 3 meters is allowed.
- Secondly, you can make an adjustment for the region of residence, that is, take not a rigid standard of 100 W/m², but a “floating” one:
That is, the formula will take a slightly different form:
Q=Stotal ×Qud / 1000
Qud - the value of the specific thermal power taken from the table shown above square meter area.
- Third - the calculation is valid for houses or apartments with an average degree of insulation of enclosing structures.
However, despite the mentioned reservations, such a calculation cannot be called accurate. Agree that it is largely based on the “geometry” of the house and its premises. But heat loss is practically not taken into account, except for the rather “blurred” ranges of specific thermal power by region (which also have very vague boundaries), and remarks that the walls should have an average degree of insulation.
But be that as it may, this method is still popular precisely for its simplicity.
It is clear that the operating power reserve of the boiler must be added to the obtained calculated value. You should not overestimate it - experts advise staying in the range from 10 to 20%. This, by the way, applies to all methods for calculating the power of heating equipment, which will be discussed below.
Calculation of the required thermal power by volume of premises
By and large, this method of calculation largely repeats the previous one. Is it true, original value here it is no longer the area, but the volume - essentially the same area, but multiplied by the height of the ceilings.
And the norms of specific thermal power adopted here are:
- for brick houses – 34 W/m³;
- For panel houses– 41 W/m³.
Even based on the proposed values (from their wording), it becomes clear that these standards were established for apartment buildings, and are used mainly to calculate the heat energy demand for premises connected to central system department or to an autonomous boiler station.
It is quite obvious that “geometry” is again being put at the forefront. And the entire system for accounting for heat losses comes down to only differences in the thermal conductivity of brick and panel walls.
In a word, this approach to calculating thermal power is also no different in accuracy.
Calculation algorithm taking into account the characteristics of the house and its individual premises
Description of the calculation method
So, the methods proposed above give only a general idea of the required amount of thermal energy for heating a house or apartment. They have a common weak point - almost complete ignorance of possible heat losses, which are recommended to be considered “average”.
But it is quite possible to carry out more accurate calculations. The proposed calculation algorithm will help with this, which is also embodied in the form of an online calculator, which will be offered below. Just before starting the calculations, it makes sense to consider step by step the very principle of their implementation.
First of all - important note. The proposed methodology involves assessing not the entire house or apartment by total area or volume, but each heated room separately. Agree that the rooms equal area, but differing, say, in the number external walls, will require and different quantities heat. It is impossible to put an equal sign between rooms that have a significant difference in the number and area of windows. And there are many such criteria for evaluating each of the rooms.
So it would be more correct to calculate the required power for each room separately. Well, then a simple summation of the obtained values will lead us to the desired indicator of the total thermal power for the entire heating system. That is, in fact, for her “heart” - the cauldron.
One more note. The proposed algorithm does not pretend to be “scientific”, that is, it is not directly based on any specific formulas established by SNiP or other governing documents. However, it has been tested by practical application and shows results with a high degree of accuracy. The differences with the results of professionally carried out thermal engineering calculations are minimal and do not in any way affect making the right choice equipment according to its rated thermal power.
The “architecture” of the calculation is as follows: the basic, already mentioned above value of specific thermal power, equal to 100 W/m², is taken, and then a whole series of correction factors is introduced, to one degree or another reflecting the amount of heat loss in a particular room.
If we express this in a mathematical formula, it will turn out something like this:
Qк= 0.1 × Sc× k1 × k2 × k3 × k4 × k5 × k6 × k7 × k8 × k9 × k10 × k11
Qк- the required thermal power required for full heating of a specific room
0.1 - conversion of 100 W to 0.1 kW, just for the convenience of obtaining the result in kilowatts.
Sk- area of the room.
k1 ÷k11- correction factors to adjust the result taking into account the characteristics of the room.
Presumably, there should be no problems with determining the area of the room. So let’s immediately move on to a detailed consideration of correction factors.
- k1 is a coefficient that takes into account the height of the ceilings in the room.
It is clear that the height of the ceilings directly affects the volume of air that the heating system must warm up. For calculation it is proposed to take following values correction factor:
- k2 is a coefficient that takes into account the number of walls of the room in contact with the street.
How larger area contact with external environment, the higher the level of heat loss. Everyone knows that a corner room is always much cooler than one with only one external wall. And some rooms of a house or apartment may even be internal, having no contact with the street.
In your mind, of course, you should take not only the number of external walls, but also their area. But our calculation is still simplified, so we will limit ourselves to only introducing a correction factor.
Odds for various cases are given in the table below:
We do not consider the case when all four walls are external. This is no longer a residential building, but just some kind of barn.
- k3 is a coefficient that takes into account the position of external walls relative to the cardinal points.
Even in winter, one should not discount the possible effects of solar energy. On a clear day, they penetrate through the windows into the rooms, thereby joining the general heat supply. In addition, the walls also receive a charge of solar energy, which leads to a reduction total number heat loss through them. But all this is true only for those walls that “see” the Sun. There is no such influence on the northern and northeastern sides of the house, for which a certain correction can also be made.
The values of the correction factor for the cardinal directions are in the table below:
- k4 is a coefficient that takes into account the direction of winter winds.
This amendment may not be mandatory, but for houses located in open areas, it makes sense to take it into account.
You might be interested in information about what they are
In almost any area there is a predominance of winter winds - this is also called the “wind rose”. Local meteorologists are required to have such a diagram - it is compiled based on the results of many years of weather observations. Quite often, the local residents themselves are well aware of which winds most often bother them in winter.
And if the wall of the room is located on the windward side, and is not protected by any natural or artificial barriers from the wind, then it will cool down much more. That is, and heat losses premises are increasing. This will be less pronounced near a wall located parallel to the direction of the wind, and to a minimum - located on the leeward side.
If you don’t want to “bother” with this factor, or there is no reliable information about the winter wind rose, then you can leave the coefficient equal to one. Or, on the contrary, take it as maximum, just in case, that is, for the most unfavorable conditions.
The values of this correction factor are in the table:
- k5 is a coefficient that takes into account the level of winter temperatures in the region of residence.
If you carry out thermal calculations According to all the rules, the assessment of heat losses is carried out taking into account the difference in temperatures indoors and outdoors. It is clear that the colder the climatic conditions of the region, the more heat required to be supplied to the heating system.
Our algorithm will also take this into account to a certain extent, but with an acceptable simplification. Depending on the level of minimum winter temperatures falling on the coldest ten-day period, the correction factor k5 is selected .
It would be appropriate to make one remark here. The calculation will be correct if temperatures that are considered normal for a given region are taken into account. There is no need to remember the abnormal frosts that happened, say, a few years ago (and that’s why, by the way, they were remembered). That is, the lowest but normal temperature for the given area should be selected.
- k6 is a coefficient that takes into account the quality of thermal insulation of walls.
It is quite clear what more efficient system insulation of walls, the lower the level of heat loss will be. Ideally, what we should strive for, thermal insulation should generally be complete, carried out on the basis of thermal calculations performed, taking into account the climatic conditions of the region and the design features of the house.
When calculating the required thermal power of the heating system, the existing thermal insulation of the walls should also be taken into account. The following gradation of correction factors is proposed:
In theory, an insufficient degree of thermal insulation or its complete absence should not be observed in a residential building. Otherwise, the heating system will be very expensive, and even without a guarantee of creating truly comfortable living conditions.
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If the reader wishes to independently assess the level of thermal insulation of his home, he can use the information and calculator that are posted in last section of this publication.
- k7 andk8 – coefficients taking into account heat loss through the floor and ceiling.
The next two coefficients are similar - their introduction into the calculation takes into account the approximate level of heat loss through the floors and ceilings of the premises. There is no need to describe in detail here - both the possible options and the corresponding values of these coefficients are shown in the tables:
For starters, the k7 coefficient, which adjusts the result depending on the characteristics of gender:
Now - the coefficient k8, which corrects for the proximity from above:
- k9 is a coefficient that takes into account the quality of windows in the room.
Here, too, everything is simple - the better the quality of the windows, the less heat loss through them. Old wooden frames, as a rule, do not have good thermal insulation characteristics. This situation is better with modern window systems equipped with double-glazed windows. But they can also have a certain gradation - according to the number of cameras in a double-glazed window and according to other design features.
For our simplified calculation, we can apply the following values of the k9 coefficient:
- k10 is a coefficient that corrects for the glazing area of the room.
The quality of windows does not yet fully reveal all the volumes of possible heat loss through them. Very great importance has a glass area. Agree, it’s difficult to compare a small window and a huge one panoramic window almost the entire wall.
To make adjustments for this parameter, you first need to calculate the so-called glazing coefficient of the room. This is not difficult - you simply find the ratio of the glazing area to the total area of the room.
kw =sw/S
kw- room glazing coefficient;
sw- total area of glazed surfaces, m²;
S- room area, m².
Anyone can measure and sum up the area of windows. And then it’s easy to find the required glazing coefficient by simple division. And it, in turn, makes it possible to go into the table and determine the value of the correction factor k10 :
| Glazing coefficient value kw | k10 coefficient value |
|---|---|
| - up to 0.1 | 0.8 |
| - from 0.11 to 0.2 | 0.9 |
| - from 0.21 to 0.3 | 1.0 |
| - from 0.31 to 0.4 | 1.1 |
| - from 0.41 to 0.5 | 1.2 |
| - over 0.51 | 1.3 |
- k11 is a coefficient that takes into account the presence of doors to the street.
The last of the considered coefficients. The room may have a door leading directly to the street, to a cold balcony, to an unheated corridor or entrance, etc. Not only is the door itself often a very serious “cold bridge” - when it is opened regularly, a fair amount of cold air will penetrate into the room each time. Therefore, an allowance should be made for this factor: such heat losses, of course, require additional compensation.
The values of the coefficient k11 are given in the table:
This coefficient should be taken into account if the doors are regularly used in winter.
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* * * * * * *
So, all correction factors have been considered. As you can see, there is nothing super complicated here, and you can safely move on to the calculations.
One more tip before starting the calculations. Everything will be much simpler if you first draw up a table, in the first column of which you sequentially indicate all the sealed rooms of the house or apartment. Next, place the data required for calculations in columns. For example, in the second column - the area of the room, in the third - the height of the ceilings, in the fourth - the orientation to the cardinal points - and so on. It’s not difficult to create such a sign if you have a plan of your residential property in front of you. It is clear that the calculated values of the required thermal power for each room will be entered in the last column.
The table can be compiled in an office application, or even simply drawn on a piece of paper. And don’t rush to part with it after making the calculations - the obtained thermal power indicators will still be useful, for example, when purchasing heating radiators or electric heating devices used as backup source heat.
To make the task of carrying out such calculations extremely simple for the reader, a special online calculator is located below. With it, with the initial data pre-collected in a table, the calculation will take literally a matter of minutes.
Calculator for calculating the required heating power for the premises of a house or apartment.
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 home. In this case, it is quite possible that it will go to two extremes: either the boiler power will not be enough - the equipment will begin to work “at full blast”, 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 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 of building structures | from 5 to 10% |
| Entry points for utilities (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 is calculated for each heated room, the obtained values are summed up, approximately 10% of the reserve is added (so that the equipment does not work at the limit of its capabilities) - and the result will show how much power the heating boiler is needed. And the values for each room will become Starting point for counting required quantity radiators.
The most simplified and most frequently used method in a non-professional environment is to adopt a norm of 100 W of thermal energy per square meter of area:
The most primitive way of calculating is the ratio of 100 W/m²
Q = S× 100
Q– required heating power for the room;
S– room area (m²);
100 — specific power per unit area (W/m²).
For example, a room 3.2 × 5.5 m
S= 3.2 × 5.5 = 17.6 m²
Q= 17.6 × 100 = 1760 W ≈ 1.8 kW
The method is obviously very simple, but very imperfect. It is worth mentioning right away that it is conditionally applicable only when standard height ceilings - approximately 2.7 m (acceptable - in the range from 2.5 to 3.0 m). From this point of view, the calculation will be more accurate not from the area, but from the volume of the room.
It is clear that in this case the power density is calculated at cubic meter. It is taken equal to 41 W/m³ for reinforced concrete panel house, or 34 W/m³ - in brick or made of other materials.
Q = S × h× 41 (or 34)
h– ceiling height (m);
41 or 34 – specific power per unit volume (W/m³).
For example, the same room in panel house, with a ceiling height of 3.2 m:
Q= 17.6 × 3.2 × 41 = 2309 W ≈ 2.3 kW
The result is more accurate, since it takes into account not only everything linear dimensions premises, but even, to a certain extent, the features of the walls.
But still, it is still far from real accuracy - many nuances are “outside the brackets”. How to perform calculations closer to real conditions is in the next section of the publication.
You may be interested in information about what they are
Carrying out calculations of the required thermal power taking into account the characteristics of the premises
The calculation algorithms discussed above can be useful for an initial “estimate,” but you should still rely on them completely with great caution. Even to a person who does not understand anything about building heating engineering, the indicated average values may certainly seem dubious - they cannot be equal, say, for the Krasnodar Territory 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, 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 - 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 the area through which heat loss occurs. In addition, the presence of two or more external walls also means corners - extremely vulnerable places from the point of view of the formation of “cold bridges”. Coefficient “a” will correct for this specific feature of the room.
The coefficient is taken equal to:
— external walls No (interior space): a = 0.8;
- external wall one: a = 1.0;
— external walls two: a = 1.2;
— external walls three: a = 1.4.
- “b” is a coefficient that takes into account the location of the external walls of the room relative to the cardinal directions.
You might be interested in information about what types of
Even on the coldest winter days solar energy still has an impact on the temperature balance in the building. It is quite natural that the side of the house that faces south receives some heat from the sun's rays, and heat loss through it is lower.
But walls and windows facing north “never see” the Sun. East End at home, although he “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 - a graphic diagram showing the prevailing wind directions in the winter and summer seasons. 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 that takes into account the climatic conditions of the 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 lowest temperatures characteristic of the coldest five-day period of the year (usually this is typical for 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 living conditions in a room depends on the quality of their thermal insulation.
The value of the coefficient for our calculations can be taken as follows:
— external walls do not have insulation: e = 1.27;
- average degree of insulation - walls made of two bricks or their surface thermal insulation is provided with other insulation materials: e = 1.0;
— insulation was carried out with high quality, based on thermal engineering calculations: e = 0.85.
Below in the course of this publication, recommendations will be given on how to determine the degree of insulation of walls and other building structures.
- coefficient "f" - correction for ceiling heights
Ceilings, especially in private homes, can have different heights. Therefore, the thermal power to warm up a particular room of the same area will also differ in this parameter.
It would not be a big mistake to accept the following values for the correction factor “f”:
— ceiling heights up to 2.7 m: f = 1.0;
— flow height from 2.8 to 3.0 m: f = 1.05;
- ceiling heights from 3.1 to 3.5 m: f = 1.1;
— ceiling heights from 3.6 to 4.0 m: f = 1.15;
- ceiling height more than 4.1 m: f = 1.2.
- « g" is a coefficient that takes into account the type of floor or room located under the ceiling.
As shown above, the floor is one of the significant sources of heat loss. This means that it is necessary to make some adjustments to account for this feature of a particular room. The correction factor “g” can be taken equal to:
- cold floor on the ground or above 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 heating 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 structure itself. 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, 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 total area room glazing
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 /SP
∑ SOK– total area of windows in the room;
SP– area of the room.
Depending on the obtained value, the correction factor “j” is determined:
— x = 0 ÷ 0.1 →j = 0,8 ;
— x = 0.11 ÷ 0.2 →j = 0,9 ;
— x = 0.21 ÷ 0.3 →j = 1,0 ;
— x = 0.31 ÷ 0.4 →j = 1,1 ;
— x = 0.41 ÷ 0.5 →j = 1,2 ;
- « k" - coefficient that corrects for the presence of an entrance door
A door to the street or to an unheated balcony is always an additional “loophole” for the cold
Door to the street or open balcony is capable of making adjustments to the thermal balance of the room - each opening of it is accompanied by the penetration of a considerable volume of cold air into the room. Therefore, it makes sense to take into account its presence - for this we introduce the coefficient “k”, which we take equal to:
- no door: k = 1,0 ;
- one door to the street or to the balcony: k = 1,3 ;
- two doors to the street or balcony: k = 1,7 .
- « l" - possible amendments to the heating radiator connection diagram
Perhaps this may seem like an insignificant detail to some, but still, why not immediately take into account the planned connection diagram for 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 with different types of insertion of supply and return pipes.
| Illustration | Radiator insert type | The value of the coefficient "l" |
|---|---|---|
 | Diagonal connection: supply from above, return from below | l = 1.0 |
 | Connection on one side: supply from above, return from below | l = 1.03 |
 | Two-way connection: both supply and return from below | l = 1.13 |
 | Diagonal connection: supply from below, return from above | l = 1.25 |
 | Connection on one side: supply from below, return from above | l = 1.28 |
 | One-way connection, both supply and return from below | l = 1.28 |
- « m" - correction factor for the peculiarities of the installation location of heating radiators
And finally, the last coefficient, which is also related to the peculiarities of connecting heating radiators. It is probably clear that if the battery is installed openly and is not blocked by anything from above or from the front, then it will give maximum heat transfer. However, such an installation is not always possible - more often the radiators are partially hidden by window sills. Other options are also possible. In addition, some owners, trying to fit heating elements into the created interior ensemble, hide them completely or partially with decorative screens - this also significantly affects the thermal output.
If there are certain “outlines” of how and where radiators will be mounted, this can also be taken into account when making calculations by introducing a special coefficient “m”:
| 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. The climatic features of the region are easy to clarify. 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 windows, 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 required total power of the heating system.
The result for each room, by the way, will help you choose the right number of heating radiators - all that remains is to divide by the specific thermal power of one section and round up.
A competent choice of boiler will allow you to save comfortable temperature indoor air in the winter season. Big choice devices allows you to most accurately select the desired model depending on the required parameters. But in order to ensure heat in the house and at the same time avoid unnecessary waste of resources, you need to know how to calculate power gas boiler for heating a private house.
A gas boiler floor type has more power Source termoresurs.ru
The main characteristics affecting the boiler power
The boiler power indicator is the main characteristic, however, the calculation can be carried out according to different formulas, depending on the device configuration and other parameters. For example, a detailed calculation may take into account the height of the building and its energy efficiency.
Varieties of boiler models
Boilers can be divided into two types depending on the purpose of application:
Single-circuit– used only for heating;
Dual-circuit– used for heating, as well as in hot water supply systems.
Units with one circuit have a simple structure, consisting of a burner and a single heat exchanger.
In dual-circuit systems, the water heating function is primarily provided. When using hot water, heating is automatically switched off for the duration of use hot water so that the system does not overload. The advantage of a dual-circuit system is its compactness. Such a heating complex takes up much less space than if the supply systems hot water and heating were used separately.
Boiler models are often divided by placement method.
Depending on their type, boilers can be installed in different ways. You can choose a wall-mounted or floor-mounted model. It all depends on the preferences of the owner of the house, the capacity and functionality of the room in which the boiler will be located. The installation method of the boiler is also affected by its power. For example, floor-standing boilers have more power compared to wall-mounted models.
In addition to fundamental differences in purposes of application and methods of placement gas boilers They also differ in their methods of control. There are models with electronic and mechanical control. Electronic systems can only operate in homes with constant access to the power grid.
On our website you can find contacts of construction companies that offer house insulation service. You can communicate directly with representatives by visiting the “Low-Rise Country” exhibition of houses.
Typical power calculations for devices
There is no single algorithm for calculating both single- and double-circuit boilers - each system must be selected separately.
Formula for a typical project
When calculating the required power to heat a house built according to standard project, that is, with a room height of no more than 3 meters, the volume of the room is not taken into account, and the power indicator is calculated as follows:
Determine the specific thermal power: Um = 1 kW/10 m 2 ;
Rm = Mind * P * Kr, where
P – a value equal to the sum of the areas of heated premises,
Kr – correction factor, which is taken in accordance with climatic zone, in which the building is located.
Some coefficient values for different regions of Russia:
Southern – 0.9;
Located in the middle zone – 1.2;
Northern – 2.0.
For the Moscow region, a coefficient value of 1.5 is taken.
This technique does not reflect the main factors influencing the microclimate in the house, and only approximately shows how to calculate the power of a gas boiler for a private home.
Some manufacturers issue recommendations, but for accurate calculations they still recommend contacting specialists Source parki48.ru
Example calculation for a single-circuit device installed in a room with an area of 100 m2, located in the Moscow region:
Рм = 1/10 * 100 * 1.5 = 15 (kW)
Calculations for dual-circuit devices
Double-circuit devices have the following operating principle. For heating, water is heated and supplied through the heating system to radiators, which release heat environment e, thus heating the rooms and cooling them. When cooling, the water flows back to be heated. Thus, water circulates around the circuit heating system, and goes through heating cycles and transfer to radiators. At the moment when the ambient temperature becomes equal to the set one, the boiler goes into standby mode for some time, i.e. Temporarily stops heating the water, then starts heating again.
For domestic needs, the boiler heats water and supplies it to the taps, and not to the heating system.
When calculating the power of a device with two circuits, another 20% of the calculated value is usually added to the resulting power.
An example of calculation for a two-circuit device that is installed in a room with an area of 100 m2; the coefficient is taken for the Moscow region:
R m = 1/10 * 100 * 1.5 = 15 (kW)
P total = 15 + 15*20% = 18 (kW)
Additional factors taken into account when installing the boiler
In construction, there is also the concept of energy efficiency of a building, that is, how much heat a building releases to the environment.
One of the indicators of heat transfer is the dissipation coefficient (Kp). This value is a constant, i.e. constant and does not change when calculating the level of heat transfer of structures made of the same materials.
It is necessary to take into account not only the power of the boiler, but also the possible heat loss of the building itself Source pechiudachi.ru
For calculations, a coefficient is taken, which, depending on the building, can be equal to different sizes and the use of which will help you understand how to calculate the power of a gas boiler for a home more accurately:
Most low level heat transfer corresponding to a K p value of 0.6 to 0.9 is assigned to buildings made of modern materials, with insulated floors, walls and roofs;
K p is equal to from 1.0 to 1.9, if the external walls of the building are insulated, the roof is insulated;
K p is equal to from 2.0 to 2.9 in houses without insulation, for example, brick houses with single masonry;
K p is equal to from 3.0 to 4.0 in non-insulated rooms, in which the level of thermal insulation is low.
Heat loss level QT calculated according to the formula:
Q T = V * P t *k/860 where
V – is the volume of the room
Pt- R temperature difference calculated by subtracting the minimum possible air temperature in the region from the desired room temperature,
k – safety factor.
The boiler power, when taking into account the dissipation coefficient, is calculated by multiplying the calculated level of heat loss by the safety factor (usually from 15% to 20%, then multiply by 1.15 and 1.20, respectively)
This technique allows you to more accurately determine productivity and, therefore, approach the issue of choosing a boiler as efficiently as possible.
What happens if you incorrectly calculate the required power
It is still worth choosing a boiler so that it matches the power required to heat the building. This will be the most the best option, since, first of all, purchasing a boiler that does not match the power level can lead to two types of problems:
A low-power boiler will always work at the limit, trying to heat the room to set temperature, and can quickly fail;
Appliance with excessive high level power costs more and even in economy mode consumes more gas than a less powerful device.
Calculator for calculating boiler power
For those who do not like to do calculations, even if they are not very complicated, a special calculator will help you calculate a boiler for heating your home - a free online application.
Interface online calculator boiler power calculation Source idn37.ru
As a rule, the calculation service requires you to fill out all the fields, which will help you make the most accurate calculations, including the power of the device and the thermal insulation of the house.
To obtain the final result, you will also need to enter the total area that will require heating.
Next, you should fill out information about the type of glazing, the level of thermal insulation of walls, floors and ceilings. As additional parameters, the height at which the ceiling is located in the room is also taken into account, and information about the number of walls interacting with the street is entered. The number of floors of the building and the presence of structures on top of the house are taken into account.
After entering the required fields, the calculation button becomes “active” and you can get the calculation by clicking on the corresponding button. To check the information received, you can use calculation formulas.
Video description
To see how to calculate the power of a gas boiler, watch the video:
Advantages of using gas boilers
Gas equipment has a number of advantages and disadvantages. The advantages include:
possibility of partial automation of the boiler operation process;
unlike other energy sources, natural gas has low cost;
The devices do not require frequent maintenance.
To the disadvantages gas systems consider the gas to be highly explosive, however proper storage gas cylinders, timely Maintenance, this risk is minimal.
On our website you can familiarize yourself with construction companies, which offer services for connecting electrical and gas equipment. You can communicate directly with representatives at the Low-Rise Country exhibition of houses.
Conclusion
Despite the apparent simplicity of the calculations, we must remember that gas equipment should be selected and installed by professionals. In this case, you will receive a trouble-free device that will work properly for many years.