How to calculate the power of a boiler for heating a house. How to choose a gas boiler: we choose a boiler according to the criteria for heating a private house How to calculate a gas boiler
Fulfilling repair work, which are associated with the replacement of heating equipment, or when designing a heating system for a new house, you must be able to calculate the heat output for the planned heating system. It is this calculation that will make it possible to make a decision that can provide optimal, efficient and economical heating of all housing. How to calculate the power of a gas boiler, and how much information is needed for this, is indicated in this review.
TMK - What is this indicator and how to work with it
However, this value in itself does not give any idea of what area of the premises can be heated using this boiler. It is also not clear how external factors will affect heat consumption, and how much heat will be spent to cover objective heat losses in each specific case.
Taking into account all the circumstances under which the heating system will operate will make it possible to determine how much heat power should be transferred to external devices in order to provide the owners with the heat they need in the house.
It is necessary to start the calculation with the simplest.
Calculation of the required heat output by area
Preliminary information about required power a gas boiler can be obtained by making a simple calculation of the power of a gas boiler by area using the formula:
Boiler power \u003d Heated area (sq.m.) x Boiler specific power / 10
The specific power of a gas boiler (UMK) is a value calculated for each region of Russia, which is:
The obtained MK is relevant for single-circuit boilers that provide only heating.
Thus, if it is necessary to heat a residential building of 100 sq.m. in the Moscow region, then the calculation of the gas boiler for the area of \u200b\u200bthe house will look like this:
100×1.5/10 = 15 kW
But do not rush to look gas boilers fifteen watts. It is necessary to determine the sources of heat loss and the total heat loss of the building or apartment. Building codes determine that heat losses occur through all the enclosures of the premises (walls, windows, doors, ceilings, floors).
General formula for determining heat loss for building envelopes
Heat loss coefficient = heat transfer coefficient of the enclosure multiplied by the total area of the enclosure and the difference between the room's internal temperature and outdoor temperature environment.
- All heat loss and heat transfer coefficients are measured in W / (m.kv * C).
- The area of enclosing structures is calculated according to the project.
- The lowest possible ambient temperatures for a particular region are published in information guides.
- The internal temperature is determined by the order of the customer of the construction or repair work.
- Determination of heat loss through walls and ceiling - the table shows the thermal conductivity of the main materials
To calculate the heat loss through the walls and ceiling, it is necessary to determine the coefficient of thermal conductivity of the building materials that make up these enclosing structures, and the thickness of each layer of a certain building material.
To calculate it, you will need the following indicators:
- a(vn) is a coefficient that determines the intensity of heat transfer from the internal air in the room to the walls and ceiling. Usually a constant value is taken - 8.7;
- a (нр) is a coefficient that determines the intensity of heat transfer from the walls and ceiling to the outside air. Usually a constant value is taken - 23 (for heated rooms).
- k - thermal conductivity of building materials from which the walls and ceiling are made;
- d - the thickness of each layer of building materials.
The formula for calculating the thermal conductivity coefficient:
The calculation is made separately for the walls and separately for the ceiling.
- K (st) - the heat transfer coefficient of glass or double-glazed window determined by the manufacturer;
- F(st) - area of glass or double-glazed window;
- K(p) - the heat transfer coefficient of the frame determined by the manufacturer;
- F(p) - frame area;
- P is the perimeter of the glass.
Calculation: K (windows) \u003d K (st) * F (st) + K (p) * F (p) + P / F (windows)
The coefficient of thermal conductivity for doors is also calculated. Only instead of the values for the materials from which the windows are made, the values for the materials from which the doors are made are substituted.
An unheated floor gives a heat loss of approximately 10%, and the calculation is made using the same formula that calculates the heat loss of walls and ceilings. The same formula for calculating the thermal conductivity of the floor.
However, there is a subtlety in the calculation of thermal conductivity for each zone of the floor. There are four zones in total and they are located in the direction of movement from the outer walls to the center of the room.
Average heat loss values for building envelopes
On average, heat loss is determined by:
- through windows and doors - up to 50% of heat;
- through walls and ceiling - 15%;
- through the floor - 10%.
Using the entire amount of information listed, you can independently draw conclusions about the state of thermal insulation of the house and, if necessary, take measures to insulate certain building envelopes.
Having received information about how much heat generated by the gas boiler will go to heat loss, it is necessary to correct the indicator that gave the calculation of the power of the gas heating boiler from the area. To do this, the preliminary power of the boiler is multiplied by the heat loss coefficient - 0.75.
Those who do not have the ability to independently make complex calculations can use the power calculator. However, before calculating the power of a gas boiler with a calculator, it is necessary to measure the building structures of the house (according to the current technical plan or directly on the object, using a laser line).
Boiler power selection - video
Calculation of the power of a gas boiler depending on the equipment
There is no such parameter in the heating system that would not affect the determination of the required heat output of a gas boiler:
- specifications the boiler itself and heating equipment;
- the use of the boiler not only for heating, but also for heating water;
- boiler draft type;
- type of use of heat of combustion of fuel.
All of the above should be taken into account in the process of finding the answer to the question of how to choose the right gas heating boiler.
Technical characteristics of the boiler and its heat output:
- the larger the boiler heat exchanger, the more heat output will be spent on heating the coolant;
- depending on what the heat exchanger is made of - cast iron, steel or copper, it is necessary to determine the operating mode of the boiler, since the listed materials have different inertia;
- a double-circuit boiler (designed not only for heating, but also for heating water) will take up to 25% of the heat output specifically for hot water supply (hot water supply);
- if the working type of boiler draft is forced, then the thermal power of such a boiler is higher than that of a boiler with natural draft;
- a condensing gas boiler produces more heat than a convection one, its efficiency is about 110%, respectively, there will be much less loss of rated heat output;
- boiler automation must regulate the temperature of the heat carrier and, accordingly, the supplied heat output.
Calculation of the power of a wall and floor gas boiler
For small residential premises or residential buildings, you can choose a wall-mounted gas boiler. These boilers are classified as low-power, but they are much more economical. In addition, the wall-mounted boiler is sold with all accessories: a pump, expansion tank, measuring instruments etc. A complete set of heating equipment provides the least loss of heat produced and the highest efficiency.
Equipment for floor boilers is determined by the designers and purchased separately. With any miscalculations in the project, the entire heating system will malfunction.
How to choose a gas boiler for a cottage
150*1.5/10=22.5kW;
Accounting for the heat loss coefficient can be taken in half of the calculated value, since the maximum indicators were taken for its calculation;
22.5 kW * 0.3 = 6.75 kW;
22.5 kW + 6.75 k. W = 29.25 k. W - the calculated thermal power of the gas boiler.
An important role in calculating the required power of a gas boiler is played by the technical characteristics of pipes and radiators. The slower the coolant cools, the more efficiency all heating system.
In any heating system using a liquid heat carrier, its “heart” is the boiler. It is here that the energy potential of the fuel (solid, gaseous, liquid) or electricity is converted into heat, which is transferred to the coolant, and is already carried by it to all heated rooms of the house or apartment. Naturally, the possibilities of any boiler are not unlimited, that is, they are limited by its technical and operational characteristics indicated in the product passport.
One of key features is the heat output of the unit. Simply put, it must be able to produce in a unit of time such an amount of heat that would be sufficient to fully heat all the premises of a house or apartment. 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 boiler power for heating a house.
A banal question - why know the required boiler power
Despite the fact that the question does seem rhetorical, it still seems necessary to give a couple of explanations. The fact is that some owners of houses or apartments still manage to make mistakes, falling into one or another extreme. That is, purchasing equipment of either obviously insufficient thermal performance, in the hope of saving money, or greatly overestimated, so that, in their opinion, it is guaranteed, with a large margin, to provide themselves with heat in any situation.
Both are completely wrong, and negatively affect 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. With the onset of winter cold weather, the boiler will operate at its full capacity, and it is not a fact that there will be a comfortable microclimate in the rooms. This means that you will have to “catch up with heat” with the help of electric heaters, which will entail considerable extra costs. And the boiler itself, functioning at the limit of its capabilities, is unlikely to last long. In any case, after a year or two, homeowners clearly realize the need to replace the unit with a more powerful one. One way or another, the cost of a mistake is quite impressive.
- Well, why not buy a boiler with a large margin, what can prevent it? Yes, of course, high-quality space heating will be provided. But now we list the "cons" of this approach:
Firstly, a boiler of greater power can cost much more in itself, and it is difficult to call such a purchase rational.
Secondly, with increasing power, the dimensions and weight of the unit almost always increase. These are unnecessary installation difficulties, “stolen” space, which is especially important if the boiler is planned to be placed, for example, in the kitchen or in another room in the living area of the house.
Thirdly, you may encounter uneconomical operation of the heating system - part of the energy spent will be spent, in fact, wasted.
Fourthly, excess power is regular long shutdowns of the boiler, which, in addition, are accompanied by cooling of the chimney and, accordingly, abundant formation of condensate.
Fifth, if powerful equipment is never properly loaded, it does not benefit him. Such a statement may seem paradoxical, but it is true - wear becomes higher, the duration of trouble-free operation is significantly reduced.
Prices for popular heating boilers
Excess boiler power will be appropriate only if it is planned to connect a water heating system to it for economic needs– boiler indirect heating. Well, or when it is planned to expand the heating system in the future. For example, in the plans of the owners - the construction of a residential extension to the house.
Methods for calculating the required boiler power
In truth, it is always better to entrust the conduct of heat engineering calculations to specialists - there are too many nuances to take into account. But, it is clear that such services are not provided free of charge, so many owners prefer to take responsibility for choosing the parameters of boiler equipment.
Let's see what methods of calculating thermal power are most often offered on the Internet. But first, let's clarify the question of what exactly should affect this parameter. So it will be easier to understand the advantages and disadvantages of each of the proposed calculation methods.
What principles are key in making calculations
So, the heating system faces two main tasks. Let us immediately clarify that there is no clear division between them - on the contrary, there is a very close relationship.
- The first is the creation and maintenance of a comfortable temperature for living in the premises. Moreover, this level of heating should apply to the entire volume of the room. Of course, due to physical laws, temperature gradation in height is still inevitable, but it should not affect the feeling of comfort in the room. It turns out that it should be able to warm up a certain volume of air.
The degree of temperature comfort is, of course, a subjective value, that is different people they can evaluate it in their own way. But still, it is generally accepted that this indicator is in the region of +20 ÷ 22 ° С. Usually, it is precisely this temperature that is used during thermal engineering calculations.
This is also indicated by the standards established by the current GOST, SNiP and SanPiN. For example, the table below shows the requirements of GOST 30494-96:
| Room type | Air temperature level, °С | |
|---|---|---|
| optimal | admissible | |
| Living spaces | 20÷22 | 18:24 |
| Residential premises for regions with minimum winter temperatures from -31 °С and below | 21÷23 | 20÷24 |
| Kitchen | 19:21 | 18:26 |
| Toilet | 19:21 | 18:26 |
| Bathroom, combined bathroom | 24÷26 | 18:26 |
| Office, recreation and study rooms | 20÷22 | 18:24 |
| Corridor | 18:20 | 16:22 |
| lobby, stairwell | 16÷18 | 14:20 |
| Storerooms | 16÷18 | 12÷22 |
| Residential premises (the rest are not standardized) | 22÷25 | 20÷28 |
- The second task is the constant compensation of possible heat losses. To create an “ideal” house in which there would be no heat leakage is a problem of problems, practically unsolvable. You can only reduce them to the ultimate minimum. And almost all elements of the building structure become leakage paths to one degree or another.
| Building element | Approximate share of total heat loss |
|---|---|
| Foundation, basement, floors of the first floor (on the ground or over an unheated basement) | from 5 to 10% |
| Joints of building structures | from 5 to 10% |
| Sections of the passage of engineering communications through building structures (sewerage, water supply, gas supply pipes, electrical or communication cables, etc.) | up to 5% |
| External walls, depending on the level of thermal insulation | from 20 to 30% |
| Windows and doors to the street | about 20÷25%, of which about half - due to insufficient sealing of boxes, poor fit of frames or canvases |
| Roof | up to 20% |
| Chimney and ventilation | up to 25÷30% |
Why were all these rather lengthy explanations given? And only in order for the reader to have complete clarity that in the calculations, willy-nilly, it is necessary to take into account both directions. That is, the "geometry" of the heated premises of the house, and the approximate level of heat loss from them. And the amount of these heat leaks, in turn, depends on a number of factors. This is the temperature difference in the street and in the house, and the quality of thermal insulation, and the features of the whole house as a whole and the location of each of its premises, and other evaluation criteria.
You might be interested in information on which are suitable
Now, armed with this preliminary knowledge, we turn to the consideration various methods calculation of the required thermal power.
Calculation of power by the area of heated premises
It is proposed to proceed from their conditional ratio, that for high-quality heating of one square meter of the area of the room it is necessary to spend 100 W of thermal energy. Thus, it will help to calculate which:
Q=Stotal / 10
Q- the required thermal power of the heating system, expressed in kilowatts.
Stot- the total area of the heated premises of the house, square meters.
However, there are caveats:
- The first - the ceiling height of the room should be on average 2.7 meters, a range of 2.5 to 3 meters is allowed.
- The second - you can make an adjustment for the region of residence, that is, take not a rigid norm of 100 W / m², but a “floating” one:
That is, the formula will take a slightly different form:
Q=Stot ×Qud / 1000
Qud - taken from the table shown above, the value of the specific heat output per square meter area.
- Third - the calculation is valid for houses or apartments with an average degree of insulation of enclosing structures.
However, despite the above reservations, such a calculation cannot be called accurate. Agree that it is largely based on the "geometry" of the house and its premises. But heat losses are practically not taken into account, except for the rather “blurred” ranges of specific thermal power by region (which are also with very vague boundaries), and remarks that the walls should have an average degree of insulation.
But be that as it may, this method is still popular, precisely for its simplicity.
It is clear that it is necessary to add the operating power reserve of the boiler to the calculated value obtained. It should not be excessively overestimated - experts advise stopping at a range of 10 to 20%. This, by the way, applies to all methods for calculating the power of heating equipment, which will be discussed below.
Calculation of the required heat output by the volume of the premises
By and large, this method of calculation largely repeats the previous one. True, the initial value here is no longer the area, but the volume - in fact, the same area, but multiplied by the height of the ceilings.
And the norms of specific thermal power here are accepted as follows:
- for brick houses - 34 W / m³;
- for panel houses - 41 W / m³.
Even based on the proposed values (from their wording), it becomes clear that these norms were established for apartment buildings, and are mainly used to calculate the heat demand for premises connected to central system branch or to an autonomous boiler station.
It is quite obvious that "geometry" is again put at the forefront. And the whole system for accounting for heat losses comes down only to differences in the thermal conductivity of brick and panel walls.
In a word, this approach to calculating thermal power also does not differ in accuracy.
Calculation algorithm taking into account the characteristics of the house and its individual premises
Description of the calculation method
So, the methods proposed above give only a general idea of the required amount of thermal energy for heating a house or apartment. They have a common vulnerability - the almost complete disregard for possible heat losses, which are recommended to be considered "average".
But it is quite possible to carry out more precise calculations. This will help the proposed calculation algorithm, which is embodied, in addition, in the form of an online calculator, which will be proposed below. Just before starting the calculations, it makes sense to consider step by step the very principle of their implementation.
First of all, an important note. The proposed methodology involves the assessment not of the entire house or apartment in terms of total area or volume, but of each heated room separately. Agree that rooms of equal area, but differing, say, in the number of external walls, will require a different amount of heat. It is impossible to put an equal sign between rooms that have a significant difference in the number and area of windows. And there are many such criteria for evaluating each of the rooms.
So it would be more correct to calculate the required power for each of the premises separately. Well, then a simple summation of the obtained values \u200b\u200bwill lead us to the desired indicator of the total heat output for the entire heating system. That is, in fact, for its "heart" - the boiler.
One more note. The proposed algorithm does not claim to be "scientific", that is, it is not directly based on any specific formulas established by SNiP or other governing documents. However, it has been field tested and shows results with a high degree of accuracy. Differences with the results of professionally carried out heat engineering calculations are minimal, and do not affect the correct choice of equipment in terms of its rated thermal power.
The “architecture” of the calculation is as follows - the base value of the specific thermal power mentioned above is taken, equal to 100 W / m², and then a whole series of correction factors is introduced, to one degree or another reflecting the amount of heat loss in a particular room.
If this is expressed mathematical formula, then it will look something like this:
Qk= 0.1 × Sk× k1 × k2 × k3 × k4 × k5 × k6 × k7 × k8 × k9× k10 × k11
Qk- the desired thermal power required for the full heating of a particular room
0.1 - translation of 100 W into 0.1 kW, just for the convenience of obtaining the result in kilowatts.
Sk- area of the room.
k1 hk11- correction factors for adjusting the result, taking into account the characteristics of the room.
With the determination of the area of \u200b\u200bthe room, presumably, there should be no problems. So let's move on to a detailed discussion of the correction factors.
- k1 is a coefficient that takes into account the height of the ceilings in the room.
It is clear that the height of the ceilings directly affects the amount of air that the heating system must warm up. For the calculation, it is proposed to accept the following values of the correction factor:
- k2 is a coefficient that takes into account the number of walls in the room that are in contact with the street.
The larger the area of contact with the external environment, the higher the level of heat loss. Everyone knows that it is always much cooler in a corner room than in a room with only one outer wall. And some rooms of a house or apartment may even be internal, not having contact with the street.
According to the mind, of course, one should take not only the number of external walls, but also their area. But our calculation is still simplified, so we restrict ourselves only to the introduction of a correction factor.
The coefficients for various cases are shown in the table below:
The case when all four walls are external is not considered. This is no longer a residential building, but just some kind of barn.
- k3 is a coefficient that takes into account the position of the outer walls relative to the cardinal points.
Even in winter, you should not discount the possible impact of the energy of the sun's rays. On a clear day, they penetrate through the windows into the premises, thereby being included in the overall heat supply. In addition, the walls receive a charge of solar energy, which leads to a decrease in the total amount of heat loss through them. But all this is true only for those walls that "see" the Sun. There is no such influence on the north and northeast side of the house, which can also be corrected.
The values of the correction factor for the cardinal points are in the table below:
- k4 is a coefficient that takes into account the direction of winter winds.
Perhaps this amendment is not mandatory, but for houses located in open areas, it makes sense to take it into account.
You may be interested in information about what they are
In almost any area there is a predominance of winter winds - this is also called the "wind rose". Local meteorologists must have such a scheme - it is compiled based on the results of many years of weather observations. Quite often, the locals themselves are well aware of which winds most often disturb them in winter.
And if the wall of the room is located on the windward side, and is not protected by any natural or artificial barriers from the wind, then it will cool much more. That is, the heat loss of the room increases. To a lesser extent, this will be expressed near the wall located parallel to the direction of the wind, and to a minimum - located on the leeward side.
If there is no desire to "bother" with this factor, or there is no reliable information about the winter wind rose, then you can leave the coefficient equal to one. Or, on the contrary, take it to the maximum, just in case, that is, for the most unfavorable conditions.
The values of this correction factor are in the table:
- k5 is a coefficient that takes into account the level of winter temperatures in the region of residence.
If carried out thermotechnical calculations according to all the rules, then the assessment of heat losses is carried out taking into account the temperature difference in the room and on the street. It is clear that the colder the climatic conditions of the region, the more heat is required to be supplied to the heating system.
In our algorithm, this will also be in some degree taken into account, but with an acceptable simplification. Depending on the level of minimum winter temperatures falling on the coldest decade, a correction factor k5 is selected .
Here it would be appropriate to make one remark. The calculation will be correct if temperatures are taken into account, which are considered normal for a given region. There is no need to recall the anomalous frosts that happened, say, a few years ago (and that's why, by the way, they are remembered). That is, the lowest, but normal temperature for the area should be selected.
- k6 is a coefficient that takes into account the quality of the thermal insulation of the walls.
It is quite clear that what more efficient system wall insulation, the lower the level of heat loss. Ideally, to which one should strive, thermal insulation in general should be complete, carried out on the basis of the performed thermal engineering calculations, taking into account the climatic conditions of the region and the design features of the house.
When calculating the required heat output of the heating system, the existing thermal insulation of the walls should also be taken into account. The following gradation of correction factors is proposed:
An insufficient degree of thermal insulation or its complete absence, in theory, should not be observed at all in a residential building. Otherwise, the heating system will be very expensive, and even without a guarantee of creating really comfortable living conditions.
You might be interested in information about the heating system
If the reader wishes to independently assess the level of thermal insulation of his home, he can use the information and calculator that are located in the last section of this publication.
- k7 andk8 - coefficients that take into account heat loss through the floor and ceiling.
The following two coefficients are similar - their introduction into the calculation takes into account the approximate level of heat loss through the floors and ceilings of the premises. There is no need to describe in detail here - both the possible options and the corresponding values of these coefficients are shown in the tables:
To begin with, the coefficient k7, which corrects the result depending on the characteristics of the floor:
Now - the coefficient k8, which corrects for the neighborhood from above:
- k9 is a coefficient that takes into account the quality of the windows in the room.
Here, too, everything is simple - the better the windows, the less heat loss through them. Old wooden frames usually do not have good thermal insulation properties. This is better with modern window systems equipped with double-glazed windows. But they can also have a certain gradation - according to the number of cameras in a double-glazed window and according to other design features.
For our simplified calculation, the following values of the coefficient k9 can be applied:
- k10 is a coefficient that corrects for the room's glazing area.
The quality of windows does not yet fully reveal all the volumes of possible heat loss through them. Highly great importance has a glazed area. Agree, it is difficult to compare a small window and a huge panoramic window almost the entire wall.
To make an adjustment for this parameter, first you need to calculate the so-called room glazing coefficient. It's easy - just find the ratio of the glazing area to the total area of the room.
kw =sw/S
kw- coefficient of glazing of the room;
sw- total area of glazed surfaces, m²;
S- room area, m².
Anyone can measure and sum the area of windows. And then it is easy to find the desired glazing coefficient by simple division. And he, in turn, makes it possible to enter the table and determine the value of the correction factor k10 :
| Value of glazing factor kw | The value of the coefficient k10 |
|---|---|
| - up to 0.1 | 0.8 |
| - from 0.11 to 0.2 | 0.9 |
| - from 0.21 to 0.3 | 1.0 |
| - from 0.31 to 0.4 | 1.1 |
| - from 0.41 to 0.5 | 1.2 |
| - over 0.51 | 1.3 |
- k11 - coefficient taking into account the presence of doors to the street.
The last of the considered coefficients. The room may have a door leading directly to the street, to a cold balcony, to an unheated corridor or entrance, etc. Not only is the door itself often a very serious "cold bridge" - if it is opened regularly, a fair amount of cold air will enter the room every time. Therefore, this factor should also be corrected: such heat losses, of course, require additional compensation.
The values of the coefficient k11 are given in the table:
This coefficient should be taken into account if the doors in winter time use regularly.
You may be interested in information about what is
* * * * * * *
So, all correction factors are considered. As you can see, there is nothing super complicated here, and you can safely proceed to the calculations.
One more tip before starting calculations. Everything will be much easier if you first draw up a table, in the first column of which you sequentially indicate all the rooms of the house or apartment to be soldered. Next, in columns, place the data that is required for calculations. For example, in the second column - the area of \u200b\u200bthe room, in the third - the height of the ceilings, in the fourth - orientation to the cardinal points - and so on. It is not difficult to make such a plate, having in front of you a plan of your residential properties. It is clear that the calculated values of the required heat output for each room will be entered in the last column.
The table can be compiled in an office application, or even simply drawn on a piece of paper. And do not rush to part with it after making the calculations - the obtained indicators of thermal power will still be useful, for example, when purchasing heating radiators or electric heaters used as a backup heat source.
To make it as easy as possible for the reader to carry out such calculations, a special online calculator is placed below. With it, with the initial data previously collected in a table, the calculation will take literally a few minutes.
Calculator for calculating the required heat output for the premises of a house or apartment.
To ensure comfortable living in the house in winter, the boiler must produce enough heat energy to fully compensate for the heat loss of the building. In addition, it is necessary to provide a certain power reserve in case of severe cold or an increase in the area of \u200b\u200bthe building. To calculate the power of the boiler, you need to take into account quite a few factors. In heat engineering, such a calculation is one of the most difficult.
There are many calculations of the heating system, namely the power of the boiler - one of the most difficult
The need to calculate the heat transfer of the boiler
Whatever materials a building is made of, it constantly releases heat to the outside. The heat loss of the house for each room may differ and depend on the materials of construction and the degree of insulation. If you take the calculations seriously, then it is better to entrust such work to specialists. Then, in accordance with the results obtained, a boiler is selected.
It is not very difficult to independently calculate the heat loss of a building, but many factors must be taken into account. The easiest way to solve the problem is with the help of a special device - a thermal imager. This is a device of small dimensions, the display of which indicates the actual heat loss of the building. At the same time, you can clearly see those places where the maximum leakage of thermal energy is observed, and take measures to correct the situation.
You can immediately install a powerful boiler without calculations
Of course, you can just take a powerful boiler and not carry out any calculations. However, in such a situation, gas costs can be very high. In addition, if the boiler is underloaded, then its service life is reduced. However, the heat generator can be loaded, for example, by using it to heat previously unheated rooms. However, not a single owner of a private house wants to overpay for wasted fuel.
If the power of the heat generator turned out to be insufficient, then it will not be possible to create comfortable living conditions in the building, and the boiler itself will operate in constant overload mode. As a result, expensive equipment will fail prematurely. Thus, only one conclusion can be drawn - you need to calculate the power of the boiler for the house, thereby making a competent selection of heating equipment.
The easiest way is to independently calculate the power of the heating boiler for the area of \u200b\u200bthe house. After that, it will be possible to say exactly which heating unit needed to heat all areas of the building.
Basic Formula
If we analyze the results of calculations carried out over several years, then one regularity is observed - for heating every 10 m 2 of an area, 1 kW of thermal energy must be spent. This statement is true for buildings with medium insulation, and the height of the ceilings in them is in the range from 2.5 to 2.7 m.
If the building meets these standards, then it will be quite simple to determine the capacity of the heating boilers, just use a simple formula:
Latest indicator for different regions of the country has the following meanings:
- Moscow region - from 1.2 to 1.5 kW.
- The middle band is from 1 to 1.2 kW.
- South of the country - from 0.7 to 0.9 kW.
- Northern territories - from 1.5 to 2 kW.
As an example, you can calculate the power of a heat generator for a 12 × 14 m house built of brick in the Moscow region. The total area of the building is 168 m 2 . The value of specific power Wsp is taken equal to 1. As a result, W = (168 × 1) / 10 = 16.8 kW. The resulting design power of the heat generator should be rounded up. However, this is not yet a complete calculation of a gas boiler for a house by area, since it is necessary to adjust the obtained indicator.
Additional Calculations
Residential buildings with average characteristics are quite rare in practice. In order for the calculation of the power of the boiler house to be as accurate as possible, additional indicators have to be taken into account. One of them has already been considered in the main formula - the specific power spent on heating 10 m 2.
It should be used as a benchmark for middle lane. At the same time, in each zone, one can see a rather serious scatter of specific capacitance values. The way out of this situation is simple - the further north the area is located in the climatic zone, the higher the coefficient should be, and vice versa. For example, for Siberia with frosts of about 35 degrees, it is customary to use Wsp = 1.8.
Another factor affecting the calculation of the boiler power is the height of the ceilings. If this parameter differs significantly from the average (2.6 m), then a correction factor must be calculated. To do this, the real value must be divided by the average.
It is equally important to take into account the thermal losses of the structure when calculating. The process of heat leakage is observed in every building. For example, if the walls are poorly insulated, then losses can reach up to 35%. Thus, during the calculations a special coefficient should be used:
- A structure made of wood, foam blocks or bricks, the age of which exceeds 15 years with high-quality insulation - K = 1.
- Buildings of other materials with poorly insulated walls - K = 1.5.
- If the roof was not insulated in the building, and not just the walls - K = 1.8.
- Modern high-quality insulated houses - K = 0.6.
Do not forget to take into account the coefficient of wood blocks This is how the required power of the heat generator is calculated in order to make right choice equipment. However, if the boiler is also planned to be used for heating water, the obtained value of its power will have to be increased by 25%. Thus, to determine the required power of the heat generator you need to use the following algorithm:
- Calculated total area buildings and is divided by 10. In this case, the indicator Wsp does not need to be taken into account.
- The calculated value is adjusted depending on the climatic zone in which the building was erected. The indicator determined at the first stage is multiplied by the coefficient of the region.
- If the actual value of the ceiling height differs significantly from the average, this must be taken into account in the calculation. First you need to split actual figure to the middle. The resulting coefficient is multiplied by the power of the heat generator, determined taking into account the correction for the climatic features of the area.
- The heat losses of the building are taken into account. The result obtained at the previous stage must be multiplied by the heat loss coefficient.
- If the boiler is also used for heating water, its capacity is increased by 25%.
The result obtained using this algorithm is different high precision, and it is suitable for choosing a boiler that runs on any type of fuel.
In accordance with the norms of SNiP
You can calculate the power of equipment for the heating system at home based on building codes and rules (SNiP). This document defines the required amount of thermal energy to heat 1 m 3 of air. The volume calculation is fairly easy to do. It is enough to determine the volume interior spaces buildings and multiply it by the rate of consumption of thermal energy.
According to SNiP, in a panel building, 41 W of heat energy must be spent to heat 1 m 3 of air.
For brick house the norm is 34 watts. After performing the calculation, the resulting power value must be converted to kilowatts. It should also be recalled that in heat engineering, calculated indicators are rounded up.
If you want to get the most accurate results, then correction factor must be taken into account:
- If a heated room is located above or below the apartment, the correction is 0.7.
- If it is unheated, the coefficient will be 1.
- If the apartment is located above the basement or under the attic - the amendment will be 0.9.
You also need to take into account the number of external walls in the room. When only one wall goes out, the coefficient will be 1.1, with two - 1.2, three - 1.3. Thus, the calculation of a boiler for heating a house can be calculated by the total volume of the building or its area. Whichever method is chosen, the process is not very complex. All necessary calculations can be carried out by anyone who does not have special knowledge.
How not to make a mistake and correctly choose a device so as not to freeze and not to thin the budget - read on. From the article you will find out which technique will be correct and necessary for you.
Calculation of heat losses at home
We say right away - there is no single method for calculating the coefficient. The setting varies depending on your climate. It is all the more important to pay more attention to this stage of preparation. Even a specialist will not determine by eye, without calculations, information on the required boiler power. Even low-power ones, such as, can heat an average apartment up to 65m². But what it exactly should be - it will become known after filling out a special questionnaire - the document is freely available, anyone can fill it out on the Internet.
Experts approached the compilation of the questionnaire responsibly. By filling in the fields, you will not be able to make a mistake. The only exception is the incorrect filling of the online form. All other calculations of the boiler for the house will be performed by the program.
So, here are the questions you need to be prepared for - specify:
1. Heat loss through walls
This parameter is affected by the area of the facade and the ventilated layer (walls are with it, and sometimes without it). The first wall covering is the paramount criterion, without which it will be too risky to choose a heating boiler. Reinforced concrete or foam concrete, mineral wool, drywall, plywood or wood - the material affects the decision of what power to buy solid fuel equipment. The thickness of the first layer of the house is also important. For thin-walled houses, buy a medium power boiler - for example,.
2. Heat loss through windows
Important condition. It is logical that more heat will "leave" with a single-chamber double-glazed window than with two-chamber ones. The area of windows is also important when calculating the power of the boiler. Before filling out the questionnaire, measure it again.
3. Heat loss through ceiling and floor
As you understand, in a room with an attic and an unheated basement, you need to install powerful equipment - like. Incorrectly selected power of the device will spoil several winter months spent in country house- heating is clearly not enough for a comfortable life.
Useful to know:
If you do everything right, your efforts will be rewarded with a profitable investment in the purchase. Consider that you have coped with the task - most likely you will receive best result for price and quality.
Why is it important to accurately determine the power of the boiler
The first thing that comes to mind is saving money on a purchase. For this alone, it is worth spending a couple of hours on calculations. Given the good work and efficient operation of the boiler, the calculation of the power of the equipment becomes even more necessary.
Here are some unhappy scenarios that will inevitably unfold if you do not take into account the above.
Remember: The correction for the region for our climate is a factor of 1.2.
An incorrect calculation of the power of a not so popular, but still occurring pellet device (for example) and a wood-fired boiler is the first choice parameter. To calculate the parameter, do not be too lazy to spend time, otherwise you cannot avoid the above problems in the lack of heat (if we are talking about weak appliances) or inefficient waste of fuel (when you pick up an expensive and too powerful boiler, like).
Determining the power of the boiler is the most important stage of work
So you got acquainted with the theoretical part of the question, having received information about the importance of calculating the power of boilers. Now it's time to move on to the practical part - the most important. As an option, a specialist responsible for the calculation of parameters and installation. But you yourself can find out what technique is really needed.
When calculating power, we start from the area of \u200b\u200bthe heated object - it is she who will help to evaluate performance. Keep in mind that with a room height of 2.7 m (and such ceilings are in almost all houses), it takes 1 kW to heat 10 m².
This ratio is approximate. It is influenced by the climate of the region and, again, the height of the ceilings, the presence basements etc.
Advice: in order to calculate the power of an ideal boiler for high ceilings, it is necessary to determine the correction factor by dividing the parameter by the standard 2.7m.
Example:
- The ceilings are 3.1m.
- We divide the parameter by 2.7 - we get 1.14.
- So, for high-quality heating of a house of 200 m² with ceilings of 3.1 m, a boiler with a capacity of 200 kW * 1.14 = 22.8 kW is useful.
- In order not to freeze for sure, we recommend rounding the parameter up. Then get 23kW. We are suitable for 24 kW.
Please note that this calculation is suitable for a single-circuit boiler. In the case of c, you need to calculate what water temperature you want to get in the cold, and choose a technique in accordance with the parameter (+ 25%, power, if you like hot water).
Step-by-step calculation of boiler power (double-circuit) for apartments
With apartments, the situation is somewhat different. Here the coefficient is less than in the house - in apartments there is no heat loss through the roof (if we are not talking about the last floor) and losses through the floor (except for the first floor).
- if another room "warms up" the apartment from above, the coefficient will be 0.7
- if there is an attic above you - 1
To calculate the parameter, we use the technique indicated above, taking into account the coefficient.
Example: The area of the apartment is 163m². Its ceilings are 2.9 m, the apartment is located in our lane.
We determine the power in five steps:
- We divide the area by the coefficient: 163m² / 10m² = 16.3 kW.
- Do not forget about the correction for the region: 16.3 kW * 1.2 = 19.56 kW.
- Since the double-circuit boiler is designed for hot water, add 25% 7.56 kW * 1.25 \u003d 9.45 kW.
- And now do not forget about the cold (experts advise adding another 10%): 9.45 kW * 1.1 \u003d 24.45 kW.
- We round up, and it turns out 25 kW. It turns out that it will suit us - a device that works on natural gas and interact with solar collectors.
Please note that in this way the power of the boilers is calculated, no matter what fuel they operate on - even gas, even electricity, even solid fuel. .
Step-by-step calculation of the power of the boiler (single-circuit) for an apartment
But what if you do not need a double-circuit boiler, and with tasks? We will make calculations, taking into account one more factor - the material of manufacture of the house. The norm of heating established at the legislative level looks like this:
- Heating 1m³ in panel house will require 41 watts.
- Heating 1m³ in brick house will require 34 watts.
We invite you to familiarize yourself with:
We remember the area of the apartment, multiply it by the height of the ceilings, we get the volume. This indicator must be multiplied by the norm - we get the power of the boiler.
Example:
- You live in a 120m² apartment with 2.6m ceilings.
- The volume will be as follows: 120m² * 2.6m = 192.4m³
- We multiply by the coefficient, we calculate the need for heat 192.4m³ * 34W = 106081W.
- We translate into kilowatts and rounding up, we get 11kW. This is the power that a thermal single-circuit unit should have. A good option is the model. A little "with a margin", the power of this technique is more than enough for a comfortable microclimate in your home.
As you can see, the task of selecting a boiler will not take more than an hour. By choosing the right heating device, you will insure yourself against uncomfortable cold weather for the whole winter, saving money on the purchase of a boiler and utilities. Correctly calculate the parameter - it is equally important for all types of heaters: coal, TT,
Before designing a heating system, installing heating equipment, it is important to choose a gas boiler that can generate the required amount of heat for the room. Therefore, it is important to choose a device of such power that its performance is as high as possible, and the resource is large.
We will talk about how to calculate the power of a gas boiler with high accuracy and taking into account certain parameters. The article presented by us describes in detail all types of heat loss through openings and building construction, formulas for their calculation are given. A specific example introduces the features of the calculation.
The correct calculation of the power of a gas boiler will not only save on consumables, but also increase the efficiency of the device. Equipment whose heat output exceeds the actual heat demand will operate inefficiently when, as an insufficiently powerful device, it cannot heat the room properly.
There is modern automated equipment that independently regulates the gas supply, which eliminates unnecessary expenses. But if such a boiler performs its work at the limit of its capabilities, then its service life is reduced.
As a result, the efficiency of the equipment decreases, parts wear out faster, and condensate forms. Therefore, there is a need to calculate the optimal power.
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