A competent choice of the boiler will save comfortable temperature indoor air in winter time of the year. A large selection of devices allows you to most accurately select the right model, depending on the required parameters. But in order to provide warmth in the house and at the same time prevent extra costs resources, you need to know how to calculate the power gas boiler for heating a private house.
The floor-mounted gas boiler has more power Source termoresurs.ru
The main characteristics affecting the power of the boiler
The boiler power indicator is the main characteristic, however, the calculation can be carried out using different formulas, depending on the configuration of the device and other parameters. For example, in a detailed calculation, they can take into account the height of the building, its energy efficiency.
Varieties of boiler models
Boilers can be divided into two types depending on the purpose of the application:
single circuit– are used only for heating;
Dual circuit- are used for heating, as well as in hot water systems.
Units with a single circuit have a simple structure, consist of a burner and a single heat exchanger.
In dual-circuit systems, the water heating function is primarily provided. When using hot water, the heating is automatically turned off for the duration of use hot water so that the system is not overloaded. advantage double-circuit system is its compactness. Such a heating complex takes much less space than if the support systems hot water and heating were used separately.
Boiler models are often divided according to the method of placement.
Boilers can be installed in different ways depending on their type. You can choose a model with a wall mount or installed on the floor. 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 way the boiler is installed 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 the purposes of application and methods of placement gas boilers They also differ in the way they are managed. There are models with electronic and mechanical control. Electronic systems can only work in homes with permanent access to the mains.
On our website you can find contacts of construction companies that offer home insulation services. You can directly communicate with representatives by visiting the exhibition of houses "Low-Rise Country".
Typical device power calculations
There is no single algorithm for calculating both single and double-circuit boilers - each of the systems must be selected separately.
Formula for a typical project
When calculating the required power for heating a house built according to standard project, that is, with a room height of not more than 3 meters, the volume of the rooms 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 \u003d Um * P * Kr, where
P - a value equal to the sum of the areas of heated premises,
Kr is a correction factor, which is taken in accordance with the climatic zone in which the building is located.
Some coefficient values for different regions of Russia:
Southern - 0.9;
located in middle lane – 1,2;
Northern - 2.0.
For the Moscow region take the value of the coefficient equal to 1.5.
This technique does not reflect the main factors affecting the microclimate in the house, and only approximately shows how to calculate the power of a gas boiler for a private house.
Some manufacturers issue memos-recommendations, but for accurate calculations they still recommend contacting specialists. Source parki48.ru
Calculation example for a single-circuit device installed in a room with an area of 100 m 2, located on the territory of the Moscow region:
Pm \u003d 1/10 * 100 * 1.5 \u003d 15 (kW)
Calculations for double-circuit devices
Double-circuit devices have the following principle of operation. For heating, water is heated and flows through the heating system to radiators, which give off heat environment e, thus heating the premises and cooling. When cooled, the water flows back for heating. So the water circulates around the circuit heating system, and goes through cycles of heating and transfer to radiators. At the moment when the ambient temperature becomes equal to the set one, the boiler goes into standby mode for a while, i.e. temporarily stops heating 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 received power.
Calculation example for a two-circuit device, which is installed in a room with an area of 100m 2; the coefficient is taken for the Moscow region:
R m \u003d 1/10 * 100 * 1.5 \u003d 15 (kW)
R final \u003d 15 + 15 * 20% \u003d 18 (kW)
Additional factors to consider when installing the boiler
In construction, there is also the concept of energy efficiency of a building, that is, how much heat a building gives off to the environment.
One of the indicators of heat transfer is the coefficient of dissipation (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 values and the use of which will help to understand how to calculate the power of a gas boiler for a house more accurately:
Most low level heat transfer, corresponding to the value of K p from 0.6 to 0.9, is assigned to buildings made of modern materials, with insulated floor, walls and roof;
K p is from 1.0 to 1.9, if the outer walls of the building are insulated, the roof is insulated;
K p is from 2.0 to 2.9 in houses without insulation, for example, brick with single masonry;
K p is from 3.0 to 4.0 in non-insulated rooms, in which there is a low level of thermal insulation.
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 is the safety factor.
The power of the boiler, taking into account the dissipation factor, is calculated by multiplying the calculated level of heat loss by the safety factor (usually from 15% to 20%, then it is necessary to multiply by 1.15 and 1.20, respectively)
This technique allows you to more accurately determine the performance and, therefore, approach the issue of choosing a boiler with the highest quality.
What happens if you calculate the required power incorrectly
It is still worth choosing a boiler so that it matches the power required to heat the building. This will be the best option, since, first of all, buying a boiler that does not match the power level can lead to two types of problems:
A low-power boiler will always work to the limit, trying to heat the room to the set temperature, and can quickly fail;
Appliance with excessive high level power is more expensive and even in economy mode consumes more gas than a less powerful device.
Boiler power calculator
For those who do not like to do calculations, even if not very complicated, a special calculator will help to calculate a boiler for heating a house, a special calculator is a free online application.
Interface online calculator boiler power calculation Source idn37.ru
As a rule, the calculation service requires you to fill in 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 get the final result, you will also need to enter the total area that will require heating.
Next, you should fill in 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, information is entered on the number of walls interacting with the street. Take into account the number of storeys of the building, the presence of structures on top of the house.
After entering the required fields, the button for performing calculations becomes “active” and you can get the calculation by clicking on the corresponding button with the mouse. To check the information received, you can use the calculation formulas.
Video description
Visually about the calculation of the power of a gas boiler, see the video:
Benefits of using gas boilers
Gas equipment has a number of advantages and disadvantages. The advantages include:
the possibility of partial automation of the boiler operation process;
unlike other energy sources, natural gas has a low cost;
devices do not require frequent maintenance.
To disadvantages gas systems consider a high explosive gas, however, with proper storage gas cylinders, timely Maintenance this risk is minimal.
On our website you can find construction companies that offer services for connecting electrical and gas equipment. You can talk directly with representatives at the exhibition of houses "Low-Rise Country".
Conclusion
Despite the apparent simplicity of the calculations, we must remember that gas equipment must be selected and installed by professionals. In this case, you will receive a trouble-free device that will work properly for many years.
Power calculation heating boiler, in particular gas boiler, it is necessary not only to select boiler and heating equipment, but also to ensure the comfortable functioning of the heating system as a whole and eliminate unnecessary operating costs.
From the point of view of physics, only four parameters are involved in the calculation of thermal power: the air temperature outside, the required temperature inside, the total volume of the premises and the degree of thermal insulation of the house, on which heat losses depend. But in fact, everything is not so simple. The outdoor temperature varies with the seasons, the indoor temperature requirements are determined by the mode of living, the total volume of the premises must first be calculated, and the heat loss depends on the materials and construction of the house, as well as the size, number and quality of windows.
Calculator of gas boiler power and gas consumption for the year
The calculator presented here for the power of a gas boiler and gas consumption for a year can greatly facilitate your task of choosing a gas boiler - just select the appropriate field values and you will get the required values.
Please note that the calculator calculates not only the optimal power of a gas boiler for heating a house, but also the average annual gas consumption. That is why the “number of residents” parameter was introduced into the calculator. It is necessary in order to take into account the average gas consumption for cooking and getting hot water for domestic needs.
This parameter is relevant only if you also use gas for the stove and water heater. If you use other appliances for this, for example, electrical ones, or even don’t cook at home and do without hot water, put zero in the “number of residents” field.
The following information was used in the calculation:
- duration heating season- 5256 h;
- duration of temporary residence (summer and weekends 130 days) - 3120 hours;
- the average temperature for the heating period is minus 2.2°C;
- the air temperature of the coldest five-day period in St. Petersburg is minus 26°C;
- soil temperature under the house during the heating period - 5 ° C;
- reduced room temperature in the absence of a person - 8.0 ° C;
- warming attic floor- a layer of mineral wool with a density of 50 kg / m³ and a thickness of 200 mm.
Creating a heating system in your own home or even in a city apartment is an extremely responsible task. At the same time, it would be completely unreasonable to purchase boiler equipment, as they say, “by eye”, that is, without taking into account all the features of housing. In this, it is quite possible to fall into two extremes: either the power of the boiler will not be enough - the equipment will work “to its fullest”, without pauses, but will not give the expected result, or, conversely, an overly expensive device will be purchased, the capabilities of which will remain completely unclaimed.
But that's not all. It is not enough to purchase the necessary heating boiler correctly - it is very important to optimally select and correctly place heat exchange devices in the premises - radiators, convectors or "warm floors". And again, relying only on your intuition or the "good advice" of your neighbors is not the most reasonable option. In a word, certain calculations are indispensable.
Of course, ideally, such heat engineering calculations should be carried out by appropriate specialists, but this often costs a lot of money. Isn't it interesting to try to do it yourself? This publication will show in detail how heating is calculated by the area of \u200b\u200bthe room, taking into account many important nuances. By analogy, it will be possible to perform, built into this page, will help you perform the necessary calculations. The technique cannot be called completely “sinless”, however, it still allows you to get a result with a completely acceptable degree of accuracy.
The simplest methods of calculation
In order for the heating system to create comfortable living conditions during the cold season, it must cope with two main tasks. These functions are closely related, and their separation is very conditional.
- The first is maintaining an optimal level of air temperature in the entire volume of the heated room. Of course, the temperature level may vary slightly with altitude, but this difference should not be significant. Quite comfortable conditions are considered to be an average of +20 ° C - it is this temperature that, as a rule, is taken as the initial temperature in thermal calculations.
In other words, the heating system must be able to heat a certain volume of air.
If we approach with complete accuracy, then for individual rooms in residential buildings the standards for the 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 premises | Air temperature, °С | Relative humidity, % | Air speed, m/s | |||
|---|---|---|---|---|---|---|
| optimal | admissible | optimal | admissible, max | optimal, max | admissible, max | |
| For the cold season | ||||||
| Living room | 20÷22 | 18÷24 (20÷24) | 45÷30 | 60 | 0.15 | 0.2 |
| The same, but for living rooms in regions with minimum temperatures from -31 ° C and below | 21÷23 | 20÷24 (22÷24) | 45÷30 | 60 | 0.15 | 0.2 |
| Kitchen | 19:21 | 18:26 | N/N | N/N | 0.15 | 0.2 |
| Toilet | 19:21 | 18:26 | N/N | N/N | 0.15 | 0.2 |
| Bathroom, combined bathroom | 24÷26 | 18:26 | N/N | N/N | 0.15 | 0.2 |
| Premises for rest and study | 20÷22 | 18:24 | 45÷30 | 60 | 0.15 | 0.2 |
| Inter-apartment corridor | 18:20 | 16:22 | 45÷30 | 60 | N/N | N/N |
| lobby, stairwell | 16÷18 | 14:20 | N/N | N/N | N/N | N/N |
| Storerooms | 16÷18 | 12÷22 | N/N | N/N | N/N | N/N |
| For the warm season (The standard is only for residential premises. For the rest - it is not standardized) | ||||||
| Living room | 22÷25 | 20÷28 | 60÷30 | 65 | 0.2 | 0.3 |
- The second is the compensation of heat losses through the structural elements of the building.
The main "enemy" of the heating system is heat loss through building structures.
Alas, heat loss is the most serious "rival" of any heating system. They can be reduced to a certain minimum, but even with the highest quality thermal insulation, it is not yet possible to completely get rid of them. Thermal energy leaks go in all directions - their approximate distribution is shown in the table:
| Building element | Approximate value of heat loss |
|---|---|
| Foundation, floors on the ground or over unheated basement (basement) premises | from 5 to 10% |
| "Cold bridges" through poorly insulated joints of building structures | from 5 to 10% |
| Places of entry of engineering communications (sewerage, water supply, gas pipes, electrical cables, etc.) | up to 5% |
| External walls, depending on the degree of insulation | from 20 to 30% |
| Poor quality windows and external doors | about 20÷25%, of which about 10% - through non-sealed joints between the boxes and the wall, and due to ventilation |
| Roof | up to 20% |
| Ventilation and chimney | up to 25 ÷30% |
Naturally, in order to cope with such tasks, the heating system must have a certain thermal power, and this potential must not only correspond to the general needs of the building (apartment), but also be correctly distributed over the premises, in accordance with their area and a number of other important factors.
Usually the calculation is carried out in the direction "from small to large". Simply put, the required amount of thermal energy for each heated room is calculated, the obtained values are summed up, approximately 10% of the reserve is added (so that the equipment does not work at the limit of its capabilities) - and the result will show how much power the heating boiler needs. And the values for each room will become Starting point to calculate the required number of radiators.
The most simplified and most commonly used method in a non-professional environment is to accept a norm of 100 watts of thermal energy for each square meter area:
The most primitive way of counting is the ratio of 100 W / m²
Q = S× 100
Q- the required thermal power for the room;
S– area of the room (m²);
100 — specific power per unit area (W/m²).
For example, room 3.2 × 5.5 m
S= 3.2 × 5.5 = 17.6 m²
Q= 17.6 × 100 = 1760 W ≈ 1.8 kW
The method is obviously very simple, but very imperfect. It should be noted right away that it is conditionally applicable only when standard height ceilings - approximately 2.7 m (permissible - in the range from 2.5 to 3.0 m). From this point of view, the calculation will be more accurate not from the area, but from the volume of the room.
It is clear that in this case the value of specific power is calculated per cubic meter. It is taken equal to 41 W / m³ for 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 panel house, with a ceiling height of 3.2 m:
Q= 17.6 × 3.2 × 41 = 2309 W ≈ 2.3 kW
The result is more accurate, since it already takes into account not only all linear dimensions rooms, 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 - in the next section of the publication.
You may be interested in information about what they are
Carrying out calculations of the required thermal power, taking into account the characteristics of the premises
The calculation algorithms discussed above are useful for the initial “estimate”, but you should still rely on them completely with very great care. Even to a person who does not understand anything in building heat engineering, the indicated average values \u200b\u200bmay seem doubtful - they cannot be equal, say, for the Krasnodar Territory and for the Arkhangelsk Region. In addition, the room - the room is different: one is located on the corner of the house, that is, it has two external walls ki, and the other on three sides is protected from heat loss by other rooms. In addition, the room may have one or more windows, both small and very large, sometimes even panoramic. And the windows themselves may differ in the material of manufacture and other design features. And this is not a complete list - just such features are visible even to the "naked eye".
In a word, there are a lot of nuances that affect the heat loss of each particular room, and it is better not to be too lazy, but to carry out a more thorough calculation. Believe me, according to the method proposed in the article, this will not be so difficult to do.
General principles and calculation formula
The calculations will be based on the same ratio: 100 W per 1 square meter. But that's just the formula itself "overgrown" with a considerable number of various correction factors.
Q = (S × 100) × a × b × c × d × e × f × g × h × i × j × k × l × m
The Latin letters denoting the coefficients are taken quite arbitrarily, in alphabetical order, and are not related to any standard quantities accepted in physics. The meaning of each coefficient will be discussed separately.
- "a" - a coefficient that takes into account the number of external walls in a particular room.
Obviously, the more external walls in the room, the larger the area through which the heat loss. In addition, the presence of two or more external walls also means corners - extremely vulnerable places in terms of the formation of "cold bridges". The coefficient "a" will correct for this specific feature of the room.
The coefficient is taken equal to:
- external walls No(indoor): a = 0.8;
- outer wall one: a = 1.0;
- external walls two: a = 1.2;
- external walls three: a = 1.4.
- "b" - coefficient taking into account the location of the external walls of the room relative to the cardinal points.
You may be interested in information about what are
Even on the coldest winter days, solar energy still has an effect on the temperature balance in the building. It is quite natural that the side of the house that faces south receives a certain amount of heat from the sun's rays, and heat loss through it is lower.
But the walls and windows facing north never “see” the Sun. The eastern part of the house, although it "grabs" the morning sun's rays, still does not receive any effective heating from them.
Based on this, we introduce the coefficient "b":
- the outer walls of the room look at North or East: b = 1.1;
- the outer walls of the room are oriented towards South or West: b = 1.0.
- "c" - coefficient taking into account the location of the room relative to the winter "wind rose"
Perhaps this amendment is not so necessary for houses located in areas protected from the winds. But sometimes the prevailing winter winds can make their own “hard adjustments” to the thermal balance of the building. Naturally, the windward side, that is, "substituted" to the wind, will lose much more body, compared to the leeward, opposite side.
Based on the results of long-term meteorological observations in any region, the so-called "wind rose" is compiled - a graphic diagram showing the prevailing wind directions in winter and summer time of the year. This information can be obtained from the local hydrometeorological service. However, many residents themselves, without meteorologists, know perfectly well where the winds mainly blow from in winter, and from which side of the house the deepest snowdrifts usually sweep.
If there is a desire to carry out calculations with higher accuracy, then the correction factor “c” can also be included in the formula, taking it equal to:
- windward side of the house: c = 1.2;
- leeward walls of the house: c = 1.0;
- wall located parallel to the direction of the wind: c = 1.1.
- "d" - correction factor that takes into account the features climatic conditions home building region
Naturally, the amount of heat loss through all the building structures of the building will greatly depend on the level of winter temperatures. It is quite clear that during the winter the thermometer indicators “dance” in a certain range, but for each region there is an average indicator of the lowest temperatures characteristic of the coldest five-day period of the year (usually this is characteristic of January). For example, below is a map-scheme of the territory of Russia, on which approximate values are shown in colors.
Usually this value is easy to check with the regional meteorological service, but you can, in principle, rely on your own observations.
So, the coefficient "d", taking into account the peculiarities of the climate of the region, for our calculations in we take equal to:
— from – 35 °С and below: d=1.5;
— from – 30 °С to – 34 °С: d=1.3;
— from – 25 °С to – 29 °С: d=1.2;
— from – 20 °С to – 24 °С: d=1.1;
— from – 15 °С to – 19 °С: d=1.0;
— from – 10 °С to – 14 °С: d=0.9;
- not colder - 10 ° С: d=0.7.
- "e" - coefficient taking into account the degree of insulation of external walls.
The total value of the heat loss of the building is directly related to the degree of insulation of all building structures. One of the "leaders" in terms of heat loss are walls. Therefore, the value of thermal power required to maintain comfortable conditions living indoors depends on the quality of their thermal insulation.
The value of the coefficient for our calculations can be taken as follows:
- external walls are not insulated: e = 1.27;
- medium degree of insulation - walls in two bricks or their surface thermal insulation with other heaters is provided: e = 1.0;
– insulation was carried out qualitatively, on the basis of heat engineering calculations: e = 0.85.
Later in the course of this publication, recommendations will be given on how to determine the degree of insulation of walls and other building structures.
- coefficient "f" - correction for ceiling height
Ceilings, especially in private homes, can have different heights. Therefore, the thermal power for heating one or another room of the same area will also differ in this parameter.
It will not be a big mistake to accept the following values of the correction factor "f":
– ceiling height up to 2.7 m: f = 1.0;
— flow height from 2.8 to 3.0 m: f = 1.05;
– ceiling height from 3.1 to 3.5 m: f = 1.1;
– ceiling height from 3.6 to 4.0 m: f = 1.15;
– ceiling height over 4.1 m: f = 1.2.
- « g "- coefficient taking into account the type of floor or room located under the ceiling.
As shown above, the floor is one of the significant sources of heat loss. So, it is necessary to make some adjustments in the calculation of this feature of a particular room. The correction factor "g" can be taken equal to:
- cold floor on the ground or over an unheated room (for example, basement or basement): g= 1,4 ;
- insulated floor on the ground or over an unheated room: g= 1,2 ;
- a heated room is located below: g= 1,0 .
- « h "- coefficient taking into account the type of room located above.
The air heated by the heating system always rises, and if the ceiling in the room is cold, then increased heat losses are inevitable, which will require an increase in the required heat output. We introduce the coefficient "h", which takes into account this feature of the calculated room:
- a "cold" attic is located on top: h = 1,0 ;
- an insulated attic or other insulated room is located on top: h = 0,9 ;
- any heated room is located above: h = 0,8 .
- « i "- coefficient taking into account the design features of windows
Windows are one of the "main routes" of heat leaks. Naturally, much in this matter depends on the quality of the window construction. Old wooden frames, which were previously installed everywhere in all houses, are significantly inferior to modern multi-chamber systems with double-glazed windows in terms of their thermal insulation.
Without words, it is clear that the thermal insulation qualities of these windows are significantly different.
But even between PVC-windows there is no complete uniformity. For example, a two-chamber double-glazed window (with three glasses) will be much warmer than a single-chamber one.
This means that it is necessary to enter a certain coefficient "i", taking into account the type of windows installed in the room:
— standard wooden windows with conventional double glazing: i = 1,27 ;
– modern window systems with single-chamber double-glazed windows: i = 1,0 ;
– modern window systems with two-chamber or three-chamber double-glazed windows, including those with argon filling: i = 0,85 .
- « j" - correction factor for the total glazing area of the room
No matter how high-quality the windows are, it will still not be possible to completely avoid heat loss through them. But it is quite clear that there is no way to compare a small window with panoramic windows almost the whole wall.
First you need to find the ratio of the areas of all the windows in the room and the room itself:
x = ∑SOK /SP
∑ SOK- the total area of windows in the room;
SP- area of the room.
Depending on the value obtained and the correction factor "j" is determined:
- x \u003d 0 ÷ 0.1 →j = 0,8 ;
- x \u003d 0.11 ÷ 0.2 →j = 0,9 ;
- x \u003d 0.21 ÷ 0.3 →j = 1,0 ;
- x \u003d 0.31 ÷ 0.4 →j = 1,1 ;
- x \u003d 0.41 ÷ 0.5 →j = 1,2 ;
- « k" - coefficient that corrects for the presence of an entrance door
The door to the street or to an unheated balcony is always an additional "loophole" for the cold
The door to the street or to an open balcony is able to make its own adjustments to the heat balance of the room - each of its opening is accompanied by the penetration of a considerable amount of cold air into the room. Therefore, it makes sense to take into account its presence - for this we introduce the coefficient "k", which we take equal to:
- no door k = 1,0 ;
- one door to the street or balcony: k = 1,3 ;
- two doors to the street or to the balcony: k = 1,7 .
- « l "- possible amendments to the connection diagram of heating radiators
Perhaps this will seem like an insignificant trifle to some, but still - why not immediately take into account the planned scheme for connecting heating radiators. The fact is that their heat transfer, and hence their participation in maintaining a certain temperature balance in the room, changes quite noticeably with different types tie-in supply and return pipes.
| Illustration | Radiator insert type | The value of the coefficient "l" |
|---|---|---|
 | Diagonal connection: supply from above, "return" from below | l = 1.0 |
 | Connection on one side: supply from above, "return" from below | l = 1.03 |
 | Two-way connection: both supply and return from the bottom | l = 1.13 |
 | Diagonal connection: supply from below, "return" from above | l = 1.25 |
 | Connection on one side: supply from below, "return" from above | l = 1.28 |
 | One-way connection, both supply and return from below | l = 1.28 |
- « m "- correction factor for the features of the installation site of heating radiators
And finally, the last coefficient, which is also associated with the features of connecting heating radiators. It is probably clear that if the battery is installed openly, is not obstructed by anything from above and from the front part, then it will give maximum heat transfer. However, such an installation is far from always possible - more often, radiators are partially hidden by window sills. Other options are also possible. In addition, some owners, trying to fit heating priors into the created interior ensemble, hide them completely or partially with decorative screens - this also significantly affects the heat output.
If there are certain “baskets” on how and where the radiators will be mounted, this can also be taken into account when making calculations by entering a special coefficient “m”:
| Illustration | Features of installing radiators | The value of the coefficient "m" |
|---|---|---|
| The radiator is located on the wall openly or is not covered from above by a window sill | m = 0.9 | |
| The radiator is covered from above by a window sill or a shelf | m = 1.0 | |
| The radiator is blocked from above by a protruding wall niche | m = 1.07 | |
| The radiator is covered from above with a window sill (niche), and from the front - with a decorative screen | m = 1.12 | |
| The radiator is completely enclosed in a decorative casing | m = 1.2 |
So, there is clarity with the calculation formula. Surely, some of the readers will immediately take up their heads - they say, it's too complicated and cumbersome. However, if the matter is approached systematically, in an orderly manner, then there is no difficulty at all.
Any good homeowner must have a detailed graphical plan of their "possessions" with dimensions, and usually oriented to the cardinal points. It is not difficult to specify the climatic features of the region. It remains only to walk through all the rooms with a tape measure, to clarify some of the nuances for each room. Features of housing - "neighborhood vertically" from above and below, location entrance doors, the proposed or already existing scheme for installing heating radiators - no one except the owners knows better.
It is recommended to immediately draw up a worksheet, where you enter all the necessary data for each room. The result of the calculations will also be entered into it. Well, the calculations themselves will help to carry out the built-in calculator, in which all the coefficients and ratios mentioned above are already “laid”.
If some data could not be obtained, then, of course, they can not be taken into account, but in this case, the “default” calculator will calculate the result, taking into account the least favorable conditions.
It can be seen with an example. We have a house plan (taken completely arbitrary).
The region with the level of minimum temperatures in the range of -20 ÷ 25 °С. Predominance of winter winds = northeasterly. The house is one-story, with an insulated attic. Insulated floors on the ground. The optimal diagonal connection of radiators, which will be installed under the window sills, has been selected.
Let's create a table like this:
| The room, its area, ceiling height. Floor insulation and "neighborhood" from above and below | The number of external walls and their main location relative to the cardinal points and the "wind rose". Degree of wall insulation | Number, type and size of windows | Existence of entrance doors (to the street or to the balcony) | Required heat output (including 10% reserve) |
|---|---|---|---|---|
| Area 78.5 m² | 10.87 kW ≈ 11 kW | |||
| 1. Hallway. 3.18 m². Ceiling 2.8 m. Warmed floor on the ground. Above is an insulated attic. | One, South, the average degree of insulation. Leeward side | Not | One | 0.52 kW |
| 2. Hall. 6.2 m². Ceiling 2.9 m. Insulated floor on the ground. Above - insulated attic | Not | Not | Not | 0.62 kW |
| 3. Kitchen-dining room. 14.9 m². Ceiling 2.9 m. Well insulated floor on the ground. Svehu - insulated attic | Two. South, west. Average degree of insulation. Leeward side | Two, single-chamber double-glazed window, 1200 × 900 mm | Not | 2.22 kW |
| 4. Children's room. 18.3 m². Ceiling 2.8 m. Well insulated floor on the ground. Above - insulated attic | Two, North - West. High degree of insulation. windward | Two, double glazing, 1400 × 1000 mm | Not | 2.6 kW |
| 5. Bedroom. 13.8 m². Ceiling 2.8 m. Well insulated floor on the ground. Above - insulated attic | Two, North, East. High degree of insulation. windward side | One, double-glazed window, 1400 × 1000 mm | Not | 1.73 kW |
| 6. Living room. 18.0 m². Ceiling 2.8 m. Well insulated floor. Top - insulated attic | Two, East, South. High degree of insulation. Parallel to wind direction | Four, double glazing, 1500 × 1200 mm | Not | 2.59 kW |
| 7. Bathroom combined. 4.12 m². Ceiling 2.8 m. Well insulated floor. Above is an insulated attic. | One, North. High degree of insulation. windward side | One. wooden frame with double glazing. 400 × 500 mm | Not | 0.59 kW |
| TOTAL: |
Then, using the calculator below, we make a calculation for each room (already taking into account a 10% reserve). With the recommended app, it won't take long. After that, it remains to sum the obtained values \u200b\u200bfor each room - this will be the required total power of the heating system.
The result for each room, by the way, will help you choose the right number of heating radiators - it remains only to divide by the specific heat output of one section and round up.
Autonomous heating for a private house is affordable, comfortable and varied. You can install a gas boiler and not depend on the vagaries of nature or failures in the central heating system. The main thing is to choose the right equipment and calculate the heat output of the boiler. If the power exceeds the heat needs of the room, then the money for installing the unit will be thrown to the wind. In order for the heat supply system to be comfortable and financially profitable, at the design stage it is necessary to calculate the power of the gas heating boiler.
The main values \u200b\u200bof calculating the heating power
The easiest way to get data on the heat output of the boiler by area of the house: taken 1 kW of power for every 10 sq. m. However, this formula has serious errors, because it does not take into account modern building technologies, the type of terrain, climatic temperature changes, the level of thermal insulation, the use of double-glazed windows, and the like.
To make a more accurate calculation of the heating power of the boiler, you need to take into account whole line important factors affecting the final result:
- dimensions of the dwelling;
- the degree of insulation of the house;
- the presence of double-glazed windows;
- thermal insulation of walls;
- building type;
- air temperature outside the window during the coldest time of the year;
- type of wiring of the heating circuit;
- the ratio of the area of \u200b\u200bbearing structures and openings;
- building heat loss.
In houses with forced ventilation the calculation of the boiler heat output must take into account the amount of energy needed to heat the air. Experts advise making a gap of 20% when using the obtained result of the thermal power of the boiler in case of unforeseen situations, severe cooling or a decrease in gas pressure in the system.
With an unreasonable increase in thermal power, it is possible to reduce the efficiency of the heating unit, increase the cost of purchasing system elements, and lead to rapid wear of components. That is why it is so important to correctly calculate the power of the heating boiler and apply it to the specified dwelling. You can get data using a simple formula W \u003d S * W beats, where S is the area of \u200b\u200bthe house, W is the factory power of the boiler, W beats is the specific power for calculations in a certain climate zone, it can be adjusted according to the characteristics of the user's region. The result must be rounded up to a large value in terms of heat leakage in the house.
For those who do not want to waste time on mathematical calculations, you can use the gas boiler power calculator online. Just keep the individual data on the features of the room and get a ready answer.
The formula for obtaining the power of the heating system
The online heating boiler power calculator makes it possible in a matter of seconds to obtain the necessary result, taking into account all of the above characteristics that affect the final result of the data obtained. In order to use such a program correctly, it is necessary to enter the prepared data into the table: the type of window glazing, the level of thermal insulation of the walls, the ratio of floor and window opening areas, the average temperature outside the house, the number of side walls, the type and area of the room. And then press the "Calculate" button and get the result of the heat loss and heat output of the boiler.
To ensure a comfortable temperature throughout the winter, the heating boiler must produce such an amount of heat energy that is necessary to replenish all the heat losses of the building / room. Plus, it is also necessary to have a small power reserve in case of abnormal cold weather or expansion of areas. We will talk about how to calculate the required power in this article.
To determine the performance of heating equipment, it is first necessary to determine the heat loss of the building / room. Such a calculation is called thermal engineering. This is one of the most complex calculations in the industry as there are many factors to consider.
Of course, the amount of heat loss is affected by the materials that were used in the construction of the house. Therefore, the building materials from which the foundation is made, walls, floor, ceiling, floors, attic, roof, window and door openings are taken into account. The type of system wiring and the presence of warm floors. In some cases, even the presence household appliances which generates heat during operation. But such precision is not always required. There are techniques that allow you to quickly estimate the required performance of a heating boiler without plunging into the wilds of heat engineering.
Calculation of the heating boiler power by area
For an approximate assessment of the required performance of a thermal unit, the area of \u200b\u200bthe premises is sufficient. In the very simple version for central Russia, it is believed that 1 kW of power can heat 10 m 2 of area. If you have a house with an area of 160m2, the boiler power for heating it is 16kW.
These calculations are approximate, because neither the height of the ceilings nor the climate is taken into account. For this, there are coefficients derived empirically, with the help of which appropriate adjustments are made.
The indicated rate - 1 kW per 10 m 2 is suitable for ceilings 2.5-2.7 m. If you have higher ceilings in the room, you need to calculate the coefficients and recalculate. To do this, divide the height of your premises by the standard 2.7 m and get a correction factor.
Calculating the power of a heating boiler by area - the easiest way
For example, the ceiling height is 3.2m. We consider the coefficient: 3.2m / 2.7m \u003d 1.18 rounded up, we get 1.2. It turns out that for heating a room of 160m 2 with a ceiling height of 3.2m, a heating boiler with a capacity of 16kW * 1.2 = 19.2kW is required. They usually round up, so 20kW.
To take into account climatic features, there are ready-made coefficients. For Russia they are:
- 1.5-2.0 for northern regions;
- 1.2-1.5 for regions near Moscow;
- 1.0-1.2 for the middle band;
- 0.7-0.9 for the southern regions.
If the house is located in the middle lane, just south of Moscow, a coefficient of 1.2 is applied (20kW * 1.2 \u003d 24kW), if in the south of Russia in the Krasnodar Territory, for example, a coefficient of 0.8, that is, less power is required (20kW * 0 ,8=16kW).
Calculation of heating and selection of a boiler is an important stage. Find the wrong power and you can get this result ...
These are the main factors to be considered. But the values found are valid if the boiler will only work for heating. If you also need to heat water, you need to add 20-25% of the calculated figure. Then you need to add a "margin" for peak winter temperatures. That's another 10%. In total we get:
- For home heating and hot water in the middle lane 24kW + 20% = 28.8kW. Then the reserve for cold weather is 28.8 kW + 10% = 31.68 kW. We round up and get 32kW. When compared with the original figure of 16kW, the difference is two times.
- House in the Krasnodar Territory. We add power for heating hot water: 16kW + 20% = 19.2kW. Now the "reserve" for the cold is 19.2 + 10% \u003d 21.12 kW. Rounding up: 22kW. The difference is not so striking, but also quite decent.
It can be seen from the examples that it is necessary to take into account at least these values. But it is obvious that in calculating the power of the boiler for a house and an apartment, there should be a difference. You can go the same way and use coefficients for each factor. But there is an easier way that allows you to make corrections in one go.
When calculating a heating boiler for a house, a coefficient of 1.5 is applied. It takes into account the presence of heat loss through the roof, floor, foundation. It is valid with an average (normal) degree of wall insulation - laying in two bricks or building materials similar in characteristics.
For apartments, different rates apply. If there is a heated room (another apartment) on top, the coefficient is 0.7, if a heated attic is 0.9, if an unheated attic is 1.0. It is necessary to multiply the boiler power found by the method described above by one of these coefficients and get a fairly reliable value.
To demonstrate the progress of calculations, we will calculate the power of a gas heating boiler for an apartment of 65m 2 with 3m ceilings, which is located in central Russia.
- We determine the required power by area: 65m 2 / 10m 2 \u003d 6.5 kW.
- We make a correction for the region: 6.5 kW * 1.2 = 7.8 kW.
- The boiler will heat the water, so we add 25% (we like it hotter) 7.8 kW * 1.25 = 9.75 kW.
- We add 10% for cold: 7.95 kW * 1.1 = 10.725 kW.
Now we round the result and get: 11 kW.
The specified algorithm is valid for the selection of heating boilers for any type of fuel. The calculation of the power of an electric heating boiler will not differ in any way from the calculation of a solid fuel, gas or liquid fuel boiler. The main thing is the performance and efficiency of the boiler, and heat losses do not change depending on the type of boiler. The whole question is how to spend less energy. And this is the area of \u200b\u200bwarming.
Boiler power for apartments
When calculating heating equipment for apartments, you can use the norms of SNiPa. The use of these standards is also called the calculation of boiler power by volume. SNiP sets the required amount of heat for heating one cubic meter air in typical buildings:
- heating 1m 3 in a panel house requires 41W;
- v brick house 34W goes to m 3.
Knowing the area of \u200b\u200bthe apartment and the height of the ceilings, you will find the volume, then, multiplying by the norm, you will find out the power of the boiler.
For example, let's calculate the required boiler power for rooms in a brick house with an area of 74m 2 with ceilings of 2.7m.
- We calculate the volume: 74m 2 * 2.7m = 199.8m 3
- We consider according to the norm how much heat will be needed: 199.8 * 34W = 6793W. Rounding up and converting to kilowatts, we get 7kW. This will be the required power that the thermal unit should produce.
It is easy to calculate the power for the same room, but already in a panel house: 199.8 * 41W = 8191W. In principle, in heating engineering they always round up, but you can take into account the glazing of your windows. If the windows have energy-saving double-glazed windows, you can round down. We believe that double-glazed windows are good and we get 8kW.
The choice of boiler power depends on the type of building - brick heating requires less heat than panel
Next, you need, as well as in the calculation for the house, to take into account the region and the need to prepare hot water. The correction for abnormal cold is also relevant. But in apartments, the location of the rooms and the number of storeys play a big role. You need to take into account the walls facing the street:
- One outer wall — 1,1
- Two - 1.2
- Three - 1.3
After you take into account all the coefficients, you will get a fairly accurate value that you can rely on when choosing equipment for heating. If you want to get an accurate heat engineering calculation, you need to order it from a specialized organization.
There is another method: to define real losses with the help of a thermal imager - a modern device that will also show the places through which heat leaks are more intense. At the same time, you can eliminate these problems and improve thermal insulation. And the third option is to use a calculator program that will calculate everything for you. You just need to select and / or enter the required data. At the output, get the estimated power of the boiler. True, there is a certain amount of risk here: it is not clear how correct the algorithms are at the heart of such a program. So you still have to at least roughly calculate to compare the results.
We hope you now have an idea of how to calculate the power of the boiler. And it doesn’t confuse you that it is, and not solid fuel, or vice versa.
You may be interested in articles about and. In order to have a general idea of the mistakes that are often encountered when planning a heating system, watch the video.
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