What power boiler is needed for the house. How to make your own calculation of the power of the heating boiler

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. In the article presented by us, all types of heat losses through openings and building structures are described in detail, 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 costs. 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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How to calculate the power of a gas boiler for the given parameters of a heated room? I know at least three different methods that give different levels of reliability of results, and today we will get acquainted with each of them.

general information

Why do we calculate the parameters specifically for gas heating?

The fact is that gas is the most economical (and, accordingly, the most popular) source of heat. A kilowatt-hour of thermal energy obtained by its combustion costs the consumer 50-70 kopecks.

For comparison, the price of a kilowatt-hour of heat for other energy carriers:

• solid fuel- 1.1-1.6 rubles per kilowatt-hour;
• Diesel fuel- 3.5 rubles / kWh;
• Electricity- 5 rubles / kWh.

In addition to efficiency, gas equipment attracts with ease of use. The boiler requires maintenance no more than once a year, does not need kindling, cleaning the ash pan and replenishing the fuel supply. Devices with electronic ignition work with remote thermostats and are able to automatically maintain a constant temperature in the house, regardless of the weather.

Does the calculation of a gas boiler for a home differ from the calculation of a solid fuel, liquid fuel or electric boiler?

In general, no. Any heat source must compensate for heat loss through the floor, walls, windows and ceiling of the building. Its thermal power has nothing to do with the energy carrier used.

In the case of a double-circuit boiler supplying the house with domestic hot water, we need a reserve of power to heat it. Excess power will ensure the simultaneous flow of water in the DHW system and the heating of the coolant for heating.

Calculation methods

Scheme 1: by area

We will be helped in this by the normative documentation of half a century ago. According to Soviet SNiP, heating should be designed at the rate of 100 watts of heat per square meter of heated space.

Let's, for example, calculate the power for a house measuring 6x8 meters:

1. The area of ​​the house is equal to the product of its overall dimensions. 6x8x48 m2;
2. With a specific power of 100 W / m2, the total power of the boiler should be 48x100 \u003d 4800 watts, or 4.8 kW.

The choice of boiler power according to the area of ​​​​the heated room is simple, understandable and ... in most cases gives the wrong result.

Because he neglects a number of important factors that affect real heat loss:

• Number of windows and doors. More heat is lost through glazing and doorways than through a main wall;
• ceiling height. In Soviet-built apartment buildings, it was standard - 2.5 meters with a minimum error. But in modern cottages you can find ceilings 3, 4 or more meters high. The higher the ceiling, the larger the heated volume;

• climate zone. With the same quality of thermal insulation, heat loss is directly proportional to the difference between indoor and outdoor temperatures.

In an apartment building, heat loss is affected by the location of the living space relative to the outer walls: end and corner rooms lose more heat. However, in a typical cottage, all rooms have common walls with the street, so the appropriate correction factor is included in the base value of the heat output.

Scheme 2: by volume, taking into account additional factors

How to do the calculation of a gas boiler for heating a private house with your own hands, taking into account all the factors I mentioned?

First and foremost: in the calculation, we take into account not the area of ​​\u200b\u200bthe house, but its volume, that is, the product of the area by the height of the ceilings.

• base value boiler power per cubic meter of heated volume - 60 watts;
• Window increases heat loss by 100 watts;
• Door adds 200 watts;
• Heat losses are multiplied by the regional coefficient. It is determined by the average temperature of the coldest month:

Image	Coefficient and climate zone
	0,6-0,9 - for regions with an average January temperature of about 0 °C (Krasnodar Territory, Crimea).
	1,2-1,3 - for the average temperature of the coldest month at -15-20 °C (Moscow and Leningrad regions).
	1,5-1,6 - for areas with an average January temperature of -25-30 ° C (Novosibirsk region, Khabarovsk Territory).
	2 - for -40 and below (Chukotka, Yakutia).

Let's again calculate the boiler power for our house measuring 6x8 meters, specifying a few additional parameters:

• House location- the city of Sevastopol (the average January temperature is +3 degrees Celsius);
• Number of windows- 5. One door leads to the street;
• Ceiling height- 3.2 meters.

1. House volume(with external walls) is equal to the product of its three dimensions: 6x8x3.2 = 153.6 cubic meters;

1. Base power for this volume - 153.6x60 \u003d 9216 W;
2. Including windows and doors it will increase by 5x100+200=700 watts. 9216+700=9916;
3. Regional coefficient for the warm climate of the Crimea, we will take equal to 0.6.

9916*0.6=6000 (with rounding) watts.

As you can see, the complicated calculation scheme gave a result that is noticeably different from the previous one. How accurate is it?

The calculation will give a reliable result for a house, the quality of insulation of which approximately corresponds to the quality of insulation of Soviet-built houses. The scheme is based on the same 100 watts per square area, recalculated taking into account the standard ceiling height of 2.5 meters at 40 W / m3 and multiplied by a factor of 1.5 to compensate for the heat loss of a private house through the roof and floor.

How to determine the need for heat in a house with non-standard insulation?

Scheme 3: by volume, taking into account the quality of insulation

The most universal formula for calculating the heat output of a boiler is Q=V*Dt*k/860.

In this formula:

• Q - heat loss of the house in kilowatts;
• V is the volume to be heated by the boiler, in cubic meters;
• Dt is the calculated temperature delta between the heated room and the air behind the outer walls;
• k - dispersion coefficient, determined by the quality of insulation of the house.

How to choose the coefficient k?

Select its value for your conditions, guided by the following table:

Image	Coefficient value and description of the building
	3-4 - building without insulation (a warehouse made of profiled sheets, a panel house with walls made of planks in one layer)
	2.0-2.9 - walls made of timber 10 cm thick or bricks 25 cm thick, wooden frames, single glazing
	1.0-1.9 - brick walls 50 cm thick, double glazing in the windows
	0.6-0.9 - facade insulated with foam plastic or mineral wool, plastic windows with triple or energy-saving double-glazed windows

How to choose the value of the calculated outdoor temperature? In calculations, it is customary to use the temperature of the coldest five-day winter for a given region. Rare extreme frosts are not taken into account: when the thermometer falls below the usual levels, auxiliary heat sources (heaters, fan heaters, etc.) can be used.

Where can I get relevant information? The instruction is quite predictable: the necessary data can be found in SNiP 23-01-99, a regulatory document on building climatology.

For the convenience of readers, I will give here a short excerpt from the text of SNiP.

City	Temperature of the coldest 5 days of winter, °C
Maykop	-22
Barnaul	-42
Blagoveshchensk	-37
Tynda	-46
Shimanovsk	-41
Arkhangelsk	-37
Astrakhan	-26
Ufa	-39
Belgorod	-28
Bryansk	-30
Ulan-Ude	-40
Vladimir	-34
Vologda	-37
Voronezh	-31
Makhachkala	-19
Irkutsk	-38
Kaliningrad	-24
Petropavlovsk-Kamchatsky	-22
Pechora	-48
Kostroma	-35
Agatha	-55
Turukhansk	-56
St. Petersburg	-30
Susuman	-57
Moscow	-32
Novosibirsk	-42
Vladivostok	-26
Komsomolsk-on-Amur	-37
Yalta	-8
Sevastopol	-11

Let's return to our example with a house in Sevastopol, once again clarifying a few details:

• Window glazing- single, in wooden large-slotted frames;
• wall material- but, about half a meter thick.

Let's start with the calculations.

1. For the calculated internal temperature, we will take the corresponding sanitary standards + 20 ° С. Given the data from the table above, the Dt parameter will be equal to 20 - -11 = 31 degrees;
2. We take the dispersion coefficient equal to 2.0: the thermal conductivity of rubble walls is much higher than that of brick walls;

1. We calculated the volume of the house earlier. It is equal to 153.6 cubes;
2. Substitute the values ​​of the variables in our formula. Q \u003d 153.6x31 * 2 / 860 \u003d 11 kW.

As you can see, the correction for significant heat losses almost doubled the calculated capacity of the gas boiler.

Two circuits

It's very simple: a 20% reserve is included in the project for the operation of the second flow. In our case, the required power will be 11x1.2 = 13.2 kW.

The heating boiler is the basis of the heating system, it is the main device, the performance of which will determine the ability of the communication network to provide the house with the amount of heat that is needed. And if you calculate the power of the heating boiler correctly and correctly, then this will eliminate the occurrence of unnecessary costs that are associated with the purchase of appliances and their operation. The boiler selected according to preliminary calculations will work with such heat transfer, which is incorporated into it by the manufacturer - this will help to maintain its technical parameters.

What is the calculation based on?

Calculating the power of a heating boiler is an important point. Power, as a rule, can be compared with the entire heat output of the heating system, which will provide a house with a certain size, with a given number of floors, and thermal properties.

To equip a one-story country or private house, you do not need a very powerful heating boiler.

So, in calculating the performance of a boiler for an autonomous house, the area is the main parameter, if we consider the heat engineering of the building in accordance with the climate of the region. So, the area of ​​\u200b\u200bthe house is the most important parameter in order to calculate the boiler for heating.

Characteristics that will affect the calculation

Those who want to calculate the boiler for home heating with maximum accuracy can use the methodology provided by SNiP II-3-79. In this case, professional calculations will take into account the following factors:

• The average temperature of the region in the coldest time.
• Insulating properties of materials that were used to build building envelopes.
• Type of wiring of the heating circuit.
• The ratio of the area of ​​\u200b\u200bbearing structures and openings.
• Separate information about each room.

How to calculate the power of a heating boiler? To perform the most accurate calculations, even information such as data on household and digital appliances is used - after all, all this also somehow releases heat into the premises.

However, we note that not every owner of the heating system requires professional calculations - it is usually customary to purchase autonomous heating circuits with devices with a power reserve.

So, the efficiency of heating boilers can be higher than the calculated values, especially since they are usually rounded off.

What is required to be taken into account?

How to calculate the power of a heating boiler, what data must be present without fail? One rule to remember is that every 10 sq.m of a cottage with insulating characteristics, the standard ceiling height limit (up to 3m) will require approximately 1 kW for heating. To the power of the boiler, which is designed to work together in heating and hot water supply, you will need to add at least 20%.

An autonomous heating circuit, which has unstable pressure in the heating boiler, will need to be equipped with a device so that its power reserve is at least 15 percent higher than the calculated value. To the power of the boiler, which provides heating and hot water, you need to add 15%.

We take into account heat loss

Note that regardless of whether the power of the electric boiler, gas boiler, diesel or wood is calculated, in any case, the operation of the heating system will be accompanied by heat loss:

• It is necessary to ventilate the premises, however, if the windows are constantly open, the house will lose about 15% of energy.
• If the walls are poorly insulated, then 35% of the heat will go away.
• Through the window openings, 10% of the heat will go away, and if the frames are of the old model, then even more.
• If the floor is not insulated, then 15% of the heat will be transferred to the basement or the ground.
• 25% of the heat will go through the roof.

The simplest formula

Thermal calculations in any case will have to be rounded off and also increased in order to provide a power reserve. That is why, in order to determine the power of a heating boiler, it will be possible to use a very simple formula:

W \u003d S * Wsp.

Here S is the total area of ​​the heated building, which takes into account residential and household rooms in sq.m.

W is the power of the heating boiler, kW.

Wud. - this is the average specific power, this parameter is used for calculations taking into account a certain climatic zone, kW / sq.m. And it is worth noting that this characteristic is based on many years of experience in the operation of various heating systems in the regions. And when we multiply the area by this indicator, we get the average power value. It will need to be adjusted based on the features listed above.

Calculation example

Consider an example using a heating boiler power calculator. Natural gas is the most affordable fuel used in Russia. For this reason, it is so common and in demand. Therefore, we will calculate the power of a gas boiler. And as an example, let's take a private house with an area of ​​140 sq.m. Territory - Krasnodar Territory. Also in the example, we take into account that our boiler will provide not only heating the house, but also plumbing fixtures with water. We will do the calculations for a system with natural circulation, the pressure here will not be maintained by the circulation pump.

Specific power - 0.85 kW/sq.m.

So, 140 sq.m / 10 sq.m = 14 - this is an intermediate coefficient of calculations. It will provide for the condition that for every 10 square meters of heated premises, 1 kW of heat will be required, which will be provided by the boiler.

14 * 0.85 = 11.9 kW.

We receive thermal energy that the house will need, which has standard thermal properties. To provide hot water for the shower, sinks - we will add another 20%.

11.9 + 11.9 * 0.2 = 14.28 kW.

We do not use a circulation pump, so we should be aware that the pressure here can be unstable. Therefore, we must add another 15% to ensure the reserve of heat energy.

14.28 + 11.9 * 0.15 = 16.07 kW.

Also be aware that there will be some heat leakage. That's why we have to round our result up. Thus, we need a heating boiler with a capacity of at least 17 kW.

As a rule, the calculation of the power of the heating boiler is carried out at the design stage of the building. After all, in order for the heating system to work effectively, specific conditions are required - the arrangement of the furnace room, the supply of rooms with a chimney and ventilation.

Compared to electrical heating appliances, own heating system is more advantageous in terms of cost savings, and in maximum convenience when heating rooms.

The efficiency and profitability of the heating system in the house depends on the correct calculations, adherence to precise rules and instructions.

Calculation of heating by the area of ​​​​the house is a laborious and complex process. Don't skimp on materials. Quality equipment and its installation affects the financial budget, but then serves the house well and comfortably.

When equipping a house with a heating system, construction work and installation of heating must proceed strictly according to the project and taking into account all safety regulations for use.

The following points should be taken into account:

• house building material,
• footage of window openings;
• climatic features of the area where the house is located;
• location of window frames according to the compass;
• what is the device of the "warm floor" system.

Subject to all the above rules and calculations for heating, some knowledge in the field of engineering is required. But there is also a simplified system - the calculation of heating by area, which can be done independently, again, adhering to the rules and complying with all norms.

Choosing a boiler requires an individual approach

If there is gas in the house, then the best option is a gas boiler. In the absence of a centralized gas pipeline, we choose an electric boiler, a heat generator using solid or liquid fuel. Taking into account regional characteristics, access to the supply of materials, it is possible to install a combined boiler. Combined generator heat will always maintain a comfortable temperature, in any emergency and force majeure situations. Here you need to start from a simple type of operation, the heat transfer coefficient.

After determining the type of boiler, it is necessary to calculate the heating according to the area of ​​\u200b\u200bthe room. The formula is simple, but it takes into account the temperature of the cold period, the heat loss coefficient for large windows and their location, the thickness of the walls and the height of the ceilings.

Each boiler has a certain power. With the wrong choice, the room will be either cold or excessively hot. Thus, if the specific power of the boiler per 10 cubic meters. taking into account the area of ​​the heated room of 100 sq.m., you can choose the most optimal heat generator.

From the formula used by engineers, Wcat = (SxWsp)/10, kW. – it follows that the boiler with a capacity of 10 kW heats a room of 100 sq.m.

The required number of sections of the heating radiator.

To make it more clear, let's solve the problem using the example of specific numbers. If we assume that room area 14 sq.m. And ceiling height 3 meters, the volume is determined by multiplication.

14 x 3 = 42 cubic meters.

In central Russia, Ukraine, Belarus, thermal power per cubic meter corresponds to 41 W. We determine: 41x 42 \u003d 1722 watts. Found out that for a room of 14 sq.m. need a 1700W heatsink. Each individual section (rib) has a power of 150 watts. By dividing the results obtained, we obtain the number of sections required for the acquisition. The calculation of heating by area is not the same everywhere. For premises over 100 sq.m. required circulation pump installation, which serves as a "forcer" of the movement of the coolant through the pipes. Its installation takes place in the opposite direction from the heating devices to the heat generator. The circulation pump increases the life of the heating system by reducing the contact of hot coolants with appliances.

When installing a heating system warm floor» the coefficient of heating of the house increases many times. You can connect the underfloor heating system to existing types of heating. A pipe is removed from the heating radiators and floor heating wiring is supplied. This is the most convenient and profitable option, taking into account the savings in time and money.

The selection of a gas boiler of optimal power is possible only after calculations have been made. The technical documentation for boiler equipment indicates its thermal power - TMK. This parameter means the power that the boiler is able to transmit to external devices (heating, ventilation, DHW preparation), taking into account its efficiency. But this value does not in any way inform the user what area can be heated using a particular boiler model.

The problem is that any building, even insulated, gives off part of the heat to the outside air through structures such as walls, ceilings, floors, windows and doors. Therefore, without a thermal calculation of the building, it is difficult not to make a mistake in choosing the right boiler.

In this article:

What parameters need to be taken into account

Heat loss of a private house

When choosing boiler equipment for heating a house, it is necessary to consider:

• climatic conditions of the region (the calculation formula includes the value of the average temperature for the coldest week of the year);
• the set air temperature inside the heated premises;
• the need for organizing hot water supply;
• heat loss from forced ventilation (if any in the house);
• number of storeys of the building;
• ceiling height;
• construction and materials of floors;
• the thickness of the outer walls and the materials from which they were built;
• geometric dimensions of the outer walls;
• floor construction (layer thickness and materials from which they are built);
• dimensions, number of windows and doors and their type (glass thickness, number of chambers, etc.).

Heat loss at home

The amount of heat loss of a building is greatly influenced by:

• type of attic (insulated, non-insulated);
• the presence or absence of a basement.

To clearly show dependence of heat loss at home on materials used in its construction, we suggest considering a small comparative table.

The table shows that a wooden house loses less heat than a brick house, respectively, and the boiler in the first case will require less power than for a brick house.

In building codes, thermal conductivity indicators for all building materials are painted.

Something similar happens with windows..

Only they are characterized not by thermal conductivity, but, on the contrary, by the heat transfer resistance coefficient: the higher the figure, the less heat the window will release from the house (in another way, this indicator is called the R-factor).

As you can see, the more chambers in the window design, the higher its resistance to heat loss. An important role is also played by the gas mixture, which fills the chambers of double-glazed windows.

How to calculate the TMK of a gas boiler

First of all - the thermal calculation of the building itself

The heat output of a heating boiler can be calculated in two ways:

1. complete;
2. simplified.

The first method involves making calculations taking into account the thermal properties of all building materialsinvolved in the construction of the house and its decoration. It can be seen from the data in the above tables how important it is to complete the calculation.

But this work is not easy, in the absence of certain experience, it is difficult to cope with it.

This is usually done by designers in design organizations. Although with a strong desire, you can arm yourself with SNiPs and try to do everything yourself.

Thermal conductivity coefficient of building materials

Thermal conductivity coefficients of common building materials

To determine the amount of heat loss through the building envelope, it is necessary to calculate the thermal conductivity of the building materials of which they are composed.

The initial data for the calculation are:

• a(vn)- coefficient that determines the intensity of heat transfer from the air in the room to the ceiling and walls. This is a constant value of 8.7.
• a(nr)- another constant coefficient equal to 23. It characterizes the intensity of heat transfer from the walls and ceiling to the outside air.
• To- thermal conductivity of building materials that make up the ceiling and walls. Data is taken from building codes. For some materials, the thermal conductivity is given in the table of building materials (see above).
• D- the thickness of the layers of building materials.

After collecting all the initial data, you can begin to calculate the heat transfer coefficient using the formula:

Kt = 1/

Kt is calculated for the ceiling and walls separately.

The principle of calculating the Kt of the floor is the same, but there are some nuances here: the correct approach requires dividing the floor area into 4 zones, located from the outer walls to the center. To simplify the calculations, heat losses through the floor structure without heating can be taken equal to 10%.

Calculation of heat loss through windows and doors

The initial data for this part of the calculation are:

• kst- the heat transfer coefficient of a double-glazed window or glass (indicated by the manufacturer).
• F Art.- the area of ​​the glazed surface of the window.
• Cr- heat transfer coefficient of the window frame (specified by the manufacturer).
• F p- the area of ​​the window frame.
• R- the perimeter of the glazed surface of the window.

The heat transfer coefficient of windows (Ko) is calculated by the formula:

Kst. x F Art. + Кр х F р + Р/F, where F is the window area.

The same formula is used to calculate the heat transfer coefficient of doors.

In this case, instead of the values ​​of glass and frames, the values ​​of the materials from which the doors are made are substituted.

To simplify the calculations, you can use the following data:

To determine heat loss, the conditional coefficient is multiplied by the total area of ​​\u200b\u200bthe house.

This method gives only an approximate result. It does not take into account the number of windows, the configuration of the house and its location. But for a preliminary assessment of heat loss, it is quite suitable.

Simplified Method

The power of the heating boiler is defined as the sum of the powers required to heat each heated room. That is, the calculations described in the previous sections are carried out for each room separately.

At the same time, designers are required to take into account the number of lamps, people in the room, and even the operation of household appliances.

Fortunately, in most cases it is possible to do without such complex and expensive heat engineering calculations. Residential buildings are usually built taking into account the climatic conditions of a particular region, so you can choose the required amount of TMK according to a simplified scheme.

This calculation is based on the assumption that the specific power of the entire house is equal to the sum of the specific power of each room. In this case, when performing calculations, they operate with experimental values ​​of the specific power of the house, depending on the region.

These tables are valid for well-insulated wooden and reinforced concrete houses with a standard ceiling height of 2.7 meters.

Boiler power per 10 sq. m is calculated by the formula:

• W \u003d S x W beats / 10, where
• W is the design power of the boiler
• S - the sum of the areas of the premises
• Wsp - specific power of the house (see table above)

Example

Typical house plan for 300 sq.m (for example)

For example, let's calculate the power of a gas boiler for a house located in the Moscow region. The total building area is 300 sq. m.

Let's take the value of specific power (according to the fourth table) equal to 1.5.

• W \u003d 300 x 1.5 / 10 \u003d 45 kW

For high ceilings

If the ceiling height differs from the standard values, in this case the power of the heating boiler is calculated by the formula:

• Mk \u003d TxKz, where
  • Mk - boiler power
  • T - estimated heat loss
  • Kz - safety factor

Heat losses T are calculated by the formula:

• T \u003d VxRxKr / 860, where
  • V is the volume of the room (in cubic meters)
  • P - the difference between the external and internal temperatures
  • Kp - dispersion coefficient

For buildings made of bricks, Kp is 2 - 2.9, for poorly insulated buildings - 3-4.

And the last thing: if you assume that the boiler will provide the house with hot water, increase the calculated power by 25%.