home Heating radiators How to choose a boiler for heating a private house based on power. Calculation options for high-power solid fuel boilers How to calculate the thermal power of a boiler

How to choose a boiler for heating a private house based on power. Calculation options for high-power solid fuel boilers How to calculate the thermal power of a boiler

Reading time: 3 min

To heat residential and office premises, equipment with an electric water heater is used. To ensure a balance of temperature and energy consumption, the electric boiler is calculated. When determining operating parameters, not only the area of the rooms is taken into account, but also the physical properties of the materials of the walls, floor and ceiling of the room.

What is the power of an electric boiler

An electric boiler is a reservoir with a heat exchanger through which tap water or a special coolant with increased thermal characteristics is pumped.

The boiler is connected to a household AC network; it heats the water using heating elements or electrodes isolated from the water. The design of the equipment includes a temperature regulator.

Power consumption depends on the degree of cooling of the coolant during circulation through the heating radiators in the building. Part of the energy is spent on heat losses in the boiler design (heating the walls or protective casings of the heating elements). An information plate is installed on the external part of the equipment, which indicates the operating parameters of the product and power consumption.

Methods for determining the power of an electric boiler

Calculation of the operating power of a heating boiler is performed to ensure a balanced heating system capable of maintaining a comfortable room temperature under various external conditions.

The equipment must ensure uniform heating of the rooms; changes in wind direction should not have a negative impact on the conditions in the rooms. Before choosing equipment, the home owner needs to know how to calculate the power of an electric boiler, taking into account the characteristics of the room.

For calculations, 2 main methods are used:

by the area of the house or rooms connected to the heating circuit and boiler;
by volume of premises.

An auxiliary technique for determining the power of a hot water supply circuit is intended to calculate additional productivity. The resulting parameter is summed with a pre-calculated value of energy consumption for heating the house.

Then the ability of the electrical wiring connected to the building is checked to withstand the maximum load when the heating elements of the boiler are operating.

Calculation of the boiler according to the area of the house

The basic method is to determine the power of an electric heating boiler based on the area of the premises. To determine the value, the base value of the power required to heat a room of 10 m² is used.

The coefficient does not depend on the climate zone; it is roughly assumed that to warm up 10 m² it is necessary to expend 1 kW of power. The coefficient does not take into account the thermal conductivity of wall materials and the height of the room, therefore, to clarify the calculation, additional correction factors determined experimentally are used.

For example, if the ceiling height is more than 2.7 m, an additional correction parameter is introduced equal to the ratio of the actual height to the value of 2.7 m. The climate coefficient depends on the location of the house, the value ranges from 0.7 for the southern regions to 2.0 - northern regions. If the heating unit is also used for hot water supply, then a power reserve of 25-30% is added to the resulting indicator.

There is another way to calculate based on the formula S*K*100, where parameter S is the area of the premises, and K is the heat loss coefficient, varying depending on the minimum air temperature threshold. The base value is 0.7, used in areas with a minimum temperature of -10°C. For every 5°C decrease in climate norm, the coefficient increases by 0.2.

The method is not used when calculating a boiler for premises with the following design features:

Availability of plastic or wooden windows with double glazing.
Use of an additional thermal insulation layer with a thickness of 150 mm, located inside or outside a brick wall (2 brick sizes thick).
Preservation of an unheated attic space and the absence of thermal insulation material on the roof trim.
Increasing the height of living rooms to 2.7 m or more.

Calculation of boiler power by volume

Calculation of the power of an electric heating boiler for the volume of residential premises is based on the heat loss coefficient, which is:

From 0.6 to 0.9 - for brick buildings with improved thermal insulation. The house uses plastic 2-chamber windows; a roof made of heat-insulating material can be used.
From 1 to 1.9 - for buildings built of brick (double masonry), with a standard roof and wooden windows.
From 2 to 2.9 - for rooms with poor thermal insulation (for example, with walls 1 brick thick).
From 3 to 4 - for buildings built of wood or made of corrugated metal sheet with a layer of heat-insulating material.

When calculating, a formula of the form is used V*K*T/860, which takes into account the volume of the house V, the correction factor K and the difference in temperature inside the house and outside the room. For the calculation, the minimum air temperature characteristic of the location of the house is taken.

The obtained value is excessive, but in the event of prolonged frosts it will be possible to maintain the temperature in the house within the specified parameters. The given method for calculating the power of an electric boiler for heating a house does not take into account the supply of additional warm liquid for washing dishes or showers.

For residential premises in panel or brick houses, calculations are carried out according to SNiP standards. The rules set the required power to heat 1 m³ of air within 41 and 34 W (for a house made of panels and sand-lime brick, respectively).

Then the owner of the premises takes measurements of the height and area, and a safety margin of 10% is added to the resulting value (in case the air temperature drops in winter). When installing energy-saving windows, it is allowed to install a boiler with a power less than the calculated one.

For corner rooms, the number of walls in contact with the street is taken into account. If only 1 wall faces the outside of the house, then a coefficient of 1.1 must be applied. Each additional wall increases the value of the correction parameter by 0.1. To reduce heat losses, it is recommended to analyze the room with a special device and then install an insulator layer.

Calculation for DHW

The calculation of an electric boiler for heating a private house, which is also used for hot water supply, takes into account the following factors:

The amount and temperature of warm water necessary to ensure the life of people living in the room.
Based on the first parameter, the volume of hot water +90°C is determined, which is then diluted with a flow of cold liquid to produce warm water.
Based on the obtained value, the electric boiler is calculated. When determining the parameters, the decrease in tap water temperature in winter is not taken into account.

For example, a residential building consumes 200 liters of warm water (Vg) heated to +40°C (Tg) every day. It is assumed that the required temperature is obtained by mixing hot and cold water. The owner plans to purchase a boiler that heats the liquid to +95°C (Tk), water is supplied to the cold water supply line at a temperature of +10°C (Tx).

The volume of hot water is determined by the formula Vg*(Tg-Tx)/(Tk-Tx)=200*(40-10)/(95-10). The calculation shows that to ensure the supply of hot water per day, it is necessary to heat 71 liters of liquid to a temperature of +95°C.

Further calculations are based on the coefficient of specific heat capacity of water (4.218 kJ per kg when heated by 1°C), the weight of the liquid and the temperature difference. The resulting value is then converted into kilowatts according to the tables; it is recommended to round the parameter upward.

For the situation described above, an additional power of about 5 kW is required. The obtained value implies heating the water in 1 hour; if the liquid is used evenly throughout the day, then it is possible to reduce additional energy costs by 2 times.

Currently, there is a fairly large selection of heating devices with which you can effectively organize an autonomous heating system. The desire of consumers to reduce dependence on centralized heat and energy services is understandable. Saving money spent on gas heating is a significant factor that residents of private houses pay attention to.

In addition, it is not always technologically possible to connect to a centralized gas supply. In such a situation, boiler technology operating on solids plays the main role. A powerful solid fuel boiler is an excellent alternative to gas equipment. Manufacturers have managed not only to improve the manufacturability of heating equipment of this type, but also to achieve a significant increase in the efficiency of solid fuel units. The high power and high efficiency of a solid fuel boiler operating on various types of fossil and organic fuels make such devices in demand and popular.

An important aspect in order to choose the right heating device for your own needs is calculating the boiler power. Let's look in detail at how to do this and what you should pay attention to.

Why is it necessary to calculate the power of a heating device?

The appearance of the heating equipment and the high technological characteristics stated in the technical passport give only a superficial idea of the technical capabilities of a solid fuel boiler. The main parameter influencing your choice is the power of the device. In pursuit of it, we sometimes make hasty conclusions and overpay, purchasing powerful units that do not meet the real requirements and assigned tasks.

Price-quality + thermal output, the ratio is of decisive importance for any heating equipment. Manufacturers offer consumers heating boilers of a variety of models, each of which corresponds to certain operating conditions. Despite this, in each individual case it is important to have an understanding of how the heating device should work and how the resource of the heating unit will be spent. The operating parameters of a solid fuel heating device, calculated taking into account the needs and design features of the room, and the correct installation of the equipment will allow the home heating system to reach optimal operating conditions.

Many consumers are wondering. How to independently calculate the power of your own solid fuel boiler, so that in the future there will be no problems with the operation of the heating system. There is nothing complicated. With a minimum of knowledge and effort, you can obtain preliminary data that gives an idea of what kind of heating device should be and how best to heat it.

Heating boiler power - theory and real facts

A heating device operating on coal, wood or other organic fuel performs a certain job related to heating the coolant. The amount of work of boiler equipment is determined by the volume of heat load that a solid fuel boiler can withstand when burning a certain amount of fuel. The ratio of the amount of fuel consumed, the amount of thermal energy released at optimal operating modes of the equipment is the boiler power.

A heating unit that is incorrectly selected for power will not be able to provide the required boiler water temperature in the heating circuit. Low-power solid fuel devices will not allow the autonomous system to fully meet your needs in terms of heating your home and ensuring the operation of the hot water supply. There will be a need to increase the power of the autonomous device. A powerful device, on the contrary, will create problems during operation. It will be necessary to make design changes to the existing heating complex to reduce the thermal load of the solid fuel heating device. Why waste precious fuel if there is no need for so much heat.

For reference: Exceeding the boiler power of the technological parameters of the heating system leads to the fact that the coolant in the circuit will disperse impulsively. Frequent switching on and off of the heating unit leads to excessive fuel consumption and a decrease in the operational capabilities of heating equipment in general.

From a theoretical point of view, calculating the optimal operating mode of boiler equipment is not difficult. It is generally accepted that 10 kW is enough to heat a living area of 10 m2. This indicator is taken taking into account the high thermal efficiency of the building and standard design features of the building (ceiling height, glazing area).

In theory, the calculation is made based on the following parameters:

area of the heated room;
specific power of heating equipment for heating is 10 kW. m, taking into account the climatic conditions of your region.

The table shows the average parameters of boiler equipment used by consumers in the Moscow region:

The thermal load parameters look optimal on paper, in theory, which is clearly not enough in relation to local conditions. The selected unit in reality should have redundant capabilities. In reality, you need to focus on equipment that can operate with a small power reserve.

On a note: The excess power of a solid fuel boiler will allow the entire heating system in the house to quickly reach optimal operating conditions. The additional resource should exceed the calculated data by 20-30%.

The actual load indicators of solid fuel units depend on a combination of various factors. The climatic conditions of the region in which you live may make adjustments when choosing a heating boiler. For the middle zone, the following power parameters of boiler equipment are considered optimal:

one-room city apartment - boiler with an output load of 4.16-5 kW;
for a two-room apartment - equipment rated at 5.85-6 kW;
for a three-room apartment it will be enough to have a unit of 8.71-10 kW;
a four-room apartment or a private residential house will require a boiler with parameters of 12-24 kW for heating.

Important! When it comes to installing solid fuel boiler equipment in private homes and suburban residential buildings, it is necessary to focus on devices with greater technological capabilities. To heat and provide hot water supply to a residential building with an area of 150 m2 or more, you will need to install a solid fuel boiler of 24 kW or more. It all depends on the intensity of the heating system and the volume of domestic needs for hot water.

It is always necessary to choose heating equipment individually, based on calculated data and your own needs.

Options for calculating the power of solid fuel units

The accuracy of your calculations depends on taking into account all the factors and indicators that we paid attention to above. For greater clarity, you can follow a number of steps that will give you an idea of how this is done.

The specific power of the heating device is indicated by the letter W. For regions of our country with a harsh climate, this parameter is 1.2-2 kW. In the southern regions, the specific heater value varies between 0.7-0.9 kW. The average value in this case is 1.2-1.5 kW.

First, we determine the area of the premises to be heated. Next, we divide the obtained area data by the specific power of the boiler installed in a house in a certain area. We divide the resulting result by 10, based on the theoretical ratio of the expended power of heating equipment to heat 10 square meters. meters.

For example: we calculate the maximum load of a coal-fired heating boiler for an average residential building with an area of 150 m2.

The living area is 150 sq. m. meters.
The specific power of the heating apparatus for heating 10 m2 is 1.5 kW.

We use the following formula for work: W = (150 x 1.5)/10. As a result, we get 22.5 kW. The obtained value is the starting point for choosing an autonomous solid fuel boiler, taking into account the technological capabilities of the heating system and your own household needs.

On a note: Having found a similar model of heating equipment, add 20-30% of the power to increase the technological capabilities of all heating equipment. The load on the hot water system and the comfortable temperature in the house depend on the number of residents in the house, provided that the boiler operates at optimal conditions.

The optimal choice of heating equipment - the nuances and subtleties of the issue

Having found out for yourself the necessary power parameters of the solid fuel boiler that will be in your home, you can begin to design and install the heating system. You should be aware that the declared data on the thermal load life of the equipment affects the cost of the unit. Low-power heating devices have limited technological capabilities and are designed mainly for heating small rooms. These can be country houses, saunas and country-type guest buildings.

If necessary, the question arises of how to increase the functionality and efficiency of a solid fuel device. In this case, there are reasonable technical and engineering solutions with which increasing the boiler’s performance will have a tangible effect.

On a note: The efficiency of the device can be significantly increased by installing an additional heat exchanger in the chimney, which will receive heat from volatile combustion waste escaping into the atmosphere. An economizer (additional heat exchanger) will give an increase of 20-30% to the rated power of the boiler equipment.

It is not advisable to use high-power solid fuel boilers for autonomous heating of residential buildings. Such equipment is cumbersome and requires a special large area for installation. Considering the size and enormous power of industrial boiler equipment, one should remember the significant consumption of fuel resources.

This technique is ideal for heating on an industrial scale. A lot of heat will be required when heating large industrial facilities and structures. Solid fuel units with a large thermal load are installed at enterprises.

conclusions

Selecting heating equipment is a complex and responsible task. You should not immediately pursue models of solid fuel units that have more power. In some cases, for heating a residential building, installing a unit with output parameters of 24-36 kW is sufficient. At a temperature outside the window of -30 0 C, such a boiler will make it possible to create a temperature inside the room of +20-22 0 C and heat the water in the hot water system to 40-45 0 C.

In each individual case, you can make a choice in favor of one type of heating technology or another.

Higher boiler power may be required in peak situations, when climatic conditions force the heating system to work in increased mode. However, such situations are not systematic, and most of the time your heating device will operate at reduced settings. If you expect a large consumption of hot water for domestic purposes, then you should immediately focus on equipment with higher power. In modern private homes, more than 50% of the power of heating equipment is used to provide hot water to the inhabitants of the house. Connecting a “warm floor” heating system also forces you to pay attention to boiler equipment with higher power.

You need to select a boiler not only based on its actual power. The operational capabilities of heating equipment, the method and quality of maintenance of boiler equipment play a role here. Using the optimal type of fuel for your heating equipment, the presence of automation will allow you to achieve normal operation of your solid fuel boiler.

The comfort of people staying indoors, especially in the winter, largely depends on the temperature of the air around them. Therefore, among the utilities installed in residential premises, the heating system takes first place. In urban conditions, issues of heating apartments are most often resolved centrally, but in private buildings their owners have to install autonomous heating systems, the main element of which is a hot water boiler. The efficiency of the entire system depends on the technical and economic characteristics of the latter.

How to calculate boiler power

Calculation of boiler power is carried out taking into account the area of the heated object

The power of a heating boiler is the main indicator characterizing its capabilities associated with optimal heating of premises during peak loads. The main thing here is to correctly calculate how much heat will be needed to heat them. Only in this case will it be possible to choose the right boiler for heating a private house in terms of power.

To calculate the power of a boiler for a home, various methods are used, in which the area or volume of heated premises is taken as a basis. More recently, the required power of a heating boiler was determined using the so-called house coefficients established for different types of houses within the limits (W/sq.m.):

130…200 – houses without thermal insulation;
90…110 – houses with a partially insulated facade;
50...70 – houses built using 21st century technologies.

By multiplying the area of the house by the corresponding house coefficient, the required power of the heating boiler was obtained.

Calculation of boiler power based on the geometric dimensions of the room

Dependence of gas boiler power on room area

Wcat = S*Wud/10, Where:

Wcat– design power of the boiler, kW;
S– total area of the heated room, sq. m.;
Wud– specific power of the boiler, which falls on every 10 sq. m. heated area.

In general, it is accepted that, depending on the region in which the room is located, the specific power of the boiler is (kW/sq.m.):

for the southern regions - 0.7...0.9;
for areas of the middle zone - 1.0...1.2;
for Moscow and the Moscow region – 1.2...1.5;
for northern regions - 1.5...2.0.

The above formula for calculating a boiler for heating a house by area is used in cases where the water heating unit will be used only for heating rooms with a height of no more than 2.5 m.

If it is assumed that a double-circuit boiler will be installed in the room, which in addition to heating should provide users with hot water, the resulting calculated power must be increased by 25%.

If the height of the heated premises exceeds 2.5 m, then the obtained result is corrected by multiplying it by the coefficient Kv. Kv = N/2.5, where H is the actual height of the room, m.

In this case, the final formula looks like this: P = (S*Wsp/10)*Kv

This method of calculating the required power that a heating boiler must have is suitable for small buildings with an insulated attic, insulated walls and windows (double glazing), etc. In other cases, the result obtained from an approximate calculation may lead to that the purchased boiler will not be able to operate normally. At the same time, excess or insufficient power contributes to a number of problems that are undesirable for the user:

reduction of technical and economic indicators of boiler operation;
failure of automation systems;
rapid wear of parts and components;
formation of condensation in the chimney;
clogging of the chimney with products of incomplete combustion of fuel, etc.;

To obtain more accurate results, it is necessary to take into account the amount of actual heat loss through individual elements of buildings (windows, doors, walls, etc.).

Refined calculation of boiler power

The power of a double-circuit boiler must be greater due to DHW

The calculation of a heating system that includes a heating boiler must be carried out individually for each facility. In addition to its geometric dimensions, it is important to take into account a number of such parameters:

presence of forced ventilation;
climate zone;
availability of hot water supply;
degree of insulation of individual elements of the facility;
the presence of an attic and basement, etc.

In general, the formula for a refined calculation of boiler power is as follows:

Wcat = Qt*Kzap, Where:

Qt– heat loss of the object, kW.
Kzap– safety factor by the amount of which it is recommended to increase the design capacity of the facility. As a rule, its value is in the range of 1.15...1.20 (15-20%).

Predicted heat losses are determined by the formulas:

Qt = V*ΔT*Kp/860, V = S*H; Where:

V– volume of the room, cubic meters;
ΔT– difference between external and internal air temperatures, °C;
Kr– dissipation coefficient, depending on the degree of thermal insulation of the object.

The dissipation coefficient is selected based on the type of building and the degree of its thermal insulation.

Objects without thermal insulation: hangars, wooden barracks, structures made of corrugated iron, etc. – Kr = 3.0...4.0.
Buildings with a low level of thermal insulation: single brick walls, wooden windows, slate or iron roof - Kp is taken to be within the range of 2.0...2.9.
Houses with an average degree of thermal insulation: two-brick walls, a small number of windows, a standard roof, etc. - Kr is 1.0...1.9.
Modern, well-insulated buildings: heated floors, double-glazed windows, etc. – Kp is in the range of 0.6...0.9.

To make it easier for consumers to find a heating boiler, many manufacturers place special calculators on their websites and dealer websites. With their help, by entering the necessary information in the appropriate fields, you can with a high degree of probability determine for what area, for example, a 24 kW boiler is designed.

Typically, such a calculator performs calculations using the following data:

average value of outside air temperature in the coldest week in the winter season;
air temperature inside the object;
presence or absence of hot water supply;
data on the thickness of external walls and ceilings;
materials from which floors and external walls are made;
ceiling height;
geometric dimensions of all external walls;
number of windows, their sizes and detailed description;
information about the presence or absence of forced ventilation.

Having processed the received data, the calculator will give the customer the required power of the heating boiler, and also indicate the type and brand of the unit that satisfies the request. An example of calculating a line of gas boilers intended for heating houses of different sizes is given in the table:

Note to column 11: Нс – wall-mounted atmospheric boiler, А – floor-standing boiler, Нд – wall-mounted turbocharged boiler.

Using the above methods, the power of a gas boiler is calculated. However, they can also be used to calculate the power characteristics of water heating units operating on other types of fuel.

Heat loss accounting

Without taking into account heat loss, it is difficult to correctly calculate the boiler power

When starting to develop an autonomous heating system, you must first find out how much heat is lost to the street during the most severe frosts through the so-called enclosing structures. These include walls, windows, floors and roofs. Only after determining the amount of heat loss will it be possible to worry about selecting a heat source of appropriate power. It should be taken into account that the loss of heat from a building in the winter season occurs not only through the enclosing structures. A significant part of the generated heat (up to 30%) is spent on heating cold air coming from the street due to natural ventilation.

The total amount of heat required to heat the room is determined by the formula:

Q = Qdesign + Qair, Where:

Qconstruct– the amount of heat lost through a similar structure, W;
Qair– the amount of heat consumed to heat the air coming from the street, W.

By summing up the values obtained as a result of calculations, the total heat load on the heating system of the entire building is determined.

All measurements are carried out on the outside of the building, necessarily taking into account its corners. Otherwise, the calculation of heat loss will be inaccurate.

There are other ways of heat leakage in rooms, for example, through a kitchen hood, open doors and windows, cracks in structures, etc. However, the amount of heat lost for these reasons practically does not exceed 5% of the total heat loss and is therefore not taken into account in calculations .

Calculation of heat loss through building envelopes

The complexity of the calculation lies in the fact that it must be carried out for each room separately, carefully inspecting, measuring and assessing the condition of each of its elements adjacent to the environment. Only in this case can you take into account all the heat leaving the house.

Based on the results of the measurements, the area S of each element of the enclosing structure is determined, which is then inserted into the basic formula for calculating the amount of lost thermal energy:

Qconstruct = 1/R*(Tv-Tn)*S*(1+Σβ), R = δ/λ; Where:

R– thermal resistance of the construction material, m sq.°C/W;
δ – thermal conductivity of the construction material, W/m°C);
λ – thickness of the construction material, m;
S– area of the external fence, sq. m.;
TV– indoor air temperature, °C;
Tn– the lowest air temperature in the winter season, °C;
β – heat loss, which depends on the orientation of the building.

If the structure consists of several materials, for example, a brick wall with insulation, the value of thermal resistance R is calculated separately for each of these materials and then summed up.

Heat losses, depending on the orientation of the building, are selected based on where the enclosing element is oriented:

to the north side – β = 0.1;
to the west or southeast – β = 0.05;
to the south or southwest – β = 0.

The calculation of heat losses through the elements of the building envelope is carried out for each room in the building, and then summing them up, the predicted value of the total heat losses in it is obtained. After this, they proceed to the calculation in the next room. As a result of the work carried out, the home owner will be able to identify ways of maximum heat leakage and eliminate the causes of their occurrence.

Calculation of heat consumed to heat ventilation air

The amount of heat that is spent on heating ventilation air reaches, in some cases, 30% of the total thermal energy losses. This is a fairly large value that is inappropriate to ignore. To calculate the amount of heat that will be forced to be spent on heating the supply air, the formula is used:

Qair = c*m* (Tv-Tn), Where:

c– heat capacity of the air mixture, the value of which is 0.28 W/kg°C;
m– mass flow of air entering the room from the street, kg.

The mass flow of air entering the room from the outside is determined by assuming that the air is renewed throughout the entire house once every hour. In this case, by adding up the volumes of all rooms, the volumetric air flow rate is obtained. Then, using the air density value, its volume is converted to mass. Here you need to take into account the fact that the density of air depends on its temperature.

Substituting all known quantities into the above formula, the amount of heat required to heat the supply air is determined.

Common mistakes

Calculation of an autonomous heating system is a complex process consisting of several interrelated, step-by-step procedures:

Calculation of heat losses of an object.
Determination of the temperature regime of individual rooms and the building as a whole.
Calculation of the power of heating radiator batteries.
Hydraulic calculation of the heating system.
Calculation of heating boiler power.
Determination of the total volume of the autonomous heating system.

Thermal calculation of a heating system is not theoretical research, but an accurate and well-founded result, the practical implementation of which will allow you to correctly select all the necessary components and set up an effective heating system that functions without problems for many years.

The main mistake that many owners of private houses make is ignoring some stages of the calculation. They believe that to solve the problem it is enough to choose a more powerful boiler, focusing only on the data of the approximate calculation of its power based on the area of the room. This approach threatens with unnecessary operating costs and often leads to the fact that the boiler will work constantly, the radiator batteries will be hot, and the room will be cold. In this case, it is necessary to return to the original state and perform a full calculation of the heating system. Only after this can we begin to eliminate the shortcomings caused by critical errors in calculations.

When choosing a solid fuel boiler, you need to consider power. It determines whether the device can create the required amount of heat for the entire house or not. It is undesirable to choose a boiler that is too powerful, because it will operate in economy mode, and this will affect the reduction in efficiency.

To make the right one, you need to know two indicators:

The amount of heat required to heat the room and heat the water.
The real power of the device.

Calculation of power depending on the volume of the room

The calculation formula is:

Q = VxΔTxK/850,

where Q – The amount of heat, defined in kW/h4;
V – room volume(unit of measurement cubic m);
ΔT is the difference between the outside temperature and the indoor temperature;
TO - correction factor, taking into account heat loss;
the number 850 is used to convert the product of the above three indicators to kW/hour.

K can have the following meanings:

3-4 – for premises that are a simplified wooden structure or a building made of corrugated sheets.
2-2,9 – for buildings with little thermal insulation. The design of such houses is simplified, the wall thickness is equal to the length of 1 brick, the windows and roof have a simple structure.
1-1,9 – for houses whose design is standard. The brickwork is double, the number of simple windows is small. The roof has a conventional roof.
0,6-0,9 – for houses with improved construction, double thermal insulation of brick walls, double glazed windows, thick floor base, roof made of good thermal insulation material.

As an example, let's take a modern house with an area of 200 square meters. m, wall height 3 m and first-class thermal insulation. The house is located in an area where in winter the temperature does not drop below -25 °C. In this case, ΔT = 20 – (-25) = 45 °C. Therefore, to heat a house you need to create Q = 200*3*45*0.9/850 = 28.58 kW/h. The figure should not be rounded, because it is not final and you need to increase it with your own hands by the amount of heat for the hot water supply. If the water is planned to be heated in a different way, then the result obtained is not adjusted, and part of the calculation is completed.

Calculation of heat for hot water supply

where c is specific heat capacity of water(the indicator is always 4200 J/kg*K);
m – mass of water in kg;
Δt temperature difference between heated water from the water supply.

Read also: Cleaning a solid fuel boiler from tar and soot

Example. The average family's need for warm water can reach 150 liters. If the boiler heats the coolant to a temperature of 80 °C, and the water from the pipeline has a temperature of 10 °C, then Δt = 80 – 10 = 70 °C.

Qв = 4200*150*70 = 44,100,000 J or 12.25 kW/h.

If 150 liters need to be heated at a time, the capacity of the indirect boiler is 150 liters, then 12.25 kW/h is added to 28.58 kW/h. This must be done because if Qzag is less than 40.83, the room will be colder than the calculated 20 °C.
If the water must be heated in portions, the volume of the indirect boiler is 50 liters, then 12.25 is divided by 3 and added with your own hands to 28.58. Qzag will be equal to 32.67 kW/h. This is the power of the device for the heating system.

Calculation by area

It is more accurate because it takes into account more factors. The calculation is made using the formula:

Q = 0.1*S*k1*k2*k3*k4*k5*k6*k7, Where:

0.1 kW is the heat norm per 1 sq. m;

S – area of the heated house;

k1 demonstrates heat loss caused by window design. Has the meaning:

1.27 – if the windows have one glass;
1.0 – if there are double-glazed windows;
0.85 – if there are windows with triple glass.

k2 demonstrates heat loss caused by window area (Sw). Is the ratio of Sw to the floor area Sf. Its meanings are:

0.8 at Sw/Sf = 0.1;
0.9 at Sw/Sf = 0.2;
1 at Sw/Sf = 0.3;
1.1 at Sw/Sf = 0.4;
1.2 at Sw/Sf = 0.5.

k3 is coefficient of heat loss through walls. It happens like this:

1.27 with very poor thermal insulation;
1 in houses with a wall of 2 bricks or insulation, the thickness of which is 15 cm;
0.854 with good thermal insulation.

k4 shows heat loss depending on the air temperature outside the house (tz). Has the following meanings:

0.7, if tз = -10 °С;
0.9 for tз = -15 °С;
1.1 for tз = -20 °С;
1.3 for tз = -25 °С;
1.5 for tз = -30 °С.

Read also: Advantages of the Popov boiler

k5 demonstrates heat loss through external walls. Is like this:

1.1 for rooms with one external wall;
1.2 for 2 external walls;
1.3 for 3 external walls;
1.4 for a building with 4 external walls.

K6 shows how much Additional heat required depending on ceiling height (H). Its meanings are:

1 for H = 2.5 m;
1.05 for H = 3.0 m;
1.1 for H = 3.5 m;
1.15 for H = 4.0 m;
1.2 for H = 4.5 m.

k7 determines heat loss depending on the type of room located above the heated room. It happens like this:

0.8 for heated rooms;
0.9 for a warm attic;
1 for a cold attic.

Example. The conditions of the problem are the same. The windows are triple glazed and make up 30% of the floor area. The number of external walls is 4. There is a cold attic upstairs.

Q = 0.1*200*0.85*1*0.854*1.3*1.4*1.05*1 = 27.74 kW/h. This figure must be increased by adding with your own hands the amount of heat required for hot water supply.

Real power of a long-burning boiler

Many devices are designed for a specific type of fuel. If other types of fuel are burned in them, their efficiency will be lower.

The power calculation will be carried out on the basis of the Viessmann Vitoligno 100-S 60 pyrolysis boiler. Its features are as follows:

Powered by wood.
In 1 hour, from 6 to 15 kg of firewood burns in the loading chamber.
Its rated power is 60 kW.
The loading chamber volume is 294 liters.
Efficiency is 87%

Let the owner plan to burn aspen wood in it. 1 kg of such firewood produces 2.82 kW/h. If a boiler burns 15 kg in 1 hour, then it emits 2.82*15*0.87 = 36.801 kW/h of heat (0.87 is efficiency). Such a device is not enough for heating a house with a 150 liter boiler, but it is quite enough for hot water supply with a 50 liter boiler. To get the figure 32.67 kW/h, you need to burn 13.31 kg of aspen firewood in 1 hour (32.67/(2.82*0.87) = 13.31). This is the case if you calculate the heat demand by volume.