The calculation of the heating system of a private house: the rules and examples of calculation

Heating a private house is a necessary element of comfortable housing. And the arrangement of the heating complex should be approached carefully, because mistakes are expensive.

Let us consider how the calculation of the heating system of a private house is performed in order to effectively compensate for heat losses in the winter months.

Heat loss of a private house

The building loses heat due to the difference in air temperature inside and outside the house. Heat loss is higher, the more significant the area of ​​enclosing structures of the building (windows, roof, walls, basement).

Also, the loss of thermal energy associated with the materials of enclosing structures and their sizes. For example, the heat loss of thin walls is more than thick.

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The heating system of a private house with two units
The heating system of a private house with two units
The option of heating in the log house
The option of heating in the log house
Air intake and heat leakage through windows and doors
Air intake and heat leakage through windows and doors
Ventilation system with fresh air supply
Ventilation system with fresh air supply
Diagram of the device for hot water supply and heating
Diagram of the device for hot water supply and heating
Selection of the boiler by type of fuel
Selection of the boiler by type of fuel
Laying options for heating circuits
Laying options for heating circuits
Outdoor heating option
Outdoor heating option

Efficient calculation of heating for a private house necessarily takes into account the materials used in the construction of walling. For example, with an equal thickness of a wall made of wood and brick, heat is conducted with different intensity - heat losses through wooden structures go slower. Some materials transmit heat better (metal, brick, concrete), others worse (wood, mineral wool, polystyrene foam).

The atmosphere inside the residential building is indirectly connected to the external air environment. The walls, window and door openings, the roof and the foundation in the winter transfer heat from the house to the outside, supplying cold instead. They account for 70-90% of the total heat loss of the cottage.

Heat loss to calculate the heating system of a private house
The walls, the roof, the windows and the doors all let the heat out in the winter. The imager will clearly show the leakage of heat

The constant leakage of thermal energy during the heating season also occurs through ventilation and sewage. When calculating the heat loss of individual housing construction, these data are usually not taken into account. But the inclusion of heat loss through the sewer and ventilation systems in the general heat calculation of the house is still the right decision.

Heat loss country house
Properly arranged heat insulation system (+) can significantly reduce heat leakage passing through building structures, door / window openings

It is impossible to calculate the autonomous heating circuit of a country house without estimating the heat loss of its enclosing structures. More precisely, it will not be possible to determine the power of the heating boiler, sufficient to heat the cottage in the most severe frosts.

Analysis of the real consumption of thermal energy through the walls will allow you to compare the cost of boiler equipment and fuel with the cost of thermal insulation of the enclosing structures. After all, the more energy efficient the house, i.e. the less heat it loses in the winter months, the lower the cost of purchasing fuel.

Thermal conductivity of structural materials
For a proper calculation of the heating system will need a coefficient of thermal conductivity of common building materials (+)

Calculation of heat loss through the walls

Using the example of a conditional two-story cottage, we calculate heat losses through its wall structures. Initial data: a square “box” with front walls 12 m wide and 7 m high; in the walls of 16 openings, the area of ​​each 2.5 m2; front wall material - solid brick ceramic; wall thickness - 2 bricks.

Heat transfer resistance. To find out this figure for the facade wall, you need to divide the thickness of the wall material by its coefficient of thermal conductivity. For a number of construction materials, thermal conductivity data are presented in the images above and below.

Heat conductivity coefficient of insulation
For accurate calculations, you need a coefficient of thermal conductivity of heat-insulating materials used in construction in the table (+)

Our conventional wall is built of ceramic brick, the coefficient of thermal conductivity - 0.56 W / m ·aboutC. Its thickness, taking into account masonry at the TsPR, is 0.51 m. By dividing the wall thickness by the thermal conductivity coefficient of the brick, we obtain the heat transfer resistance of the wall:

0.51: 0.56 = 0.91 W / m2 × oFROM

The result of the division is rounded to two decimal places; there is no need for more accurate data on the resistance to heat transfer.

The area of ​​the outer walls.Since the square building was chosen as an example, the area of ​​its walls is determined by multiplying the width by the height of one wall, then by the number of external walls:

12 · 7 · 4 = 336 m2

So, we know the area of ​​the facade walls.But what about the openings of windows and doors, occupying together 40 m2 (2.5 · 16 = 40 m2a) front wall, do you need to take them into account? Indeed, how to correctly calculate the autonomous heating in a wooden house without taking into account the heat transfer resistance of window and door structures.

How to calculate heat loss through the walls
Heat conductivity coefficient of heat-insulating materials used for insulation of load-bearing walls (+)

If you need to calculate the heat loss of a large building or a warm house (energy efficient) - yes, taking into account the heat transfer coefficients of window frames and entrance doors when calculating will be correct.

However, for low-rise buildings IZHS constructed from traditional materials, door and window openings can be neglected. Those. do not take away their area from the total area of ​​the facade walls.

Total heat loss of the walls.We find out the heat loss of the wall from its one square meter when the difference in temperature of air inside and outside the house is one degree. To do this, we divide the unit by the heat transfer resistance of the wall, calculated earlier:

1: 0.91 = 1.09 W / m2·aboutFROM

Knowing the heat loss from the square meter of the perimeter of the external walls, one can determine the heat loss at certain street temperatures.For example, if the temperature in a cottage is +20aboutC, and on the street -17aboutC, the temperature difference will be 20 + 17 = 37aboutC. In such a situation, the total heat loss of the walls of our conditional home will be:

0.91 (heat transfer resistance per square meter of the wall) · 336 (front wall area) · 37 (temperature difference between the room and the street atmosphere) = 11313 W

Thermal insulation materials - thermal conductivity
Heat conductivity coefficient of heat-insulating materials used for floor / wall insulation, for dry floor screeding and leveling walls

Let us recalculate the obtained value of heat loss in kilowatt-hours, they are more convenient for perception and subsequent calculations of the power of the heating system.

Wall heat loss in kilowatt-hours.First, find out how much heat energy goes through the walls in one hour at a temperature difference of 37aboutFROM.

We remind you that the calculation is carried out for a house with structural characteristics conditionally selected for demonstration-demonstrative calculations:

11313 (the value of heat loss obtained earlier) · 1 (hour): 1000 (number of watts per kilowatt) = 11.313 kW · h.

Thermal conductivity of building materials and thermal insulation
Heat conductivity coefficient of building materials used for wall and floor insulation (+)

To calculate the heat loss per day, the resulting value of heat loss per hour is multiplied by 24 hours:

11.313 · 24 = 271.512 kW · h

For clarity, let's find out the heat loss for the full heating season:

7 (the number of months in the heating season) · 30 (the number of days in the month) · 271.512 (daily heat loss of walls) = 57017.52 kW · h

So, the calculated heat loss from a house with the above-selected characteristics of the building envelope will be 57017.52 kWh for the seven months of the heating season.

Accounting for the effects of ventilation of a private house

Calculation of heat ventilation losses during the heating season as an example will be carried out for a conventional square-shaped cottage, with a wall 12 meters wide and 7 meters high. Without furniture and interior walls, the internal volume of the atmosphere in this building will be:

12 · 12 · 7 = 1008 m3

At air temperature +20aboutC (norm in the heating season) its density is equal to 1.2047 kg / m3and the specific heat capacity is 1,005 kJ / (kg ·aboutFROM). Calculate the mass of the atmosphere in the house:

1008 (volume of home atmosphere) · 1.2047 (air density at t +20aboutC) = 1214.34 kg

Table of thermal conductivity of related materials
The table with the value of the coefficient of thermal conductivity of materials that may be required when carrying out accurate calculations (+)

Suppose a fivefold change in air volume in the premises of the house.Note that the exact need for fresh air intake depends on the number of residents of the cottage. With an average temperature difference between the house and the street during the heating season, equal to 27aboutC (20aboutWith home, -7aboutFrom the external atmosphere) for the day for heating the incoming cold air need thermal energy:

5 (number of air changes in the rooms) · 27 (temperature difference between the room and outdoor atmosphere) · 1214.34 (air density at t +20aboutС) · 1,005 (specific heat capacity of air) = 164755,58 kJ

We translate kilojoules to kilowatt-hours:

164,755.58: 3,600 (number of kilojoules per kilowatt-hour) = 45.76 kWh

Having found out the cost of thermal energy for heating the air in the house with its fivefold replacement through the inlet ventilation, it is possible to calculate the “air” heat losses during the seven-month heating season:

7 (number of “heated” months) · 30 (average number of days in a month) · 45.76 (daily costs of heat energy for heating the supply air) = 9609.6 kW · h

Ventilation (infiltration) energy costs are inevitable, since the renewal of air in the rooms of the cottage is vital. The heating needs of a replaceable air atmosphere in a house need to be calculated, summed up with heat loss through walling structures and taken into account when choosing a heating boiler.There is another type of thermal energy, the latter - sewer heat loss.

Energy costs for DHW preparation

If during warm months cold water comes from the tap to the cottage, then during the heating season it is ice cold, with a temperature not exceeding +5aboutC. Bathing, washing dishes and washing impossible without heating water. The water collected in the toilet tank contacts through the walls with the home atmosphere, taking some heat. What happens to the water heated by burning not free fuel and spent on domestic needs? It is drained into the sewer.

Boiler with boiler
Double-circuit boiler with indirect heating boiler, used both for heating the heat carrier and for supplying hot water to the circuit built for it

Consider an example. A family of three, suppose it consumes 17 m3water monthly. 1000 kg / m3- the density of water, and 4,183 kJ / kg ·aboutC is its specific heat. The average temperature of heating water intended for domestic needs, let it be +40aboutC. Accordingly, the difference of the average temperature between the cold water entering the house (+5aboutC) and heated in a boiler (+30aboutC) it turns out 25aboutFROM.

To calculate the sewage heat loss we consider:

17 (monthly volume of water consumption) · 1000 (water density) · 25 (temperature difference between cold and heated water) · 4.183 (specific heat capacity of water) = 1777775 kJ

To convert kilojoules to clearer kilowatt hours:

1777775: 3600 = 493.82 kWh

Thus, for the seven-month period of the heating season, the thermal energy in the amount of:

493.82 · 7 = 3456.74 kW · h

The consumption of thermal energy for heating water for hygienic needs is small compared to heat loss through walls and ventilation. But this, too, energy costs, loading the heating boiler or boiler and causing fuel consumption.

Calculation of the power of the heating boiler

The boiler in the heating system is designed to compensate for the heat loss of the building. And also, in the case of a dual-circuit system or when the boiler is equipped with an indirect heating boiler, to warm the water for hygienic needs.

Having calculated the daily heat losses and the flow of warm water “to the sewage system”, it is possible to accurately determine the required boiler capacity for a cottage of a certain area and the characteristics of the building envelope.

Boiler
Single-circuit boiler produces only the heating medium heating for the heating system

To determine the power of the heating boiler, it is necessary to calculate the cost of heat energy at home through the facade walls and to heat the alternating air atmosphere in the interior. Required data on heat loss in kilowatt-hours per day - in the case of a conditional house, calculated as an example, is:

271.512 (daily heat losses by external walls) + 45.76 (daily heat losses for supply air heating) = 317.272 kWh

Accordingly, the required heating capacity of the boiler will be:

317.272: 24 (hours) = 13.22 kW

However, such a boiler will be under a constantly high load, reducing its service life. And in especially frosty days, the calculated boiler capacity will not be enough, since with a high temperature difference between the room and street atmospheres the building heat loss will sharply increase.

Therefore, the boiler selected by the average calculation of the cost of thermal energy, with severe frosts can not cope. It would be rational to increase the required power of the boiler equipment by 20%:

13.22 · 0.2 + 13.22 = 15.86 kW

To calculate the required power of the second circuit of the boiler, heating water for washing dishes, bathing, etc., it is necessary to divide the monthly heat consumption of the “sewer” heat loss by the number of days in the month and by 24 hours:

493.82: 30: 24 = 0.68 kW

According to the results of calculations, the optimal boiler power for the example cottage is 15.86 kW for the heating circuit and 0.68 kW for the heating circuit.

The choice of radiators

Traditionally, the power of the heating radiator is recommended to choose the area of ​​the heated room, and with 15-20% overestimate of power needs, just in case. For example, consider how correct the method of selecting a radiator "10 m2 of area - 1.2 kW".

Ways to connect radiators
Thermal power of radiators depends on the method of their connection, which must be taken into account when calculating the heating system

Baseline: corner room on the first level of a two-story house IZHS; outer wall of double-row masonry ceramic bricks; the width of the room is 3 m, the length is 4 m, the ceiling height is 3 m. According to a simplified selection scheme, it is proposed to calculate the area of ​​the room, we assume:

3 (width) · 4 (length) = 12 m2

Those. the necessary power of the heating radiator with a 20% surcharge is 14.4 kW. And now we calculate the power parameters of the heating radiator on the basis of the heat loss of the room.

In fact, the area of ​​the room affects the loss of heat energy less than the area of ​​its walls, going out one side to the outside of the building (facade). Therefore, we will consider exactly the area of ​​"street" walls in the room:

3 (width) · 3 (height) + 4 (length) · 3 (height) = 21 m2

Knowing the area of ​​the walls that transmit heat "to the street", we calculate the heat loss when the difference between room and outdoor temperature is 30about(in the house +18aboutC, outside -12aboutC), and immediately in kilowatt-hours:

0.91 (heat transfer m2 of room walls facing the street) · 21 (area of ​​"street" walls) · 30 (temperature difference inside and outside the house): 1000 (watts per kilowatt) = 0.57 kW

Installation of radiators
According to building standards, heating devices are located in places of maximum heat loss. For example, radiators are installed under window openings, heat guns - over the entrance to the house. In the corner rooms, batteries are installed on blank walls exposed to the maximum winds.

It turns out that to compensate for heat losses through the facade walls of this structure, at 30aboutthe temperature difference in the house and on the street is enough heating capacity of 0.57 kW · h. Increase the required power by 20, even by 30% - we get 0.74 kWh.

Thus, the actual power needs of heating can be significantly lower than the “1.2 kW per square meter of the floor space” trading scheme. Moreover, the correct calculation of the required capacity of heating radiators will reduce the amount of coolant inheating system, which will reduce the load on the boiler and fuel costs.

Conclusions and useful video on the topic

Preservation of heat in the premises of the house - the main task of the heating system in the winter months. However, the heat is constantly not enough. Where the heat leaves the house - the answers are provided by a visual video:

The video describes the procedure for calculating the heat loss at home through the building envelope. Knowing the heat loss, you can accurately calculate the power of the heating system:

The choice of heating boiler power depends on the state of the house and on the quality of the insulation of its enclosing structures. The principle of "kilowatt per 10 squares of the area" works in the cottage of the average condition of the facades, roof and foundation. Detailed video about the principles of selection of power characteristics of the heating boiler, see below:

Heat production is becoming more expensive annually - fuel prices are rising. It is impossible to relate to the energy costs of the cottage, it is completely unprofitable. On the one hand, each new heating season is more expensive and more expensive for a homeowner. On the other hand, weatherization of the walls, foundation and roofing of a country house costs good moneyHowever, the less heat leaves the building, the cheaper it will be to heat it.

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