Heating a private house is a necessary element of comfortable housing. Agree that the arrangement of the heating complex should be approached carefully, because mistakes are expensive. But you have never done such calculations and do not know how to perform them correctly?
We will help you  in our article we will examine in detail how the calculation of the heating system of a private house is done in order to effectively compensate for heat losses in the winter months.
Let us give concrete examples, adding material photos and useful video tips, as well as uptodate tables with indicators and coefficients necessary for calculations.
The content of the article:

Heat loss of a private house
 Calculation of heat loss through the walls
 Accounting for the effects of ventilation of a private house
 Energy costs for preparing DHW
 Calculation of the power of the heating boiler
 The choice of radiators
 Conclusions and useful video on the topic
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 heat loss 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 main purpose of the calculation of heating is the competent choice of a heating unit that can compensate for heat loss during the cold period of the year.
To select the necessary power equipment, heat losses are accumulated through the building envelope.
The calculations take into account the leakage of heat through the loosely fitting window sashes and door leafs, as well as the energy required for heating the incoming air
For rooms with organized mechanical ventilation, carrying out mixing of fresh air mass from the outside, the need for energy consumption for its heating is taken into account.
If it is planned to use a doublecircuit boiler as the main unit of heating and water heating for the DHW system, the energy necessary for this task is taken into account in the calculations.
Competently performed calculations necessarily take into account the type of fuel and its energy efficiency.
All calculations are adjusted with reference to the method of arrangement of heating circuits, with hidden installation of the system, consideration should be given to the heating of building structures.
When calculating for an open heating scheme that communicates directly with the atmosphere through an open expansion tank, energy losses are necessarily taken into account when the coolant cools.
The heating system of a private house with two units
The option of heating in a log house
Air intake and heat leakage through windows and doors
Ventilation system with fresh air supply
Diagram of the device for hot water supply and heating
Selection of the boiler by type of fuel
Options for laying heating circuits
Outdoor heating option
Effective heating calculation for a private house, it is necessary to take into account the materials used in the construction of walling.
For example, with 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 with 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 7090% of the total heat loss of the cottage.
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 the right solution.
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 evaluating the heat loss of its enclosing structures. More precisely, it will not work determine the power of the heating boiler, sufficient to heat the cottage in the most fierce frosts.
Analysis of the actual consumption of thermal energy through the walls will allow you to compare the cost of boiler equipment and fuel with the cost of insulation of the enclosing structures.
After all, the more energyefficient the house, i.e. the less heat it loses in the winter months, the lower the cost of purchasing fuel.
For proper calculation of the heating system will require coefficient of thermal conductivity common building materials.
Table of values of thermal conductivity of various building materials, most often used when erected
Calculation of heat loss through the walls
Using the example of a conditional twostory cottage, we calculate heat losses through its wall structures.
Initial data:
 square “box” with front walls 12 m wide and 7 m high;
 in the walls of 16 openings, the area of each 2.5 m^{2};
 front wall material  solid brick ceramic;
 wall thickness  2 bricks.
Next, we will calculate the group of indicators, of which the total value of heat loss through the walls is formed.
Heat Resistance Indicator
To find out the heat transfer resistance index for a facade wall, it is necessary to divide the wall material thickness by its thermal conductivity coefficient.
For a number of construction materials, thermal conductivity data are presented in the images above and below.
For accurate calculations, the thermal conductivity coefficient of the heatinsulating materials used in construction is required.
Our conventional wall is built of ceramic brick, the coefficient of thermal conductivity  0.56 W / m ·^{about}WITH. Its thickness, taking into account the laying on the TsPR, is 0.51 m. Dividing the wall thickness by the thermal conductivity coefficient of a brick, we obtain the resistance to heat transfer of the wall:
0.51: 0.56 = 0.91 W / m^{2 × o}WITH
The result of the division is rounded to two decimal places, there is no need for more accurate data on heat transfer resistance.
Exterior wall area
Since a 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 m^{2}
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 m^{2}a) front wall, do you need to take them into account?
Indeed, how to correctly calculate independent heating in a wooden house excluding heat transfer resistance of window and door structures.
The coefficient of thermal conductivity of thermal insulation materials used for insulation of loadbearing 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 lowrise 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 wall heat loss
We find out the heat loss of the wall from its one square meter at a difference of one and two degrees inside and outside the house.
To do this, we divide the unit by the heat transfer resistance of the wall, calculated earlier:
1: 0.91 = 1.09 W / m^{2}·^{about}WITH
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 +20 ^{about}C, and on the street 17 ^{about}C, the temperature difference will be 20 + 17 = 37 ^{about}WITH. In this situation, the total heat loss of the walls of our conditional home will be:
0.91 · 336 · 37 = 11313 W,
Where: 0.91  heat transfer resistance per square meter wall; 336  the area of the front walls; 37  the temperature difference between room and outdoor atmosphere.
Heat conductivity coefficient of heatinsulating materials used for floor / wall insulation, for dry floor screeding and leveling walls
Let us recalculate the obtained value of heat losses in kilowatthours, they are more convenient for perception and subsequent calculations of the power of the heating system.
Wall heat loss in kilowatthours
First, find out how much heat energy goes through the walls in one hour at a temperature difference of 37 ^{about}WITH.
We remind you that the calculation is carried out for a house with design characteristics that are conditionally selected for demonstration and demonstration calculations:
11313 · 1: 1000 = 11.313 kW · h,
Where: 11313 is the heat loss value obtained earlier; 1 hour; 1000 is the number of watts per kilowatt.
The coefficient of thermal conductivity of building materials used for wall and ceiling 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 · 30 · 271.512 = 57017.52 kW · h,
Where: 7  the number of months in the heating season; 30  the number of days in the month; 271,512  daily heat loss of the walls.
So, the calculated heat loss from a house with the abovechosen 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 ventilation heat losses during the heating season as an example will be carried out for a conditional cottage of square shape, with a wall 12 meters wide and 7 meters high.
Excluding furniture and interior walls, the internal volume of the atmosphere in this building will be:
12 · 12 · 7 = 1008 m^{3}
At air temperature +20 ^{about}C (norm in the heating season) its density is equal to 1.2047 kg / m^{3}and the specific heat 1,005 kJ / (kg ·^{about}WITH).
Calculate the mass of the atmosphere in the house:
1008 · 1.2047 = 1214.34 kg,
Where: 1008  the volume of the home atmosphere; 1.2047  air density at t +20 ^{about}WITH .
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 of air volume in the premises of the house. Note that the exact supply demand fresh air depends on the number of tenants of the cottage.
With an average temperature difference between the house and the street during the heating season, equal to 27 ^{about}C (20 ^{about}With home, 7 ^{about}From the external atmosphere) for the day for heating the incoming cold air need thermal energy:
5 · 27 · 1214.34 · 1,005 = 164755,58 kJ,
Where: 5  the number of air changes in the premises; 27  temperature difference between room and street atmosphere; 1214.34  air density at t +20 ^{about}WITH; 1,005  specific heat of air.
We translate kilojoules to kilowatthours, dividing the value by the number of kilojoules per kilowatthour (3600):
164755.58: 3600 = 45.76 kWh
Having found out the cost of thermal energy for heating the air in the house when it is replaced five times through fresh air ventilation, it is possible to calculate the “air” heat losses during the sevenmonth heating season:
7 · 30 · 45.76 = 9609.6 kW · h,
Where: 7  the number of "heated" months; 30 is the average number of days in a month; 45.76  the daily cost of thermal energy for heating the supply air.
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 preparing DHW
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 no higher than +5 ^{about}WITH. 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.
Dualcircuit 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 m^{3} water monthly. 1000 kg / m^{3}  the density of water, and 4,183 kJ / kg ·^{about}C is its specific heat.
The average temperature of heating water intended for domestic needs, let it be +40 ^{about}WITH. Accordingly, the difference of the average temperature between the cold water entering the house (+5 ^{about}C) and heated in a boiler (+30 ^{about}C) it turns out 25 ^{about}WITH.
For the calculation of sewage heat loss we consider:
17 · 1000 · 25 · 4.183 = 1777775 kJ,
Where: 17  the monthly volume of water consumption; 1000 is the density of water; 25  temperature difference between cold and heated water; 4.183 — specific heat capacity of water;
To convert kilojoules to clearer kilowatt hours:
1777775: 3600 = 493.82 kWh
Thus, for the sevenmonth 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 dual circuit system or at equipment of a boiler with a boiler of indirect heating, for water heating 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.
Singlecircuit 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 kilowatthours per day  in the case of a conditional house, calculated as an example, is:
271.512 + 45.76 = 317.272 kWh,
Where: 271,512  daily heat losses by external walls; 45.76  daily heat losses for heating the supply air.
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 choose a boiler on average calculation of the cost of heat energy is not worth it  it can not cope with severe frosts.
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 “sewer” heat losses by the number of days in a month and by 24 hours:
493.82: 30: 24 = 0.68 kW
According to the results of calculations, the optimal boiler power for an example cottage is 15.86 kW for the heating circuit and 0.68 kW for the heating circuit.
The choice of radiators
Traditionally heating radiator power It is recommended to choose by the area of the heated room, and with 1520% overestimation of power needs, just in case.
For example, consider how correct the method of selecting a radiator "10 m2 of area  1.2 kW".
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 twostory house IZHS; outer wall of doublerow masonry ceramic bricks; room width 3 m, length 4 m, ceiling height 3 m.
Under a simplified selection scheme, it is proposed to calculate the area of the room, we consider:
3 (width) · 4 (length) = 12 m^{2}
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 m^{2}
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 30^{about} (in the house +18 ^{about}C, outside 12 ^{about}C), and immediately in kilowatthours:
0.91 · 21 · 30: 1000 = 0.57 kW,
Where: 0.91  heat transfer resistance m2 room walls, facing the street; 21  the area of "street" walls; 30  temperature difference inside and outside the house; 1000 is the number of watts in kilowatts.
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, the batteries are installed on blank walls exposed to the maximum exposure to winds.
It turns out that to compensate for heat losses through the facade walls of this structure, at 30^{about} the 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 power of heating radiators will reduce the volume heating medium in the heating system, which will reduce the load on the boiler and fuel costs.
Conclusions and useful video on the topic
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 enclosing structures. Knowing the heat loss, you can accurately calculate the power of the heating system:
Detailed video about the principles of selection of power characteristics of the heating boiler, see below:
Heat generation annually rises in price  rising fuel prices. And the heat is constantly not enough. It is impossible to treat indifferently to the energy consumption 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, the foundation and the roof of a country house costs good money. However, the less heat leaves the building, the cheaper it will be to heat it..
Preservation of heat in the premises of the house  the main task of the heating system in the winter months. The choice of power of the heating boiler depends on the state of the house and the quality of the insulation of its enclosing structures. The principle of "kilowatt per 10 squares of area" works in the cottage of the average condition of the facades, roofing and basement.
Did you independently calculate the heating system for your home? Or noticed the inconsistency of the calculations given in the article? Share your practical experience or the amount of theoretical knowledge, leaving a comment in the block under this article.