Heating a private house is a necessary element of comfortable housing. Agree that the arrangement of the heating complex should be approached carefully, as 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 consider in detail how the calculation of the heating system of a private house is done to effectively compensate for heat losses in the winter months.
We give specific examples, supplementing the article with visual photos and useful video tips, as well as relevant tables with indicators and coefficients needed for calculations.
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 the building envelope (windows, roofs, walls, foundations).
Also, the loss of thermal energy is associated with the materials of the building envelope and their size. For example, the heat loss of thin walls is greater than thick.
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The heating system of a private house with two units
Option for heating in a log house
Air and heat leakage through windows and doors
Fresh air ventilation system
DHW and heating circuit diagram
Boiler selection by fuel type
Options for laying heating circuits
Outdoor heating option
An effective calculation of heating for a private house necessarily takes into account the materials used in the construction of building envelopes.
For example, with an equal thickness of a wall made of wood and brick, heat is carried out with different intensities - heat loss through wooden structures is slower. Some materials let heat pass better (metal, brick, concrete), others worse (wood, mineral wool, polystyrene foam).
The atmosphere inside a residential building is indirectly related to the external air environment. Walls, openings of windows and doors, roof and foundation in winter transfer heat from the house to the outside, supplying cold in return. They account for 70-90% of the total heat loss of the cottage.
Walls, roof, windows and doors - everything lets heat out in the winter. The thermal imager clearly shows heat leaks
A constant leak of thermal energy during the heating season also occurs through ventilation and sewage.
When calculating the heat loss of an individual housing construction, these data are usually not taken into account. But the inclusion of heat losses through the sewer and ventilation systems in the general thermal calculation of the house is still the right decision.
Significantly arranged thermal 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 be possible to determine the capacity of the heating boiler, sufficient to heat the cottage in the most severe frosts.
Analysis of the actual consumption of thermal energy through the walls will allow you to compare the costs of boiler equipment and fuel with the costs of thermal insulation of building envelopes.
After all, the more energy-efficient the house, i.e. the less heat it loses in the winter months, the lower the cost of acquiring fuel.
For a competent calculation of the heating system, the thermal conductivity coefficient of common building materials will be required.
The table of values of the coefficient of thermal conductivity of various building materials, most often used in the construction of
Calculation of heat loss through walls
Using the conditional two-story cottage as an example, we calculate the heat loss through its wall structures.
Initial data:
- square “box” with front walls 12 m wide and 7 m high;
- within the walls of 16 openings, the area of each 2.5 m2;
- material of front walls - solid brick ceramic;
- wall thickness - 2 bricks.
Next, we will calculate the group of indicators from which the total value of heat loss through the walls is added.
Heat transfer resistance
To find out the heat transfer resistance index for a facade wall, it is necessary to divide the thickness of the wall material by its thermal conductivity coefficient.
For a number of structural materials, data on the coefficient of thermal conductivity are presented in the images above and below.
For accurate calculations, the coefficient of thermal conductivity indicated in the table of thermal insulation materials used in construction will be required.
Our conditional wall is built of solid ceramic bricks, the thermal conductivity of which is 0.56 W / maboutC. Its thickness, taking into account the masonry on the central distribution center, is 0.51 m. Dividing the wall thickness by the thermal conductivity of the brick, we obtain the wall heat transfer resistance:
0.51: 0.56 = 0.91 W / m2 × oFROM
We round off the division result to two decimal places; there is no need for more accurate data on heat transfer resistance.
External Wall Area
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 front walls. But what about the openings of windows and doors, occupying together 40 m2 (2.5 · 16 = 40 m2) of the front wall, should they be taken into account?
Indeed, how to correctly calculate autonomous heating in a wooden house without taking into account the heat transfer resistance of window and door structures.
Thermal conductivity coefficient of heat-insulating materials used for insulation of load-bearing walls
If it is necessary to calculate the heat loss of a large-area building or a warm house (energy efficient) - yes, taking into account the heat transfer coefficients of window frames and entrance doors will be correct in the calculation.
However, for low-rise buildings IZHS built from traditional materials, door and window openings can be neglected. Those. do not take away their area from the total area of the front walls.
Common wall heat loss
We find out the heat loss of the wall from its one square meter when the temperature difference between the air inside and outside the house is one degree.
To do this, 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 per square meter of the perimeter of the external walls, you can determine the heat loss at certain street temperatures.
For example, if the temperature in the cottage is +20 aboutC, and on the street -17 aboutC, the temperature difference will be 20 + 17 = 37 aboutC. In such a 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 of the wall; 336 - area of front walls; 37 - temperature difference between indoor and outdoor atmosphere.
Thermal conductivity coefficient of heat-insulating materials used for floor / wall insulation, for dry floor screed and wall alignment
We recalculate the resulting 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 we find out how much thermal energy will go through the walls in one hour with a temperature difference of 37 aboutFROM.
We remind you that the calculation is carried out for a house with structural characteristics, conditionally selected for demonstration and demonstration calculations:
113131: 1000 = 11.313 kWh,
Where: 11313 - the amount of heat loss obtained earlier; 1 hour; 1000 is the number of watts per kilowatt.
Thermal conductivity coefficient of building materials used for insulation of walls and floors
To calculate the heat loss per day, the resulting heat loss per hour is multiplied by 24 hours:
11.31324 = 271.512 kWh
For clarity, we find out the loss of thermal energy for the full heating season:
7 · 30 · 271.512 = 57017.52 kWh,
Where: 7 - the number of months in the heating season; 30 - the number of days in a month; 271,512 - daily heat loss of the walls.
So, the estimated heat loss of a house with the characteristics of the enclosing structures selected above will amount to 57017.52 kWh for seven months of the heating season.
Taking into account the effects of private house ventilation
As an example, we will calculate the ventilation heat loss during the heating season for a conditional cottage of a square shape, with a wall of 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 m3
At an air temperature of +20 aboutC (norm in the heating season) its density is 1.2047 kg / m3and the specific heat is 1.005 kJ / (kgaboutFROM).
We calculate the mass of the atmosphere in the house:
10081.2047 = 1214.34 kg,
Where: 1008 - the volume of the home atmosphere; 1.2047 - air density at t +20 aboutFROM .
A table with the value of the coefficient of thermal conductivity of materials that may be required for accurate calculations
Suppose a five-fold change in air volume in the premises of the house. Note that the exact need for the supply volume of fresh air depends on the number of residents of the cottage.
With an average temperature difference between the house and the street in the heating season, equal to 27 aboutC (20 aboutC home, -7 aboutWith the external atmosphere) per day for heating the supply of cold air you need thermal energy:
5.271214.34-1.005 = 164755.58 kJ,
Where: 5 - the number of air changes in the premises; 27 - temperature difference between indoor and outdoor atmosphere; 1214.34 - air density at t +20 aboutFROM; 1.005 - specific heat of air.
We convert kilojoules into kilowatt hours, dividing the value by the number of kilojoules in one kilowatt hour (3600):
164755.58: 3600 = 45.76 kWh
Having ascertained the cost of thermal energy for heating the air in the house during its five-fold replacement through the supply ventilation, it is possible to calculate the “air” heat loss for the seven-month heating season:
7 · 30 · 45.76 = 9609.6 kWh,
Where: 7 - the number of "heated" months; 30 - the average number of days in a month; 45.76 - daily heat energy costs for heating the supply air.
Ventilation (infiltration) energy costs are inevitable, since air renewal in the cottage is vital.
The heating needs of the replaceable air atmosphere in the house must be calculated, summed with heat losses through the building envelope and taken into account when choosing a heating boiler. There is another type of heat energy consumption, the latter - sewer heat loss.
Energy costs for DHW preparation
If during the warmer months cold water comes from the tap to the cottage, then in the heating season it is icy, with a temperature not exceeding +5 aboutC. Bathing, washing dishes and washing are not possible without heating the water.
Water collected in the toilet bowl contacts the atmosphere through the walls, taking a little heat. What happens to water heated by burning non-free fuel and spent on household needs? It is poured into the sewer.
A double-circuit boiler with an indirect heating boiler, used both for heating the coolant and for supplying hot water to the circuit constructed for it
Consider an example. A family of three, suppose to spend 17 m3 water monthly. 1000 kg / m3 - the density of water, and 4.183 kJ / kgaboutC is its specific heat.
The average temperature of heating water intended for domestic needs, let it be +40 aboutC. Accordingly, the difference in average temperature between cold water entering the house (+5 aboutC) and heated in a boiler (+30 aboutC) it turns out 25 aboutFROM.
To calculate sewer heat loss, we consider:
17 · 1000 · 25 · 4.183 = 1777775 kJ,
Where: 17 - monthly volume of water consumption; 1000 is the density of water; 25 - temperature difference between cold and heated water; 4,183 - specific heat of water;
To convert kilojoules to more understandable kilowatt hours:
1777775: 3600 = 493.82 kWh
Thus, for a seven-month period of the heating season, heat energy in the amount of:
493.827 = 3456.74 kWh
The consumption of thermal energy for heating water for hygienic needs is small, in comparison with heat loss through walls and ventilation. But this is also energy consumption, loading a heating boiler or boiler and causing fuel consumption.
Calculation of the power of the 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 equipping a boiler with an indirect heating boiler, for heating water for hygienic needs.
By calculating the daily heat loss and the consumption of warm water “for sewage”, it is possible to accurately determine the necessary boiler capacity for a cottage of a certain area and the characteristics of the enclosing structures.
The single-circuit boiler only produces heating medium for the heating system
To determine the power of the heating boiler, it is necessary to calculate the cost of thermal energy of the house through the facade walls and the heating of the replaceable air atmosphere of the interior.
Data on heat losses in kilowatt hours per day is required - in the case of a conditional house, calculated as an example, this is:
271.512 + 45.76 = 317.272 kWh,
Where: 271.512 - daily heat loss by external walls; 45.76 - daily heat loss for heating the supply air.
Accordingly, the necessary heating capacity of the boiler will be:
317.272: 24 (hours) = 13.22 kW
However, such a boiler will be under constantly high load, reducing its service life. And on especially frosty days, the rated capacity of the boiler will not be enough, because with a high temperature difference between indoor and outdoor atmospheres, the heat loss of the building will increase sharply.
Therefore, it is not worth choosing a boiler according to an average calculation of thermal energy costs - it may not be able to cope with severe frosts.
It will be rational to increase the required capacity of boiler equipment by 20%:
13.22.2 + 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 losses by the number of days in the month and by 24 hours:
493.82: 30: 24 = 0.68 kW
According to the calculation results, the optimal boiler power for the cottage example is 15.86 kW for the heating circuit and 0.68 kW for the heating circuit.
The choice of radiators
Traditionally, it is recommended that the power of the heating radiator be selected according to the area of the heated room, and with a 15-20% overstatement of power requirements, just in case.
As an example, let us consider how correct the method of choosing a radiator is “10 m2 of area - 1.2 kW”.
The heat output of radiators depends on how they are connected, which must be taken into account when calculating the heating system
Initial data: corner room on the first level of a two-story house IZHS; external wall of double-row ceramic brick masonry; room width 3 m, length 4 m, ceiling height 3 m.
According to the simplified selection scheme, it is proposed to calculate the area of the room, we consider:
3 (width) · 4 (length) = 12 m2
Those. the required power of the heating radiator with a 20% premium is 14.4 kW. And now we calculate the power parameters of the heating radiator based on the heat loss of the room.
In fact, the area of a room affects the loss of thermal energy less than the area of its walls that extend on one side of the building (front).
Therefore, we will consider exactly the area of "street" walls available in the room:
3 (width) · 3 (height) + 4 (length) · 3 (height) = 21 m2
Knowing the area of the walls that transfer heat “to the street”, we calculate the heat loss with a difference in room and street temperature of 30about (in the house +18 aboutC, outside -12 aboutC), and immediately in kilowatt hours:
0.91 · 21 · 30: 1000 = 0.57 kW,
Where: 0.91 - heat transfer resistance m2 of 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 per kilowatt.
According to building standards, heating appliances are located in places of maximum heat loss. For example, radiators are installed under the window openings, heat guns - above the entrance to the house. In corner rooms, batteries are installed on dull walls that are subject to maximum winds.
It turns out that to compensate for heat loss through the facade walls of this design, at 30about the temperature difference in the house and on the street is enough heating with a capacity of 0.57 kWh. We increase the required power by 20, even by 30% - we get 0.74 kWh.
Thus, the real power requirements of heating can be significantly lower than the trading scheme “1.2 kW per square meter of floor space”.
Moreover, the correct calculation of the required capacity of heating radiators will reduce the amount of coolant in the heating system, which will reduce the load on the boiler and fuel costs.
Where the heat goes from home - the video provides the answers:
In the video, the procedure for calculating the heat loss of a house through the building envelope is considered. Knowing the heat loss, you can accurately calculate the power of the heating system:
For a detailed video on the principles of selecting the power characteristics of a heating boiler, see below:
Heat production rises annually - fuel prices are rising. And the heat is constantly not enough. You can’t be indifferent to the energy consumption of the cottage - it is completely unprofitable.
On the one hand, each new heating season costs the homeowner more and more expensive. On the other hand, insulation of walls, foundations and suburban roofs costs good money. However, the less heat that leaves the building, the cheaper it will be to heat it..
The preservation of heat in the premises of the house is the main task of the heating system in the winter months. The choice of heating boiler power depends on the condition of the house and on the quality of insulation of its enclosing structures. The principle of “kilowatts per 10 squares of area” works in a cottage of an average state of facades, roofs and foundations.
Have you independently calculated a heating system for your home? Or did you notice a mismatch in the calculations given in the article? Share your practical experience or the volume of theoretical knowledge by leaving a comment in the block under this article.