The first step in organizing the heating of a private home is calculating heat loss. The purpose of this calculation is to find out how much heat escapes out through walls, floors, roofing and windows (commonly known as building envelopes) during the most severe frosts in a given area. Knowing how to calculate heat loss according to the rules, you can get a fairly accurate result and begin selecting a heat source based on power.

Basic formulas

To get a more or less accurate result, you need to perform calculations according to all the rules; a simplified method (100 W of heat per 1 m² of area) will not work here. The total heat loss of a building during the cold season consists of 2 parts:

• heat loss through enclosing structures;
• losses of energy used for heating ventilation air.

The basic formula for calculating the thermal energy consumption through external fences is as follows:

Q = 1/R x (t in - t n) x S x (1+ ∑β). Here:

• Q is the amount of heat lost by a structure of one type, W;
• R - thermal resistance of the construction material, m²°C / W;
• S—external fence area, m²;
• t in - temperature internal air, °C;
• t n - most low temperature environment, °C;
• β - additional heat loss, depending on the orientation of the building.

The thermal resistance of the walls or roof of a building is determined based on the properties of the material from which they are made and the thickness of the structure. To do this, use the formula R = δ / λ, where:

• λ—reference value of the thermal conductivity of the wall material, W/(m°C);
• δ is the thickness of the layer of this material, m.

If a wall is built from 2 materials (for example, brick with mineral wool insulation), then the thermal resistance is calculated for each of them, and the results are summed up. Outdoor temperature selected according to regulatory documents, and according to personal observations, internal - as necessary. Additional heat losses are coefficients determined by the standards:

1. When a wall or part of the roof is turned to the north, northeast or northwest, then β = 0.1.
2. If the structure faces southeast or west, β = 0.05.
3. β = 0 when the outer fence faces the south or southwest.

Calculation order

To take into account all the heat leaving the house, it is necessary to calculate the heat loss of the room, each separately. To do this, measurements are taken of all fences adjacent to the environment: walls, windows, roof, floor and doors.

Important point: measurements should be taken according to outside, capturing the corners of the building, otherwise the calculation of the heat loss of the house will give an underestimated heat consumption.

Windows and doors are measured by the opening they fill.

Based on the measurement results, the area of ​​each structure is calculated and substituted into the first formula (S, m²). The value R is also inserted there, obtained by dividing the thickness of the fence by the thermal conductivity coefficient building material. In the case of new windows made of metal-plastic, the R value will be told to you by a representative of the installer.

As an example, it is worth calculating heat loss through enclosing walls made of brick 25 cm thick, with an area of ​​5 m² at an ambient temperature of -25°C. It is assumed that the temperature inside will be +20°C, and the plane of the structure faces north (β = 0.1). First you need to take the thermal conductivity coefficient of brick (λ) from the reference literature; it is equal to 0.44 W/(m°C). Then, using the second formula, the resistance to heat transfer is calculated brick wall 0.25 m:

R = 0.25 / 0.44 = 0.57 m²°C / W

To determine the heat loss of a room with this wall, all initial data must be substituted into the first formula:

Q = 1 / 0.57 x (20 - (-25)) x 5 x (1 + 0.1) = 434 W = 4.3 kW

If the room has a window, then after calculating its area, the heat loss through the translucent opening should be determined in the same way. The same actions are repeated regarding floors, roofing and front door. At the end, all the results are summed up, after which you can move on to the next room.

Heat metering for air heating

When calculating the heat loss of a building, it is important to take into account the amount of thermal energy consumed by the heating system to heat the ventilation air. The share of this energy reaches 30% of total losses, so it is unacceptable to ignore it. You can calculate the ventilation heat loss of a house through the heat capacity of the air using a popular formula from a physics course:

Q air = cm (t in - t n). In it:

• Q air - heat consumed by the heating system for heating supply air, W;
• t in and t n - the same as in the first formula, °C;
• m is the mass flow of air entering the house from outside, kg;
• c is the heat capacity of the air mixture, equal to 0.28 W / (kg °C).

Here all quantities are known, except for the mass air flow rate during ventilation of premises. In order not to complicate the task for yourself, it is worth agreeing with the condition that air environment updated throughout the house once per hour. Then the volumetric air flow rate can be easily calculated by adding the volumes of all rooms, and then you need to convert it into mass air flow through density. Since the density of the air mixture changes depending on its temperature, you need to take suitable value from the table:

m = 500 x 1.422 = 711 kg/h

Heating such a mass of air by 45°C will require the following amount of heat:

Q air = 0.28 x 711 x 45 = 8957 W, which is approximately equal to 9 kW.

At the end of the calculations, the results of heat losses through external fences are summed up with ventilation heat losses, which gives the total heat load on the building’s heating system.

The presented calculation methods can be simplified if the formulas are entered into Excel program in the form of tables with data, this will significantly speed up the calculation.

You can calculate the heating of a private house yourself by taking some measurements and substituting your values ​​into the necessary formulas. Let's tell you how it's done.

Calculating heat loss at home

Several critical parameters of the heating system and, first of all, the boiler power depend on the calculation of heat loss at home.

The calculation sequence is as follows:

We calculate and record in a column the area of ​​windows, doors, external walls, floors, and ceilings of each room. Opposite each value we write down the coefficient from which our house is built.

If you haven't found required material c, then look at the extended version of the table, which is called the thermal conductivity coefficients of materials (soon on our website). Next, using the formula below, we calculate the heat loss of each structural element of our house.

Q = S * ΔT / R,

Where Q– heat loss, W
S— structure area, m2
Δ T— temperature difference between indoors and outdoors for the coldest days °C

R— value of thermal resistance of the structure, m2 °C/W

R layer = V / λ

Where V- layer thickness in m,

λ — thermal conductivity coefficient (see table on materials).

We sum up the thermal resistance of all layers. Those. for walls, plaster and wall material and external insulation (if any) are taken into account.

Let's add it all up Q for windows, doors, external walls, floors, ceilings

To the resulting amount we add 10-40% of ventilation losses. They can also be calculated using the formula, but when good windows and moderate ventilation, you can safely set 10%.

We divide the result by the total area of ​​the house. Precisely general, because Indirectly, heat will also be wasted in corridors where there are no radiators. The calculated value of specific heat loss can vary between 50-150 W/m2. The highest heat losses are in rooms on the upper floors, the lowest in the middle ones.

After graduation installation work, check walls, ceilings and other structural elements to make sure there are no heat leaks anywhere.

The table below will help you more accurately determine the indicators of materials.

Deciding on the temperature regime

This stage is directly related to the choice of boiler and method of heating the premises. If you intend to install " heated floors", Maybe, The best decision– condensing boiler and low temperature regime 55C in the supply and 45C in the return. This mode ensures maximum boiler efficiency and, accordingly, the best gas savings. In the future, if you want to use high-tech heating methods, ( , solar collectors) you won’t have to redo the heating system for new equipment, because It is designed specifically for low temperature conditions. Additional advantages - the air in the room does not dry out, the intensity of the flow is lower, and less dust collects.

If you choose a traditional boiler, it is better to choose a temperature regime that is as close as possible to European standards - 75C - at the boiler outlet, 65C - return flow, 20C - room temperature. This mode is provided in the settings of almost all imported boilers. In addition to choosing a boiler, the temperature regime affects the calculation of radiator power.

Selection of radiator power

When calculating heating radiators in a private home, the material of the product does not play a role. This is a matter of taste of the owner of the house. Only the radiator power indicated in the product data sheet is important. Manufacturers often indicate inflated figures, so the calculation results will be rounded up. The calculation is made for each room separately. Simplifying the calculations somewhat for a room with 2.7 m ceilings, we present simple formula:

K=S * 100/P

Where TO— the required number of radiator sections

S– room area

P– power indicated in the product passport

Calculation example: For a room with an area of ​​30 m2 and a power of one section of 180 W, we obtain: K= 30 x 100/180

K=16.67 rounded 17 sections

The same calculation can be applied for cast iron batteries, assuming that

1 rib (60 cm) = 1 section.

Hydraulic calculation of the heating system

The point of this calculation is to choose the right pipe diameter and characteristics. Due to the complexity of the calculation formulas, for a private house it is easier to select pipe parameters from the table.

Here is the total power of the radiators for which the pipe supplies heat.

Pipe diameter	Min. radiator power kW	Max. radiator power kW
Metal-plastic pipe 16 mm	2,8	4,5
Metal-plastic pipe 20 mm	5	8
Metal-plastic pipe 25 mm	8	13
Metal-plastic pipe 32 mm	13	21
Polypropylene pipe 20 mm	4	7
Polypropylene pipe 25 mm	6	11
Polypropylene pipe 32 mm	10	18
Polypropylene pipe 40 mm	16	28

Calculating the volume of the heating system

This value is necessary to select the correct volume expansion tank. It is calculated as the sum of the volume in radiators, pipelines and boiler. Reference information on radiators and pipelines is given below, on the boiler - indicated in its passport.

Coolant volume in the radiator:

• aluminum section - 0.450 liters
• bimetallic section - 0.250 liters
• new cast iron section- 1,000 liter
• old cast iron section - 1,700 liters

Coolant volume in 1 lm. pipes:

• ø15 (G ½") - 0.177 liters
• ø20 (G ¾") - 0.310 liters
• ø25 (G 1.0″) - 0.490 liters
• ø32 (G 1¼") - 0.800 liters
• ø15 (G 1½") - 1,250 liters
• ø15 (G 2.0″) - 1,960 liters

Installation of a heating system for a private house - selection of pipes

It is made with pipes made of different materials:

Steel

• They have a lot of weight.
• They require proper skills, special tools and equipment for installation.
• Subject to corrosion
• May accumulate static electricity.

Copper

• Withstands temperatures up to 2000 C, pressure up to 200 atm. (in a private house completely unnecessary advantages)
• Reliable and durable
• Have a high cost
• Mounted with special equipment, silver solder

Plastic

• Antistatic
• Corrosion resistant
• Inexpensive
• Have minimal hydraulic resistance
• No special skills required for installation

Summarize

A correctly made calculation of the heating system of a private house ensures:

• Comfortable warmth in the rooms.
• Sufficient amount of hot water.
• Silence in the pipes (without gurgling or growling).
• Optimal boiler operating modes
• Correct load on the circulation pump.
• Minimum installation costs

Energy efficient building renovation can help you save money thermal energy and improve the comfort of life. The greatest savings potential lies in good thermal insulation of the external walls and roof. The easiest way to evaluate the possibilities of effective repair is the consumption of thermal energy. If more than 100 kWh of electricity (10 m³) is consumed per year natural gas) on square meter heated area, including wall area, then energy-saving renovations can be beneficial.

Heat loss through the outer shell

The basic concept of an energy-saving building is a continuous layer of thermal insulation over the heated surface of the house contour.

1. Roof. With a thick layer of insulation, heat loss through the roof can be reduced;

Important! IN wooden structures Thermal sealing of the roof is difficult, since the wood swells and can be damaged by high humidity.

1. Walls. As with a roof, heat loss is reduced when a special coating is used. When internal thermal insulation walls there is a risk that condensation will collect behind the insulation if the humidity in the room is too high;

1. Floor or basement. For practical reasons thermal insulation produced from inside the building;
2. Thermal bridges. Thermal bridges are unwanted cooling fins (thermal conductors) on the outside of a building. For example, a concrete floor, which is also a balcony floor. Many thermal bridges are found in the soil area, parapets, window and door frames. There are also temporary thermal bridges if the wall parts are secured with metal elements. Thermal bridges can account for a significant portion of heat loss;
3. Window. Over the past 15 years, the thermal insulation of window glass has improved 3 times. Today's windows have a special reflective layer on the glass, which reduces radiation loss; these are single- and double-glazed windows;
4. Ventilation. A typical building has air leaks, especially around windows, doors and the roof, which provide the necessary air exchange. However, during the cold season, this causes significant heat loss in the house from the heated air escaping. Good modern buildings are quite airtight, and it is necessary to regularly ventilate the premises by opening the windows for a few minutes. To reduce heat loss due to ventilation, comfortable ventilation systems. This type of heat loss is estimated at 10-40%.

Thermographic surveys in a poorly insulated building provide insight into how much heat is lost. This is very good tool for quality control of repairs or new construction.

Methods for assessing heat loss at home

Exist complex techniques calculations that take into account various physical processes: convection exchange, radiation, but they are often unnecessary. Simplified formulas are usually used, and if necessary, you can add 1-5% to the result. The orientation of the building is taken into account in new buildings, but solar radiation also does not significantly affect the calculation of heat loss.

Important! When applying formulas for calculating thermal energy losses, the time spent by people in a particular room is always taken into account. The smaller it is, the smaller temperature indicators must be taken as a basis.

1. Average values. The most approximate method does not have sufficient accuracy. There are tables compiled for individual regions, taking into account climatic conditions and average building parameters. For example, for a specific area, the power value in kilowatts required to heat 10 m² of room area with 3 m high ceilings and one window is indicated. If the ceilings are lower or higher, and there are 2 windows in the room, the power indicators are adjusted. This method does not take into account the degree of thermal insulation of the house at all and will not save thermal energy;
2. Calculation of heat loss from the building envelope. The area is summed up external walls minus the sizes of window and door areas. Additionally there is a roof area with a floor. Further calculations are carried out using the formula:

Q = S x ΔT/R, where:

• S – found area;
• ΔT – difference between internal and external temperatures;
• R – resistance to heat transfer.

The results obtained for the walls, floor and roof are combined. Ventilation losses are then added.

Important! Such a calculation of heat loss will help determine the boiler power for the building, but will not allow you to calculate the number of radiators per room.

1. Calculation of heat loss by room. When using a similar formula, losses are calculated for all rooms of the building separately. Then the heat loss for ventilation is determined by determining the volume of the air mass and approximate quantity once a day during her indoor shift.

Important! When calculating ventilation losses, it is necessary to take into account the purpose of the room. Increased ventilation is required for the kitchen and bathroom.

An example of calculating heat loss in a residential building

The second calculation method is used, only for external structures Houses. Up to 90 percent of thermal energy is lost through them. Accurate results are important to select the right boiler for your boiler effective heat without excessive heating of the premises. It is also an indicator of the economic efficiency of the selected materials for thermal protection, showing how quickly you can recoup the costs of their purchase. The calculations are simplified, for a building without a multilayer thermal insulation layer.

The house has an area of ​​10 x 12 m and a height of 6 m. The walls are 2.5 bricks thick (67 cm), covered with plaster, a layer of 3 cm. The house has 10 windows 0.9 x 1 m and a door 1 x 2 m.

Calculation of heat transfer resistance of walls:

1. R = n/λ, where:

• n – wall thickness,
• λ – thermal conductivity (W/(m °C).

This value is looked up in the table for your material.

1. For brick:

Rkir = 0.67/0.38 = 1.76 sq.m °C/W.

1. For plaster coating:

Rpc = 0.03/0.35 = 0.086 sq.m °C/W;

1. Total value:

Rst = Rkir + Rsht = 1.76 + 0.086 = 1.846 sq.m °C/W;

Calculation of the area of ​​external walls:

1. Total area of ​​external walls:

S = (10 + 12) x 2 x 6 = 264 sq.m.

1. Area of ​​windows and doorway:

S1 = ((0.9 x 1) x 10) + (1 x 2) = 11 sq.m.

1. Adjusted wall area:

S2 = S – S1 = 264 – 11 = 253 sq.m.

Heat losses for walls will be determined:

Q = S x ΔT/R = 253 x 40/1.846 = 6810.22 W.

Important! The ΔT value is taken arbitrarily. For each region, you can find the average value of this value in the tables.

On next stage Heat loss through the foundation, windows, roof, and door is calculated in the same way. When calculating the heat loss index for the foundation, a smaller temperature difference is taken. Then you need to sum up all the received numbers and get the final one.

To determine the possible energy consumption for heating, you can present this figure in kWh and calculate it in heating season.

If you use only the number for the walls, you get:

• per day:

6810.22 x 24 = 163.4 kWh;

• per month:

163.4 x 30 = 4903.4 kWh;

• for a heating season of 7 months:

4903.4 x 7 =34,323.5 kWh.

When heating is gas, gas consumption is determined based on its calorific value and the efficiency of the boiler.

Heat losses due to ventilation

1. Find the air volume of the house:

10 x 12 x 6 = 720 m³;

1. The mass of air is found by the formula:

M = ρ x V, where ρ is the air density (taken from the table).

M = 1, 205 x 720 = 867.4 kg.

1. It is necessary to determine the number of times the air in the entire house is changed per day (for example, 6 times), and calculate heat loss for ventilation:

Qв = nxΔT xmx С, where С – specific heat for air, n is the number of times the air is replaced.

Qв = 6 x 40 x 867.4 x 1.005 = 209217 kJ;

1. Now we need to convert to kWh. Since there are 3600 kilojoules in one kilowatt-hour, then 209217 kJ = 58.11 kWh

Some calculation methods suggest taking heat losses for ventilation from 10 to 40 percent of total heat losses, without calculating them using formulas.

To make it easier to calculate heat loss at home, there are online calculators where you can calculate the result for each room or the entire house. Simply enter your data in the fields provided.

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Every building, regardless of design features, transmits thermal energy through the fences. Heat loss in environment needs to be restored using a heating system. The sum of heat losses with a normalized reserve is the required power of the heat source that heats the house. To create in a home comfortable conditions, the calculation of heat loss is carried out taking into account various factors: the structure of the building and the layout of the premises, orientation to the cardinal points, wind direction and the average mildness of the climate in cold period, physical qualities of building and thermal insulation materials.

Based on the results of the thermal engineering calculation, a heating boiler is selected, the number of battery sections is specified, the power and length of the heated floor pipes are calculated, a heat generator is selected for the room - in general, any unit that compensates for heat loss. By by and large, it is necessary to determine heat losses in order to heat the house economically - without excess power reserves of the heating system. Calculations are performed manually or choose a suitable computer program into which the data is inserted.

How to perform the calculation?

First, it’s worth understanding the manual technique to understand the essence of the process. To find out how much heat a house loses, the losses through each building envelope are determined separately and then added up. The calculation is performed in stages.

1. Form a base of initial data for each room, preferably in the form of a table. The first column records the pre-calculated area of ​​door and window blocks, external walls, ceilings, and floors. The thickness of the structure is entered in the second column (this is design data or measurement results). In the third - the thermal conductivity coefficients of the corresponding materials. Table 1 contains standard values ​​that will be needed in further calculations:

The higher λ, the more heat is lost through the meter thick surface.

2. Determine the thermal resistance of each layer: R = v/ λ, where v is the thickness of the building or thermal insulation material.

3. Calculate the heat loss of each structural element according to the formula: Q = S*(T in -T n)/R, where:

• Tn – outside temperature, °C;
• T in – indoor temperature, °C;
• S – area, m2.

Of course, throughout heating season the weather varies (for example, the temperature ranges from 0 to -25°C), and the house is heated to the desired level of comfort (for example, up to +20°C). Then the difference (T in -T n) varies from 25 to 45.

To make the calculation, you need the average temperature difference for the entire heating season. To do this, in SNiP 23-01-99 “Building climatology and geophysics” (Table 1), the average temperature of the heating period for a particular city is found. For example, for Moscow this figure is -26°. In this case the average difference is 46°C. To determine the heat consumption through each structure, the heat losses of all its layers are added up. So, for walls, plaster, masonry material, external thermal insulation, and cladding are taken into account.

4. Calculate the total heat losses, defining them as the sum Q of external walls, floors, doors, windows, ceilings.

5. Ventilation. From 10 to 40% of infiltration (ventilation) losses are added to the addition result. If you install high-quality double-glazed windows in your house and do not abuse ventilation, the infiltration coefficient can be taken as 0.1. Some sources indicate that the building does not lose heat at all, since leaks are compensated by solar radiation and household heat emissions.

Manual counting

Initial data. Cottage area 8x10 m, height 2.5 m. The walls are 38 cm thick and made of ceramic bricks, the inside is finished with a layer of plaster (thickness 20 mm). The floor is made of 30mm edged boards, insulated with mineral wool (50 mm), sheathed chipboard sheets(8 mm). The building has a basement, the temperature in which in winter is 8°C. The ceiling is covered with wooden panels and insulated with mineral wool (thickness 150 mm). The house has 4 windows 1.2x1 m, an oak entrance door 0.9x2x0.05 m.

Assignment: determine the total heat loss of a house based on the assumption that it is located in the Moscow region. The average temperature difference during the heating season is 46°C (as mentioned earlier). The room and the basement have a difference in temperature: 20 – 8 = 12°C.

1. Heat loss through external walls.

Total area (minus windows and doors): S = (8+10)*2*2.5 – 4*1.2*1 – 0.9*2 = 83.4 m2.

Thermal resistance is determined brickwork and plaster layer:

• R clade. = 0.38/0.52 = 0.73 m2*°C/W.
• R pieces = 0.02/0.35 = 0.06 m2*°C/W.
• R total = 0.73 + 0.06 = 0.79 m2*°C/W.
• Heat loss through the walls: Q st = 83.4 * 46/0.79 = 4856.20 W.

2. Heat loss through the floor.

Total area: S = 8*10 = 80 m2.

The thermal resistance of a three-layer floor is calculated.

• R boards = 0.03/0.14 = 0.21 m2*°C/W.
• R chipboard = 0.008/0.15 = 0.05 m2*°C/W.
• R insulation = 0.05/0.041 = 1.22 m2*°C/W.
• R total = 0.03 + 0.05 + 1.22 = 1.3 m2*°C/W.

We substitute the values ​​of the quantities into the formula for finding heat loss: Q floor = 80*12/1.3 = 738.46 W.

3. Heat loss through the ceiling.

Square ceiling surface equal to floor area S = 80 m2.

When determining the thermal resistance of the ceiling, in this case they do not take into account wooden boards: They are secured with gaps and do not act as a barrier to the cold. The thermal resistance of the ceiling coincides with the corresponding insulation parameter: R sweat. = R insulation = 0.15/0.041 = 3.766 m2*°C/W.

Amount of heat loss through the ceiling: Q sweat. = 80*46/3.66 = 1005.46 W.

4. Heat loss through windows.

Glazing area: S = 4*1.2*1 = 4.8 m2.

For the manufacture of windows, a three-chamber PVC profile(occupies 10% of the window area), as well as a double-chamber double-glazed window with a glass thickness of 4 mm and a distance between glasses of 16 mm. Among technical characteristics the manufacturer indicated the thermal resistance of the glass unit (R st.p. = 0.4 m2*°C/W) and profile (R prof. = 0.6 m2*°C/W). Taking into account the dimensional fraction of each structural element, the average thermal resistance of the window is determined:

• R approx. = (R st.p.*90 + R prof.*10)/100 = (0.4*90 + 0.6*10)/100 = 0.42 m2*°C/W.
• Based on the calculated result, heat loss through the windows is calculated: Q approx. = 4.8*46/0.42 = 525.71 W.

Door area S = 0.9*2 = 1.8 m2. Thermal resistance R dv. = 0.05/0.14 = 0.36 m2*°C/W, and Q dv. = 1.8*46/0.36 = 230 W.

The total amount of heat loss at home is: Q = 4856.20 W + 738.46 W + 1005.46 W + 525.71 W + 230 W = 7355.83 W. Taking into account infiltration (10%), losses increase: 7355.83 * 1.1 = 8091.41 W.

To accurately calculate how much heat a building loses, they use online calculator heat loss This computer program, into which not only the data listed above is entered, but also various additional factors that influence the result. The advantage of the calculator is not only the accuracy of the calculations, but also an extensive reference data base.