Heat loss at home, calculation of heat loss. Simple calculation of heat loss of buildings

The first step in organizing the heating of a private house is the calculation of heat loss. The purpose of this calculation is to find out how much heat escapes to the outside through walls, floors, roofs and windows ( common name- enclosing structures) at the most severe frosts in this locality. Knowing how to calculate heat loss according to the rules, you can get a fairly accurate result and start selecting a heat source by power.

Basic Formulas

To get a more or less accurate result, it is necessary 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;
loss of energy used to heat the ventilation air.

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

Q \u003d 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 is the thermal resistance of the construction material, m²°C / W;
S is the area of the outer fence, m²;
t in - internal air temperature, ° С;
t n - most low temperature environment, °С;
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. For this, the formula R = δ / λ is used, where:

λ is the 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 the wall is built from 2 materials (for example, a brick with a mineral wool insulation), then the thermal resistance is calculated for each of them, and the results are summarized. The outdoor temperature is selected as regulatory documents, and according to personal observations, internal - by necessity. Additional heat losses are the coefficients defined by the standards:

When the wall or part of the roof is turned to the north, northeast or northwest, then β = 0.1.
If the structure is facing southeast or west, β = 0.05.
β = 0 when the outer fence faces 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 made of all fences adjacent to the environment: walls, windows, roofs, floors and doors.

An important point: measurements should be carried out 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 results of measurements, the area of \u200b\u200beach structure is calculated and substituted into the first formula (S, m²). The value of R is also inserted there, obtained by dividing the thickness of the fence by the thermal conductivity of the building material. In the case of new metal-plastic windows, the value of R will be prompted by a representative of the installer.

As an example, it is worthwhile to calculate the heat loss through the enclosing walls made of bricks 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 is facing north (β = 0.1). First you need to take from the reference literature the coefficient of thermal conductivity of the brick (λ), it is equal to 0.44 W / (m ° C). Then, according to the second formula, the resistance to heat transfer is calculated brick wall 0.25 m:

R \u003d 0.25 / 0.44 \u003d 0.57 m² ° C / W

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

Q \u003d 1 / 0.57 x (20 - (-25)) x 5 x (1 + 0.1) \u003d 434 W \u003d 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 for floors, roofs and front door. At the end, all the results are summarized, 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 heat energy consumed by the heating system for heating the ventilation air. The share of this energy reaches 30% of the total losses, so it is unacceptable to ignore it. You can calculate the ventilation heat loss at home through the heat capacity of the air using the popular formula from the physics course:

Q air \u003d cm (t in - t n). In it:

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

Here, all quantities are known, except for the mass air flow during ventilation of rooms. In order not to complicate your task, you should agree with the condition that the air environment is updated throughout the house 1 time per hour. Then it is not difficult to calculate the volumetric air flow by adding the volumes of all rooms, and then you need to convert it into mass air through density. Since the density of the air mixture varies with its temperature, you need to take the appropriate 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 \u003d 0.28 x 711 x 45 \u003d 8957 W, which is approximately equal to 9 kW.

Upon completion of the calculations, the results of heat losses through the external fences are added to the 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 the Excel program in the form of tables with data, this will significantly speed up the calculation.

Below is a pretty simple heat loss calculation buildings, which, nevertheless, will help to accurately determine the power required for heating your warehouse, shopping center or other similar building. This will make it possible at the design stage to preliminarily estimate the cost of heating equipment and subsequent heating costs, and, if necessary, adjust the project.

Where does the heat go? Heat escapes through walls, floors, roofs and windows. In addition, heat is lost during ventilation of the premises. To calculate heat loss through building envelope, use the formula:

Q - heat loss, W

S – construction area, m2

T - temperature difference between indoor and outdoor air, °C

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

The calculation scheme is as follows - we calculate the heat loss individual elements, summarize and add the heat loss during ventilation. All.

Suppose we want to calculate the heat loss for the object shown in the figure. The height of the building is 5 ... 6 m, width - 20 m, length - 40 m, and thirty windows measuring 1.5 x 1.4 meters. Indoor temperature 20 °C, outside temperature -20 °C.

We consider the area of enclosing structures:

floor: 20 m * 40 m = 800 m2

roof: 20.2 m * 40 m = 808 m2

window: 1.5 m * 1.4 m * 30 pcs = 63 m2

walls:(20 m + 40 m + 20 m + 40 m) * 5 m = 600 m2 + 20 m2 (accounting pitched roof) = 620 m2 - 63 m2 (windows) = 557 m2

Now let's see the thermal resistance of the materials used.

The value of thermal resistance can be taken from the table of thermal resistances or calculated based on the value of the thermal conductivity coefficient using the formula:

R - thermal resistance, (m2 * K) / W

? - coefficient of thermal conductivity of the material, W / (m2 * K)

d – material thickness, m

The value of thermal conductivity coefficients for different materials can be seen.

floor: concrete screed 10 cm and mineral wool with a density of 150 kg/m3. 10 cm thick.

R (concrete) = 0.1 / 1.75 = 0.057 (m2*K)/W

R (mineral wool) \u003d 0.1 / 0.037 \u003d 2.7 (m2 * K) / W

R (floor) \u003d R (concrete) + R (mineral wool) \u003d 0.057 + 2.7 \u003d 2.76 (m2 * K) / W

roof:

R (roof) = 0.15 / 0.037 = 4.05 (m2*K)/W

window: the value of thermal resistance of windows depends on the type of double-glazed window used
R (windows) \u003d 0.40 (m2 * K) / W for single-chamber glass wool 4–16–4 at? T \u003d 40 ° С

walls: panels from mineral wool 15 cm thick
R (walls) = 0.15 / 0.037 = 4.05 (m2*K)/W

Let's calculate the heat loss:

Q (floor) \u003d 800 m2 * 20 ° C / 2.76 (m2 * K) / W \u003d 5797 W \u003d 5.8 kW

Q (roof) \u003d 808 m2 * 40 ° C / 4.05 (m2 * K) / W \u003d 7980 W \u003d 8.0 kW

Q (windows) \u003d 63 m2 * 40 ° C / 0.40 (m2 * K) / W \u003d 6300 W \u003d 6.3 kW

Q (walls) \u003d 557 m2 * 40 ° C / 4.05 (m2 * K) / W \u003d 5500 W \u003d 5.5 kW

We get that the total heat loss through the building envelope will be:

Q (total) = 5.8 + 8.0 + 6.3 + 5.5 = 25.6 kWh

Now about ventilation losses.

To heat 1 m3 of air from a temperature of -20 °C to +20 °C, 15.5 W will be required.

Q (1 m3 of air) \u003d 1.4 * 1.0 * 40 / 3.6 \u003d 15.5 W, here 1.4 is the air density (kg / m3), 1.0 - specific heat air (kJ / (kg K)), 3.6 - conversion factor to watts.

It remains to determine the number required air. It is believed that with normal breathing, a person needs 7 m3 of air per hour. If you use a building as a warehouse and 40 people work on it, then you need to heat 7 m3 * 40 people = 280 m3 of air per hour, this will require 280 m3 * 15.5 W = 4340 W = 4.3 kW. And if you have a supermarket and on average there are 400 people on the territory, then air heating will require 43 kW.

Final result:

For heating the proposed building, a heating system of the order of 30 kWh is required, and a ventilation system with a capacity of 3000 m3 / h with a heater with a power of 45 kW / h.

Calculation of heat loss at home is a necessary step in the design of a heating system. Executed by complex formulas. Incorrectly leads to insufficient heating of the room (if the heat loss indicators are underestimated) or to overpayments for the system and for heating (if the indicators are too high).

The calculation of heat supply must be carried out on highest level

Initial data for calculating the heat loss of a house

To calculate correctly, you need to have basic set data. Only with them it is possible to work.

Heated area (you will need it in the future to calculate the volume of heated air);
Floor plan of the building (involved, among other things, in determining the installation locations of heating units);
Section of the building (sometimes not required);
The type of local climate is taken into account in the calculation. You can find out from the NSS - 2. 04. 02 - 2000 "Construction climatology". The resulting coefficient is taken into account in the calculation;
The geographical position of the building, the location of the heated volume relative to the north, south, west and east;
Building materials from which the walls and floor are made;
The structure of enclosing structures (walls, floors). You need a profile listing the layers of materials, their location and thickness;
every kind of building material, etc.;
Type and design of doors from the room, their profile, section;
The materials from which the doors are made with clarification specific gravity each, the location and thickness of the layers and the coefficient of thermal conductivity. Those. the same information is required as for wall materials;
The calculation of the heat output of the heating system is impossible without information on the windows, if any. It is required to take into account their dimensions, geometry, type of double-glazed window, sometimes materials. A profile and data similar to doors may also be required;
Roof data: structure, type, height, profile listing the type of materials and thickness, position of the layers. Characteristics of building materials - thermal conductivity, quantity, etc.;
Window sill height. It is considered as the distance from the surface of the top layer of the floor (not facing, but a clean layer) to the underside of the board;
The presence or absence of heating batteries;
In the presence of a "warm floor" - its profile, building material of the coating over communications with a listing of the thickness of the layers, their location, thermal conductivity, etc .;
Building material and type of pipeline.

Defined data for the walls of a residential building

Think about what the future functions of the room are, based on this, draw a conclusion about the desired temperature regime (for example, in warehouses the temperature can be lower than in those where the staff is constantly located, in greenhouses, flower bases have even more specific heating requirements).

The next step is to determine temperature regime premises. It is carried out by periodically measuring temperatures. The desired temperatures to be maintained are determined. The heating scheme and the proposed (or desired) installation sites for the risers are selected. The source of heat supply is determined.

When calculating heat losses, the architecture of the building, in particular, its shape and geometry, also plays an important role. Since 2003, SNiP has taken into account the indicator of the shape of the structure. It is calculated as the ratio of the area of the shell (walls, floor and ceiling) to the volume that it surrounds. Until 2003, the parameter was not taken into account, which led to the fact that energy was significantly overused.

Progress of work: calculating the percentage of allowable heat loss for a country house made of timber, logs, bricks, panels

Before proceeding directly to work, the performer conducts some field surveys at the facility. The premises are examined and measured, the wishes and information from the customer are taken into account. This process involves certain steps:

Natural measurement of premises;
Specification according to the customer's data;
Study of the heating system, if any;
Ideas for improving or correcting an error in heating (in an existing system);
Study of the hot water supply system;
Development of ideas for using it for heating or reducing heat loss (for example, using Valtec equipment (Valtek);
Calculation of heat losses and others necessary for the development of a heating system plan.

After these stages, the contractor provides the necessary technical documentation. It includes floor plans, profiles, where each heater is displayed and general device systems, materials according to the specifics and type of equipment used.

Calculations: where are the greatest heat losses in a frame insulated house and how to reduce them using a device

Most important process in the design of heating - calculations of the future system. Calculation of heat losses through enclosing structures is carried out, additional losses and heat gains are determined, required amount heaters of the selected type, etc. The calculation of the heat loss coefficient of the house should be done by an experienced person.

The equation heat balance plays an important role in determining heat losses and developing ways to compensate for them. is given below:

V is the volume of the room, calculated taking into account the area of \u200b\u200bthe room and the height of the ceilings. T is the difference between the outside and inside temperatures of the building. K is the heat loss coefficient.

The heat balance formula does not give the most accurate indicators, therefore it is rarely used.

The main value that is used in the calculation is − thermal load for heaters. To determine it, the values of heat losses and are used. allows you to calculate the amount of heat that the heating system will produce, has the form:

Volume heat loss () is multiplied by 1.2. This is a reserve thermal coefficient - a constant that helps to compensate for some random heat losses (long-term opening of doors or windows, etc.).

Calculating heat loss is quite difficult. On average, different building envelopes contribute to the loss of different amounts of energy. 10% is lost through the roof, 10% - through the floor, foundation, 40% - walls, 20% each - windows and poor insulation, ventilation system, etc. Specific thermal characteristic various materials is not the same. Therefore, the formula contains coefficients that allow you to take into account all the nuances. The table below shows the values of the coefficients needed to calculate the amount of heat.

The heat loss formula is as follows:

In the formula, the specific heat loss is 100 watts per square meter. m. Pl - the area of \u200b\u200bthe room, also participating in the definition. Now a formula can be applied to calculate the amount of heat required to release the boiler.

Count correctly and your house will be warm

An example of calculating the heat loss coefficient in a private house: a formula for success

The formula for calculating heat for space heating is easily applicable to any building. As an example, consider a hypothetical building with simple glazing, wooden walls and a window-to-floor ratio of 20%. It is located in the temperate climate zone, where the minimum outside temperature is 25 degrees. It has 4 walls, 3 m high. Above the heated room is cold attic. The value of the coefficients is found out according to the table K1 - 1.27, K2 - 1.25, K3 - 1, K4 - 1.1, K5 - 1.33, K6 - 1, K7 - 1.05. The area of the premises is 100 sq.m. The formula of the heat balance equation is not complicated and is within the power of every person.

Since the formula is known, the amount of heat required to heat a room can be calculated as follows:

Tp \u003d 100 * 100 * 1.27 * 1.25 * 1 * 1.1 * 1.33 * 1 * 1.05 \u003d 24386.38 W \u003d 24.386 kW

And in order to calculate the thermal energy for heating, the boiler power formula is used as follows:

Mk \u003d 1.2 * 24.386 \u003d 29.2632 kW.

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At further stages, the number of required heating elements and the load on each of them, as well as the energy consumption for heating, are determined. The calculation of heat loss at home in our time of saving is very relevant.

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

According to the results thermotechnical calculation they choose a heating boiler, specify the number of battery sections, consider the power and length of the underfloor heating pipes, select a heat generator in the room - in general, any unit that compensates for heat loss. By and large, it is necessary to determine heat losses in order to heat the house economically - without an extra supply of power from the heating system. Calculations are performed manually or a suitable computer program is selected into which data are substituted.

How to make a calculation?

First, you should deal with the manual technique - to understand the essence of the process. To find out how much heat a house loses, determine the losses through each building envelope separately, and then add them up. The calculation is carried out in stages.

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

The higher λ, the more heat escapes through the meter thickness of the given surface.

2. The heat resistance of each layer is determined: R = v/ λ, where v is the thickness of the building or heat-insulating material.

3. Calculate the heat loss of each structural element according to the formula: Q \u003d S * (T in -T n) / R, where:

T n - outdoor temperature, ° C;
T in - indoor temperature, ° C;
S is the area, m2.

Of course, during the heating period, 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 a calculation, you need the average temperature difference for the entire heating season. To do this, in SNiP 23-01-99 "Construction climatology and geophysics" (table 1) find the average temperature of the heating period for a particular city. 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. So, for walls, plaster, masonry material, external thermal insulation, and cladding are taken into account.

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

5. Ventilation. From 10 to 40% of infiltration (ventilation) losses are added to the result of addition. If high-quality double-glazed windows are installed in the house, and ventilation is not abused, the infiltration coefficient can be taken as 0.1. In some sources, it is indicated that the building does not lose heat at all, since leakages are compensated for by solar radiation and domestic heat emissions.

Counting by hand

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

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

1. Heat loss through external walls.

Total area (excluding windows and doors): S \u003d (8 + 10) * 2 * 2.5 - 4 * 1.2 * 1 - 0.9 * 2 \u003d 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 walls: Q st \u003d 83.4 * 46 / 0.79 \u003d 4856.20 W.

2. Heat loss through the floor.

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

The heat 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 \u200b\u200bof the quantities into the formula for finding heat losses: Q floor \u003d 80 * 12 / 1.3 \u003d 738.46 W.

3. Heat loss through the ceiling.

The area of the ceiling surface is equal to the area of the floor S = 80 m2.

When determining the thermal resistance of the ceiling, in this case, wooden panels are not taken into account: they are fixed with gaps and are not a barrier to cold. The thermal resistance of the ceiling coincides with the corresponding parameter of the insulation: R pot. = R ins. = 0.15/0.041 = 3.766 m2*°C/W.

The amount of heat loss through the ceiling: Q sweat. \u003d 80 * 46 / 3.66 \u003d 1005.46 W.

4. Heat loss through windows.

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

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

R ok. \u003d (R st.p. * 90 + R prof. * 10) / 100 \u003d (0.4 * 90 + 0.6 * 10) / 100 \u003d 0.42 m2 * ° C / W.
Based on the calculated result, the heat losses through the windows are calculated: Q approx. \u003d 4.8 * 46 / 0.42 \u003d 525.71 W.

Door area S = 0.9 * 2 = 1.8 m2. Thermal resistance R dv. \u003d 0.05 / 0.14 \u003d 0.36 m2 * ° C / W, and Q ext. \u003d 1.8 * 46 / 0.36 \u003d 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%), the losses increase: 7355.83 * 1.1 = 8091.41 W.

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

Can be ordered in specialized firm. True, it is not cheap, and it will be impossible to check the results. It is quite another matter if you learn to analyze heat losses in the house yourself. Then no one will have to pay, and you will be one hundred percent sure of your calculations.

The amount of heat lost by a building in a certain unit of time is called heat loss. This value is not constant. It depends on the temperature, as well as the heat-shielding properties of the enclosing structures (these include walls, windows, ceilings, etc.). Significant heat losses also occur due to drafts - the air entering the room is scientifically called infiltration. And a great way to deal with them is the installation of modern double-glazed windows. Calculation of heat loss must take into account all these factors.

All construction and Decoration Materials differ in their characteristics and, consequently, thermal properties. Their structure is often heterogeneous, consists of several layers, and sometimes has closed air gaps. You can calculate the heat loss of this entire structure by adding up the indicators for each of the layers.

The main characteristic of the materials in our calculations will be the indicator. It will show how much heat the structure will lose (for example, 1 m 2) at a certain temperature difference.

We have the following formula: R=DT/Q

· DT - indicator of temperature difference;

Q - the amount of W / m 2 of heat that the structure loses;

· R - heat transfer resistance coefficient.

All these indicators are easy to calculate using SNiP. They contain information regarding most traditional building materials. As for modern designs(double-glazed windows, drywall and others), the required data can be obtained from the manufacturer.

Thus, it is possible to calculate the heat loss for each Special attention should be given to the outer walls, attic floors, areas above cold basements and unheated floors. Additional heat losses occur through doors and windows (especially those facing north and east), as well as external gates in the absence of a vestibule.

Calculation of heat loss of the building is carried out in relation to the most unfavorable period of the year. In other words, the coldest and windiest week is taken. Summing up the heat losses in this way, it is possible to determine the required power of all heaters in the room, necessary for its comfortable heating. These calculations will also help to identify the “weak link” in the thermal insulation system and take additional measures.

You can also make a calculation on the basis of general, averaged indicators. For example, for one and two-story buildings at minimum temperature air -25°С heat per one square meter 213 watts will be required. For buildings with quality, this figure drops to 173 W, or even less.

Based on the foregoing, we can say that you should not save on high-quality thermal insulation. In the context of a constant increase in energy prices, competent insulation and ventilation of structures lead to significant benefits.