Heat loss calculation: building heat loss indicators and calculator. Simple calculation of building heat loss House heat loss coefficient

So that your house does not turn out to be a bottomless pit for heating costs, we suggest studying the basic directions of thermal engineering research and calculation methodology. Without preliminary calculation thermal permeability and moisture accumulation, the whole essence of housing construction is lost.

Physics of thermal processes

Different areas of physics have much in common in describing the phenomena they study. So it is in heat engineering: the principles describing thermodynamic systems clearly echo the foundations of electromagnetism, hydrodynamics and classical mechanics. After all, we are talking about the description of the same world, so it is not surprising that models of physical processes are characterized by some common features in many areas of research.

The essence of thermal phenomena is easy to understand. The temperature of a body or the degree of its heating is nothing but a measure of the intensity of oscillations of the elementary particles of which this body is composed. Obviously, when two particles collide, the one with the higher energy level will transfer energy to the particle with lower energy, but never vice versa. However, this is not the only way of energy exchange; transfer is also possible through thermal radiation quanta. Wherein basic principle necessarily conserved: a quantum emitted by a less heated atom is not able to transfer energy to a hotter elementary particle. It is simply reflected from it and either disappears without a trace, or transfers its energy to another atom with less energy.

Thermodynamics is good because the processes occurring in it are absolutely visual and can be interpreted under the guise of various models. The main thing is to follow the basic postulates, such as the law of energy transfer and thermodynamic equilibrium. So if your presentation complies with these rules, you will easily understand the method of thermal engineering calculations from and to.

The concept of resistance to heat transfer

The ability of a material to transfer heat is called thermal conductivity. In the general case, it is always higher, the greater the density of the substance and the better its structure is adapted to transmit kinetic vibrations.

The quantity inversely proportional to thermal conductivity is thermal resistance. For each material, this property takes on unique values depending on the structure, shape, and a number of other factors. For example, the efficiency of heat transfer in the thickness of materials and in the zone of their contact with other media may differ, especially if there is at least a minimal layer of substance between the materials in another state of aggregation. Quantitatively, thermal resistance is expressed as the temperature difference divided by the heat flow rate:

R t \u003d (T 2 - T 1) / P

R t - thermal resistance of the section, K / W;
T 2 - temperature of the beginning of the section, K;
T 1 - temperature of the end of the section, K;
P is the heat flux, W.

In the context of calculating heat losses, thermal resistance plays a decisive role. Any enclosing structure can be represented as a plane-parallel barrier to the heat flow. Its total thermal resistance is the sum of the resistances of each layer, while all the partitions are folded into a spatial structure, which is, in fact, a building.

R t \u003d l / (λ S)

R t - thermal resistance of the circuit section, K / W;
l is the length of the section of the thermal circuit, m;
λ is the thermal conductivity of the material, W/(m K);
S is the cross-sectional area of the site, m 2.

Factors affecting heat loss

Thermal processes correlate well with electrical processes: the temperature difference acts as a voltage, the heat flux can be considered as a current strength, but you don’t even need to come up with your own term for resistance. The concept of least resistance, which appears in heat engineering as cold bridges, is also fully true.

If we consider an arbitrary material in a section, it is quite easy to establish the path of the heat flow both at the micro- and at the macro level. Let us take as the first model concrete wall, in which, due to technological necessity, through fastenings are made with steel rods of arbitrary section. Steel conducts heat somewhat better than concrete, so we can distinguish three main heat fluxes:

through the concrete
through steel bars
from steel bars to concrete

The last heat flow model is the most interesting. Since the steel rod warms up faster, the temperature difference between the two materials will be observed closer to the outer part of the wall. Thus, steel not only "pumps" heat outward by itself, it also increases the thermal conductivity of adjacent masses of concrete.

In porous media, thermal processes proceed in a similar way. Almost all Construction Materials consist of a branched web of solid matter, the space between which is filled with air. Thus, a solid, dense material serves as the main conductor of heat, but due to the complex structure, the path along which heat spreads turns out to be larger than the cross section. Thus, the second factor that determines thermal resistance is the heterogeneity of each layer and the building envelope as a whole.

The third factor affecting the thermal conductivity, we can name the accumulation of moisture in the pores. Water has a thermal resistance 20-25 times lower than that of air, so if it fills the pores, the overall thermal conductivity of the material becomes even higher than if there were no pores at all. When water freezes, the situation becomes even worse: thermal conductivity can increase up to 80 times. The source of moisture is usually room air and atmospheric precipitation. Accordingly, the three main methods of dealing with such a phenomenon are outdoor waterproofing walls, the use of vapor protection and the calculation of moisture accumulation, which is necessarily carried out in parallel with the forecasting of heat losses.

Differentiated calculation schemes

The simplest way to determine the amount of heat loss in a building is to sum up the values of heat flow through the structures that form this building. This technique fully takes into account the difference in the structure various materials, as well as the specifics of the heat flow through them and at the junctions of one plane to another. Such a dichotomous approach greatly simplifies the task, because different enclosing structures can differ significantly in the design of thermal protection systems. Accordingly, in a separate study, it is easier to determine the amount of heat loss, because for this various ways calculations:

For walls, heat leakage is quantitatively equal to the total area multiplied by the ratio of the temperature difference to the thermal resistance. In this case, the orientation of the walls to the cardinal points is necessarily taken into account to take into account their heating in daytime, as well as the permeability building structures.
For overlaps, the technique is the same, but the presence of attic space and mode of operation. Also for room temperature a value of 3-5 °C is taken higher, the calculated humidity is also increased by 5-10%.
Heat losses through the floor are calculated zonal, describing the belts along the perimeter of the building. This is due to the fact that the temperature of the soil under the floor is higher near the center of the building compared to the foundation part.
The heat flux through the glazing is determined by the nameplate data of the windows; the type of adjoining windows to the walls and the depth of the slopes must also be taken into account.

Q = S (ΔT / Rt)

Q is heat loss, W;
S - wall area, m 2;
ΔT - temperature difference inside and outside the room, ° С;
R t - resistance to heat transfer, m 2 ° C / W.

Calculation example

Before moving on to the demo, let's answer the question last question: how to correctly calculate the integral thermal resistance of complex multilayer structures? This, of course, can be done manually, since not many types of load-bearing bases and insulation systems are used in modern construction. However, take into account the presence of decorative finishes, interior and facade plaster, as well as the influence of all transients and other factors is quite difficult, it is better to use automated calculations. One of the best online resources for such tasks is smartcalc.ru, which additionally plots the dew point offset depending on climatic conditions.

For example, let's take an arbitrary building, having studied the description of which the reader will be able to judge the set of initial data necessary for the calculation. Available cottage correct rectangular shape with dimensions of 8.5x10 m and a ceiling height of 3.1 m, located in the Leningrad region. The house has an uninsulated floor on the ground with boards on logs with an air gap, the floor height is 0.15 m higher than the ground planning mark on the site. The wall material is a slag monolith 42 cm thick with internal cement-lime plaster up to 30 mm thick and external slag-cement plaster of the "fur coat" type up to 50 mm thick. The total area of glazing is 9.5 m 2 , the windows used are a double-glazed window in a heat-saving profile with an average thermal resistance of 0.32 m 2 °C / W. The cover was made on wooden beams: from below it is plastered on shingles, filled with blast-furnace slag and covered with clay screed from above, above the ceiling there is a cold-type attic. The task of calculating heat loss is the formation of a thermal protection system for walls.

First of all, heat losses through the floor are determined. Since their share in the total heat outflow is the smallest, and also due to the large number of variables (density and type of soil, freezing depth, massiveness of the foundation, etc.), heat loss is calculated using a simplified method using the reduced heat transfer resistance. Along the perimeter of the building, starting from the line of contact with the ground, four zones are described - encircling strips 2 meters wide. For each of the zones, its own value of the reduced resistance to heat transfer is taken. In our case, there are three zones with an area of 74, 26 and 1 m 2. Do not be confused by the total area of the zones, which is 16 m 2 more than the area of the building, the reason for this is the double recalculation of the intersecting strips of the first zone in the corners, where heat losses are much higher compared to sections along the walls. Applying heat transfer resistance values of 2.1, 4.3 and 8.6 m 2 °C / W for zones one through three, we determine the heat flow through each zone: 1.23, 0.21 and 0.05 kW respectively.

Walls

Using the terrain data, as well as the materials and thickness of the layers that form the walls, you need to fill in the appropriate fields on the smartcalc.ru service mentioned above. According to the results of the calculation, the resistance to heat transfer is equal to 1.13 m 2 ° C / W, and the heat flux through the wall is 18.48 W per square meter. With a total wall area (excluding glazing) of 105.2 m 2, the total heat loss through the walls is 1.95 kW / h. In this case, heat loss through the windows will be 1.05 kW.

Covering and roofing

Calculation of heat loss through attic floor can also be performed in the online calculator by selecting the desired type of enclosing structures. As a result, the resistance of the overlap to heat transfer is 0.66 m 2 · ° С / W, and the heat loss is 31.6 W per square meter, that is, 2.7 kW from the entire area of the building envelope.

Total total heat loss according to calculations, they are 7.2 kWh. Given the rather low quality of the building structures, this figure is obviously much lower than the real one. In fact, such a calculation is idealized, it does not take into account special coefficients, blowing, the convection component of heat transfer, losses through ventilation and entrance doors. In fact, due to poor quality installation windows, lack of protection at the junction of the roof to the mauerlat and poor waterproofing of the walls from the foundation, real heat losses can be 2 or even 3 times more than the calculated ones. However, even basic thermal engineering studies help determine whether the structures of the house under construction will comply with sanitary standards at least as a first approximation.

Finally, let's give one important recommendation: If you really want to get a complete understanding of the thermal physics of a particular building, you must use an understanding of the principles described in this overview and specialized literature. For example, a reference manual by Elena Malyavina “Heat loss of a building” can be a very good help in this matter, where the specifics of heat engineering processes are explained in great detail, links are given to the necessary regulations, as well as examples of calculations and all the necessary background information.

I figured out the loss of overlap (floors on the ground without insulation) even STRONGLY a lot
with a thermal conductivity of concrete of 1.8, it turns out 61491 kWh season
I think the average temperature difference should not be taken as 4033 * 24, since the earth is still warmer than atmospheric air
For floors, the temperature difference will be less, the air outside is -20 degrees and the ground under the floors can be +10 degrees. That is, at a temperature in the house of 22 degrees, to calculate the heat loss in the walls, the temperature difference will be 42 degrees, and for the floors at the same time it will be only 12 degrees.
I also made such a calculation for myself last year in order to choose an economically justified thickness of insulation. But I made a more complex calculation. I found in the internet for my city statistics on temperatures for the previous year, and in increments of every four hours. Ie I consider that during four hours the temperature is constant. For each temperature, he determined how many hours a year this temperature had and calculated the losses for each temperature for the season, of course, divided into articles, walls, attic, floor, windows, ventilation. For the floor, I took a temperature difference of a constant 15 degrees, like (I have a basement). I did it all in an excel spreadsheet. I set the thickness of the insulation and immediately see the result.
I have walls silicate brick 38 cm. The house is two-story plus a basement, the area with a basement is 200 sq. m. The results are as follows:
Styrofoam 5 cm. Savings for the season will be 25919 rubles, a simple payback period (without inflation) is 12.8 years.
Styrofoam 10 cm. Savings for the season will be 30,017 rubles, a simple payback period (without inflation) is 12.1 years.
Styrofoam 15 cm. Savings per season will be 31,690 rubles, a simple payback period (without inflation) is 12.5 years.
Now let's think of a slightly different number. compare 10 cm and the payback to them of an additional 5 cm (up to 15)
So, additional savings at +5 cm is about 1,700 rubles per season. and the additional costs for warming are approximately 31,500 rubles, that is, these additional. 5 cm of insulation will pay off only after 19 years. It is not worth it, although before the calculations I was determined to make 15 cm in order to reduce the operating costs of gas, but now I see that the sheepskin skin is not worth the candle, add. saving 1700 rubles a year, it's not serious
For comparison, to the first five cm, we additionally add another 5 cm, then add. savings will be 4100 per year, add. costs 31500, payback 7.7 years, this is already normal. I will do 10 cm thinner, but I don’t want to, not seriously like that.
Yes, according to my calculations, I got the following results
brick wall 38 cm plus 10 cm foam.
energy saving windows.
Ceiling 20 cm. air layer, losses mean even less, but for now I don’t take it into account), the floor of foam boards or whatever else is 10 cm plus ventilation.
The total losses for the year are 41,245 kW. h, it is approximately 4,700 cubic meters of gas per year or so 17500 rub/ year (1460 rubles / month) It seems to me that it turned out okay. I also want to make a self-made heat exchanger for ventilation, otherwise I estimated 30-33% of all heat losses, these are ventilation losses, something needs to be decided with this., I don’t want to sit in a corked box.

Not all materials used in construction are capable of providing proper level heat saving of a private house. Through the walls, roof, floor, window openings there is a constant leakage of heat. Having determined with the help of a thermal imager which structural elements of the building act as “weak links”, it is possible to significantly reduce heat loss in a private house by means of complex or fragmentary insulation.

Insulate windows

Insulation of windows at home is most often performed according to Swedish technology, for which all window sashes are removed from the frames, then a groove is selected along the perimeter of the frame with a cutter, into which a tubular sealant made of silicone (with a diameter of 2 to 7 mm) is filled - this allows you to reliably seal the window porches. Small gaps in the frames, the gaps between the double-glazed window and the frame are filled with sealant after preliminary washing, cleaning and drying the windows.

Window insulation can also be performed using a heat-saving film, which is fixed to the window with a self-adhesive strip. window frame. Letting light into the room, the film reliably shields heat flows due to metallized spraying, returning back to the room about 60% of the heat. Significant heat losses through windows are often associated with a violation of the geometry of the frame, gaps between the frame and slopes, sagging and skewed sashes, poor functioning of fittings - to eliminate these problems, qualified adjustment or repair of windows is required.

Insulate the walls

The most significant heat loss - about 40% - occurs through the walls of buildings, so thoughtful insulation of the main walls of a private house will drastically improve its heat-saving parameters. Wall insulation can be done from the inside or/and outside - the method of insulation depends on the material used in the construction of the house. Brick and foam concrete houses are most often insulated from the outside, but a heat insulator can also be laid from the inside of these buildings. Wooden houses almost never insulated from the side interior spaces, to avoid the greenhouse effect in the rooms. Outside, houses are insulated from a bar, sometimes from a log house.

Insulation of the walls of the house can be performed using the "wet" or "wet" technology. hinged facade- the main difference between these methods lies in the principle of installation facade cladding. When arranging a “wet” facade, a dense heat insulator (polystyrene foam, polystyrene foam) is attached to the wall, and then decorative trim using adhesives. When installing a hinged facade, after installing a heater (mineral or glass wool), a crate is mounted, and then facing modules are fixed in its profiles. An obligatory element of the "pie" of the walls is a vapor barrier film, which removes condensate from the insulation layer, protects it from getting wet and prevents the loss of insulating properties.

Insulate the roof

The roof of the house is another surface through which heat constantly escapes from the house. Depending on the material used in the arrangement roof deck, the roof can be more or less warm. Capital insulation, as a rule, require metal roofs from corrugated board and metal tiles. Roofs made of ondulin, flexible and ceramic tiles have low thermal conductivity, so for them the insulating "pie" can be thinner than in the case of metal. Similar to the technology for insulating other surfaces of the house, a vapor barrier must be included in the “pie” of the roof, and one or two ventilation gaps are provided for effective ventilation of the under-roof space.

Insulate the floor

Unlike walls and window openings, heat leakage through the floor of a private house is small - approximately 10%, and subject to the arrangement of insulation, it will be reduced to a minimum. As a heater for floors, the same polystyrene, polystyrene or mineral wool, but it is also possible to use expanded clay, foamed concrete, cement-bonded mixtures and peat mats. An additional insulation measure in country house installation of warm floors can act: water, cable or infrared.

Similar to the device for insulating walls and roofs, an obligatory component of the "pie" of the floor is vapor barrier membrane, which shields moisture-saturated steam seeping from inner space home outside. Thus, the heat-insulating layer is reliably protected from getting wet.

Any construction of the house, begins with drawing up the project of the house. Already at this stage, you should think about warming your home, because. there are no buildings and houses with zero heat loss, which we pay for in the cold winter, during the heating season. Therefore, it is necessary to carry out the insulation of the house outside and inside, taking into account the recommendations of the designers.

What and why to insulate?

During the construction of houses, many do not know, and do not even realize that in a private house built, during the heating season, up to 70% of the heat will go to heat the street.

Concerned about saving family budget and the problem of home insulation, many are wondering: what and how to insulate ?

This question is very easy to answer. It is enough to look at the screen of the thermal imager in winter, and you will immediately notice through which structural elements the heat escapes into the atmosphere.

If you do not have such a device, then it does not matter, below we will describe the statistics that show where and in what percentage the heat leaves the house, as well as post a video of the thermal imager from a real project.

When insulating a house it is important to understand that heat escapes not only through floors and roofs, walls and foundations, but also through old windows and doors that will need to be replaced or insulated during the cold season.

Distribution of heat losses in the house

All experts recommend insulation of private houses , apartments and industrial premises not only from the outside, but also from the inside. If this is not done, then the warmth that is “dear” to us, in the cold season, will simply quickly disappear into nowhere.

Based on statistics and data from specialists, according to which, if the main heat leaks are identified and eliminated, it will already be possible to save 30% or more percent on heating in winter.

So, let's analyze in what directions, and in what percentage our heat leaves the house.

The largest heat loss occurs through:

Heat loss through the roof and floors

As is known, warm air always rises to the top, so it heats the uninsulated roof of the house and ceilings, through which 25% of our heat leaks.

To produce house roof insulation and reduce heat loss to a minimum, you need to use roof insulation with a total thickness of 200mm to 400mm. The technology of insulating the roof of the house can be seen by enlarging the picture on the right.

Heat loss through walls

Many will probably wonder: why is the heat loss through the uninsulated walls of the house (about 35%) more than through the uninsulated roof of the house, because all the warm air rises to the top?

Everything is very simple. Firstly, the area of the walls is much larger than the area of \u200b\u200bthe roof, and secondly, different materials have different thermal conductivity. Therefore, during construction country houses, you need to take care of the house wall insulation. For this, insulation for walls with a total thickness of 100 to 200 mm is suitable.

For proper insulation the walls of the house, you must have knowledge of technology and a special tool. wall insulation technology brick house can be seen by zooming in on the picture on the right.

Heat loss through floors

Strange as it may seem, but not insulated floors in the house take from 10 to 15% of the heat (the figure may be more if your house is built on piles). This is due to ventilation under the house during the cold period of winter.

To minimize heat loss through insulated floors in the house, you can use insulation for floors with a thickness of 50 to 100mm. This will be enough to walk barefoot on the floor in the cold winter season. The technology of home floor insulation can be seen by enlarging the picture on the right.

Heat loss through windows

Window- perhaps this is the very element that is almost impossible to insulate, because. then the house will become like a dungeon. The only thing that can be done to reduce heat loss by up to 10% is to reduce the number of windows in the design, insulate the slopes and install at least double-glazed windows.

Heat loss through doors

The last element in the design of the house, through which up to 15% of heat escapes, is the doors. This is due to the constant opening entrance doors, through which heat constantly escapes. For reducing heat loss through doors to a minimum, it is recommended to install double doors, seal them with sealing rubber and install thermal curtains.

Benefits of an insulated home

Payback in the first heating season
Savings on air conditioning and heating at home
Cool indoors in summer
Excellent additional soundproofing walls and floors, ceilings and floors
Protection of house structures from destruction
Increased indoor comfort
It will be possible to turn on the heating much later

The results of the insulation of a private house

It is very profitable to warm the house , and in most cases even necessary, because this is due large quantity advantages over non-insulated houses, and allows you to save your family budget.

Having carried out external and internal insulation at home, your a private house becomes like a thermos. Heat will not fly away from it in winter and heat will not come in in summer, and all costs for the complete insulation of the facade and roof, basement and foundation will pay off within one heating season.

For optimal choice heater for home , we recommend that you read our article: The main types of insulation for the house, which discusses in detail the main types of insulation used in the insulation of a private house outside and inside, their pros and cons.

Video: Real project - where does the heat go in the house

Energy-efficient renovation of the building will save thermal energy and improve the comfort of life. The greatest savings potential lies in the good thermal insulation of the outer walls and roof. The easiest way to assess opportunities effective repair is the consumption of thermal energy. If more than 100 kWh of electricity (10 m³ of natural gas) is consumed per year for 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.

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

Important! AT wooden structures the thermal insulation of the roof is difficult, as the wood swells and can be damaged by high humidity.

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

Floor or basement. For practical reasons, thermal insulation is made from inside the building;
thermal bridges. Thermal bridges are unwanted cooling fins (heat conductors) on the outside of a building. For example, a concrete floor, which is also a balcony floor. Many thermal bridges are found in soil areas, parapets, window and door frames. There are also temporary thermal bridges if the wall details are fixed metal elements. Thermal bridges can account for a significant portion of heat loss;
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 losses, these are single- and double-glazed windows;
Ventilation. A typical building has air leaks, especially around windows, doors and on the roof, which provides the necessary air exchange. However, during the cold season, this causes significant heat loss from the house from the outgoing heated air. Good ones modern buildings are sufficiently airtight, and it is necessary to regularly ventilate the premises by opening windows for a few minutes. To reduce heat loss through ventilation, comfort panels are increasingly being installed. ventilation systems. This type of heat loss is estimated at 10-40%.

Thermographic surveys in a poorly insulated building give an idea of how much heat is being wasted. This is very good tool for quality control of repair or new construction.

Ways to assess heat loss at home

There are complex calculation methods that take into account various physical processes: convection exchange, radiation, but they are often redundant. Simplified formulas are usually used, and if necessary, 1-5% can be added to the result. Building orientation 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 heat losses, the time spent by people in a particular room is always taken into account. The smaller it is, the lower temperature indicators should be taken as a basis.

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 is indicated, which is required to heat 10 m² of room area with 3 m high ceilings and one window. 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;
Calculation of heat loss of the enclosing contour of the building. Summarized area external walls minus the dimensions of the areas of windows and doors. Additionally, there is a roof area with a floor. Further calculations are carried out according to the formula:

Q = S x ΔT/R, where:

S is the found area;
ΔT is the difference between indoor and outdoor temperatures;
R is the resistance to heat transfer.

The result obtained for the walls, floor and roof is combined. Then ventilation losses are added.

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

Calculation of heat loss by rooms. When using a similar formula, losses are calculated for all rooms of the building separately. Then, heat losses for ventilation are found by determining the volume of air mass and the approximate number of times a day it is changed in the room.

Important! When calculating ventilation losses, it is necessary to take into account the purpose of the room. The kitchen and bathroom need enhanced ventilation.

An example of calculating the heat loss of a residential building

The second calculation method is used, only for the external structures of the house. Through them, up to 90 percent of thermal energy is lost. Accurate results are important to select the right boiler to deliver efficient heat without overheating the rooms. It is also an indicator of the economic efficiency of the selected materials for thermal protection, showing how quickly you can recoup the cost 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 (67 cm) thick, covered with plaster, with a layer of 3 cm. The house has 10 windows 0.9 x 1 m and a door 1 x 2 m.

Calculation of resistance to heat transfer of walls:

R = n/λ, where:

n - wall thickness,
λ is the specific thermal conductivity (W/(m °C).

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

For brick:

Rkir \u003d 0.67 / 0.38 \u003d 1.76 sq.m ° C / W.

For plaster coating:

Rpcs \u003d 0.03 / 0.35 \u003d 0.086 sq.m ° C / W;

Total value:

Rst \u003d Rkir + Rsht \u003d 1.76 + 0.086 \u003d 1.846 sq.m ° C / W;

Calculation of the area of external walls:

Total area of external walls:

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

Area of windows and doorway:

S1 \u003d ((0.9 x 1) x 10) + (1 x 2) \u003d 11 sq.m.

Adjusted wall area:

S2 = S - S1 = 264 - 11 = 253 sq.m.

Heat losses for walls will be determined by:

Q \u003d S x ΔT / R \u003d 253 x 40 / 1.846 \u003d 6810.22 W.

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

On the next step heat losses through the foundation, windows, roof, door are calculated in an identical 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 consumption of electricity for heating, you can present this figure in kWh and calculate it for the heating season.

If you use only the number for the walls, it turns out:

per day:

6810.22 x 24 = 163.4 kWh;

per month:

163.4 x 30 = 4903.4 kWh;

for the heating season of 7 months:

4903.4 x 7 \u003d 34,323.5 kWh.

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

Heat losses for ventilation

Find the air volume of the house:

10 x 12 x 6 = 720 m³;

The mass of air is found by the formula:

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

M \u003d 1, 205 x 720 \u003d 867.4 kg.

It is necessary to determine the figure, how many times the air in the whole house is replaced per day (for example, 6 times), and calculate the heat loss for ventilation:

Qv = nxΔT xmx C, where C is the specific heat capacity for air, n is the number of times the air is replaced.

Qv \u003d 6 x 40 x 867.4 x 1.005 \u003d 209217 kJ;

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 the total heat losses, without calculating them using formulas.

To facilitate the calculation of heat loss at home, there are online calculators where you can calculate the result for each room or the entire house. You simply enter your data in the proposed fields.