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Heat loss during the heating season. Ideal home: calculation of heat loss at home. Calculating the area of ​​external walls

Comfort is a tricky thing. Sub-zero temperatures come, it immediately becomes chilly, and uncontrollably drawn to home improvement. "Global warming" begins. And there is one “but” here - even after calculating the heat loss of the house and installing the heating “according to the plan”, you can stay face to face with the quickly leaving heat. The process is not visually noticeable, but it feels great through woolen socks and big heating bills. The question remains - where did the "precious" heat go?

Natural heat losses are well hidden behind bearing structures or “well-made” insulation, where there should not be gaps by default. But is it? Let's look at the issue of thermal leakage for different structural elements.

Cold places on the walls

Up to 30% of all heat loss at home falls on the walls. IN modern construction they are multilayer structures made of materials with different thermal conductivity. Calculations for each wall can be carried out individually, but there are errors common to all, through which heat leaves the room, and cold enters the house from the outside.

The place where the insulating properties are weakened is called the “cold bridge”. For walls it is:

• Masonry joints

The optimal masonry seam is 3mm. It is achieved more often adhesive compositions fine texture. When the volume of the solution between the blocks increases, the thermal conductivity of the entire wall increases. Moreover, the temperature of the masonry seam can be 2-4 degrees colder than the base material (brick, block, etc.).

Masonry joints as a "thermal bridge"

• Concrete lintels over openings.

One of the highest thermal conductivity coefficients among building materials (1.28 - 1.61 W / (m * K)) for reinforced concrete. This makes it a source of heat loss. The issue is not completely resolved by cellular or foam concrete lintels. The temperature difference between the reinforced concrete beam and the main wall is often close to 10 degrees.

It is possible to isolate the jumper from the cold with continuous external insulation. And inside the house - by assembling a box from the Civil Code under the eaves. This creates an additional air layer for warmth.

• Mounting holes and fasteners.

Connecting an air conditioner, TV antenna leaves holes in the overall insulation. Through metal fasteners and the passage hole must be tightly sealed with insulation.

And if possible, do not withdraw metal mounts outside, fixing them inside the wall.

Insulated walls also have defects with heat loss.

Installation of damaged material (with chips, squeezing, etc.) leaves vulnerable areas for heat leakage. This is clearly seen when examining the house with a thermal imager. Bright spots show gaps in the outer insulation.

During operation, it is important to monitor general condition insulation. An error in the choice of glue (not special for thermal insulation, but tiled) can give out cracks in the structure after 2 years. Yes, and the main insulation materials also have their drawbacks. For example:

• Mineral wool - does not rot, and is not interesting to rodents, but is very sensitive to moisture. Therefore, its good service life in external insulation is about 10 years - then damage appears.
• Styrofoam - has good insulating properties, but is easily amenable to rodents, and is not resistant to force and ultraviolet radiation. The insulation layer after installation requires immediate protection (in the form of a structure or a layer of plaster).

When working with both materials, it is important to observe a clear fit of the locks of the insulation boards and the cross arrangement of the sheets.

• Polyurethane foam - creates seamless insulation, is convenient for uneven and curved surfaces, but is vulnerable to mechanical damage, and collapses under UV rays. It is desirable to cover plaster mixture- fastening frames through a layer of insulation violates the overall insulation.

Experience! Heat loss can increase during operation, because all materials have their own nuances. It is better to periodically assess the condition of the insulation and repair damage immediately. A crack on the surface is a “high-speed” road to the destruction of the insulation inside.

Foundation heat loss

Concrete is the predominant material in foundation construction. Its high thermal conductivity and direct contact with the ground give up to 20% of heat loss around the entire perimeter of the building. The foundation conducts heat especially strongly from the basement and improperly installed underfloor heating on the ground floor.

Heat loss is also increased by excess moisture not removed from the house. It destroys the foundation, creating loopholes for the cold. Many heat-insulating materials are also sensitive to humidity. For example, mineral wool, which often goes to the foundation from general insulation. It is easily damaged by moisture, and therefore requires a dense protective frame. Expanded clay also loses its thermal insulation properties permanently wet ground. Its structure creates air cushion and well compensates for the pressure of soils during freezing, but the constant presence of moisture minimizes beneficial features expanded clay in insulation. That is why the creation of working drainage - required condition long life of the foundation and heat preservation.

In terms of importance, this also includes the waterproofing protection of the base, as well as a multi-layer blind area, at least a meter wide. At column foundation or heaving soil, the blind area around the perimeter is insulated to protect the soil at the base of the house from freezing. The blind area is insulated with expanded clay, sheets of expanded polystyrene or polystyrene.

Sheet materials for foundation insulation are best chosen with groove connection, and treat it with a special silicone compound. The tightness of the locks blocks access to the cold and guarantees complete protection of the foundation. In this matter, seamless spraying of polyurethane foam has an indisputable advantage. In addition, the material is elastic and does not crack when the soil is heaving.

For all types of foundations, you can use the developed insulation schemes. An exception may be the foundation on piles, due to its design. Here, when processing the grillage, it is important to take into account the heaving of the soil and choose a technology that does not destroy the piles. This is a complex calculation. Practice shows that a house on stilts protects the well-insulated floor of the first floor from the cold.

Attention! If the house has a basement, and it is often flooded, then this must be taken into account with the insulation of the foundation. Since the insulation / insulator in this case will clog moisture in the foundation, and destroy it. Accordingly, the heat will be lost even more. The first thing to do is solve the problem of flooding.

Vulnerabilities of the floor

An uninsulated ceiling gives off a significant part of the heat to the foundation and walls. This is especially noticeable when the underfloor heating is not properly installed - the heating element cools down faster, increasing the cost of heating the room.

In order for the heat from the floor to go into the room, and not out into the street, you need to make sure that the installation goes according to all the rules. The main ones are:

• Protection. A damper tape (or foil polystyrene sheets up to 20 cm wide and 1 cm thick) is attached to the walls around the entire perimeter of the room. Before this, the gaps are necessarily eliminated, and the surface of the wall is leveled. The tape is fixed as tightly as possible to the wall, isolating the heat transfer. When there are no air pockets, there are no heat leaks.
• Indent. From outer wall to the heating circuit should be at least 10 cm. If the warm floor is mounted closer to the wall, then it starts to heat the street.
• Thickness. The characteristics of the necessary screen and insulation for underfloor heating are calculated individually, but it is better to add 10-15% of the margin to the figures obtained.
• Finishing. The screed over the floor should not contain expanded clay (it isolates heat in concrete). Optimal Thickness screeds 3-7 cm. The presence of a plasticizer in a mixture of concrete improves thermal conductivity, and hence heat transfer to the room.

Serious insulation is relevant for any floor, and not necessarily heated. Poor thermal insulation turns the floor into a large "radiator" for the ground. Should it be heated in winter?

Important! Cold floors and dampness appear in the house when ventilation of the underground space is not working or not done (vents are not organized). No heating system compensates for such a shortcoming.

Places of adjoining building structures

Compounds violate the integral properties of materials. Therefore, corners, joints and junctions are so vulnerable to cold and moisture. The junctions of concrete panels are the first to dampen, and fungus and mold appear there. The temperature difference between the corner of the room (the place where the structures are joined) and the main wall can range from 5-6 degrees, to sub-zero temperatures and condensation inside the corner.

Clue! At the places of such connections, the masters recommend making an increased layer of insulation from the outside.

Heat often escapes through interfloor overlap when the slab is laid on the entire thickness of the wall and its edges go out into the street. Here, heat losses of both the first and second floors increase. Drafts are formed. Again, if there is a warm floor on the second floor, external insulation should be designed for this.

Heat leakage through ventilation

The heat from the room is removed through equipped ventilation ducts that provide healthy air exchange. Ventilation, working "on the contrary", tightens the cold from the street. This happens when there is a lack of air in the room. For example, when the switched on fan in the hood takes too much air from the room, due to which it begins to be drawn in from the street through other exhaust channels(without filters and heating).

Questions of how not to bring a large amount of heat outside, and how not to let cold air into the house, have long had their own professional solutions:

1. IN ventilation system recuperators are installed. They return up to 90% of heat to the house.
2. Settling down supply valves. They "prepare" the outdoor air in front of the room - it is cleaned and warmed. The valves come with manual adjustment or automatic, which focuses on the difference in temperature outside and inside the room.

Comfort is worth good ventilation. With normal air exchange, mold does not form, and a healthy microclimate is created for living. That is why a well-insulated house with a combination of insulating materials must necessarily have working ventilation.

Outcome! To reduce heat loss through ventilation ducts it is necessary to eliminate errors in the redistribution of air in the room. In well-functioning ventilation, only warm air leaves the house, some of the heat from which can be returned back.

Heat loss through windows and doors

Through door and window openings, the house loses up to 25% of heat. Weak spots for doors, this is a leaky seal that can be easily re-glued to a new one and a thermal insulation that has strayed inside. It can be replaced by removing the cover.

Vulnerabilities for wood and plastic doors similar to "cold bridges" in similar window designs. Therefore, we will consider the general process using their example.

What gives out "window" heat loss:

• Explicit gaps and drafts (in the frame, around the window sill, at the junction of the slope and the window). Poor sash fit.
• Damp and moldy internal slopes. If the foam and plaster have lagged behind the wall over time, then the moisture from the outside gets closer to the window.
• Cold glass surface. For comparison - energy-saving glass (at -25 ° outside, and inside the room + 20 °) has a temperature of 10-14 degrees. And, of course, it does not freeze.

The sashes may not fit snugly when the window is not adjusted and the rubber bands around the perimeter have worn out. The position of the flaps can be adjusted independently, as well as change the seal. It is better to replace it completely every 2-3 years, and preferably with a “native” production seal. Seasonal cleaning and lubrication of rubber bands maintains their elasticity during temperature changes. Then the sealant does not let the cold through for a long time.

Slots in the frame itself (relevant for wooden windows) are filled silicone sealant, better transparent. When it hits the glass, it's not so noticeable.

The joints of the slopes and the window profile are also sealed with sealant or liquid plastic. In a difficult situation, you can use self-adhesive polyethylene foam - "insulating" adhesive tape for windows.

Important! It is worth making sure that in the decoration of the external slopes, the insulation (polystyrene, etc.) completely covers the seam polyurethane foam and the distance to the middle of the window frame.

Modern ways to reduce heat loss through glass:

• Use of PVI films. They reflect wave radiation and reduce heat loss by 35-40%. Films can be glued to an already installed double-glazed window if there is no desire to change it. It is important not to confuse the sides of the glass and the polarity of the film.
• Installation of glass with low emission characteristics: k- and i-glass. Double-glazed windows with k-glasses transmit the energy of short waves of light radiation into the room, accumulating the body in it. Long-wave radiation no longer leaves the room. As a result, the glass on the inner surface has a temperature twice as high as that of conventional glass. i-glass holds thermal energy in the house by reflecting up to 90% of the heat back into the room.
• The use of silver-coated glasses, which are 2x chamber double-glazed windows save 40% more heat (compared to conventional glasses).
• The choice of double-glazed windows with an increased number of glasses and the distance between them.

Healthy! Reduce heat loss through glass - organized air curtains above the windows (can be in the form warm skirting boards) or protective shutters for the night. Especially relevant when panoramic glazing and extreme sub-zero temperatures.

Causes of heat leakage in the heating system

Heat loss also applies to heating, where heat leakage occurs more often for two reasons.

• A powerful radiator without a protective screen heats the street.

• Not all radiators fully warm up.

Compliance with simple rules reduces heat loss and prevents the heating system from working “idle”:

1. A reflective screen should be installed behind each radiator.
2. Before starting the heating, once a season, it is necessary to bleed the air from the system and see if all the radiators are fully warmed up. The heating system can become clogged due to accumulated air or debris (delamations, poor-quality water). Once every 2-3 years, the system must be completely flushed.

The note! When refilling, it is better to add anti-corrosion inhibitors to the water. It will support metal elements systems.

Heat loss through the roof

Heat initially tends to the top of the house, which makes the roof one of the most vulnerable elements. It accounts for up to 25% of all heat losses.

A cold attic or residential attic is insulated equally tightly. The main heat losses occur at the junctions of materials, it does not matter whether it is insulation or structural elements. So, the often overlooked bridge of cold is the border of the walls with the transition to the roof. It is desirable to process this area together with the Mauerlat.

The main insulation also has its own nuances, related more to the materials used. For example:

1. Mineral wool insulation should be protected from moisture and it is advisable to change every 10 - 15 years. Over time, it cakes and begins to let heat through.
2. Ecowool, which has excellent properties of "breathing" insulation, should not be near hot springs - when heated, it smolders, leaving gaps in the insulation.
3. When using polyurethane foam, it is necessary to equip ventilation. The material is vapor-tight, and it is better not to accumulate excess moisture under the roof - other materials are damaged, and a gap appears in the insulation.
4. Slabs in multilayer thermal insulation must be laid in a checkerboard pattern and must be closely adjacent to the elements.

Practice! In overhead structures, any gap can remove a lot of expensive heat. Here it is important to focus on dense and continuous insulation.

Conclusion

It is useful to know the places of heat loss not only in order to equip a house and live in comfortable conditions, but also not to overpay for heating. Proper insulation in practice pays off in 5 years. The term is long. But after all, we are not building a house for two years.

Related videos

Today, many families choose for themselves Vacation home as a place of permanent residence or year-round recreation. However, its content, and in particular the payment utilities, are quite costly, while most homeowners are not oligarchs at all. One of the most significant expenses for any homeowner is the cost of heating. To minimize them, it is necessary to think about energy saving even at the stage of building a cottage. Let's consider this question in more detail.

« The problems of energy efficiency of housing are usually remembered from the perspective of urban housing and communal services, however, the owners individual houses this topic is sometimes much closer,- considers Sergey Yakubov , deputy director of sales and marketing, a leading manufacturer of roofing and facade systems in Russia. - The cost of heating a house can be much more than half the cost of maintaining it in the cold season and sometimes reach tens of thousands of rubles. However, with a competent approach to the thermal insulation of a residential building, this amount can be significantly reduced.».

Actually, you need to heat the house in order to constantly maintain in it comfortable temperature no matter what's going on outside. In this case, it is necessary to take into account heat losses both through the building envelope and through ventilation, because. heat leaves with heated air, which is replaced by cooled air, as well as the fact that a certain amount of heat is emitted by people in the house, Appliances, incandescent lamps, etc.

To understand how much heat we need to get from our heating system and how much money we have to spend on it, let's try to evaluate the contribution of each of the other factors to the heat balance using the example of a brick building located in the Moscow region two-story house with a total area of ​​150 m2 (to simplify the calculations, we assumed that the dimensions of the cottage in terms of approximately 8.7x8.7 m and it has 2 floors 2.5 m high).

Heat loss through building envelope (roof, walls, floor)

The intensity of heat loss is determined by two factors: the temperature difference inside and outside the house and the resistance of its enclosing structures to heat transfer. By dividing the temperature difference Δt by the heat transfer resistance coefficient Ro of walls, roofs, floors, windows and doors and multiplying by their surface area S, we can calculate the intensity of heat loss Q:

Q \u003d (Δt / R o) * S

The temperature difference Δt is not constant, it changes from season to season, during the day, depending on the weather, etc. However, our task is simplified by the fact that we need to estimate the need for heat in total for the year. Therefore, for an approximate calculation, we may well use such an indicator as the average annual air temperature for the selected area. For the Moscow region it is +5.8°C. If we take +23°C as a comfortable temperature in the house, then our average difference will be

Δt = 23°C - 5.8°C = 17.2°C

Walls. The area of ​​​​the walls of our house (2 square floors 8.7x8.7 m high 2.5 m) will be approximately equal to

S \u003d 8.7 * 8.7 * 2.5 * 2 \u003d 175 m 2

However, the area of ​​windows and doors must be subtracted from this, for which we will calculate the heat loss separately. Suppose we have one front door, standard size 900x2000 mm, i.e. area

S doors \u003d 0.9 * 2 \u003d 1.8 m 2,

and windows - 16 pieces (2 on each side of the house on both floors) with a size of 1500x1500 mm, the total area of ​​\u200b\u200bwhich will be

S windows \u003d 1.5 * 1.5 * 16 \u003d 36 m 2.

Total - 37.8 m 2. Remaining area brick walls -

S walls \u003d 175 - 37.8 \u003d 137.2 m 2.

The heat transfer resistance coefficient of a 2-brick wall is 0.405 m2°C/W. For simplicity, we will neglect the resistance to heat transfer of the layer of plaster covering the walls of the house from the inside. Thus, the heat dissipation of all the walls of the house will be:

Q walls \u003d (17.2 ° C / 0.405 m 2 ° C / W) * 137.2 m 2 \u003d 5.83 kW

Roof. For simplicity of calculations, we will assume that the resistance to heat transfer roofing cake equal to the heat transfer resistance of the insulation layer. For light mineral wool insulation 50-100 mm thick, most often used for roof insulation, it is approximately equal to 1.7 m 2 °C / W. We will neglect the heat transfer resistance of the attic floor: let's assume that the house has an attic, which communicates with other rooms and heat is distributed evenly between all of them.

Square gable roof with a slope of 30 ° will be

Roof S \u003d 2 * 8.7 * 8.7 / Cos30 ° \u003d 87 m 2.

Thus, its heat dissipation will be:

Roof Q \u003d (17.2 ° C / 1.7 m 2 ° C / W) * 87 m 2 \u003d 0.88 kW

Floor. The heat transfer resistance of a wooden floor is approximately 1.85 m2°C/W. Having made similar calculations, we obtain heat dissipation:

Q floor = (17.2°C / 1.85m 2 °C/W) * 75 2 = 0.7 kW

Doors and windows. Their resistance to heat transfer is approximately equal to 0.21 m 2 °C / W, respectively (double wooden door) and 0.5 m 2 °C / W (ordinary double-glazed window, without additional energy-efficient "gadgets"). As a result, we get heat dissipation:

Q door = (17.2°C / 0.21W/m 2 °C) * 1.8m 2 = 0.15 kW

Q windows \u003d (17.2 ° C / 0.5 m 2 ° C / W) * 36 m 2 \u003d 1.25 kW

Ventilation. According to building codes, the air exchange coefficient for a dwelling should be at least 0.5, and preferably 1, i.e. in an hour, the air in the room should be completely updated. Thus, with a ceiling height of 2.5 m, this is approximately 2.5 m 3 of air per hour per square meter. This air must be heated from outdoor temperature (+5.8°C) to room temperature (+23°C).

The specific heat capacity of air is the amount of heat required to raise the temperature of 1 kg of a substance by 1 ° C - approximately 1.01 kJ / kg ° C. At the same time, the air density in the temperature range of interest to us is approximately 1.25 kg/m3, i.e. the mass of 1 cubic meter of it is 1.25 kg. Thus, to heat the air by 23-5.8 = 17.2 ° C for each square meter of area, you will need:

1.01 kJ / kg ° C * 1.25 kg / m 3 * 2.5 m 3 / hour * 17.2 ° C = 54.3 kJ / hour

For a house of 150 m2, this will be:

54.3 * 150 \u003d 8145 kJ / h \u003d 2.26 kW

Summarize

Heat loss through	Temperature difference, °C	Area, m2	Heat transfer resistance, m2°C/W	Heat loss, kW
Walls	17,2	175	0,41	5,83
Roof	17,2	87	1,7	0,88
Floor	17,2	75	1,85	0,7
doors	17,2	1,8	0,21	0,15
Window	17,2	36	0,5	0,24
Ventilation	17,2	-	-	2,26
Total:				11,06

Let's breathe now!

Suppose a family of two adults with two children lives in a house. The nutritional norm for an adult is 2600-3000 calories per day, which is equivalent to a heat dissipation power of 126 watts. The heat dissipation of a child will be estimated at half the heat dissipation of an adult. If everyone who lived at home is in it 2/3 of the time, then we get:

(2*126 + 2*126/2)*2/3 = 252W

Let's say that there are 5 rooms in the house, lit by ordinary incandescent lamps with a power of 60 W (not energy-saving), 3 per room, which are turned on for an average of 6 hours a day (i.e. 1/4 of the total time). Approximately 85% of the power consumed by the lamp is converted into heat. In total we get:

5*60*3*0.85*1/4=191W

The refrigerator is a very efficient heating device. Its heat dissipation is 30% of the maximum power consumption, i.e. 750 W.

Other household appliances (let it be washing and dishwasher) releases about 30% of the maximum power input as heat. The average power of these devices is 2.5 kW, they work for about 2 hours a day. Total we get 125 watts.

A standard electric stove with an oven has a power of approximately 11 kW, however, the built-in limiter regulates the operation of the heating elements so that their simultaneous consumption does not exceed 6 kW. However, it is unlikely that we will ever use more than half of the burners at the same time or all the heating elements of the oven at once. Therefore, we will proceed from the fact that the average operating power of the stove is approximately 3 kW. If she works 3 hours a day, then we get 375 watts of heat.

Each computer (and there are 2 in the house) emits approximately 300 W of heat and works 4 hours a day. Total - 100 watts.

TV is 200 W and 6 hours a day, i.e. per circle - 50 watts.

In total we get: 1.84 kW.

Now we calculate the required heat output of the heating system:

Heating Q = 11.06 - 1.84 = 9.22 kW

heating costs

Actually, above we calculated the power that will be needed to heat the coolant. And we will heat it, of course, with the help of a boiler. Thus, heating costs are fuel costs for this boiler. Since we are considering the most general case, we will make a calculation for the most universal liquid (diesel) fuel, since gas pipelines are far from being everywhere (and the cost of their summing up is a figure with 6 zeros), but solid fuel it is necessary, firstly, to bring it somehow, and secondly, to throw it into the boiler furnace every 2-3 hours.

To find out what volume V of diesel fuel per hour we have to burn to heat the house, we need specific heat its combustion q (the amount of heat released during the combustion of a unit mass or volume of fuel, for diesel fuel - approximately 13.95 kWh / l) multiplied by the boiler efficiency η (approximately 0.93 for diesel) and then the required power of the heating system Qheating ( 9.22 kW) divided by the resulting figure:

V = heating Q / (q * η) = 9.22 kW / (13.95 kW * h / l) * 0.93) = 0.71 l / h

With an average cost of diesel fuel for the Moscow Region of 30 rubles per liter per year, it will take us

0.71 * 30 rub. * 24 hours * 365 days = 187 thousand rubles. (rounded).

How to save?

The natural desire of any homeowner is to reduce heating costs even at the construction stage. Where does it make sense to invest money?

First of all, you should think about the insulation of the facade, which, as we saw earlier, accounts for the bulk of all heat loss at home. In the general case, external or internal additional insulation can be used for this. However internal insulation much less efficient: when installing thermal insulation from the inside, the boundary between the warm and cold areas “moves” inside the house, i.e. moisture will condense in the thickness of the walls.

There are two ways to insulate facades: “wet” (plaster) and by installing a hinged ventilated facade. Practice shows that due to the need for constant repairs, “wet” insulation, taking into account operating costs, ends up being almost twice as expensive as a ventilated facade. The main disadvantage of the plaster facade is high price its service and content. " The initial costs for the arrangement of such a facade are lower than for a hinged ventilated one, by only 20-25%, a maximum of 30%,- explains Sergey Yakubov ("Metal Profile"). - However, considering the cost of Maintenance, which must be done at least once every 5 years, after the first five years, the plaster facade will be equal in cost to the ventilated one, and in 50 years (the service life of the ventilated facade) it will be 4-5 times more expensive».

What is a hinged ventilated facade? This is an external "screen" attached to a light metal frame, which is attached to the wall with special brackets. A light insulation is placed between the wall of the house and the screen (for example, Isover "VentFacade Bottom" with a thickness of 50 to 200 mm), as well as a wind and hydroprotective membrane (for example, Tyvek Housewrap). Various materials can be used as exterior cladding, but in individual construction most commonly used steel siding. " The use of modern high-tech materials in the production of siding, such as steel coated with Colorcoat Prisma ™, allows you to choose almost any design solution, - says Sergey Yakubov. - This material has excellent resistance to both corrosion and mechanical stress. The warranty period for it is 20 years with a real life of 50 years or more. Those. provided that steel siding is used, the entire facade structure will last 50 years without repair».

An additional layer of facade insulation made of mineral wool has a heat transfer resistance of approximately 1.7 m2°C/W (see above). In construction, to calculate the heat transfer resistance of a multi-layer wall, add up the corresponding values ​​for each of the layers. As we remember, our main bearing wall in 2 bricks has a heat transfer resistance of 0.405 m2°C/W. Therefore, for a wall with a ventilated facade, we get:

0.405 + 1.7 = 2.105 m 2 °C / W

Thus, after insulation, the heat dissipation of our walls will be

Q facade \u003d (17.2 ° C / 2.105 m 2 ° C / W) * 137.2 m 2 \u003d 1.12 kW,

which is 5.2 times less than the same indicator for an uninsulated facade. Impressive, isn't it?

Again we calculate the required heat output of the heating system:

Q heating-1 = 6.35 - 1.84 = 4.51 kW

Diesel fuel consumption:

V 1 \u003d 4.51 kW / (13.95 kW * h / l) * 0.93) \u003d 0.35 l / h

Amount for heating:

0.35 * 30 rub. * 24 hours * 365 days = 92 thousand rubles.

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 energy level higher will transfer energy to a particle with less energy, but never vice versa. However, this is not the only way of energy exchange; transfer is also possible through thermal radiation quanta. At the same time, the basic principle is necessarily preserved: a quantum emitted by a less heated atom is not able to transfer energy to a hotter one. 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 matter between the materials in a different 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, as a rule, is room air and atmospheric precipitation. Accordingly, the three main methods of combating this phenomenon are external waterproofing of walls, the use of vapor protection and the calculation of moisture accumulation, which is necessarily carried out in parallel with predicting heat loss.

Differentiated calculation schemes

The simplest way to determine the size of the heat losses of a building is to sum up the values ​​of the 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 total area multiplied by the ratio of temperature difference to 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- 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 demo, let's answer the 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 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 shift 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. There is a one-story house of the correct rectangular shape with dimensions of 8.5x10 m and a ceiling height of 3.1 m, located in 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 ceiling is made on wooden beams: it is plastered on shingles from below, 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 a large number variables (density and type of soil, freezing depth, massiveness of the foundation, etc.), the calculation of heat losses is carried out according to a simplified method using the reduced resistance to heat transfer. 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. Let it not bother you total amount areas of zones, which more area building on 16 m 2, 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 overlap resistance to heat transfer is 0.66 m 2 ° C / W, and heat loss is 31.6 W s square meter, that is, 2.7 kW from the entire area of ​​\u200b\u200bthe 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 of windows, lack of protection at the junction of the roof to the Mauerlat and poor waterproofing of the walls from the foundation real heat loss 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, we 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 review 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 regulatory documents, as well as examples of calculations and all the necessary background information.

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.

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 a preliminary calculation of 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. At the same time, the basic principle is necessarily preserved: 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 clear and can be interpreted under the guise of various models. The main thing is to observe 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 matter between the materials in a different state of aggregation. Quantitatively, thermal resistance is expressed as the temperature difference divided by the heat flow rate:

Rt = (T2 - T1) / P

Where:

• Rt - thermal resistance of the section, K / W;
• T2 - temperature of the beginning of the section, K;
• T1 - temperature of the end of the section, K;
• P - 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.

Rt = l / (λ S)

Where:

• Rt - thermal resistance of the circuit section, K / W;
• l - length of the section of the thermal chain, m;
• λ - coefficient of thermal conductivity of the material, W/(m K);
• S - cross-sectional area of ​​the site, m2.

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. As the first model, we will take a 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 flows:

• 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 building 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, as a rule, is room air and atmospheric precipitation. Accordingly, the three main methods of combating this phenomenon are external waterproofing of walls, the use of vapor protection and the calculation of moisture accumulation, which must be carried out in parallel with predicting heat loss.

Differentiated calculation schemes

The simplest way to determine the size of the heat loss of a building is to sum up the values ​​of the heat flow through the structures that form this building. This technique fully takes into account the difference in the structure of different 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 various calculation methods are provided for this:

• For walls, heat leakage is quantitatively equal to the total area multiplied by the ratio of the temperature difference to the thermal resistance. At the same time, the orientation of the walls to the cardinal points is necessarily taken into account to take into account their heating in the daytime, as well as the ventilation of building structures.
• For floors, the methodology is the same, but the presence of an attic space and its mode of operation are taken into account. Also, a value 3–5 °С higher is taken as room temperature, 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 (∆T / Rt)

Where:

• Q - heat losses, W;
• S - wall area, m2;
• ΔT - temperature difference inside and outside the room, ° С;
• Rt - resistance to heat transfer, m2 °C / W.

Calculation example

Before moving on to the demo, let's answer the 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, it is rather difficult to take into account the presence of decorative finishes, interior and facade plaster, as well as the influence of all transient processes and other factors, 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 shift 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. There is a one-story house of a regular 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 m2; the windows used are double-glazed windows in a heat-saving profile with an average thermal resistance of 0.32 m2 °C/W.

The ceiling is made on wooden beams: it is plastered on shingles from below, 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.

Floor

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 m2. Do not be confused by the total area of ​​the zones, which is 16 m2 larger 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. Using heat transfer resistance values ​​of 2.1, 4.3, and 8.6 m2 °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 calculation results, the heat transfer resistance is equal to 1.13 m2 °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 m2, the total heat loss through the walls is 1.95 kWh. In this case, heat loss through the windows will be 1.05 kW.

Covering and roofing

The calculation of heat loss through the 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 floor to heat transfer is 0.66 m2 °C/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.

The total total heat loss according to the calculations is 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 of windows, the 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. Nevertheless, even basic thermal engineering studies help to determine whether the structures of the house under construction will comply with sanitary standards, at least in the first approximation.

Finally, we 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 review 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 thermal processes are explained in great detail, links are given to the necessary regulatory documents, as well as examples of calculations and all the necessary background information.published

If you have any questions on this topic, ask them to specialists and readers of our project.

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 that we pay for 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 you know, 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 in cold period winters.

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 construction 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 sound insulation of walls and 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 big amount advantages over non-insulated houses, and allows you to save your family budget.

Having carried out the external and internal insulation of the house, 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