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

Calculating heat loss at home

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

The calculation sequence is as follows:

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

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

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

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

Where V- layer thickness in m,

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

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

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

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

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

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

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

Deciding on the temperature regime

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

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

Selection of radiator power

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

Where TO— the required number of radiator sections

S– room area

P– power indicated in the product passport

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

K=16.67 rounded 17 sections

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

1 rib (60 cm) = 1 section.

Hydraulic calculation of the heating system

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

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

Calculating the volume of the heating system

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

Coolant volume in the radiator:

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

Coolant volume in 1 lm. pipes:

• ø15 (G ½") - 0.177 liters
• ø20 (G ¾") - 0.310 liters
• ø25 (G 1.0″) - 0.490 liters
• ø32 (G 1¼") - 0.800 liters
• ø15 (G 1½") - 1,250 liters
• ø15 (G 2.0″) - 1.960 liters

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

It is made with pipes made of different materials:

Steel

• They have a lot of weight.
• Requires proper skill special tools and installation equipment.
• Subject to corrosion
• May accumulate static electricity.

Copper

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

Plastic

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

Let's sum it up

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

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

Comfort is a fickle thing. Sub-zero temperatures arrive, you immediately feel chilly, and are 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 plan,” you can be left face to face with the quickly disappearing heat. The process is not visually noticeable, but is perfectly felt through woolen socks and large heating bills. The question remains: where did the “precious” heat go?

Natural heat loss is well hidden behind load-bearing structures or “well-made” insulation, where there should be no gaps by default. But is this true? Let's look at the issue of heat leaks for different elements designs.

Cold spots on the walls

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

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

• Masonry joints

The optimal masonry seam is 3mm. It is achieved more often with adhesive compositions of fine texture. When the volume of mortar 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.

Reinforced concrete has one of the highest thermal conductivity coefficients among building materials (1.28 - 1.61 W/(m*K)). 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.

You can insulate the lintel from the cold with continuous external insulation. And inside the house - by assembling a box from HA under the cornice. This creates an additional air layer for heat.

• Mounting holes and fasteners.

Connecting an air conditioner or TV antenna leaves gaps in the overall insulation. The through metal fasteners and the passage hole must be tightly sealed with insulation.

And if possible, do not move metal fasteners outside, fixing them inside the wall.

Insulated walls also have heat loss defects

Installation of damaged material (with chips, compression, etc.) leaves vulnerable areas for heat leaks. This is clearly visible when examining a house with a thermal imager. Bright spots indicate gaps in the external insulation.

During operation, it is important to monitor the general condition of the insulation. An error in choosing an adhesive (not a special one for thermal insulation, but a tile one) can cause cracks in the structure within 2 years. Yes, and the main insulation materials also have their disadvantages. 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.
• Foam plastic - has good insulating properties, but is easily susceptible 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 ensure a precise 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 is destroyed by UV rays. It is advisable to cover it plaster mixture— fastening frames through a layer of insulation violates the overall insulation.

Experience! Heat losses 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 “fast” road to destruction of the insulation inside.

Heat loss from the foundation

Concrete is the predominant material in foundation construction. His high thermal conductivity and direct contact with the ground result in up to 20% heat loss along the entire perimeter of the building. The foundation conducts heat particularly strongly from the basement and improperly installed heated floors on the first floor.

Heat loss is also increased by excess moisture that is not removed from the house. It destroys the foundation, creating openings for the cold. Many thermal insulation materials are also sensitive to humidity. For example, mineral wool, which is often transferred 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 on constantly moist soil. Its structure creates an air cushion and well compensates for ground pressure during freezing, but the constant presence of moisture minimizes beneficial properties expanded clay insulation. That is why the creation of working drainage is prerequisite long life of the foundation and heat retention.

This also includes in importance the waterproofing protection of the base, as well as a multi-layer blind area, not wide less than a meter. At columnar 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.

It is better to choose sheet materials for foundation insulation with a groove connection, and treat it with a special silicone compound. The tightness of the locks blocks access to the cold and guarantees continuous protection of the foundation. In this matter, seamless spraying of polyurethane foam has an undeniable advantage. In addition, the material is elastic and does not crack when the soil heaves.

For all types of foundations, you can use the developed insulation schemes. An exception may be a 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 is protected from the cold by a properly insulated floor of the first floor.

Attention! If the house has a basement and it often floods, then this must be taken into account when insulating the foundation. Since the insulation/insulator in this case will clog moisture in the foundation and destroy it. Accordingly, heat will be lost even more. The first thing that needs to be resolved is the flooding issue.

Vulnerable areas of the floor

An uninsulated ceiling transfers a significant portion of the heat to the foundation and walls. This is especially noticeable if the heated floor is installed incorrectly - heating element cools down faster, increasing the cost of heating the room.

To ensure that the heat from the floor goes into the room and not outside, you need to make sure that the installation follows all the rules. The main ones:

• 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 cracks must be eliminated and the wall surface leveled. The tape is fixed as tightly as possible to the wall, isolating heat transfer. When there are no air pockets, there are no heat leaks.
• Indent. There should be at least 10 cm from the outer wall to the heating circuit. If the heated floor is installed closer to the wall, then it begins to heat the street.
• Thickness. The characteristics of the required screen and insulation for underfloor heating are calculated individually, but it is better to add a 10-15% margin to the obtained figures.
• Finishing. The screed on top of the floor should not contain expanded clay (it insulates heat in the concrete). Optimal thickness screeds 3-7 cm. The presence of a plasticizer in the concrete mixture improves thermal conductivity, and therefore heat transfer into the room.

Serious insulation is important for any floor, and not necessarily with heating. Poor thermal insulation turns the floor into a large “radiator” for the ground. Is it worth heating it in winter?!

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

Junction points of building structures

The compounds disrupt the integrity of the materials. Therefore, corners, joints and abutments are so vulnerable to cold and moisture. The joints of concrete panels become damp first, and fungus and mold appear there. The temperature difference between the corner of the room (the junction of the structures) and the main wall can range from 5-6 degrees to subzero temperatures and condensation inside the corner.

Clue! At the sites of such connections, craftsmen recommend making an increased layer of insulation on the outside.

Heat often escapes through interfloor covering, when the slab is laid over the entire thickness of the wall and its edges face the street. Here the heat loss of both the first and second floors increases. Drafts form. Again, if there is a heated floor on the second floor, the external insulation should be designed for this.

Heat leaks through ventilation

Heat is removed from the room through equipped ventilation ducts, ensuring healthy air exchange. Ventilation that works “in reverse” draws in the cold from the street. This happens when there is a shortage of air in the room. For example, when the fan in the hood is turned on, it takes too much air from the room, due to which it begins to be drawn in from the street through other exhaust ducts (without filters and heating).

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

1. Recuperators are installed in the ventilation system. They return up to 90% of the heat to the house.
2. Supply valves are being installed. They "prepare" street air before the room - it is cleaned and warmed. The valves come with manual or automatic adjustment, which is based on the difference in temperature outside and inside the room.

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

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

Heat loss through windows and doors

A house loses up to 25% of heat through door and window openings. The weak points for doors are a leaky seal, which can be easily replaced with a new one, and thermal insulation that has become loose inside. It can be replaced by removing the casing.

Vulnerable spots for wooden and plastic doors similar to “cold bridges” in similar window designs. That's why general process Let's look at their example.

What indicates “window” heat loss:

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

The sashes may not fit tightly when the window is not adjusted and the rubber bands around the perimeter are worn out. The position of the valves can be adjusted independently, as well as the seal can be changed. It is better to completely replace it once every 2-3 years, and preferably with a “native” made seal. Seasonal cleaning and lubrication of rubber bands maintains their elasticity during temperature changes. Then the seal does not let the cold in 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 is 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” tape for windows.

Important! It is worth making sure that in the finishing of external slopes the insulation (foam plastic, 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 glass unit 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-glass 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 inner surface has a temperature twice as high as that of ordinary glass. i-glass retains thermal energy in the house by reflecting up to 90% of the heat back into the room.
• The use of silver-coated glass, which in 2-chamber double-glazed windows saves 40% more heat (compared to conventional glass).
• Selection 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 (possibly in the form of warm baseboards) or protective shutters at night. This is especially true for panoramic glazing and severe sub-zero temperatures.

Causes of heat leakage in the heating system

Heat loss also applies to heating, where heat leaks often occur for two reasons.

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

• Not all radiators warm up completely.

Following simple rules reduces heat loss and prevents the heating system from running 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 check whether all radiators are fully warmed up. The heating system can become clogged due to accumulated air or debris (delaminations, poor quality water). Once every 2-3 years the system must be completely flushed.

Note! When refilling, it is better to add anti-corrosion inhibitors to the water. This will support the metal elements of the system.

Heat loss through the roof

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

Cold attic space or residential attic are insulated equally tightly. The main heat losses occur at the junctions of materials, it does not matter whether it is insulation or structural elements. Thus, an often overlooked bridge of cold is the boundary of the walls with the transition to the roof. It is advisable to treat this area together with the Mauerlat.

Basic 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 it every 10 to 15 years. Over time, it cakes and begins to let in heat.
2. Ecowool, which has excellent “breathable” insulation properties, should not be located near hot springs - when heated, it smolders, leaving holes in the insulation.
3. When using polyurethane foam, it is necessary to arrange ventilation. The material is vapor proof and excess moisture It is better not to accumulate under the roof - other materials are damaged and a gap appears in the insulation.
4. Plates in multi-layer thermal insulation must be laid in a checkerboard pattern and must adhere closely to the elements.

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

Conclusion

It is useful to know the places of heat loss not only for arranging a house and living in comfortable conditions, but also so as not to overpay for heating. Proper insulation in practice it pays for itself in 5 years. The term is long. But we’re not building a house for two years.

Related videos

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

Heat loss through the outer shell

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

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

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

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

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

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

Methods for assessing heat loss at home

There are complex calculation methods that take into account various physical processes: convection exchange, radiation, but they are often unnecessary. Simplified formulas are usually used, and if necessary, you can add 1-5% to the result. 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 thermal energy 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.

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

Q = S x ΔT/R, where:

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

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

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

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

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

An example of calculating heat loss in a residential building

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

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

Calculation of heat transfer resistance of walls:

1. R = n/λ, where:

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

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

1. For brick:

Rkir = 0.67/0.38 = 1.76 sq.m °C/W.

1. For plaster coating:

Rpc = 0.03/0.35 = 0.086 sq.m °C/W;

1. Total value:

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

Calculation of the area of ​​external walls:

1. Total area of ​​external walls:

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

1. Area of ​​windows and doorway:

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

1. Adjusted wall area:

S2 = S – S1 = 264 – 11 = 253 sq.m.

Heat losses for walls will be determined:

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

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

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

To determine possible consumption 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, you get:

• per day:

6810.22 x 24 = 163.4 kWh;

• per month:

163.4 x 30 = 4903.4 kWh;

• for a heating season of 7 months:

4903.4 x 7 =34,323.5 kWh.

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

Heat losses due to ventilation

1. Find the air volume of the house:

10 x 12 x 6 = 720 m³;

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

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

M = 1, 205 x 720 = 867.4 kg.

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

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

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

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

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

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

Video

To prevent your home from becoming a bottomless pit for heating costs, we suggest you study basic directions 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

Various areas of physics have many similarities in the description of the phenomena that they study. The same is true in thermal engineering: the principles that describe thermodynamic systems clearly echo the fundamentals of electromagnetism, hydrodynamics and classical mechanics. After all, we are talking about describing the same world, so it is not surprising that models of physical processes are characterized by some general 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 more than a measure of the intensity of vibrations of the elementary particles of which this body consists. Obviously, when two particles collide, the one with a 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. In this case, 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.

The good thing about thermodynamics is that the processes occurring in it are absolutely clear and can be interpreted as various models. The main thing is to comply with basic postulates, such as the law of energy transfer and thermodynamic equilibrium. So if your presentation follows these rules, you will easily understand the technique thermotechnical calculations from and to.

Concept of heat transfer resistance

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

A 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 between the materials there is at least a minimal layer of substance in a different state of aggregation. Quantitatively, thermal resistance is expressed as the temperature difference divided by the heat flow rate:

R t = (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—heat flow, W.

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

R t = l / (λ·S)

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

Factors influencing heat loss

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

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

• through the thickness of concrete
• through steel rods
• from steel bars to concrete

The last heat flow model is the most interesting. Since the steel rod heats up faster, closer to the outside of the wall there will be a temperature difference between the two materials. Thus, the steel not only “pumps” heat outward on its own, it also increases the thermal conductivity of the adjacent masses of concrete.

In porous media, thermal processes proceed in a similar way. Almost everything building materials consist of a branched web of solid matter, the space between which is filled with air. Thus, the main conductor of heat is a solid, dense material, but due to its 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 enclosing structure as a whole.

The third factor affecting thermal conductivity is 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 precipitation. Accordingly, the three main methods of combating this phenomenon are external waterproofing of walls, the use of vapor barrier and calculation of moisture accumulation, which must be carried out in parallel with the prediction of heat loss.

Differentiated calculation schemes

The simplest way to determine the amount of heat loss from a building is to sum up the heat flow through the structures that make up the building. This technique fully takes into account the difference in structure various materials, as well as the specifics of the heat flow through them and at the junctions of one plane to another. This dichotomous approach greatly simplifies the task, because different enclosing structures can differ significantly in the design of thermal protection systems. Accordingly, with 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 thermal resistance. In this case, the orientation of the walls to the cardinal points must be taken into account to take into account their heating in daytime, as well as airflow building structures.
• For floors, the technique is the same, but the presence of an attic space and its operating mode are taken into account. Also, the room temperature is taken to be 3-5 °C higher, the calculated humidity is also increased by 5-10%.
• Heat loss through the floor is calculated zonally, describing the zones around the perimeter of the building. This is due to the fact that the temperature of the soil under the floor is higher at the center of the building compared to the foundation part.
• The heat flow through the glazing is determined by the passport data of the windows; you also need to take into account the type of connection of the windows to the walls and the depth of the slopes.

Q = S (Δ T / R t)

• Q—heat losses, W;
• S—wall area, m2;
• ΔT—temperature difference inside and outside the room, ° C;
• R t - heat transfer resistance, m 2 °C/W.

Calculation example

Before moving on to the demo example, we will answer the last question: how to correctly calculate the integral thermal resistance of complex multilayer structures? This, of course, can be done manually; fortunately, not many types are used in modern construction. load-bearing foundations and insulation systems. However, take into account the presence decorative finishing, interior and facade plaster, as well as the influence of all transient processes and other factors is quite complex, it is better to use automated calculations. One of the best online resources for such tasks is smartcalc.ru, which additionally creates a diagram of dew point displacement depending on climatic conditions.

For example, let’s take an arbitrary building, after studying 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 rectangular shape dimensions 8.5x10 m and ceiling height 3.1 m, located in the Leningrad region. The house has an uninsulated floor on the ground with boards on joists with air gap, the floor height is 0.15 m higher than the ground level 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 “fur coat” plaster up to 50 mm thick. The total glazing area is 9.5 m2; the windows are double-chamber 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: the bottom is plastered over shingles, filled with blast furnace slag and covered with a clay screed on top; above the ceiling there is a cold-type attic. The task of calculating heat loss is to form a thermal protection system for walls.

The first step is to determine the heat loss through the floor. Since their share in the total heat outflow is the smallest, and also because large number variables (density and type of soil, freezing depth, massiveness of the foundation, etc.), the calculation of heat loss is carried out 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 stripes 2 meters wide. For each zone, its own value of the reduced heat transfer resistance is taken. In our case, there are three zones with an area of ​​74, 26 and 1 m2. Don’t 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 stripes of the first zone in the corners, where heat loss is significantly higher compared to areas along the walls. Using 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 data about the terrain, 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 turns out to be 1.13 m 2 °C/W, and the heat flow through the wall is 18.48 W per square meter. With a total wall area (minus glazing) of 105.2 m2, 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.

Ceiling and roofing

Calculation of heat loss through attic floor You can also do it in the online calculator by selecting the desired type of enclosing structures. As a result, the heat transfer resistance of the ceiling is 0.66 m 2 °C/W, and the heat loss is 31.6 W per square meter, that is, 2.7 kW from the entire area of ​​the enclosing structure.

Total total heat loss according to calculations is 7.2 kWh. Given the rather low quality of the building's construction, 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, airflow, 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 and the mauerlat and poor waterproofing of the walls from the foundation, real heat losses can be 2 or even 3 times greater than calculated. However, even basic thermal engineering studies help determine whether the designs of the house being built will comply sanitary standards at least to a first approximation.

Finally, let's give one important recommendation: If you really want to gain a complete understanding of the thermal physics of a particular building, you must use an understanding of the principles described in this overview and specialist literature. For example, Elena Malyavina’s reference guide “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 to the necessary regulatory documents are given, as well as examples of calculations and all the necessary background information are given.

Conventionally, heat loss in a private home can be divided into two groups:

• Natural - heat loss through walls, windows or the roof of a building. These are losses that cannot be completely eliminated, but they can be minimized.
• “Heat leaks” are additional heat losses that can most often be avoided. These are various visually invisible errors: hidden defects, installation errors, etc., which cannot be detected visually. A thermal imager is used for this.

Below we present to your attention 15 examples of such “leaks”. These are real problems that are most often encountered in private homes. You will see what problems may be present in your home and what you should pay attention to.

Poor quality wall insulation

Insulation does not work as effectively as it could. The thermogram shows that the temperature on the wall surface is distributed unevenly. That is, some areas of the wall heat up more than others (than brighter color, the higher the temperature). This means that the heat loss is no greater, which is not correct for an insulated wall.

In this case, the bright areas are an example of ineffective insulation. It is likely that the foam in these places is damaged, poorly installed or missing altogether. Therefore, after insulating a building, it is important to make sure that the work is done efficiently and that the insulation works effectively.

Poor roof insulation

Joint between wooden beam and mineral wool is not compacted enough. This causes the insulation to not work effectively and causes additional heat loss through the roof that could be avoided.

The radiator is clogged and gives off little heat

One of the reasons why the house is cold is that some sections of the radiator do not heat up. This can be caused by several reasons: construction debris, air accumulation or manufacturing defects. But the result is the same - the radiator operates at half its heating capacity and does not warm the room enough.

The radiator “warms” the street

Another example of inefficient radiator operation.

There is a radiator installed inside the room, which heats up the wall very much. As a result, part of the heat it generates goes outside. In fact, the heat is used to warm the street.

Laying heated floors close to the wall

The underfloor heating pipe is laid close to the outer wall. The coolant in the system is cooled more intensively and has to be heated more often. The result is an increase in heating costs.

Cold influx through cracks in windows

There are often cracks in windows that appear due to:

• insufficient pressing of the window to the window frame;
• wear of rubber seals;
• poor-quality window installation.

Cold air constantly enters the room through the cracks, causing drafts that are harmful to health and increasing heat loss in the building.

Cold influx through cracks in doors

Also, cracks appear in balconies and entrance doors.

Bridges of cold

“Cold bridges” are areas of a building with lower thermal resistance compared to other areas. That is, they transmit more heat. For example, these are corners, concrete lintels over windows, junctions of building structures, and so on.

Why are cold bridges harmful?

• Increases heat loss in the building. Some bridges lose more heat, others less. It all depends on the characteristics of the building.
• Under certain conditions, condensation forms in them and fungus appears. Such potentially dangerous areas must be prevented and eliminated in advance.

Cooling the room through ventilation

Ventilation works in reverse. Instead of removing air from the room to the outside, cold street air is drawn into the room from the street. This, as in the example with windows, provides drafts and cools the room. In the example given, the temperature of the air that enters the room is -2.5 degrees, at a room temperature of ~20-22 degrees.

Cold influx through the sunroof

And in this case, the cold enters the room through the hatch into the attic.

Cold flow through the air conditioner mounting hole

Cold flow into the room through the air conditioner mounting hole.

Heat loss through walls

The thermogram shows “heat bridges” associated with the use of materials with weaker resistance to heat transfer during the construction of the wall.

Heat loss through the foundation

Often when insulating the wall of a building, they forget about another important area - the foundation. Heat loss also occurs through the building's foundation, especially if the building has a basement or a heated floor is installed inside.

Cold wall due to masonry joints

Masonry joints between bricks are numerous cold bridges and increase heat loss through the walls. The example above shows that the difference between the minimum temperature (masonry joint) and maximum (brick) is almost 2 degrees. Thermal resistance the walls have been lowered.

Air leaks

Cold bridge and air leak under the ceiling. It occurs due to insufficient sealing and insulation of the joints between the roof, wall and floor slab. As a result, the room is additionally cooled and drafts appear.

Conclusion

All this typical mistakes, which are found in most private homes. Many of them can be easily eliminated and can significantly improve the energy status of the building.

Let's list them again:

1. Heat leaks through walls;
2. Ineffective operation of thermal insulation of walls and roofs - hidden defects, poor-quality installation, damage, etc.;
3. Cold inflows through air conditioner mounting holes, cracks in windows and doors, ventilation;
4. Ineffective operation of radiators;
5. Bridges of cold;
6. The influence of masonry joints.

15 hidden heat leaks in a private home that you didn't know about