## Recommendation points

- Physics of heat engineering processes
- Heat transfer resistance
- Factors affecting heat loss
- Differentiated calculation schemes
- Calculation example

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

## Physics of heat engineering 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 resonate with the basics of electromagnetism, hydrodynamics and classical mechanics. After all, we are talking about describing the same world, so it is not surprising that the 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 more than a measure of the intensity of vibrations of the elementary particles that make up this body. 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; transmission is also possible by means of quanta of thermal radiation. 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 simply reflects from it and either disappears without a trace, or transfers its energy to another atom with less energy.

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

## Heat transfer resistance

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 oscillations.

Comparison of energy efficiency of different building materials

Thermal resistance is a quantity inversely proportional to thermal conductivity. 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 minimum interlayer of matter in a different aggregate state. Thermal resistance is quantitatively expressed as the temperature difference divided by the heat flow rate:

R_{t}= (T_{2}– T_{1}) / PWhere:

- R
_{t}– thermal resistance of the site, K / W;- T
_{2}– temperature of the beginning of the section, K;- T
_{1}– temperature of the end of the section, K;- P – heat flux, W.
In the context of calculating heat loss, thermal resistance plays a decisive role. Any enclosing structure can be represented as a plane-parallel obstacle to the heat flow path. Its total thermal resistance is the sum of the resistances of each layer, while all partitions are added to a spatial structure, which is, in fact, a building.

R_{t}= l / (? S)Where:

- R
_{t}– thermal resistance of the circuit section, K / W;- l is the length of the heat circuit section, m;
- ? – coefficient of thermal conductivity of the material, W / (m · K);
- S – cross-sectional area of the site, m
^{2}.## Factors affecting heat loss

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

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

- through the thickness of concrete
- through steel rods
- from steel bars to concrete
Heat loss through cold bridges in concrete

The last heat flow model is most interesting. Since the steel bar heats up faster, there will be a temperature difference between the two materials closer to the outside 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 solids, the space between which is filled with air. Thus, the main conductor of heat is a solid, dense material, but due to the complex structure, the path along which the heat propagates 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..

Reducing heat loss and shifting the dew point into the insulation with external wall insulation

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 atmospheric precipitation. Accordingly, the three main methods of dealing with this phenomenon are external waterproofing of walls, the use of steam 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 establish the amount of heat loss in a building is to sum the values of the heat flow through the structures that constitute the building. This technique fully takes into account the difference in the structure of various materials, as well as the specifics of the heat flow through them and in the nodes of the abutment 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, with a separate study, it is easier to determine the amount of heat loss, because for this, various calculation methods are provided:

- For walls, heat leaks are quantitatively equal to the total area multiplied by the ratio of the temperature difference to the thermal resistance. In this case, the orientation of the walls to the cardinal points must be taken into account to take into account their heating in the daytime, as well as the blow-through of building structures.
- For floors, the technique is the same, but it takes into account the presence of the attic space and its mode of operation. Also, the room temperature is taken as a value 3-5 ° C higher, the calculated humidity is also increased by 5-10%.
- Heat loss through the floor is calculated zonally, 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 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 abutment of the windows to the walls and the depth of the slopes.

Q = S (?T / R_{t})Where:

- Q — heat loss, W;
- S – wall area, m
^{2};- ?T – temperature difference inside and outside the room, ° С;
- R
_{t }– resistance to heat transfer, m^{2}° С / W.## Calculation example

Before moving on to the demo example, 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 there are not many types of load-bearing bases and insulation systems 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 transients and other factors; it is better to use automated calculations. One of the best network resources for such tasks is smartcalc.ru, which additionally draws a dew point shift diagram 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 required for the calculation. There is a one-storey house of a regular rectangular shape with dimensions of 8.5×10 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. Wall material – 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. Total glazing area – 9.5 m

^{2}, a double-glazed unit in a heat-saving profile with an average thermal resistance of 0.32 m was used as windows^{2}° С / W. The overlap was made on wooden beams: the bottom was plastered along the shingles, filled with blast furnace slag and covered with a clay screed on top, above the ceiling there was a cold-type attic. The task of calculating heat loss is the formation of a system of thermal protection of walls.

FloorThe first step is to determine the heat loss through the floor. 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.), the calculation of heat loss is carried out according to a simplified method using the reduced heat transfer resistance. Along the perimeter of the building, starting from the line of contact with the ground surface, four zones are described – encircling stripes 2 meters wide. For each of the zones, its own value of the reduced resistance to heat transfer is taken. In our case, there are three zones with an area of 74, 26 and 1 m

^{2}. Do not be confused by the total sum of the areas of zones, which is more than the area of the building by 16 m^{2}, the reason for this is the double recalculation of the intersecting strips of the first zone in the corners, where the heat loss is much higher compared to the sections along the walls. Applying heat transfer resistance values of 2.1, 4.3 and 8.6 m^{2}° С / W for zones one through three, we determine the heat flux through each zone: 1.23, 0.21 and 0.05 kW, respectively.

WallsUsing 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 heat transfer resistance turns out to be equal to 1.13 m

^{2}° С / 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}total heat loss through the walls is 1.95 kW / h. In this case, heat loss through the windows will be 1.05 kW.

Overlap and roofThe 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 floor resistance to heat transfer is 0.66 m

^{2}° С / W, and 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. With a sufficiently low quality of building structures, this indicator is obviously much lower than the real one. In fact, such a calculation is idealized, it does not take into account special coefficients, airflow, 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 abutment 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 higher than the calculated one. Nevertheless, even basic heat engineering studies help to determine whether the structures of a house under construction will meet sanitary standards at least in the first approximation..

Heat loss at home

Finally, we will 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 special literature. For example, Elena Malyavina’s reference book “Heat loss of a building” can be a very good help in this matter, where the specificity of heat engineering processes is explained in great detail, links to the necessary regulatory documents are given, as well as examples of calculations and all the necessary reference information.