Creating comfortable conditions for living or working is the primary task of construction. A significant part of the territory of our country is located in northern latitudes with a cold climate. Therefore, maintaining a comfortable temperature in buildings is always important. With the growth of energy tariffs, the reduction of energy consumption for heating comes to the fore.
Climatic characteristics
The choice of wall and roof structure depends primarily on the climatic conditions of the construction area. To determine them, it is necessary to refer to SP131.13330.2012 "Construction climatology". The following quantities are used in the calculations:
- the temperature of the coldest five-day period with a security of 0.92, denoted by Tn;
- average temperature, denoted by Tot;
- duration, denoted by ZOT.
On the example for Murmansk, the values have the following values:
- Тн=-30 deg;
- Tot=-3.4 deg;
- ZOT=275 days.
In addition, it is necessary to set the design temperature inside the room TV, it is determined in accordance with GOST 30494-2011. For housing, you can take TV=20 degrees.
20 - (-3, 4)) x 275=6435.
Key indicators
For the right choice of building envelope materials, it is necessary to determine what thermal characteristics they should have. The ability of a substance to conduct heat is characterized by its thermal conductivity, denoted by the Greek letter l (lambda) and is measured in W / (m x deg.). The ability of a structure to retain heat is characterized by its resistance to heat transfer R and is equal to the ratio of thickness to thermal conductivity: R=d/l.
If the structure consists of several layers, the resistance is calculated for each layer and then summed up.
Heat transfer resistance is the main indicator of outdoor construction. Its value must exceed the standard value. When performing a thermal engineering calculation of the building envelope, we must determine the economically justified composition of the walls and roof.
Thermal conductivity values
Insulation qualitydetermined primarily by thermal conductivity. Each certified material undergoes laboratory tests, as a result of which this value is determined for operating conditions "A" or "B". For our country, most regions correspond to the operating conditions "B". When performing a heat engineering calculation of the enclosing structures of a house, this value should be used. The thermal conductivity values are indicated on the label or in the material passport, but if they are not available, you can use the reference values \u200b\u200bfrom the Code of Practice. The values for the most popular materials are listed below:
- Ordinary brickwork - 0.81 W(m x deg.).
- Silicate brick masonry - 0.87 W(m x deg.).
- Gas and foam concrete (density 800) - 0.37 W(m x deg.).
- Softwood - 0.18 W(m x deg.).
- Extruded Styrofoam - 0.032 W(m x deg.).
- Mineral wool boards (density 180) - 0.048 W(m x deg.).
Regular value of heat transfer resistance
The calculated value of heat transfer resistance should not be less than the base value. The base value is determined according to Table 3 SP50.13330.2012 "Thermal protection of buildings". The table defines coefficients for calculating the basic values of heat transfer resistance for all enclosing structures and types of buildings. Continuing the started thermotechnical calculation of enclosing structures, an example of calculation can be presented as follows:
- Rsten=0.00035x6435 + 1.4=3.65 (m x deg/W).
- Rpokr=0, 0005х6435 +2, 2=5, 41 (m x deg/W).
- Rchard=0.00045x6435 + 1.9=4.79 (m x deg/W).
- Rockna=0.00005x6435 + 0.3=0.62 (m x deg/W).
Thermotechnical calculation of the external enclosing structure is performed for all structures that close the "warm" contour - the floor on the ground or the floor of the technical underground, the outer walls (including windows and doors), the combined cover or the floor of the unheated attic. Also, the calculation must be performed for internal structures, if the temperature difference in adjacent rooms is more than 8 degrees.
Thermotechnical calculation of walls
Most walls and ceilings are multi-layered and heterogeneous in their design. The thermotechnical calculation of enclosing structures of a multilayer structure is as follows:
R=d1/l1 +d2/l2 +dn/ln, where n is the parameters of the n-th layer.
If we consider a brick plastered wall, we get the following construction:
- outer layer of plaster 3 cm thick, thermal conductivity 0.93 W(m x deg.);
- solid clay brick masonry 64 cm, thermal conductivity 0.81 W(m x deg.);
- Inner layer of plaster 3 cm thick, thermal conductivity 0.93 W(m x deg.).
The formula for thermal engineering calculation of enclosing structures is as follows:
R=0.03/0.93 + 0.64/0.81 + 0.03/0.93=0.85(m x deg/W).
The resulting value is significantly less than the previously determined base resistance valueheat transfer of the walls of a residential building in Murmansk 3, 65 (m x deg / W). The wall does not meet the regulatory requirements and needs to be insulated. For wall insulation, we use mineral wool boards with a thickness of 150 mm and a thermal conductivity of 0.048 W (m x deg.).
Having selected the insulation system, it is necessary to perform a verification thermal calculation of the enclosing structures. An example calculation is shown below:
R=0.15/0.048 + 0.03/0.93 + 0.64/0.81 + 0.03/0.93=3.97(m x deg/W).
The resulting calculated value is greater than the base value - 3.65 (m x deg / W), the insulated wall meets the requirements of the standards.
Calculation of overlaps and combined coverings is performed similarly.
Heat engineering calculation of floors in contact with the ground
Often in private houses or public buildings the floors of the first floors are made on the ground. The resistance to heat transfer of such floors is not standardized, but at a minimum the design of the floors must not allow dew to fall. The calculation of structures in contact with the ground is carried out as follows: the floors are divided into strips (zones) 2 meters wide, starting from the outer boundary. Up to three such zones are allocated, the remaining area belongs to the fourth zone. If the floor design does not provide for effective insulation, then the heat transfer resistance of the zones is taken as follows:
- 1 zone – 2, 1 (m x deg/W);
- 2 zone – 4, 3 (m x deg/W);
- 3 zone – 8, 6 (m x deg/W);
- 4 zone – 14, 3 (m x deg/W).
It is easy to see that the farther the floor is from the outer wall, the higher its resistance to heat transfer. Therefore, they are often limited to warming the perimeter of the floor. At the same time, the heat transfer resistance of the insulated structure is added to the heat transfer resistance of the zone. An example of the calculation of floors on the ground will be considered below. Let's take the floor area 10 x 10, equal to 100 square meters.
- The area of 1 zone will be 64 square meters.
- The area of Zone 2 will be 32 square meters.
- The area of Zone 3 will be 4 square meters.
Average floor heat transfer resistance on the ground:Rfloor=100 / (64/2, 1 + 32/4, 3 + 4/8, 6)=2.6 (m x deg/ Tue).
Having completed the insulation of the floor perimeter with a polystyrene foam plate 5 cm thick, a strip 1 meter wide, we obtain the average value of heat transfer resistance:
Рpol=100 / (32/2, 1 + 32/(2, 1+0, 05/0, 032) + 32/4, 3 + 4/8, 6)=4, 09 (m x deg/W).
It is important to note that not only floors are calculated in this way, but also wall structures in contact with the ground (walls of a recessed floor, warm basement).
Thermotechnical calculation of doors
The base value of the heat transfer resistance of entrance doors is calculated somewhat differently. To calculate it, you will first need to calculate the heat transfer resistance of the wall according to the sanitary and hygienic criterion (non-fallingdew): Rst=(Tv - Tn) / (DTn x av).
Here ДТн - the temperature difference between the inner surface of the wall and the air temperature in the room, is determined according to the Code of Rules and for housing is 4.0.
av - the heat transfer coefficient of the inner surface of the wall, according to the joint venture is 8, 7. The base value of the doors is taken equal to 0, 6xRst.
For the selected door design, it is required to perform a verification thermotechnical calculation of enclosing structures. Entrance door calculation example:
Rdv=0.6 x (20-(-30))/(4 x 8.7)=0.86 (m x deg/W).
This design value will correspond to a door insulated with a mineral wool board 5 cm thick.
Complex requirements
Calculations of walls, floors or coverings are performed to check the element-by-element requirements of the regulations. The set of rules also establishes a complete requirement that characterizes the quality of insulation of all enclosing structures as a whole. This value is called "specific heat-shielding characteristic". Not a single thermotechnical calculation of enclosing structures can do without its verification. An example of a JV calculation is shown below.
| Design name | Square | R | A/R |
| Walls | 83 | 3, 65 | 22, 73 |
| Covering | 100 | 5, 41 | 18, 48 |
| Basement ceiling | 100 | 4, 79 | 20, 87 |
| Windows | 15 | 0, 62 | 24, 19 |
| Doors | 2 | 0, 8 | 2, 5 |
| Amount | 88, 77 |
Kob \u003d 88, 77 / 250 \u003d 0.35, which is less than the normalized value of 0.52. In this case, the area and volume are taken for a house measuring 10 x 10 x 2.5 m. Heat transfer resistances are equal to the base values.
The normalized value is determined in accordance with the joint venture, depending on the heated volume of the house.
In addition to the complex requirement, to draw up an energy passport, they also perform a heat engineering calculation of enclosing structures, an example of a passport is given in the Appendix to SP50.13330.2012.
Uniformity coefficient
All above calculations are applicable for homogeneous structures. Which is quite rare in practice. To take into account inhomogeneities that reduce the resistance to heat transfer, a correction factor for thermal uniformity, r, is introduced. It takes into account the change in heat transfer resistance introduced by window and door openings, external corners, inhomogeneous inclusions (for example, lintels, beams, reinforcing belts), cold bridges, etc.
The calculation of this coefficient is quite complicated, so in a simplified form, you can use approximate values from the reference literature. For example, for brickwork - 0.9, three-layer panels - 0.7.
Effective insulation
When choosing a home insulation system, it is easy to make sure that it is almost impossible to meet modern thermal protection requirements without the use of effective insulation. So, if you use a traditional clay brick, you will need masonry several meters thick, which is not economically feasible. At the same time, the low thermal conductivity of modern insulation based on expanded polystyrene or stone wool allows you to limit yourself to thicknesses of 10-20 cm.
For example, to achieve a base heat transfer resistance value of 3.65 (m x deg/W), you would need:
- 3m thick brick wall;
- laying of foam concrete blocks 1, 4 m;
- mineral wool insulation 0.18 m.
