Thermal calculation of the floor online. Example of thermal engineering calculation of an external wall

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A long time ago, buildings and structures were built without thinking about what heat-conducting qualities the enclosing structures have. In other words, the walls were simply made thick. And if you ever happened to be in old merchant houses, then you might notice that the outer walls of these houses are made of ceramic brick, the thickness of which is about 1.5 meters. This thickness brick wall ensured and still provides quite a comfortable stay of people in these houses even in the most severe frosts.

At present, everything has changed. And now it is not economically profitable to make the walls so thick. Therefore, materials have been invented that can reduce it. One of them: heaters and gas silicate blocks. Thanks to these materials, for example, the thickness brickwork can be reduced up to 250 mm.

Now walls and ceilings are most often made of 2 or 3 layers, one layer of which is a material with good thermal insulation properties. And in order to determine the optimal thickness of this material, a thermal calculation is carried out and the dew point is determined.

How the calculation is made to determine the dew point, you can find on the next page. Here, the heat engineering calculation will be considered using an example.

Required regulatory documents

For the calculation, you will need two SNiPs, one joint venture, one GOST and one allowance:

SNiP 23-02-2003 (SP 50.13330.2012). " Thermal protection buildings". Updated version of 2012.
SNiP 23-01-99* (SP 131.13330.2012). "Construction climatology". Updated edition from 2012.
SP 23-101-2004. "Design of thermal protection of buildings".
GOST 30494-96 (replaced by GOST 30494-2011 since 2011). "Residential and public buildings. Indoor microclimate parameters".
Benefit. E.G. Malyavin "Heat loss of the building. Reference guide".

Calculated parameters

In the process of performing a heat engineering calculation, the following are determined:

thermal characteristics building materials enclosing structures;
reduced heat transfer resistance;
compliance of this reduced resistance with the standard value.

Example. Thermal engineering calculation of a three-layer wall without an air gap

Initial data

1. The climate of the area and the microclimate of the room

Construction area: Nizhny Novgorod.

Purpose of the building: residential.

The calculated relative humidity of the indoor air from the condition of no condensation on the inner surfaces of the outer fences is - 55% (SNiP 23-02-2003 p.4.3. Table 1 for normal humidity conditions).

The optimum air temperature in the living room in cold period years t int = 20°C (GOST 30494-96 Table 1).

Estimated outdoor temperature text, determined by the temperature of the coldest five-day period with a security of 0.92 = -31 ° С (SNiP 23-01-99 table 1 column 5);

The duration of the heating period with an average daily outdoor temperature of 8°С is equal to z ht = 215 days (SNiP 23-01-99 table 1 column 11);

The average outdoor temperature during the heating period t ht = -4.1 ° C (SNiP 23-01-99 table. 1 column 12).

2. Wall construction

The wall consists of the following layers:

Brick decorative (besser) 90 mm thick;
insulation (mineral wool board), in the figure its thickness is indicated by the sign "X", since it will be found in the calculation process;
silicate brick 250 mm thick;
plaster (complex mortar), an additional layer to obtain a more objective picture, since its influence is minimal, but there is.

3. Thermophysical characteristics of materials

The values of the characteristics of the materials are summarized in the table.

Note (*): These characteristics can also be found from manufacturers of thermal insulation materials.

Calculation

4. Determining the thickness of the insulation

To calculate the thickness of the heat-insulating layer, it is necessary to determine the heat transfer resistance of the enclosing structure based on the requirements sanitary norms and energy saving.

4.1. Determination of the norm of thermal protection according to the condition of energy saving

Determination of degree-days of the heating period according to clause 5.3 of SNiP 23-02-2003:

D d = ( t int - tht) z ht = (20 + 4.1)215 = 5182°С×day

Note: also degree-days have the designation - GSOP.

The normative value of the reduced resistance to heat transfer should be taken not less than the normalized values determined by SNIP 23-02-2003 (Table 4) depending on the degree-day of the construction area:

R req \u003d a × D d + b \u003d 0.00035 × 5182 + 1.4 \u003d 3.214m 2 × °С/W,

where: Dd - degree-day of the heating period in Nizhny Novgorod,

a and b - coefficients taken according to table 4 (if SNiP 23-02-2003) or according to table 3 (if SP 50.13330.2012) for walls residential building(column 3).

4.1. Determination of the norm of thermal protection according to the condition of sanitation

In our case, it is considered as an example, since this indicator is calculated for industrial buildings with excess sensible heat of more than 23 W / m 3 and buildings intended for seasonal operation (in autumn or spring), as well as buildings with an estimated internal air temperature of 12 ° С and below the given resistance to heat transfer of enclosing structures (with the exception of translucent ones).

Determination of the normative (maximum allowable) resistance to heat transfer according to the condition of sanitation (formula 3 SNiP 23-02-2003):

where: n \u003d 1 - coefficient taken from table 6 for outer wall;

t int = 20°C - value from the initial data;

t ext \u003d -31 ° С - value from the initial data;

Δt n \u003d 4 ° С - normalized temperature difference between the temperature of the indoor air and the temperature of the inner surface of the building envelope, is taken according to table 5 in this case for the outer walls of residential buildings;

α int \u003d 8.7 W / (m 2 × ° С) - heat transfer coefficient of the inner surface of the building envelope, taken according to table 7 for external walls.

4.3. Thermal protection rate

From the above calculations for the required heat transfer resistance, we choose R req from the condition of energy saving and denote it now R tr0 \u003d 3.214 m 2 × °С/W .

5. Determining the thickness of the insulation

For each layer of a given wall, it is necessary to calculate the thermal resistance using the formula:

where: δi - layer thickness, mm;

λ i - calculated coefficient of thermal conductivity of the layer material W/(m × °С).

1 layer ( decorative brick): R 1 \u003d 0.09 / 0.96 \u003d 0.094 m 2 × °С/W .

3rd layer (silicate brick): R 3 = 0.25 / 0.87 = 0.287 m 2 × °С/W .

4th layer (plaster): R 4 = 0.02 / 0.87 = 0.023 m 2 × °С/W .

Determination of the minimum allowable (required) thermal resistance thermal insulation material(formula 5.6 E.G. Malyavin "Heat loss of the building. Reference manual"):

where: R int = 1/α int = 1/8.7 - resistance to heat transfer on the inner surface;

R ext \u003d 1/α ext \u003d 1/23 - resistance to heat transfer on the outer surface, α ext is taken according to table 14 for external walls;

ΣR i = 0.094 + 0.287 + 0.023 - the sum of thermal resistances of all layers of the wall without a layer of insulation, determined taking into account the coefficients of thermal conductivity of materials taken in column A or B (columns 8 and 9 of Table D1 SP 23-101-2004) in accordance with the humidity conditions of the wall, m 2 ° С /W

The thickness of the insulation is (formula 5.7):

where: λ ut - coefficient of thermal conductivity of the insulation material, W / (m ° C).

Determination of the thermal resistance of the wall from the condition that the total thickness of the insulation will be 250 mm (formula 5.8):

where: ΣR t, i - the sum of thermal resistances of all layers of the fence, including the insulation layer, of the accepted structural thickness, m 2 ·°С / W.

From the result obtained, it can be concluded that

R 0 \u003d 3.503m 2 × °С/W> R tr0 = 3.214m 2 × °С/W→ therefore, the thickness of the insulation is selected right.

Influence of the air gap

In the case when in a three-layer masonry, mineral wool, glass wool or other slab insulation, it is necessary to install an air ventilated layer between the outer masonry and the insulation. The thickness of this layer should be at least 10 mm, and preferably 20-40 mm. It is necessary in order to drain the insulation, which gets wet from condensate.

This air layer is not a closed space, therefore, if it is present in the calculation, it is necessary to take into account the requirements of clause 9.1.2 of SP 23-101-2004, namely:

a) structural layers located between the air gap and outer surface(in our case, this is a decorative brick (besser)), they are not taken into account in the heat engineering calculation;

b) on the surface of the structure facing towards the layer ventilated by the outside air, the heat transfer coefficient α ext = 10.8 W/(m°C) should be taken.

Note: the influence of the air gap is taken into account, for example, in the heat engineering calculation of plastic double-glazed windows.

Creation comfortable conditions for living or labor activity is the primary goal of construction. A significant part of the territory of our country is located in northern latitudes with a cold climate. Therefore, maintaining comfortable temperature in buildings is always relevant. With the growth of energy tariffs, the reduction of energy consumption for heating comes to the fore.

Climate characteristics

The choice of wall and roof construction 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 is denoted by Tn;
average temperature, denoted by Tot;
duration, denoted ZOT.

On the example for Murmansk, the values have the following values:

Tn=-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 \u003d 20 degrees.

To perform a heat engineering calculation of enclosing structures, pre-calculate the value of GSOP (degree-day of the heating period):
GSOP = (Tv - Tot) x ZOT.
In our example, GSOP \u003d (20 - (-3.4)) x 275 \u003d 6435.

Basic indicators

For right choice materials of enclosing structures, 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 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 outdoor structure. Its value must exceed normative 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

The quality of thermal insulation is determined primarily by thermal conductivity. Each certified material passes laboratory research, 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 given 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.).
Wood conifers- 0.18 W (m x deg.).
Extruded polystyrene foam - 0.032 W (m x deg.).
Mineral wool slabs (density 180) - 0.048 W (m x deg.).

Standard value of resistance to heat transfer

The calculated value of the heat transfer resistance should not be less than base value. The base value is determined according to Table 3 SP50.13330.2012 "buildings". The table defines the coefficients for calculating the basic values of heat transfer resistance for all enclosing structures and types of buildings. Continuing the started thermal engineering calculation of enclosing structures, an example of calculation can be presented as follows:

Рsten \u003d 0.00035x6435 + 1.4 \u003d 3.65 (m x deg / W).
Рpocr \u003d 0.0005x6435 + 2.2 \u003d 5.41 (m x deg / W).
Rcherd \u003d 0.00045x6435 + 1.9 \u003d 4.79 (m x deg / W).
Rockna \u003d 0.00005x6435 + 0.3 \u003d x deg / W).

The 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 carried out for internal structures if the temperature difference in adjacent rooms is more than 8 degrees.

Thermal engineering calculation of walls

Most walls and ceilings are multi-layered and heterogeneous in their design. The thermotechnical calculation of the enclosing structures of a multilayer structure is as follows:
R= d1/l1 +d2/l2 +dn/ln,
where n are the parameters of the nth layer.

If we consider a brick plastered wall, we get the following design:

outer layer of plaster 3 cm thick, thermal conductivity 0.93 W (m x deg.);
masonry of solid clay bricks 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 the thermotechnical calculation of enclosing structures is as follows:

R \u003d 0.03 / 0.93 + 0.64 / 0.81 + 0.03 / 0.93 \u003d 0.85 (m x deg / W).

The obtained value is significantly less than the previously determined base value of the resistance to heat transfer of the walls of a residential building in Murmansk 3.65 (m x deg/W). The wall does not satisfy regulatory requirements and needs to be warmed up. For wall insulation, we use 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 thermotechnical calculation of the enclosing structures. An example calculation is shown below:

R \u003d 0.15 / 0.048 + 0.03 / 0.93 + 0.64 / 0.81 + 0.03 / 0.93 \u003d 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.

The calculation of overlaps and combined coverings is carried out in a similar way.

Thermal 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 out. 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 structure 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);
zone 2 - 4.3 (m x deg / W);
zone 3 - 8.6 (m x deg / W);
4 zone - 14.3 (m x deg / W).

It is easy to see that the farther the floor area is from outer wall, the higher its resistance to heat transfer. Therefore, they are often limited to warming the perimeter of the floor. In this case, the heat transfer resistance of the insulated structure is added to the heat transfer resistance of the zone.
The calculation of the resistance to heat transfer of the floor must be included in the overall heat engineering calculation of enclosing structures. 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 sq. m.
The area of zone 2 will be 32 sq. m.
The area of the 3rd zone will be 4 sq. m.

The average value of the resistance to heat transfer of the floor on the ground:
Rpol \u003d 100 / (64 / 2.1 + 32 / 4.3 + 4 / 8.6) \u003d 2.6 (m x deg / W).

Having performed 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 the heat transfer resistance:

Rpol \u003d 100 / (32 / 2.1 + 32 / (2.1 + 0.05 / 0.032) + 32 / 4.3 + 4 / 8.6) \u003d 4.09 (m x deg / W).

It is important to note that not only floors are calculated in this way, but also the structures of walls in contact with the ground (walls of a recessed floor, a warm basement).

Thermotechnical calculation of doors

The basic value of heat transfer resistance is calculated somewhat differently entrance doors. To calculate it, you will first need to calculate the heat transfer resistance of the wall according to the sanitary and hygienic criterion (non-dew):
Rst \u003d (Tv - Tn) / (DTn x av).

Here DTN is the temperature difference between the inner surface of the wall and the air temperature in the room, determined by the Code of Rules and for housing is 4.0.
av - 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. An example of the calculation of the front door:

Рdv \u003d 0.6 x (20-(-30)) / (4 x 8.7) \u003d 0.86 (m x deg / W).

This design value will correspond to a door insulated with a 5 cm thick mineral wool board.

Complex requirements

Wall, floor or roof calculations 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 SP calculation is shown below.

Kob = 88.77 / 250 = 0.35, which is less than the normalized value of 0.52. In this case, the area and volume are taken for a house with dimensions of 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, in order to draw up an energy passport, a thermal engineering calculation of building envelopes is also performed; an example of a passport is given in the appendix to SP50.13330.2012.

Uniformity coefficient

All the above calculations are applicable for homogeneous structures. Which is quite rare in practice. To take into account the inhomogeneities that reduce the resistance to heat transfer, a correction factor for thermal engineering uniformity, r, is introduced. It takes into account the change in heat transfer resistance introduced by window and doorways, external corners, inhomogeneous inclusions (for example, lintels, beams, reinforcing belts), etc.

The calculation of this coefficient is quite complicated, therefore, 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 modern thermal protection requirements are met without using effective insulation almost impossible. 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:

brick wall 3 m thick;
masonry from foam concrete blocks 1.4 m;
mineral wool insulation 0.18 m.

To keep the house warm in the most very coldy, it is necessary to choose the right thermal insulation system - for this, a thermal engineering calculation of the outer wall is performed. The result of the calculations shows how effective the actual or projected method of insulation is.

How to make a thermal calculation of the outer wall

First you need to prepare the initial data. On the design parameter influenced by the following factors:

the climatic region in which the house is located;
the purpose of the premises is a residential building, an industrial building, a hospital;
mode of operation of the building - seasonal or year-round;
the presence in the design of door and window openings;
indoor humidity, the difference between indoor and outdoor temperatures;
number of floors, floor features.

After collecting and recording the initial information, the thermal conductivity coefficients of the building materials from which the wall is made are determined. The degree of heat absorption and heat transfer depends on how damp the climate is. In this regard, to calculate the coefficients, moisture maps compiled for Russian Federation. After that, all the numerical values \u200b\u200bnecessary for the calculation are entered into the appropriate formulas.

Thermal engineering calculation of the outer wall, an example for a foam concrete wall

As an example, the heat-shielding properties of a wall made of foam blocks, insulated with expanded polystyrene with a density of 24 kg / m3 and plastered on both sides with a lime-sand mortar are calculated. Calculations and selection of tabular data are carried out on the basis of building rules. Initial data: construction area - Moscow; relative humidity - 55%, average temperature in the house tv = 20 ° C. The thickness of each layer is set: δ1, δ4 = 0.01m (plaster), δ2 = 0.2m (foam concrete), δ3 = 0.065m (expanded polystyrene "SP Radoslav" ).
The purpose of the heat engineering calculation of the outer wall is to determine the required (Rtr) and actual (Rf) resistance to heat transfer.
Calculation

According to Table 1 of SP 53.13330.2012, under given conditions, the humidity regime is assumed to be normal. The required value of Rtr is found by the formula:
Rtr=a GSOP+b,
where a, b are taken according to Table 3 of SP 50.13330.2012. For a residential building and an outer wall, a = 0.00035; b = 1.4.
GSOP - degree-days of the heating period, they are found according to the formula (5.2) SP 50.13330.2012:
GSOP=(tin-tot)zot,
where tv \u003d 20O C; tot is the average outdoor temperature during the heating season, according to Table 1 SP131.13330.2012 tot = -2.2°C; zot = 205 days (duration heating season according to the same table).
Substituting the tabular values, they find: GSOP = 4551O C * day; Rtr \u003d 2.99 m2 * C / W
According to table 2 SP50.13330.2012 for normal humidity choose the coefficients of thermal conductivity of each layer of the "pie": λB1=0.81W/(m°C), λB2=0.26W/(m°C), λB3=0.041W/(m°C), λB4=0.81W/ (m°C).
According to the formula E.6 of SP 50.13330.2012, the conditional resistance to heat transfer is determined:
R0cond=1/αint+δn/λn+1/αext.
where αext \u003d 23 W / (m2 ° С) from clause 1 of Table 6 of SP 50.13330.2012 for external walls.
Substituting the numbers, get R0usl = 2.54 m2 ° C / W. It is refined using the coefficient r = 0.9, which depends on the homogeneity of structures, the presence of ribs, reinforcement, cold bridges:
Rf=2.54 0.9=2.29m2 °C/W.

The result obtained shows that the actual thermal resistance is less than required, so the wall design needs to be reconsidered.

Thermotechnical calculation of the outer wall, the program simplifies calculations

Simple computer services speed up computational processes and the search for the required coefficients. It is worth familiarizing yourself with the most popular programs.

"TeReMok". Initial data are entered: type of building (residential), internal temperature 20O, humidity regime - normal, area of residence - Moscow. In the next window, the calculated value of the standard resistance to heat transfer opens - 3.13 m2 * ° C / W.
Based on the calculated coefficient, a thermal engineering calculation of the outer wall of foam blocks (600 kg / m3), insulated with extruded polystyrene foam Flurmat 200 (25 kg / m3) and plastered with cement-lime mortar, is carried out. Choose from the menu the right materials, putting down their thickness (foam block - 200 mm, plaster - 20 mm), leaving the cell with the thickness of the insulation unfilled.
By pressing the "Calculation" button, the desired thickness of the heat insulator layer is obtained - 63 mm. The convenience of the program does not eliminate its disadvantage: it does not take into account the different thermal conductivity of the masonry material and mortar. Thanks to the author can be said at this address http://dmitriy.chiginskiy.ru/teremok/
The second program is offered by the site http://rascheta.net/. Its difference from the previous service is that all thicknesses are set independently. The coefficient of thermal engineering homogeneity r is introduced into the calculation. It is selected from the table: for foam concrete blocks with wire reinforcement in horizontal joints r = 0.9.
After filling in the fields, the program issues a report on the actual thermal resistance of the selected design, whether it meets climatic conditions. In addition, a sequence of calculations is provided with formulas, normative sources, and intermediate values.

When building a house or carrying out thermal insulation work, it is important to evaluate the effectiveness of the insulation of the outer wall: a thermal calculation performed independently or with the help of a specialist allows you to do this quickly and accurately.

When determining the need for additional insulation of a house, it is important to know the heat loss of its structures, in particular. An online wall thermal conductivity calculator will help you make calculations quickly and accurately.

In contact with

Why do you need a calculation

Thermal conductivity given element buildings - the property of a building to conduct heat through a unit of its area with a temperature difference between inside and outside the room of 1 deg. WITH.

The heat engineering calculation of enclosing structures performed by the service mentioned above is necessary for the following purposes:

for selection heating equipment and the type of system that allows not only to compensate for heat loss, but also to create a comfortable temperature inside the living quarters;
to determine the need for additional insulation of the building;
when designing and constructing a new building to select a wall material that provides the least heat loss in certain climatic conditions;
to create a comfortable temperature indoors not only during the heating period, but also in summer in hot weather.

Attention! Performing independent thermotechnical calculations wall structures, use the methods and data described in such normative documents, as SNiP II 03 79 "Construction heat engineering" and SNiP 23-02-2003 "Thermal protection of buildings".

What does thermal conductivity depend on?

Heat transfer depends on factors such as:

The material from which the building is built various materials differ in their ability to conduct heat. Yes, concrete different kinds bricks contribute to a large loss of heat. On the contrary, galvanized logs, beams, foam and gas blocks, with a smaller thickness, have lower thermal conductivity, which ensures the preservation of heat inside the room and much lower costs for insulation and heating of the building.
Wall thickness - than given value more, the less heat transfer occurs through its thickness.
Humidity of the material - the greater the moisture content of the raw material from which the structure is erected, the more it conducts heat and the faster it collapses.
The presence of air pores in the material - air-filled pores prevent accelerated heat loss. If these pores are filled with moisture, heat loss increases.
The presence of additional insulation - lined with a layer of insulation outside or inside the wall in terms of heat loss, have values \u200b\u200bmany times less than non-insulated ones.

In construction, along with the thermal conductivity of walls, such a characteristic as thermal resistance (R) has become widespread. It is calculated taking into account the following indicators:

coefficient of thermal conductivity of the wall material (λ) (W/m×0С);
construction thickness (h), (m);
the presence of a heater;
moisture content of the material (%).

The lower the thermal resistance value, the more the wall is subject to heat loss.

The thermotechnical calculation of enclosing structures according to this characteristic is performed according to the following formula:

R=h/λ; (m2×0С/W)

Example of thermal resistance calculation:

Initial data:

the load-bearing wall is made of dry pine timber 30 cm (0.3 m) thick;
thermal conductivity coefficient is 0.09 W/m×0С;
result calculation.

Thus, the thermal resistance of such a wall will be:

R=0.3/0.09=3.3 m2×0С/W

The values obtained as a result of the calculation are compared with the normative ones in accordance with SNiP II 03 79. At the same time, such an indicator as the degree-day of the period during which the heating season continues is taken into account.

If the obtained value is equal to or greater than the standard value, then the material and thickness of the wall structures are selected correctly. Otherwise, the building should be insulated to achieve the standard value.

In the presence of a heater, its thermal resistance is calculated separately and summarized with the same value of the main wall material. Also, if the material of the wall structure has high humidity, apply the appropriate coefficient of thermal conductivity.

For a more accurate calculation of the thermal resistance of this design, similar values \u200b\u200bof windows and doors facing the street are added to the result obtained.

Valid values

When performing a heat engineering calculation of the outer wall, the region in which the house will be located is also taken into account:

For the southern regions with warm winters and small temperature differences, it is possible to build walls of small thickness from materials with an average degree of thermal conductivity - ceramic and clay fired single and double, and of high density. The thickness of the walls for such regions can be no more than 20 cm.
At the same time for northern regions it is more expedient and cost-effective to build enclosing wall structures of medium and large thickness from materials with high thermal resistance - logs, gas and foam concrete of medium density. For such conditions, wall structures up to 50–60 cm thick are erected.
For regions with temperate climate and alternating temperature regime in winter they are suitable with high and medium thermal resistance - gas and foam concrete, timber, medium diameter. Under such conditions, the thickness of wall enclosing structures, taking into account heaters, is no more than 40–45 cm.

Important! The most accurate calculation of the thermal resistance of wall structures is the heat loss calculator, which takes into account the region where the house is located.

Heat transfer of various materials

One of the main factors affecting the thermal conductivity of the wall is the building material from which it is built. This dependence is explained by its structure. So, materials with a low density have the lowest thermal conductivity, in which the particles are arranged quite loosely and there is a large number of pores and voids filled with air. These include various types of wood, light porous concrete - foam, gas, slag concrete, as well as hollow silicate bricks.

Materials with high thermal conductivity and low thermal resistance include various types of heavy concrete, monolithic silicate brick. This feature is explained by the fact that the particles in them are located very close to each other, without voids and pores. This contributes to faster heat transfer in the thickness of the wall and a large heat loss.

Table. Thermal conductivity coefficients of building materials (SNiP II 03 79)

Calculation of a sandwich structure

The thermotechnical calculation of the outer wall, consisting of several layers, is carried out as follows:

according to the formula described above, the value of thermal resistance of each of the layers of the "wall cake" is calculated;
the values of this characteristic of all layers are added together, obtaining the total thermal resistance of the wall multilayer structure.

Based on this technique, it is possible to calculate the thickness. To do this, it is necessary to multiply the thermal resistance missing to the norm by the thermal conductivity coefficient of the insulation - as a result, the thickness of the insulation layer will be obtained.

With the help of the TeReMOK program, the thermotechnical calculation is performed automatically. In order for the wall thermal conductivity calculator to perform calculations, it is necessary to enter the following initial data into it:

type of building - residential, industrial;
wall material;
construction thickness;
region;
required temperature and humidity inside the building;
presence, type and thickness of insulation.

Useful video: how to independently calculate the heat loss in the house

Thus, the thermotechnical calculation of enclosing structures is very important both for a house under construction and for a building that has already been built for a long time. In the first case, the correct heat calculation will save on heating, in the second case, it will help to choose the insulation that is optimal in thickness and composition.

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