Thermotechnical calculation of enclosing structures of buildings. Thermal calculation with an example How to make a thermal calculation of a building

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 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 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

The quality of thermal insulation is determined 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 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.).
• Coniferous wood - 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 meet the regulatory requirements and needs to be insulated. 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 are carried out 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 the 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 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 heaters based on expanded polystyrene or stone wool allows us to limit ourselves 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.

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 have noticed that the outer walls of these houses are made of ceramic bricks, the thickness of which is about 1.5 meters. This thickness brick wall provided 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 of brickwork can be reduced 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 of buildings". Updated edition from 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 during the cold season 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 of sanitary standards 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 the walls of a 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 sensible heat excesses 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 ° C and below the reduced heat transfer resistance 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 = 0.09 / 0.96 = 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 the outer surface (in our case, this is a decorative brick (besser)) 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.

Thermal engineering calculation allows you to determine the minimum thickness of building envelopes so that there are no cases of overheating or freezing during the operation of the building.

Enclosing structural elements of heated public and residential buildings, with the exception of the requirements of stability and strength, durability and fire resistance, economy and architectural design, must primarily meet thermal engineering standards. Choice of guardrails depending on constructive solution, climatological characteristics of the building area, physical properties, humidity and temperature conditions in the building, as well as in accordance with the requirements of resistance to heat transfer, air permeability and vapor permeability.

What is the meaning of calculation?

1. If, when calculating the cost of a future building, only strength characteristics, then, of course, the cost will be less. However, this is a visible savings: subsequently, much more money will be spent on heating the room.
2. Properly selected materials will create an optimal microclimate in the room.
3. When planning a heating system, a heat engineering calculation is also necessary. In order for the system to be cost-effective and efficient, it is necessary to have an understanding of the real possibilities of the building.

Thermal requirements

It is important that the external structures comply with the following thermal requirements:

• They had sufficient heat-shielding properties. In other words, it should not be allowed summer time overheating of the premises, and in winter - excessive heat loss.
• The air temperature difference between the internal elements of the fences and the premises should not be higher than the standard value. Otherwise, excessive cooling of the human body by heat radiation to these surfaces and moisture condensation of the internal air flow on the enclosing structures may occur.
• In the event of a change in heat flow, temperature fluctuations inside the room should be minimal. This property called thermal stability.
• It is important that the air tightness of the fences does not cause strong cooling of the premises and does not worsen the heat-shielding properties of the structures.
• Fences must have a normal humidity regime. Since waterlogging of fences increases heat loss, causes dampness in the room, and reduces the durability of structures.

In order for the structures to meet the above requirements, they perform a thermal calculation, and also calculate the heat resistance, vapor permeability, air permeability and moisture transfer according to the requirements of regulatory documentation.

Thermotechnical qualities

From the thermal characteristics of the outdoor structural elements buildings depends on:

• Moisture regime of structural elements.
• The temperature of internal structures, which ensures that there is no condensation on them.
• Constant humidity and temperature in the premises, both in the cold and in the warm season.
• The amount of heat lost by a building winter period time.

So, based on all of the above, the heat engineering calculation of structures is considered an important stage in the process of designing buildings and structures, both civil and industrial. Designing begins with the choice of structures - their thickness and sequence of layers.

Tasks of thermal engineering calculation

So, the heat engineering calculation of enclosing structural elements is carried out in order to:

1. Compliance of structures with modern requirements for thermal protection of buildings and structures.
2. Ensuring a comfortable microclimate in the interior.
3. Ensuring optimal thermal protection of fences.

Basic parameters for calculation

To determine the heat consumption for heating, as well as to make a heat engineering calculation of the building, it is necessary to take into account many parameters that depend on the following characteristics:

• Purpose and type of building.
• Geographic location of the building.
• The orientation of the walls to the cardinal points.
• Dimensions of structures (volume, area, number of storeys).
• Type and size of windows and doors.
• Characteristics of the heating system.
• The number of people in the building at the same time.
• The material of the walls, floor and ceiling of the last floor.
• The presence of a hot water system.
• Type of ventilation systems.
• Other design features buildings.

Thermal engineering calculation: program

To date, many programs have been developed that allow you to make this calculation. As a rule, the calculation is carried out on the basis of the methodology set out in the regulatory and technical documentation.

These programs allow you to calculate the following:

• Thermal resistance.
• Heat loss through structures (ceiling, floor, door and window openings, and walls).
• The amount of heat required to heat the infiltrating air.
• Selection of sectional (bimetallic, cast iron, aluminum) radiators.
• Selection of panel steel radiators.

Thermotechnical calculation: calculation example for external walls

For the calculation, it is necessary to determine the following main parameters:

• t in \u003d 20 ° C is the temperature of the air flow inside the building, which is taken to calculate the fences according to the minimum values ​​​​of the most optimum temperature relevant building and structure. It is accepted in accordance with GOST 30494-96.

• According to the requirements of GOST 30494-96, the humidity in the room should be 60%, as a result, a normal humidity regime will be provided in the room.
• In accordance with Appendix B of SNiPa 23-02-2003, the humidity zone is dry, which means that the operating conditions of the fences are A.
• t n \u003d -34 ° C is the temperature of the outdoor air flow in the winter period, which is taken according to SNiP based on the coldest five-day period, which has a security of 0.92.
• Z ot.per = 220 days is the duration of the heating period, which is taken according to SNiP, while the average daily ambient temperature is ≤ 8 °C.
• T from.per. = -5.9 °C is the ambient temperature (average) during the heating season, which is accepted according to SNiP, at a daily ambient temperature ≤ 8 °C.

Initial data

In this case, the thermotechnical calculation of the wall will be carried out in order to determine the optimal thickness of the panels and the heat-insulating material for them. Sandwich panels will be used as external walls (TU 5284-001-48263176-2003).

Comfortable conditions

Consider how the thermal engineering calculation of the outer wall is performed. First you need to calculate the required heat transfer resistance, focusing on comfortable and sanitary conditions:

R 0 tr \u003d (n × (t in - t n)) : (Δt n × α in), where

n = 1 is a factor that depends on the position of the external structural elements in relation to the outside air. It should be taken according to SNiP 23-02-2003 from Table 6.

Δt n \u003d 4.5 ° C is the normalized temperature difference between the internal surface of the structure and internal air. Accepted according to SNiP data from table 5.

α in \u003d 8.7 W / m 2 ° C is the heat transfer of internal enclosing structures. Data are taken from table 5, according to SNiP.

We substitute the data in the formula and get:

R 0 tr \u003d (1 × (20 - (-34)) : (4.5 × 8.7) \u003d 1.379 m 2 ° C / W.

Energy Saving Conditions

When performing a thermal engineering calculation of the wall, based on the conditions of energy saving, it is necessary to calculate the required heat transfer resistance of the structures. It is determined by GSOP (heating degree-day, °C) using the following formula:

GSOP = (t in - t from.per.) × Z from.per, where

t in is the temperature of the air flow inside the building, °C.

Z from.per. and t from.per. is the duration (days) and temperature (°C) of the period with an average daily air temperature ≤ 8 °C.

Thus:

GSOP = (20 - (-5.9)) × 220 = 5698.

Based on the conditions of energy saving, we determine R 0 tr by interpolation according to SNiP from table 4:

R 0 tr \u003d 2.4 + (3.0 - 2.4) × (5698 - 4000)) / (6000 - 4000)) \u003d 2.909 (m 2 ° C / W)

R 0 = 1/ α in + R 1 + 1/ α n, where

d is the thickness of the thermal insulation, m.

l = 0.042 W/m°C is the thermal conductivity of the mineral wool board.

α n \u003d 23 W / m 2 ° C is the heat transfer of external structural elements, taken according to SNiP.

R 0 \u003d 1 / 8.7 + d / 0.042 + 1/23 \u003d 0.158 + d / 0.042.

Insulation thickness

The thickness of the heat-insulating material is determined based on the fact that R 0 \u003d R 0 tr, while R 0 tr is taken under energy saving conditions, thus:

2.909 = 0.158 + d/0.042, whence d = 0.116 m.

We select the brand of sandwich panels from the catalog with optimum thickness thermal insulation material: DP 120, while the total thickness of the panel should be 120 mm. The heat engineering calculation of the building as a whole is carried out in a similar way.

The need to perform the calculation

Designed on the basis of a competently executed heat engineering calculation, building envelopes can reduce heating costs, the cost of which is regularly increasing. In addition, saving heat is considered an important environmental task, because it is directly related to a decrease in fuel consumption, which leads to a decrease in the impact negative factors on the environment.

In addition, it is worth remembering that improperly performed thermal insulation can lead to waterlogging of structures, which will result in the formation of mold on the surface of the walls. The formation of mold, in turn, will lead to spoilage interior decoration(peeling of wallpaper and paint, destruction of the plaster layer). In particularly advanced cases, radical intervention may be necessary.

Often construction companies tend to use in their activities modern technologies and materials. Only a specialist can understand the need to use one or another material, both separately and in combination with others. It is the heat engineering calculation that will help determine the most optimal solutions that will ensure the durability of structural elements and minimal financial costs.

The purpose of thermotechnical calculation is to calculate the thickness of the insulation for a given thickness of the bearing part of the outer wall, which meets sanitary and hygienic requirements and energy saving conditions. In other words, we have external walls with a thickness of 640 mm made of silicate bricks and we are going to insulate them with polystyrene foam, but we don’t know what thickness the insulation must be chosen in order to comply with building codes.

The thermotechnical calculation of the outer wall of the building is carried out in accordance with SNiP II-3-79 "Construction Heat Engineering" and SNiP 23-01-99 "Construction Climatology".

Table 1

Thermal performance of the building materials used (according to SNiP II-3-79*)

1-internal plaster (cement-sand mortar) - 20 mm

2-brick wall (silicate brick) - 640 mm

3-insulation (polystyrene foam)

4-thin-layer plaster (decorative layer) - 5 mm

When performing a heat engineering calculation, a normal humidity regime in the premises was adopted - operating conditions ("B") in accordance with SNiP II-3-79 v.1 and adj. 2, i.e. the thermal conductivity of the materials used is taken according to column "B".

We calculate the required heat transfer resistance of the fence, taking into account sanitary and hygienic and comfortable conditions according to the formula:

R 0 tr \u003d (t in - t n) * n / Δ t n * α in (1)

where t in is the design temperature of the internal air °С, taken in accordance with GOST 12.1.1.005-88 and design standards

relevant buildings and structures, we accept equal to +22 ° С for residential buildings in accordance with Appendix 4 to SNiP 2.08.01-89;

t n is the estimated winter temperature of the outside air, °С, equal to the average temperature of the coldest five-day period, with a security of 0.92 according to SNiP 23-01-99 for the city of Yaroslavl is taken equal to -31°С;

n is the coefficient accepted according to SNiP II-3-79* (table 3*) depending on the position of the outer surface of the enclosing structure in relation to the outside air and is taken equal to n=1;

Δ t n - normative and temperature difference between the temperature of the internal air and the temperature of the inner surface of the enclosing structure - is set according to SNiP II-3-79 * (table 2 *) and is taken equal to Δ t n = 4.0 ° С;

R 0 tr \u003d (22- (-31)) * 1 / 4.0 * 8.7 \u003d 1.52

We determine the degree-day of the heating period by the formula:

GSOP \u003d (t in - t from.per) * z from.per. (2)

where t in - the same as in the formula (1);

t from.per - average temperature, ° С, of the period with an average daily air temperature below or equal to 8 ° С according to SNiP 23-01-99;

z from.per - duration, days, of the period with an average daily air temperature below or equal to 8 ° C according to SNiP 23-01-99;

GSOP \u003d (22-(-4)) * 221 \u003d 5746 ° C * day.

Let us determine the reduced resistance to heat transfer Ro tr according to the conditions of energy saving in accordance with the requirements of SNiP II-3-79* (Table 1b*) and sanitary and hygienic and comfortable conditions. Intermediate values ​​are determined by interpolation.

table 2

Heat transfer resistance of enclosing structures (according to SNiP II-3-79*)

The resistance to heat transfer of enclosing structures R(0) is taken as the largest of the values ​​calculated earlier:

R 0 tr \u003d 1.52< R 0 тр = 3,41, следовательно R 0 тр = 3,41 (м 2 *°С)/Вт = R 0 .

We write an equation for calculating the actual heat transfer resistance R 0 of the enclosing structure using the formula in accordance with the given design scheme and determine the thickness δ x of the design layer of the fence from the condition:

R 0 \u003d 1 / α n + Σδ i / λ i + δ x / λ x + 1 / α in \u003d R 0

where δ i is the thickness of the individual layers of the fence, except for the calculated one, in m;

λ i - thermal conductivity coefficients of individual layers of the fence (except for the calculated layer) in (W / m * ° C) are taken according to SNiP II-3-79 * (Appendix 3 *) - for this calculation table 1;

δ x - thickness of the design layer of the outer fence, m;

λ x - coefficient of thermal conductivity of the calculated layer of the outer fence in (W / m * ° C) are taken according to SNiP II-3-79 * (Appendix 3 *) - for this calculation table 1;

α in - the heat transfer coefficient of the inner surface of the enclosing structures is taken according to SNiP II-3-79 * (table 4 *) and is taken equal to α in \u003d 8.7 W / m 2 * ° С.

α n - heat transfer coefficient (for winter conditions) of the outer surface of the building envelope is taken according to SNiP II-3-79 * (table 6 *) and is taken equal to α n \u003d 23 W / m 2 * ° С.

The thermal resistance of a building envelope with sequentially located homogeneous layers should be determined as the sum of the thermal resistances of individual layers.

For external walls and ceilings, the thickness of the heat-insulating layer of the fence δ x is calculated from the condition that the value of the actual reduced resistance to heat transfer of the enclosing structure R 0 must not be less than the normalized value R 0 tr calculated by formula (2):

R 0 ≥ R 0 tr

Expanding the value of R 0 , we get:

R0 = 1 / 23 + (0,02/ 0,93 + 0,64/ 0,87 + 0,005/ 0.93) + δx / 0,041 + 1/ 8,7

Based on this, we determine the minimum value of the thickness of the heat-insulating layer

δ x \u003d 0.041 * (3.41 - 0.115 - 0.022 - 0.74 - 0.005 - 0.043)

δx = 0.10 m

We take into account the thickness of the insulation (polystyrene foam) δ x = 0.10 m

Determine the actual resistance to heat transfer calculated enclosing structures R 0, taking into account the accepted thickness of the heat-insulating layer δ x = 0.10 m

R0 = 1 / 23 + (0,02/ 0,93 + 0,64/ 0,87 + 0,005/ 0,93 + 0,1/ 0,041) + 1/ 8,7

R 0 \u003d 3.43 (m 2 * ° C) / W

Condition R 0 ≥ R 0 tr observed, R 0 = 3.43 (m 2 * ° C) / W ≥ R 0 tr \u003d 3.41 (m 2 * ° C) / W

Example of thermotechnical calculation of enclosing structures

1. Initial data

Technical task. In connection with the unsatisfactory heat and humidity regime of the building, it is necessary to insulate its walls and mansard roof. To this end, perform calculations of thermal resistance, heat resistance, air and vapor permeability of the building envelope with an assessment of the possibility of moisture condensation in the thickness of the fences. Determine the required thickness of the heat-insulating layer, the need to use wind and vapor barriers, the order of the layers in the structure. Develop a design solution that meets the requirements of SNiP 23-02-2003 "Thermal protection of buildings" for building envelopes. Perform calculations in accordance with the set of rules for the design and construction of SP 23-101-2004 "Design of thermal protection of buildings".

General characteristics of the building. A two-story residential building with an attic is located in the village. Sviritsa Leningrad region. The total area of ​​external enclosing structures - 585.4 m 2; total wall area 342.5 m 2; the total area of ​​windows is 51.2 m 2; roof area - 386 m 2; basement height - 2.4 m.

The structural scheme of the building includes bearing walls, reinforced concrete floors from multi-hollow panels, 220 mm thick and a concrete foundation. The outer walls are made of brickwork and plastered inside and out with a mortar layer of about 2 cm.

The roof of the building has a truss structure with a steel seam roof, made along the crate with a step of 250 mm. Insulation 100 mm thick is made of mineral wool boards laid between the rafters

The building is provided with stationary electric-thermal storage heating. The basement has a technical purpose.

climatic parameters. According to SNiP 23-02-2003 and GOST 30494-96, we take the estimated average temperature of the indoor air equal to

t int= 20 °С.

According to SNiP 23-01-99 we accept:

1) the estimated temperature of the outside air in the cold season for the conditions of the village. Sviritsa Leningrad region

t ext= -29 °С;

2) the duration of the heating period

z ht= 228 days;

3) the average outdoor temperature for the heating period

t ht\u003d -2.9 ° С.

Heat transfer coefficients. The values ​​of the heat transfer coefficient of the inner surface of the fences are accepted: for walls, floors and smooth ceilings α int\u003d 8.7 W / (m 2 ºС).

The values ​​of the heat transfer coefficient of the outer surface of the fences are accepted: for walls and coatings α ext=23; attic floors α ext\u003d 12 W / (m 2 ºС);

Normalized resistance to heat transfer. Degree-days of the heating period G d are determined by formula (1)

G d\u003d 5221 ° С day.

Since the value G d differs from table values, standard value R req determined by formula (2).

According to SNiP 23-02-2003 for the obtained degree-day value, the normalized resistance to heat transfer R req, m 2 ° С / W, is:

For external walls 3.23;

Coverings and ceilings over driveways 4.81;

Fencing over unheated undergrounds and basements 4.25;

windows and balcony doors 0,54.

2. Thermotechnical calculation of external walls

2.1. Resistance of external walls to heat transfer

Exterior walls are made of hollow ceramic bricks and have a thickness of 510 mm. The walls are plastered from the inside with lime-cement mortar 20 mm thick, from the outside - with cement mortar of the same thickness.

The characteristics of these materials - density γ 0, dry thermal conductivity coefficient  0 and vapor permeability coefficient μ - are taken from Table. Clause 9 of the application. In this case, in the calculations we use the coefficients of thermal conductivity of materials  W for operating conditions B, (for wet operating conditions), which are obtained by formula (2.5). We have:

For lime-cement mortar

γ 0 \u003d 1700 kg / m 3,

 W\u003d 0.52 (1 + 0.168 4) \u003d 0.87 W / (m ° C),

μ=0.098 mg/(m h Pa);

For brickwork from hollow ceramic bricks on cement-sand mortar

γ 0 \u003d 1400 kg / m 3,

 W\u003d 0.41 (1 + 0.207 2) \u003d 0.58 W / (m ° C),

μ=0.16 mg/(m h Pa);

For cement mortar

γ 0 \u003d 1800 kg / m 3,

 W\u003d 0.58 (1 + 0.151 4) \u003d 0.93 W / (m ° C),

μ=0.09 mg/(m h Pa).

The heat transfer resistance of a wall without insulation is

R o \u003d 1 / 8.7 + 0.02 / 0.87 + 0.51 / 0.58 + 0.02 / 0.93 + 1/23 \u003d 1.08 m 2 ° C / W.

In the presence of window openings that form the slopes of the wall, the coefficient of thermal uniformity of brick walls, 510 mm thick, is taken r = 0,74.

Then the reduced resistance to heat transfer of the walls of the building, determined by formula (2.7), is equal to

R r o \u003d 0.74 1.08 \u003d 0.80 m 2 ° C / W.

The obtained value is much lower than the normative value of heat transfer resistance, therefore, it is necessary to install external thermal insulation and subsequent plastering with protective and decorative plaster compositions with fiberglass reinforcement.

In order for the thermal insulation to dry out, the plaster layer covering it must be vapor-permeable, i.e. porous with low density. We choose a porous cement-perlite mortar having the following characteristics:

γ 0 \u003d 400 kg / m 3,

 0 \u003d 0.09 W / (m ° C),

 W\u003d 0.09 (1 + 0.067 10) \u003d 0.15 W / (m ° C),

 \u003d 0.53 mg / (m h Pa).

The total resistance to heat transfer of the added layers of thermal insulation R t and plaster lining R w must be at least

R t+ R w \u003d 3.23 / 0.74-1.08 \u003d 3.28 m 2 ° C / W.

Preliminarily (with subsequent clarification), we accept the thickness of the plaster lining as 10 mm, then its resistance to heat transfer is equal to

R w \u003d 0.01 / 0.15 \u003d 0.067 m 2 ° C / W.

When used for thermal insulation of mineral wool boards manufactured by CJSC Mineralnaya Vata, Facade Butts brand  0 \u003d 145 kg / m 3,  0 \u003d 0.033,  W \u003d 0.045 W / (m ° C) the thickness of the heat-insulating layer will be

δ=0.045 (3.28-0.067)=0.145 m.

Rockwool boards are available in thicknesses from 40 to 160 mm in 10 mm increments. We accept a standard thickness of thermal insulation of 150 mm. Thus, the slabs will be laid in one layer.

Checking compliance with energy saving requirements. The calculation scheme of the wall is shown in fig. 1. The characteristics of the layers of the wall and the total resistance of the wall to heat transfer, excluding vapor barrier, are given in Table. 2.1.

Table 2.1

Characterization of the layers of the wall andtotal resistance of the wall to heat transfer

The heat transfer resistance of the walls of the building after insulation will be:

R o = 1/8.7+4.32+1/23=4.48 m 2 °C/W.

Taking into account the coefficient of thermal engineering uniformity of the outer walls ( r= 0.74) we get the reduced resistance to heat transfer

R o r\u003d 4.48 0.74 \u003d 3.32 m 2 ° C / W.

Received value R o r= 3.32 exceeds the standard R req= 3.23, since the actual thickness of the heat-insulating plates is greater than the calculated one. This situation meets the first requirement of SNiP 23-02-2003 for the thermal resistance of the wall - R o ≥ R req .

Verification of compliance with the requirements forsanitary and hygienic and comfortable conditions in the room. Estimated difference between the temperature of the indoor air and the temperature of the inner surface of the wall Δ t 0 is

Δ t 0 =n(t int – t ext)/(R o r ·α int)=1.0(20+29)/(3.32 8.7)=1.7 ºС.

According to SNiP 23-02-2003, for the outer walls of residential buildings, a temperature difference of not more than 4.0 ºС is permissible. Thus, the second condition (Δ t 0 ≤Δ t n) done.

P
check the third condition ( τ int >t grew), i.e. is it possible to condense moisture on the inner surface of the wall at the estimated outdoor temperature t ext\u003d -29 ° С. Inner surface temperature τ int enclosing structure (without heat-conducting inclusion) is determined by the formula

τ int = t int –Δ t 0 \u003d 20–1.7 \u003d 18.3 ° С.

The elasticity of water vapor in the room e int is equal to

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