Thermotechnical calculation of the walls of the house online. Example of thermal engineering calculation of an external wall

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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, except for the requirements of stability and strength, durability and fire resistance, efficiency and architectural design, must meet, first of all, 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, during the calculation of the cost of a future building, only strength characteristics are taken into account, then, naturally, 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 on 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 external structural elements of buildings depends:

• 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. Collateral during indoor areas comfortable microclimate.
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 optimal 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 thermal engineering calculation is performed outer wall. 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.

In this way:

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

Thickness thermal insulation material is determined based on the fact that R 0 \u003d R 0 tr, while R 0 tr is taken under the conditions of energy saving, thus:

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

We select the brand of sandwich panels according to the catalog with the optimal thickness of the heat-insulating material: DP 120, while the total thickness of the panel should be 120 mm. The same way thermotechnical calculation of the building as a whole is carried out.

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, which will ensure the durability of structural elements and minimal financial costs.

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 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 internal air from the condition of no condensation on the internal surfaces of the external 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 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 adopted according to table 6 for the 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, 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 of a heat-insulating material (formula 5.6 by E.G. Malyavin "Heat loss of a 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.

AT modern conditions more and more people think about rational use resources. Electricity, water, materials. To save all this in the world came for a long time and everyone understands how to do it. But the main amount in the bills for payment is heating, and not everyone understands how to reduce the expense for this item.

What is thermal engineering calculation?

Thermal engineering calculation is performed in order to select the thickness and material of the building envelope and bring the building in line with thermal protection standards. The main regulatory document regulating the ability of a structure to resist heat transfer is SNiP 23-02-2003 "Thermal protection of buildings".

The main indicator of the enclosing surface in terms of thermal protection was the reduced resistance to heat transfer. This is a value that takes into account the heat-shielding characteristics of all layers of the structure, taking into account cold bridges.

A detailed and competent heat engineering calculation is quite laborious. When building private houses, the owners try to take into account the strength characteristics of materials, often forgetting about heat conservation. This can lead to rather disastrous consequences.

Why is the calculation performed?

Before starting construction, the customer can choose whether he will take into account the thermal characteristics or ensure only the strength and stability of the structures.

The cost of insulation will definitely increase the estimate for the construction of the building, but will reduce the cost of further operation. Individual houses are built for decades, perhaps they will serve the next generations. During this time, the cost of an effective insulation will pay off several times.

What does the owner get if the calculations are done correctly:

• Savings on space heating. The heat losses of the building are reduced, respectively, the number of radiator sections with a classical heating system and the capacity of the underfloor heating system will decrease. Depending on the heating method, the owner's costs for electricity, gas or hot water become smaller;
• Savings on repairs. At proper insulation a comfortable microclimate is created in the room, condensation does not form on the walls, and microorganisms dangerous to humans do not appear. The presence of a fungus or mold on the surface requires repair, and a simple cosmetic one will not bring any results and the problem will arise again;
• Security for residents. Here, as well as in the previous paragraph, we are talking about dampness, mold and fungus, which can cause various diseases in people who are constantly in the room;
• respect for environment. There is a shortage of resources on the planet, so reducing the consumption of electricity or blue fuel has a positive effect on the ecological situation.

Normative documents for performing the calculation

The reduced resistance and its compliance with the normalized value is the main goal of the calculation. But for its implementation, you will need to know the thermal conductivity of the materials of the wall, roof or ceiling. Thermal conductivity is a value that characterizes the ability of a product to conduct heat through itself. The lower it is, the better.

During the calculation of heat engineering, they rely on the following documents:

• SP 50.13330.2012 "Thermal protection of buildings". The document was reissued on the basis of SNiP 23-02-2003. The main standard for calculation;
• SP 131.13330.2012 "Construction climatology". New edition of SNiP 23-01-99*. This document allows you to determine the climatic conditions of the settlement in which the object is located;
• SP 23-101-2004 "Design of thermal protection of buildings" in more detail than the first document in the list, reveals the topic;
• GOST 30494-96 (replaced by GOST 30494-2011 since 2011) Residential and public buildings;
• Manual for students of construction universities E.G. Malyavin “Heat loss of the building. Reference manual".

Thermal engineering calculation is not complicated. It can be performed by a person without special education according to the template. The main thing is to approach the issue very carefully.

An example of calculating a three-layer wall without an air gap

Let's take a closer look at an example of a heat engineering calculation. First you need to decide on the source data. As a rule, you choose the materials for the construction of the walls yourself. We will calculate the thickness of the insulation layer based on the materials of the wall.

Initial data

The data is individual for each construction object and depends on the location of the object.

1. Climate and microclimate

1. Construction area: Vologda.
2. Purpose of the object: residential.
3. Relative air humidity for a room with a normal humidity regime is 55% (item 4.3. Table 1).
4. The temperature inside the residential premises tint is set by regulatory documents (Table 1) and is equal to 20 degrees Celsius.

text is the estimated outside air temperature. It is set by the temperature of the coldest five days of the year. The value can be found in, table 1, column 5. For a given area, the value is -32ᵒС.

zht = 231 days - the number of days in the period when additional space heating is needed, that is, the average daily temperature outside is less than 8ᵒС. The value is looked up in the same table as the previous one, but in column 11.

tht = -4.1ᵒС – average outside air temperature during the heating period. The value is in column 12.

2. Wall materials

All layers should be taken into account (even a layer of plaster, if any). This will allow the most accurate calculation of the design.

In this embodiment, consider a wall consisting of the following materials:

1. a layer of plaster, 2 centimeters;
2. an inner verst made of ordinary solid ceramic brick with a thickness of 38 centimeters;
3. a layer of Rockwool mineral wool insulation, the thickness of which is selected by calculation;
4. outer verst of front ceramic brick, 12 centimeters thick.

3. Thermal conductivity of adopted materials

All properties of materials must be presented in the passport from the manufacturer. Many companies represent full information about products on their websites. The characteristics of the selected materials are summarized in a table for convenience.

Calculation of the thickness of the insulation for the wall

1. Energy saving condition

Calculation of the value of degree-days of the heating period (GSOP) is carried out according to the formula:

Dd = (tint - tht) zht.

All letter designations, presented in the formula, are deciphered in the source data.

Dd \u003d (20-(-4.1)) * 231 \u003d 5567.1 ᵒС * day.

The normative resistance to heat transfer is found by the formula:

The coefficients a and b are taken according to table 4, column 3.

For initial data a=0.00045, b=1.9.

Rreq = 0.00045*5567.1+1.9=3.348 m2*ᵒС/W.

2. Calculation of the norm of thermal protection based on the conditions of sanitation

This indicator is not calculated for residential buildings and is given as an example. The calculation is carried out with an excess of sensible heat exceeding 23 W / m3, or the operation of the building in spring and autumn. Also, calculations are required at a design temperature of less than 12ᵒС indoors. Formula 3 is used:

The coefficient n is taken according to table 6 of the SP "Thermal protection of buildings", αint according to table 7, Δtn according to the fifth table.

Rreq = 1*(20+31)4*8.7 = 1.47 m2*ᵒС/W.

Of the two values ​​obtained in the first and second paragraph, the largest is selected, and further calculation is carried out on it. In this case, Rreq = 3.348 m2*ᵒС/W.

3. Determination of the thickness of the insulation

The heat transfer resistance for each layer is obtained by the formula:

where δ is the layer thickness, λ is its thermal conductivity.

a) plaster R pcs \u003d 0.02 / 0.87 \u003d 0.023 m2 * ᵒС / W;
b) ordinary brick R row.brick. \u003d 0.38 / 0.48 \u003d 0.79 m2 * ᵒС / W;
c) facing brick Rut = 0.12 / 0.48 = 0.25 m2 * ᵒС / W.

The minimum heat transfer resistance of the entire structure is determined by the formula (, formula 5.6):

Rint = 1/αint = 1/8.7 = 0.115 m2*ᵒС/W;
Rext = 1/αext = 1/23 = 0.043 m2*ᵒС/W;
∑Ri = 0.023+0.79+0.25 = 1.063 m2*ᵒC/W, i.e. the sum of the numbers obtained in point 3;

R_tr ^ ut \u003d 3.348 - (0.115 + 0.043 + 1.063) \u003d 2.127 m2 * ᵒС / W.

The thickness of the insulation is determined by the formula (formula 5.7):

δ_tr^ut \u003d 0.038 * 2.127 \u003d 0.081 m.

The value found is the minimum. The insulation layer is taken not less than this value. In this calculation, we finally accept the thickness of the mineral wool insulation as 10 centimeters, so that you do not have to cut the purchased material.

For calculations of heat losses of the building, which are performed for design heating systems, it is necessary to find the actual value of the resistance to heat transfer with the found thickness of the insulation.

Rо = Rint+Rext+∑Ri = 1/8.7 + 1/23 + 0.023 + 0.79 + 0.1/0.038 + 0.25 = 3.85 m2*ᵒС/W > 3.348 m2*ᵒС/W.

The condition is met.

Influence of the air gap on the heat-shielding characteristics

When constructing a wall protected slab insulation a ventilated layer is possible. It allows you to remove condensate from the material and prevent it from getting wet. The minimum thickness of the gap is 1 centimeter. This space is not closed and has direct communication with the outside air.

In the presence of an air-ventilated layer, only those layers that are up to it from the side are taken into account in the calculation. warm air. For example, a wall pie consists of plaster, internal masonry, insulation, an air gap and external masonry. Only plaster, internal masonry and insulation are taken into account. The outer layer of masonry goes after the ventilation gap, therefore it is not taken into account. In this case outdoor masonry performs only an aesthetic function and protects the insulation from external influences.

Important: when considering structures where the airspace is closed, it is taken into account in the calculation. For example, in the case of window fillings. The air between the panes plays a role effective insulation.

Teremok program

To perform the calculation using a personal computer, specialists often use the program for thermal calculation "Teremok". It exists online and as an application for operating systems.

The program performs calculations based on all necessary normative documents. Working with the application is extremely simple. It allows you to work in two modes:

• calculation of the required layer of insulation;
• verification of an already thought-out design.

The database contains all the necessary characteristics for settlements our country, you just need to choose the right one. It is also necessary to choose the type of construction: external wall, mansard roof, ceiling over a cold basement or attic.

When you press the continue button, a new window appears that allows you to "assemble" the structure. Many materials are available in the program memory. They are divided into three groups for ease of search: structural, heat-insulating and heat-insulating-structural. You only need to set the layer thickness, the program will indicate the thermal conductivity itself.

With absence necessary materials you can add them yourself, knowing the thermal conductivity.

Before making calculations, you must select the type of calculation above the plate with the wall structure. Depending on this, the program will give either the thickness of the insulation, or report on the compliance of the enclosing structure with the standards. After the calculations are completed, you can generate a report in text format.

"Teremok" is very convenient to use and even a person without a technical education is able to deal with it. For specialists, it significantly reduces the time for calculations and preparation of a report in electronic form.

The main advantage of the program is the fact that it is able to calculate the thickness of the insulation not only of the outer wall, but of any structure. Each of the calculations has its own characteristics, and it is quite difficult for a non-professional to understand all of them. To build a private house, it is enough to master this application, and you do not have to delve into all the difficulties. Calculation and verification of all enclosing surfaces will take no more than 10 minutes.

Thermal engineering calculation online (calculator overview)

Thermal engineering calculation can be done on the Internet online. Not bad, as in my opinion, is the service: rascheta.net. Let's take a quick look at how to work with it.

By going to the site online calculator, the first step is to choose the standards by which the calculation will be made. I choose the 2012 rulebook as it is a newer document.

Next, you need to specify the region in which the object will be built. If your city is not available, choose the nearest one. Big city. After that, we indicate the type of buildings and premises. Most likely you will calculate a residential building, but you can choose public, administrative, industrial and others. And the last thing you need to choose is the type of enclosing structure (walls, ceilings, coatings).

Estimated average temperature, relative humidity and coefficient of thermotechnical homogeneity are left the same if you do not know how to change them.

In the calculation options, set all two checkboxes except the first one.

In the table, we indicate the wall cake starting from the outside - we select the material and its thickness. On this, in fact, the whole calculation is completed. Below the table is the result of the calculation. If any of the conditions is not met, we change the thickness of the material or the material itself until the data complies with regulatory documents.

If you want to see the calculation algorithm, then click on the "Report" button at the bottom of the site page.

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 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.

Climate characteristics

The choice of wall and roof construction depends primarily on climatic conditions 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.

Main characteristics

For right choice building envelope materials, it is necessary to determine which thermal characteristics they must 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.).
• 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 must not be less than the 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 performed 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 homes 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 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 calculated 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 issuing 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, outside corners, inhomogeneous inclusions (for example, jumpers, 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 you modern requirements thermal protection without the use of effective insulation is almost impossible. So, if you use a traditional clay brick, you will need masonry several meters thick, which is not economically feasible. However, low thermal conductivity modern heaters 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.