Thermal calculation allows us to determine minimum thickness enclosing structures to ensure that there are no cases of overheating or freezing during the operation of the structure.

Enclosing structural elements of heated public and residential buildings, with the exception of the requirements of stability and strength, durability and fire resistance, efficiency and architectural design, must first of all meet thermal engineering standards. Enclosing elements are selected 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 point of the calculation?

1. If, when calculating the cost of a future building, only the strength characteristics are taken into account, then, naturally, the cost will be less. However, this is a visible saving: subsequently, significantly 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 thermal engineering calculation is also required. For the system to be cost-effective and efficient, it is necessary to have an understanding of real possibilities building.

Thermal requirements

It is important that external structures meet the following thermal requirements:

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

In order for structures to meet the above requirements, thermal engineering calculations are performed, and heat resistance, vapor permeability, air permeability and moisture transfer are calculated according to the requirements of regulatory documentation.

Thermal qualities

From the thermal characteristics of external structural elements buildings depends:

• Humidity conditions 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 warm seasons.
• The amount of heat lost by a building in winter period time.

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

Problems of thermal engineering calculations

So, the thermal engineering calculation of enclosing structural elements is carried out with the aim of:

1. Compliance of structures with modern requirements for thermal protection of buildings and structures.
2. Provisions for interior spaces 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 thermal engineering calculation of the building, it is necessary to take into account many parameters depending on the following characteristics:

• Purpose and type of building.
• Geographical location of the building.
• Orientation of walls according to cardinal directions.
• Dimensions of structures (volume, area, number of floors).
• Type and sizes of windows and doors.
• Characteristics of the heating system.
• The number of people in the building at the same time.
• Material of walls, floors and ceilings of the last floor.
• Availability of hot water supply system.
• Type of ventilation systems.
• Other design features buildings.

Thermal engineering calculation: program

To date, many programs have been developed 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.

Thermal engineering calculation: example calculation for external walls

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

• t in = 20°C is the temperature of the air flow inside the building, which is taken to calculate fences based on 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 the room will be provided with normal humidity conditions.
• In accordance with Appendix B of SNiP 02/23/2003, the humidity zone is dry, which means that the operating conditions for the fences are A.
• t n = -34 °C is the temperature of the external air flow in the winter, which is accepted according to SNiP based on the coldest five-day period, which has a probability of 0.92.
• Z from.per = 220 days - this is the duration heating season, which is accepted according to SNiP, while the average daily temperature environment≤ 8 °C.
• T from.trans. = -5.9 °C is the ambient temperature (average) during the heating period, which is accepted according to SNiP, with a daily ambient temperature ≤ 8 °C.

Initial data

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

Comfortable conditions

Let's look at how thermal engineering calculations are performed outer wall. First, you should calculate the required heat transfer resistance, focusing on comfortable and sanitary conditions:

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

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

Δt n = 4.5 °C - this is the normalized temperature difference inner surface designs and internal air. Accepted according to SNiP data from Table 5.

α in = 8.7 W/m 2 °C is the heat transfer of internal enclosing structures. The data is taken from table 5, according to SNiP.

We substitute the data into the formula and get:

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

Energy saving conditions

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

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

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

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

Thus:

GSOP = (20 - (-5.9)) ×220 = 5698.

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

R 0 tr = 2.4 + (3.0 - 2.4) × (5698 - 4000)) / (6000 - 4000)) = 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 = 23 W/m 2 °C is the heat transfer of external structural elements, accepted according to SNiP.

R0 = 1/8.7 + d/0.042+1/23 = 0.158 + d/0.042.

Insulation thickness

Thickness thermal insulation material is determined based on the fact that R 0 = 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 the optimal thickness of the thermal insulation material: DP 120, while the total thickness of the panel should be 120 mm. Thermal engineering calculations of the building as a whole are carried out in a similar way.

The need to perform a calculation

Designed on the basis of thermal engineering calculations, carried out competently, enclosing structures can reduce heating costs, the cost of which regularly increases. In addition, saving heat is considered an important environmental task, because it is directly related to reducing fuel consumption, which leads to a reduction in the impact of 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 in their activities they strive to use modern technologies and materials. Only a specialist can understand the need to use a particular material, both separately and in combination with others. It is the thermotechnical calculation that will help determine the most optimal solutions, which will ensure the durability of structural elements and minimal financial costs.

To keep your home warm during the most very coldy, it is necessary to choose the right thermal insulation system - for this, a thermal technical calculation of the outer wall is performed. The result of the calculations shows how effective the real or designed insulation method is.

How to make a thermal engineering calculation of an external wall

First, you should prepare the initial data. The following factors influence the calculated parameter:

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

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

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

As an example, the heat-protective 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 lime-sand mortar are calculated. Calculations and selection of tabular data are based on building regulations.Initial data: construction area - Moscow; relative humidity - 55%, average temperature in the house tв = 20О С. 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 thermal engineering calculation of an external wall is to determine the required (Rtr) and actual (Rph) heat transfer resistance.
Calculation

1. According to Table 1 SP 53.13330.2012, under given conditions, the humidity regime is assumed to be normal. The required value of Rtr is found using the formula:
   Rtr=a GSOP+b,
   where a, b are taken according to table 3 SP 50.13330.2012. For a residential building and an external wall a = 0.00035; b = 1.4.
   GSOP – degree-days of the heating period, they are found using formula (5.2) SP 50.13330.2012:
   GSOP=(tv-tot)zot,
   where tв=20О С; tot – average outside air temperature during the heating period, according to Table 1 SP131.13330.2012 tot = -2.2°C; zfrom = 205 days. (duration heating season according to the same table).
   Substituting the table values, they find: GSOP = 4551О С*day; Rtr = 2.99 m2*C/W
2. According to Table 2 SP50.13330.2012 for normal humidity, the thermal conductivity coefficients of each layer of the “pie” are selected: λB1 = 0.81 W/(m°C), λB2 = 0.26 W/(m°C), λB3 = 0.041 W/(m° C), λB4=0.81 W/(m°C).
   Using formula E.6 SP 50.13330.2012, the conditional heat transfer resistance is determined:
   R0condition=1/αint+δn/λn+1/αext.
   where αext = 23 W/(m2°C) from clause 1 of table 6 SP 50.13330.2012 for external walls.
   Substituting the numbers, we get R0cond=2.54m2°C/W. It is clarified using the coefficient r=0.9, depending on the homogeneity of the structures, the presence of ribs, reinforcement, and cold bridges:
   Rf=2.54 0.9=2.29m2 °C/W.

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

Thermal calculation of an external 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.

1. "TeReMok". The initial data is entered: type of building (residential), internal temperature 20O, humidity regime - normal, area of ​​residence - Moscow. The next window opens the calculated value of the standard heat transfer resistance - 3.13 m2*оС/W.
   Based on the calculated coefficient, a thermal engineering calculation is made of an external wall made of foam blocks (600 kg/m3), insulated with extruded polystyrene foam “Flurmat 200” (25 kg/m3) and plastered with cement-lime mortar. Select from the menu necessary materials, indicating their thickness (foam block - 200 mm, plaster - 20 mm), leaving the cell with the thickness of the insulation unfilled.
   By clicking the “Calculation” button, the required thickness of the heat insulation layer is obtained – 63 mm. The convenience of the program does not eliminate its shortcoming: it does not take into account different thermal conductivity masonry material and mortar. Thanks to the author you can say at this address http://dmitriy.chiginskiy.ru/teremok/
2. 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 uniformity 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 out the fields, the program issues a report on what the actual thermal resistance of the selected structure is and whether it meets the climatic conditions. In addition, a sequence of calculations with formulas, normative sources and intermediate values ​​is provided.

When building a house or carrying out thermal insulation works It is important to assess the effectiveness of insulation of an external wall: a thermal calculation performed independently or with the help of a specialist allows you to do this quickly and accurately.

A long time ago, buildings and structures were built without thinking about what thermal conductivity qualities the enclosing structures had. In other words, the walls were simply made thick. And if you have 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 a completely comfortable stay for people in these houses, even in the most severe frosts.

Nowadays 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. Some of them: insulation 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 optimal thickness of this material, a thermal engineering calculation is carried out and the dew point is determined.

You can find out how to calculate the dew point on the next page. Thermal engineering calculations will also be considered here using an example.

Required regulatory documents

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

• SNiP 23-02-2003 (SP 50.13330.2012). " Thermal protection buildings". Updated edition from 2012.
• SNiP 23-01-99* (SP 131.13330.2012). "Building 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 a building. Reference manual".

Calculated parameters

In the process of performing thermal engineering calculations, the following is determined:

• thermal characteristics of building materials of 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. Local climate and indoor microclimate

Construction area: Nizhny Novgorod.

Purpose of the building: residential.

The calculated relative humidity of the internal air under the condition of no condensation on the internal surfaces of external fences is equal to - 55% (SNiP 23-02-2003 clause 4.3. Table 1 for normal humidity conditions).

The optimal air temperature in the living room is cold period year t int = 20°С (GOST 30494-96 table 1).

Estimated outside air temperature t ext, determined by the temperature of the coldest five-day period with a probability of 0.92 = -31°C (SNiP 23-01-99 table 1 column 5);

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

Average outside air temperature for the heating period t ht = -4.1°C (SNiP 23-01-99 table 1 column 12).

2. Wall design

The wall consists of the following layers:

• Decorative brick (besser) 90 mm thick;
• insulation (mineral wool board), in the figure its thickness is indicated by the sign “X”, since it will be found during the calculation process;
• sand-lime brick thickness 250 mm;
• plaster (complex mortar), an additional layer to obtain a more objective picture, since its influence is minimal, but it exists.

3. Thermophysical characteristics of materials

The values ​​of the material characteristics are summarized in the table.

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

Calculation

4. Determination of insulation thickness

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

4.1. Determination of the thermal protection standard based on energy saving conditions

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

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

Note: degree days are also designated GSOP.

The standard value of the reduced heat transfer resistance should be taken no less than the standardized values ​​determined according to SNIP 23-02-2003 (Table 4) depending on the degree-day of the construction area:

R req = a×D d + b = 0.00035 × 5182 + 1.4 = 3.214m2 × °C/W,

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

a and b - coefficients accepted 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 thermal protection standards based on sanitation conditions

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 use (autumn or spring), as well as buildings with a design internal air temperature of 12 ° C and a lower heat transfer resistance of enclosing structures (with the exception of translucent ones).

Determination of the standard (maximum permissible) resistance to heat transfer according to sanitation conditions (formula 3 SNiP 02/23/2003):

where: n = 1 - coefficient adopted according to Table 6 for the outer wall;

t int = 20°С - value from the original data;

t ext = -31°С - value from the original data;

Δt n = 4°С - the normalized temperature difference between the temperature of the internal air and the temperature of the internal surface of the enclosing structure, taken according to Table 5 in this case for the external walls of residential buildings;

α int = 8.7 W/(m 2 ×°C) - heat transfer coefficient of the internal surface of the enclosing structure, taken according to Table 7 for external walls.

4.3. Thermal protection standard

From the above calculations, for the required heat transfer resistance we select R req from the energy saving condition and now denote it R tr0 = 3.214 m 2 × °C/W .

5. Determination of insulation thickness

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 is the calculated thermal conductivity coefficient of the layer material W/(m × °C).

1 layer ( decorative brick): R 1 = 0.09/0.96 = 0.094 m 2 × °C/W .

Layer 3 (sand-lime brick): R 3 = 0.25/0.87 = 0.287 m2 × °C/W .

4th layer (plaster): R 4 = 0.02/0.87 = 0.023 m2 × °C/W .

Determination of the minimum permissible (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 - heat transfer resistance on the inner surface;

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

ΣR i = 0.094 + 0.287 + 0.023 - the sum of the thermal resistances of all layers of the wall without a layer of insulation, determined taking into account the thermal conductivity coefficients of materials adopted 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 °C /W

The thickness of the insulation is equal to (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 is the sum of the thermal resistances of all layers of the fence, including the insulation layer, of the accepted structural thickness, m 2 °C/W.

From the obtained result we can conclude that

R 0 = 3.503 m 2 × °C/W> R tr0 = 3.214m 2 × °C/W→ therefore, the thickness of the insulation is selected Right.

Effect of air gap

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

This air gap 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) the layers of the structure located between the air gap and the outer surface (in our case, this is decorative brick (besser)) are not taken into account in the thermal engineering calculation;

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

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

In the climatic conditions of northern latitudes, a correctly made thermal calculation of a building is extremely important for builders and architects. The obtained indicators will provide the necessary information for design, including about the materials used for construction, additional insulation, floors and even finishing.

In general, heat calculation affects several procedures:

• taking into account by designers when planning the layout of rooms, load-bearing walls and fencing;
• creation of a heating system and ventilation structure project;
• selection of building materials;
• analysis of the operating conditions of the building.

All this is connected by single values ​​​​obtained as a result of settlement operations. In this article we will tell you how to make a thermal calculation of the outer wall of a building, and also give examples of using this technology.

Objectives of the procedure

A number of goals are relevant only for residential buildings or, conversely, industrial premises, but most of the problems solved are suitable for all buildings:

• Maintaining comfortable climatic conditions inside the rooms. The term “comfort” includes both the heating system and natural conditions heating the surface of walls, roofs, using all heat sources. The same concept includes the air conditioning system. Without proper ventilation, especially in production, the premises will be unsuitable for work.
• Saving electricity and other heating resources. The following meanings apply here:
  • specific heat capacity of the materials and cladding used;
  • climate outside the building;
  • heating power.

It is extremely uneconomical to carry out heating system, which simply will not be used to the proper extent, but will be difficult to install and expensive to maintain. The same rule can be applied to expensive building materials.

Thermal engineering calculation - what is it?

Heat calculation allows you to set the optimal (two boundaries - minimum and maximum) thickness of the enclosing walls and load-bearing structures which will provide long-term operation without freezing and overheating of ceilings and partitions. In other words, this procedure allows you to calculate the real or expected, if it is carried out at the design stage, the thermal load of the building, which will be considered the norm.

The analysis is based on the following data:

• room design - presence of partitions, heat-reflecting elements, ceiling height, etc.;
• features of the climate regime in a given area - maximum and minimum temperature limits, the difference and rapidity of temperature changes;
• the location of the building in the cardinal directions, that is, taking into account the absorption of solar heat, at what time of day there is maximum susceptibility of heat from the sun;
• mechanical influences and physical properties construction site;
• indicators of air humidity, the presence or absence of protection of walls from moisture penetration, the presence of sealants, including sealing impregnations;
• work natural or artificial ventilation, the presence of the “greenhouse effect”, vapor permeability and much more.

At the same time, the assessment of these indicators must comply with a number of standards - the level of resistance to heat transfer, air permeability, etc. Let's consider them in more detail.

Requirements for thermal engineering calculations of the premises and related documentation

State inspection bodies that govern the organization and regulation of construction, as well as checking the implementation of safety regulations, have drawn up SNiP No. 23-02-2003, which sets out in detail the standards for carrying out measures for the thermal protection of buildings.

The document proposes engineering solutions that will ensure the most economical consumption of heat energy, which is spent on heating premises (residential or industrial, municipal) during the heating period. These recommendations and requirements have been developed taking into account ventilation, air conversion, and the location of heat entry points.

SNiP is a bill at the federal level. Regional documentation is presented in the form of TSN - territorial construction standards.

Not all buildings are within the jurisdiction of these codes. In particular, those buildings that are heated irregularly or are constructed without heating are not checked according to these requirements. Heat calculations are mandatory for the following buildings:

• residential - private and apartment buildings;
• public, municipal - offices, schools, hospitals, kindergartens, etc.;
• industrial – factories, concerns, elevators;
• agricultural - any heated buildings for agricultural purposes;
• warehouses – barns, warehouses.

The text of the document specifies standards for all those components that are included in thermal analysis.

Design requirements:

• Thermal insulation. This is not only the preservation of heat in the cold season and the prevention of hypothermia and freezing, but also protection from overheating in the summer. Isolation, therefore, must be two-way - preventing influences from the outside and the release of energy from within.
• The permissible value of the temperature difference between the atmosphere inside the building and the thermal regime of the interior of the enclosing structures. This will lead to the accumulation of condensation on the walls, as well as negative influence on the health of people in the premises.
• Thermal stability, that is, temperature stability, preventing sudden changes in the heated air.
• Breathability. Balance is important here. On the one hand, the building cannot be allowed to cool down due to active heat transfer; on the other hand, it is important to prevent the occurrence of the “greenhouse effect”. It happens when synthetic, “non-breathable” insulation is used.
• No dampness. High humidity– this is not only the reason for the appearance of mold, but also an indicator due to which serious losses of heat energy occur.

How to do thermal engineering calculations of the walls of a house - basic parameters

Before proceeding with direct heat calculations, you need to collect detailed information about the construction. The report will include answers to the following points:

• The purpose of the building is residential, industrial or public premises, specific purpose.
• The geographic latitude of the area where the facility is or will be located.
• Climatic features of the area.
• The direction of the walls is to the cardinal points.
• Dimensions entrance structures And window frames- their height, width, permeability, type of windows - wooden, plastic, etc.
• Power of heating equipment, layout of pipes, batteries.
• The average number of residents or visitors, workers, if these are industrial premises that are located inside the walls at the same time.
• Building materials from which floors, ceilings and any other elements are made.
• Presence or absence of supply hot water, the type of system that is responsible for this.
• Features of ventilation, both natural (windows) and artificial - ventilation shafts, air conditioning.
• The configuration of the entire building - the number of floors, the total and individual area of ​​​​the premises, the location of the rooms.

Once this data has been collected, the engineer can begin calculations.

We offer you three methods that are most often used by specialists. You can also use a combined method, when facts are taken from all three possibilities.

Options for thermal calculation of enclosing structures

Here are three indicators that will be taken as the main ones:

• building area from the inside;
• volume outside;
• specialized thermal conductivity coefficients of materials.

Heat calculation by area of ​​premises

Not the most economical, but the most frequent, especially in Russia, method. It involves primitive calculations based on the area indicator. This does not take into account climate, band, minimum and maximum temperature values, humidity, etc.

Also, the main sources of heat loss are not taken into account, such as:

• Ventilation system – 30-40%.
• Roof slopes – 10-25%.
• Windows and doors – 15-25%.
• Walls – 20-30%.
• Floor on the ground – 5-10%.

These inaccuracies are due to failure to take into account the majority important elements lead to the fact that the heat calculation itself may have a strong error in both directions. Usually engineers leave a “reserve”, so they have to install this heating equipment, which is not fully used or threatens severe overheating. There are often cases when heating and air conditioning systems are installed at the same time, because they cannot correctly calculate heat loss and heat gain.

“Bigger” indicators are used. Disadvantages of this approach:

• expensive heating equipment and materials;
• uncomfortable indoor microclimate;
• additional installation automated control over temperature conditions;
• possible freezing of walls in winter.

Q=S*100 W (150 W)

• Q is the amount of heat required for a comfortable climate in the entire building;
• W S – heated area of ​​the room, m.

The value of 100-150 Watts is a specific indicator of the amount of thermal energy required to heat 1 m2.

If you choose this method, then listen to the following tips:

• If the height of the walls (to the ceiling) is no more than three meters, and the number of windows and doors per surface is 1 or 2, then multiply the result by 100 W. Typically, all residential buildings, both private and apartment buildings, use this value.
• If the design contains two window openings or a balcony, loggia, then the indicator increases to 120-130 W.
• For industrial and warehouse premises, a coefficient of 150 W is more often taken.
• When choosing heating devices(radiators), if they are located near a window, it is worth increasing their designed power by 20-30%.

Thermal calculation of enclosing structures according to the volume of the building

Typically this method is used for those buildings where high ceilings– more than 3 meters. That is, industrial facilities. The disadvantage of this method is that air conversion is not taken into account, that is, the fact that it is always warmer at the top than at the bottom.

Q=V*41 W (34 W)

• V – external volume of the building in cubic meters;
• 41 W is the specific amount of heat required to heat one cubic meter of a building. If construction is carried out using modern building materials, then the figure is 34 W.

• Glass in windows:
  • double package – 1;
  • binding – 1.25.
• Insulation materials:
  • new modern developments – 0.85;
  • standard brickwork in two layers – 1;
  • small wall thickness – 1.30.
• Air temperature in winter:
  • -10 – 0,7;
  • -15 – 0,9;
  • -20 – 1,1;
  • -25 – 1,3.
• Percentage of windows compared to total surface area:
  • 10% – 0,8;
  • 20% – 0,9;
  • 30% – 1;
  • 40% – 1,1;
  • 50% – 1,2.

All these errors can and should be taken into account, however, they are rarely used in real construction.

An example of thermal engineering calculation of the external building envelope by analyzing the insulation used

If you are building a residential building or cottage yourself, we strongly recommend that you think through everything down to the smallest detail in order to ultimately save money, create an optimal climate inside, and ensure long-term operation of the facility.

To do this, you need to solve two problems:

• make the correct heat calculation;
• install a heating system.

Example data:

• corner living room;
• one window – 8.12 sq. m;
• region – Moscow region;
• wall thickness – 200 mm;
• area according to external parameters – 3000*3000.

It is necessary to find out how much power is needed to heat 1 square meter of space. The result will be Qsp = 70 W. If the insulation (wall thickness) is smaller, the values ​​will also be lower. Let's compare:

• 100 mm – Qsp = 103 W.
• 150 mm – Qsp = 81 W.

This indicator will be taken into account when installing heating.

Software for heating system design

By using computer programs from the ZVSOFT company you can calculate all the materials spent on heating, as well as make a detailed floor plan of communications with radiators displayed, specific heat capacity, energy consumption, nodes.

The company offers basic CAD for design work of any complexity - . In it you can not only design a heating system, but also create detailed diagram for the construction of the entire house. This can be realized thanks to the large functionality, number of tools, as well as work in two- and three-dimensional space.

You can install an add-on to the base software. This program is designed to design all engineering systems, including for heating. Using easy line tracing and the function of layering plans, you can design several communications on one drawing - water supply, electricity, etc.

Before building a house, do a thermal engineering calculation. This will help you not make a mistake with the choice of equipment and the purchase of building materials and insulation.

The purpose of the thermal engineering calculation is to calculate the thickness of the insulation for a given thickness of the load-bearing part of the outer wall, which meets sanitary and hygienic requirements and energy saving conditions. In other words, we have external walls 640 mm thick made of sand-lime brick and we are going to insulate them with polystyrene foam, but we don’t know what thickness of insulation we need to choose in order to comply with building standards.

Thermal engineering calculations of the outer wall of a building are carried out in accordance with SNiP II-3-79 “Building Heat Engineering” and SNiP 23-01-99 “Building Climatology”.

Table 1

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

Scheme no.	Material	Characteristics of the material in a dry state		Design coefficients (subject to operation according to Appendix 2) SNiP II-3-79*
		Density γ 0, kg/m 3	Thermal conductivity coefficient λ, W/m*°С	Thermal conductivity λ, W/m*°С		Heat absorption (with a period of 24 hours) S, m 2 *°C/W
	Cement-sand mortar (item 71)	1800	0.57	0.76	0.93		11.09
	Brickwork made of solid silicate brick (GOST 379-79) on cement-sand mortar (item 87)	1800	0.88	0.76	0.87	9.77	10.90
	Expanded polystyrene (GOST 15588-70) (item 144)		0.038	0.038	0.041	0.41	0.49
	Cement-sand mortar - thin layer plaster(pos. 71)	1800	0.57	0.76	0.93		11.09

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

2-brick wall (sand-lime brick) - 640 mm

3-insulation (expanded polystyrene)

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

When performing thermal engineering calculations, the normal humidity regime in the premises was adopted - operating conditions (“B”) in accordance with SNiP II-3-79 t.1 and adj. 2, i.e. We take the thermal conductivity of the materials used according to column “B”.

Let's 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 = (t in – t n) * n / Δ t n *α in (1)

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

corresponding buildings and structures, we take equal to +22 °C for residential buildings in accordance with Appendix 4 to SNiP 2.08.01-89;

t n – calculated winter temperature outside air, °C, equal to the average temperature of the coldest five-day period, supply 0.92 according to SNiP 23-01-99 for the city of Yaroslavl is taken equal to -31 °C;

n – 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 - standard and temperature difference between the temperature of the internal air and the temperature of the internal surface of the enclosing structure - is established according to SNiP II-3-79* (Table 2*) and is taken equal to Δ t n = 4.0 °C;

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

Let us determine the degree-day of the heating period using the formula:

GSOP= (t in – t from.trans.)*z from.trans. (2)

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

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

GSOP=(22-(-4))*221=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, hygienic and comfortable conditions. Intermediate values ​​are determined by interpolation.

table 2

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

Buildings and premises	Degree-days of the heating period, ° C*days	Reduced resistance to heat transfer of walls, not less than R 0 tr (m 2 *°C)/W
Public administrative and domestic, with the exception of rooms with damp or wet conditions	5746	3,41

We take the heat transfer resistance of enclosing structures R(0) as the greatest of the values ​​​​calculated earlier:

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

Let us 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 enclosure from the condition:

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

where δ i is the thickness of individual layers of the fence other than the calculated one in m;

λ i – thermal conductivity coefficients of individual fencing layers (except for the design 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 in m;

λ x – thermal conductivity coefficient of the design 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 internal surface of enclosing structures is taken according to SNiP II-3-79* (Table 4*) and is taken equal to α in = 8.7 W/m 2 *°C.

α n - heat transfer coefficient (for winter conditions) of the outer surface of the enclosing structure is taken according to SNiP II-3-79* (Table 6*) and is taken equal to α n = 23 W/m 2 *°C.

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

For external walls and ceilings, the thickness of the thermal insulation 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 be no less than the standardized 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) + δ x / 0,041 + 1/ 8,7

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

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

δ x = 0.10 m

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

Determine the actual heat transfer resistance calculated enclosing structures R 0, taking into account the accepted thickness of the thermal insulation 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 = 3.43 (m 2 *°C)/W

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