Thermotechnical calculation with an example. Thermal engineering calculation of structures: what is it and how is it carried out

In order for the dwelling to be warm in the most severe frosts, it is necessary to choose the right thermal insulation system - for this, a heat engineering calculation is performed outer wall.The result of the calculations shows how effective the actual or projected method of insulation is.

How to make a thermal calculation of the outer wall

First you need to prepare the initial data. The following factors influence the design parameter:

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

After collecting and recording the initial information, the thermal conductivity coefficients 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, moisture maps compiled for Russian Federation. After that, all the numerical values ​​\u200b\u200bnecessary for the calculation are entered into the appropriate formulas.

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

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

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

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

Thermotechnical calculation of the outer wall, the program simplifies calculations

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

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

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

If you are going to build
a small brick cottage, then of course you will have questions: “What
should the wall be thick?”, “Do I need insulation?”, “Which side to put
heater? etc. and so on.

In this article, we will try to
figure it out and answer all your questions.

Thermal engineering calculation
enclosing structure is needed, first of all, in order to find out which
thickness should be your outer wall.

First, you need to decide how much
floors will be in your building and depending on this, the calculation is made
building envelopes by bearing capacity (not in this article).

Based on this calculation, we determine
the number of bricks in your building's masonry.

For example, it turned out 2 clay
bricks without voids, brick length 250 mm,
mortar thickness 10 mm, total 510 mm (brick density 0.67
will be useful to us later). You decide to cover the outer surface
facing tiles, thickness 1 cm (when buying, be sure to find out
density), and the inner surface with ordinary plaster, layer thickness 1.5
cm, also do not forget to find out its density. In total 535mm.

In order for the building to
collapsed, of course, enough, but unfortunately in most cities
Russian winters are cold and therefore such walls will freeze through. And not to
the walls are frozen, need another layer of insulation.

The thickness of the insulation layer is calculated
in the following way:

1. On the Internet you need to download SNiP
II 3-79* —
"Construction heat engineering" and SNiP 23-01-99 - "Construction climatology".

2. We open SNiP construction
climatology and find your city in table 1 *, and look at the value at the intersection
column "Air temperature of the coldest five-day period, ° С, security
0.98" and strings with your city. For the city of Penza, for example, t n \u003d -32 o C.

3. Estimated indoor air temperature
take

t in = 20 o C.

Heat transfer coefficient for interior wallsa c \u003d 8.7 W / m 2 ˚С

Heat transfer coefficient for external walls in winter conditionsa n \u003d 23W / m 2 ˚С

Normative temperature difference between the temperature of the internal
air and the temperature of the inner surface of the enclosing structuresΔ t n \u003d 4 o C.

4. Next
we determine the required resistance to heat transfer according to the formula # G0 (1a) from building heat engineering
GSOP = (t in - t from.per.) z from.per , GSOP=(20+4.5) 207=507.15 (for the city
Penza).

According to formula (1) we calculate:

(where sigma is directly the thickness
material, and lambda density. Itook as a heater
polyurethane foampanels with a density of 0.025)

We take the thickness of the insulation equal to 0.054 m.

Hence the wall thickness will be:

d = d 1 + d 2 + d 3 + d 4 =

0,01+0,51+0,054+0,015=0,589
m.

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Initial data

Place of construction - Omsk

z ht = 221 days

t ht = -8.4ºС.

t ext = -37ºС.

t int = + 20ºС;

air humidity: = 55%;

Operating conditions of enclosing structures - B. Heat transfer coefficient of the inner surface of the fence A i nt \u003d 8.7 W / m 2 ° С.

a ext \u003d 23 W / m 2 ° C.

The necessary data on the structural layers of the wall for thermal calculation are summarized in the table.

1. Determination of degree-days of the heating period according to the formula (2) SP 23-101-2004:

D d \u003d (t int - t ht) z th \u003d (20–(8.4)) 221 \u003d 6276.40

2. The normalized value of the heat transfer resistance of the outer walls according to the formula (1) SP 23-101-2004:

R reg \u003d a D d + b \u003d 0.00035 6276.40+ 1.4 \u003d 3.6 m 2 ° C / W.

3. Reduced resistance to heat transfer R 0 r external brick walls with effective insulation residential buildings is calculated by the formula

R 0 r = R 0 arb r,

where R 0 conv - heat transfer resistance of brick walls, conditionally determined by formulas (9) and (11) without taking into account heat-conducting inclusions, m 2 ·°С / W;

R 0 r - reduced resistance to heat transfer, taking into account the coefficient of thermal uniformity r, which for walls is 0.74.

The calculation is carried out from the condition of equality

hence,

R 0 conditional \u003d 3.6 / 0.74 \u003d 4.86 m 2 ° C / W

R 0 conv \u003d R si + R k + R se

R k \u003d R reg - (R si + R se) \u003d 3.6- (1 / 8.7 + 1/23) \u003d 3.45 m 2 ° C / W

4. Thermal resistance of outer brick wall layered structure can be represented as the sum of the thermal resistances of the individual layers, i.e.

R to \u003d R 1 + R 2 + R ut + R 4

5. Determine the thermal resistance of the insulation:

R ut \u003d R k + (R 1 + R 2 + R 4) \u003d 3.45– (0.037 + 0.79) \u003d 2.62 m 2 ° С / W.

6. Find the thickness of the insulation:

Ri

\u003d R ut \u003d 0.032 2.62 \u003d 0.08 m.

We accept the thickness of the insulation 100 mm.

The final wall thickness will be (510+100) = 610 mm.

We perform a check taking into account the accepted thickness of the insulation:

R 0 r \u003d r (R si + R 1 + R 2 + R ut + R 4 + R se) \u003d 0.74 (1 / 8.7 + 0.037 + 0.79 + 0.10 / 0.032 + 1/23 ) \u003d 4.1m 2 ° C / W.

Condition R 0 r \u003d 4.1> \u003d 3.6m 2 ° C / W is performed.

Checking compliance with sanitary and hygienic requirements

building thermal protection

1. Check the condition :

∆t = (t int- t ext)/ R 0r a int \u003d (20-(37)) / 4.1 8.7 \u003d 1.60 ºС

According to Table. 5SP 23-101-2004 ∆ t n = 4 °C, therefore, the condition ∆ t = 1,60< ∆t n = 4 ºС is fulfilled.

2. Check the condition :

] = 20 – =

20 - 1.60 = 18.40ºС

3. According to Appendix Sp 23-101–2004 for indoor air temperature t int = 20 ºС and relative humidity = 55% dew point temperature t d = 10.7ºС, therefore, the condition τsi = 18.40> t d= performed.

Conclusion. The enclosing structure satisfies regulatory requirements thermal protection of the building.

4.2 Thermotechnical calculation of attic roofing.

Initial data

Determine the thickness of the attic floor insulation, consisting of insulation δ = 200 mm, vapor barrier, prof. sheet

Attic floor:

Combined coverage:

Place of construction - Omsk

The length of the heating period z ht = 221 days.

Average design temperature of the heating period t ht = -8.4ºС.

The temperature of the cold five-day t ext = -37ºС.

The calculation was made for a five-story residential building:

indoor air temperature t int = + 20ºС;

air humidity: = 55%;

the humidity regime of the room is normal.

Operating conditions of enclosing structures - B.

Heat transfer coefficient of the inner surface of the fence A i nt \u003d 8.7 W / m 2 ° С.

Heat transfer coefficient of the outer surface of the fence a ext \u003d 12 W / m 2 ° C.

Name of material Y 0 , kg / m³ δ , m λ , mR, m 2 ° С / W

1. Determination of degree-days of the heating period according to the formula (2) SP 23-101-2004:

D d \u003d (t int - t ht) z th \u003d (20 -8.4) 221 \u003d 6276.4 ° C day

2. Rationing the value of the resistance to heat transfer of the attic floor according to the formula (1) SP 23-101-2004:

R reg \u003d a D d + b, where a and b are selected according to table 4 of SP 23-101-2004

R reg \u003d a D d + b \u003d 0.00045 6276.4+ 1.9 \u003d 4.72 m² ºС / W

3. Thermal engineering calculation is carried out from the condition that the total thermal resistance R 0 is equal to the normalized R reg , i.e.

4. From formula (8) SP 23-100-2004 we determine the thermal resistance of the building envelope R k (m² ºС / W)

R k \u003d R reg - (R si + R se)

Rreg = 4.72m² ºС / W

R si \u003d 1 / α int \u003d 1 / 8.7 \u003d 0.115 m² ºС / W

R se \u003d 1 / α ext \u003d 1/12 \u003d 0.083 m² ºС / W

R k \u003d 4.72– (0.115 + 0.083) \u003d 4.52 m² ºС / W

5. The thermal resistance of the building envelope (attic floor) can be represented as the sum of the thermal resistances of individual layers:

R k \u003d R cb + R pi + R tss + R ut → R ut \u003d R c + (R cb + R pi + R cs) \u003d R c - (d / λ) \u003d 4.52 - 0.29 \u003d 4 .23

6. Using the formula (6) SP 23-101-2004, we determine the thickness of the insulating layer:

d ut = R ut λ ut = 4.23 0.032= 0.14 m

7. We accept the thickness of the insulating layer 150mm.

8. We consider the total thermal resistance R 0:

R 0 \u003d 1 / 8.7 + 0.005 / 0.17 + 0.15 / 0.032 + 1 / 12 \u003d 0.115 + 4.69 + 0.083 \u003d 4.89m² ºС / W

R 0 ≥ R reg 4.89 ≥ 4.72 satisfies the requirement

Condition check

1. Check the fulfillment of the condition ∆t 0 ≤ ∆t n

The value of ∆t 0 is determined by the formula (4) SNiP 23-02-2003:

∆t 0 = n (t int - t ext) / R 0 a int 6

∆t 0 \u003d 1 (20 + 37) / 4.89 8.7 \u003d 1.34ºС

According to Table. (5) SP 23-101-2004 ∆t n = 3 ºС, therefore, the condition ∆t 0 ≤ ∆t n is fulfilled.

2. Check the fulfillment of the condition τ >t d

Value τ we calculate according to the formula (25) SP 23-101-2004

tsi = t int– [n(t int–text)]/(R o a int)

τ \u003d 20- 1 (20 + 26) / 4.89 8.7 \u003d 18.66 ºС

3. According to Appendix R SP 23-01-2004 for indoor air temperature t int = +20 ºС and relative humidity φ = 55% dew point temperature t d = 10.7 ºС, therefore, the condition τ >t d is executed.

Conclusion: attic floor meets regulatory requirements.

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

In general, heat calculation affects several procedures:

• consideration by designers when planning the location of rooms, bearing walls and fences;
• creation of a project for a heating system and ventilation facilities;
• 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, as well as give examples of the use of this technology.

Tasks of the procedure

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

• Preservation of comfortable climatic conditions inside the rooms. The term "comfort" includes both the heating system and the natural conditions for heating the surface of walls, roofs, and the use of 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 resources for heating. The following values ​​take place here:
  • specific heat capacity of the materials and skins used;
  • climate outside the building;
  • heating power.

It is extremely uneconomical to install a heating system that simply will not be used to the right extent, but will be difficult to install and expensive to maintain. The same rule can be attributed to expensive building materials.

Thermotechnical calculation - what is it

The heat calculation allows you to set the optimal (two boundaries - minimum and maximum) thickness of the walls of enclosing and supporting structures, which will ensure long-term operation without freezing and overheating of floors and partitions. In other words, this procedure allows you to calculate the real or assumed, 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:

• the design of the room - the presence of partitions, heat-reflecting elements, ceiling height, etc .;
• features of the climatic regime in a given area - maximum and minimum temperature limits, difference and rapidity of temperature changes;
• the location of the structure on the cardinal points, that is, taking into account absorption solar heat, at what time of the day is the maximum susceptibility of heat from the sun;
• mechanical influences and physical properties building object;
• indicators of air humidity, the presence or absence of protection of walls from moisture penetration, the presence of sealants, including sealing impregnations;
• the work of natural or artificial ventilation, the presence of a "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 us consider them in more detail.

Requirements for the heat engineering calculation of the premises and related documentation

State inspection bodies that manage the organization and regulation of construction, as well as checking the implementation of safety precautions, compiled SNiP No. 23-02-2003, which details the norms for carrying out measures for the thermal protection of buildings.

The document proposes engineering solutions that will provide the most economical consumption heat energy that is spent on heating premises (residential or industrial, municipal) during the heating season. These guidelines and requirements have been developed with regard to 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 building codes.

Not all buildings fall within the jurisdiction of these vaults. In particular, those buildings that are heated irregularly or are completely constructed without heating are not checked according to these requirements. Mandatory heat calculation is 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;
• storage - barns, warehouses.

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

Design requirements:

• Thermal insulation. This is not only the preservation of heat in the cold season and the prevention of hypothermia, freezing, but also protection against overheating in the summer. Isolation, therefore, must be mutual - prevention of influences from outside and return 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 building envelope. This will lead to the accumulation of condensation on the walls, as well as to negative influence on the health of people in the room.
• Heat resistance, that is, temperature stability, preventing sudden changes in the heated air.
• Breathability. Balance is important here. On the one hand, it is impossible to allow the building to cool down due to active heat transfer, on the other hand, it is important to prevent the appearance of the "greenhouse effect". It happens when synthetic, "non-breathing" insulation is used.
• Absence of dampness. High humidity is not only a reason for the appearance of mold, but also an indicator due to which serious losses of heat energy occur.

How to make a thermal calculation of the walls of the house - the main parameters

Before proceeding with the direct heat calculation, you need to collect detailed information about the building. The report will include responses to the following items:

• The purpose of the building is residential, industrial or public premises, a specific purpose.
• Geographic latitude of the area where the object is located or will be located.
• Climatic features of the area.
• The direction of the walls to the cardinal points.
• Dimensions input structures And window frames- their height, width, permeability, type of windows - wooden, plastic, etc.
• The power of heating equipment, the layout of pipes, batteries.
• The average number of residents or visitors, workers, if these are industrial premises that are inside the walls at a 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 ​​\u200b\u200bthe premises, the location of the rooms.

When these data are collected, the engineer can proceed to the calculation.

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

Variants of thermal calculation of enclosing structures

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

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

Heat calculation by area

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 the 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 the neglect of the majority important elements lead to the fact that the heat calculation itself can have a strong error in both directions. Usually, engineers leave a "reserve", so you have to install such heating equipment that is not fully activated or threatens severe overheating. It is not uncommon for a heating and air conditioning system to be installed at the same time, since they cannot correctly calculate heat losses and heat gains.

Use "aggregated" indicators. Cons of this approach:

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

Q=S*100W (150W)

• 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 m.

If you choose this method, then heed the following tips:

• If the height of the walls (to the ceiling) is not more than three meters, and the number of windows and doors per surface is 1 or 2, then multiply the result by 100 watts. Usually all residential buildings, both private and multi-family, use this value.
• If the design contains two window openings or a balcony, a loggia, then the figure increases to 120-130 watts.
• For industrial and warehouse premises, a factor of 150 W is more often taken.
• When choosing heaters (radiators), if they are located near the window, it is worth adding their projected power by 20-30%.

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

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

Q=V*41W (34W)

• V is the 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 watts.

• 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 the total surface:
  • 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 a thermotechnical calculation of the external enclosing structures of a building by analyzing the used insulation

If you are building a residential building or a cottage on your own, then we strongly recommend that you think through everything to the smallest detail in order to ultimately save money and create an optimal climate inside, ensuring 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 square meters;
• 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 the room. The result will be Qsp = 70 W. If the insulation (wall thickness) is less, then the values ​​\u200b\u200bare also lower. Compare:

• 100 mm - Qsp \u003d 103 W.
• 150 mm - Qsp \u003d 81 W.

This indicator will be taken into account when laying heating.

Heating system design software

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

The firm offers basic CAD for design work any complexity. In it, you can not only design a heating system, but also create detailed diagram for the construction of the whole house. This can be realized due to the large functionality, the 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 for the design of all engineering systems, including heating. With the help of easy line tracing and the plan layering function, you can design several communications on one drawing - water supply, electricity, etc.

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

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. Such a thickness of the brick wall provided and still provides quite a comfortable stay for 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. Some 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 optimal thickness of this material, a thermotechnical 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. Help Guide" .

Calculated parameters

In the process of performing a heat engineering calculation, the following are 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. 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 sanitary norms and energy saving.

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

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

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

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

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

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

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

a and b - coefficients taken according to table 4 (if SNiP 23-02-2003) or according to table 3 (if SP 50.13330.2012) for 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 excess sensible heat of more than 23 W / m 3 and buildings intended for seasonal operation (in autumn or spring), as well as buildings with an estimated internal air temperature of 12 ° С and below the given resistance to heat transfer of enclosing structures (with the exception of translucent ones).

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

where: n \u003d 1 - coefficient 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 \u003d 0.09 / 0.96 \u003d 0.094 m 2 × °С/W .

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

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

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

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

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

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

The thickness of the insulation is (formula 5.7):

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

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

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

From the result obtained, it can be concluded that

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

Influence of the air gap

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

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

a) structural layers located between the air gap and 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.