How to calculate the thermal calculation of the wall. Thermotechnical calculation with an example. Influence of the air gap
It is required to determine the thickness of the insulation in a three-layer brick outer wall in a residential building located in Omsk. Wall construction: inner layer - brickwork of ordinary clay bricks 250 mm thick and density 1800 kg/m 3 , outer layer - brickwork of facing brick thickness 120 mm and density 1800 kg/m 3 ; between the outer and inner layers there is an effective insulation made of expanded polystyrene with a density of 40 kg / m 3; the outer and inner layers are interconnected by fiberglass flexible ties with a diameter of 8 mm, located at a step of 0.6 m.
1. Initial data
The purpose of the building is a residential building
Construction area - Omsk
Estimated indoor air temperature t int= plus 20 0 С
Estimated outdoor temperature text= minus 37 0 С
Estimated indoor air humidity - 55%
2. Determination of the normalized resistance to heat transfer
Determined according to table 4 depending on degree-days heating period. Degree-days of the heating period, D d , °С×day, determined by formula 1, based on the average outdoor temperature and the duration of the heating period.
According to SNiP 23-01-99 * we determine that in Omsk the average outdoor temperature of the heating period is equal to: t ht \u003d -8.4 0 С, duration of the heating period z ht = 221 days The degree-day value of the heating period is:
D d = (t int - tht) z ht \u003d (20 + 8.4) × 221 \u003d 6276 0 C day.
According to Table. 4. normalized resistance to heat transfer Rreg exterior walls for residential buildings corresponding to the value D d = 6276 0 С day equals Rreg \u003d a D d + b \u003d 0.00035 × 6276 + 1.4 \u003d 3.60 m 2 0 C / W.
3. Choice of constructive solution outer wall
The constructive solution of the outer wall was proposed in the task and is a three-layer fence with an inner layer of brickwork 250 mm thick, with an outer layer of brickwork 120 mm thick, between the outer and inner layers there is a polystyrene foam insulation. The outer and inner layers are interconnected by flexible fiberglass ties with a diameter of 8 mm, located at 0.6 m increments.
4. Determining the thickness of the insulation
The thickness of the insulation is determined by formula 7:
d ut \u003d (R reg ./r - 1 / a int - d kk / l kk - 1 / a ext) × l ut
where Rreg. – normalized resistance to heat transfer, m 2 0 C / W; r- coefficient of heat engineering uniformity; a int is the heat transfer coefficient of the inner surface, W / (m 2 × ° C); a ext is the heat transfer coefficient of the outer surface, W / (m 2 × ° C); d kk- the thickness of the brickwork, m; l kk- the calculated coefficient of thermal conductivity of brickwork, W/(m×°С); l ut- the calculated coefficient of thermal conductivity of the insulation, W/(m×°С).
The normalized resistance to heat transfer is determined: R reg \u003d 3.60 m 2 0 C / W.
The thermal uniformity coefficient for a brick three-layer wall with fiberglass flexible ties is about r=0.995, and may not be taken into account in the calculations (for information - if steel flexible connections are used, then the coefficient of thermal engineering uniformity can reach 0.6-0.7).
The heat transfer coefficient of the inner surface is determined from Table. 7 a int \u003d 8.7 W / (m 2 × ° C).
The heat transfer coefficient of the outer surface is taken according to table 8 a e xt \u003d 23 W / (m 2 × ° C).
The total thickness of the brickwork is 370 mm or 0.37 m.
The design coefficients of thermal conductivity of the materials used are determined depending on the operating conditions (A or B). Operating conditions are determined in the following sequence:
According to the table 1 determine the humidity regime of the premises: since the estimated temperature of the indoor air is +20 0 С, the calculated humidity is 55%, the humidity regime of the premises is normal;
According to Appendix B (map of the Russian Federation), we determine that the city of Omsk is located in a dry zone;
According to the table 2 , depending on the humidity zone and the humidity regime of the premises, we determine that the operating conditions of the enclosing structures are BUT.
App. D determine the coefficients of thermal conductivity for operating conditions A: for expanded polystyrene GOST 15588-86 with a density of 40 kg / m 3 l ut \u003d 0.041 W / (m × ° С); for brickwork from ordinary clay bricks on cement-sand mortar density 1800 kg/m 3 l kk \u003d 0.7 W / (m × ° С).
Let us substitute all the determined values into formula 7 and calculate the minimum thickness of the polystyrene foam insulation:
d ut \u003d (3.60 - 1 / 8.7 - 0.37 / 0.7 - 1/23) × 0.041 \u003d 0.1194 m
We round the resulting value up to the nearest 0.01 m: d ut = 0.12 m. We perform a verification calculation according to formula 5:
R 0 \u003d (1 / a i + d kk / l kk + d ut / l ut + 1 / a e)
R 0 \u003d (1 / 8.7 + 0.37 / 0.7 + 0.12 / 0.041 + 1/23) \u003d 3.61 m 2 0 C / W
5. Limitation of temperature and moisture condensation on the inner surface of the building envelope
Δt o, °С, between the temperature of the internal air and the temperature of the internal surface of the enclosing structure should not exceed the normalized values Δtn, °С, established in table 5, and defined as follows
Δt o = n(t int – text)/(R 0 a int) \u003d 1 (20 + 37) / (3.61 x 8.7) \u003d 1.8 0 C i.e. less than Δt n , = 4.0 0 C, determined from table 5.
Conclusion: t The thickness of the expanded polystyrene insulation in a three-layer brick wall is 120 mm. At the same time, the heat transfer resistance of the outer wall R 0 \u003d 3.61 m 2 0 C / W, which is greater than the normalized resistance to heat transfer Rreg. \u003d 3.60 m 2 0 C / W on the 0.01m 2 0 C/W. Estimated temperature difference Δt o, °С, between the temperature of the internal air and the temperature of the internal surface of the enclosing structure does not exceed the standard value Δtn,.
Example of thermotechnical calculation of translucent enclosing structures
Translucent enclosing structures (windows) are selected according to the following method.
Rated resistance to heat transfer Rreg determined according to table 4 of SNiP 23-02-2003 (column 6) depending on the degree-days of the heating period D d. However, the type of building and D d accept as in the previous example thermotechnical calculation opaque enclosing structures. In our case D d = 6276 0 From days, then for the window of an apartment building Rreg \u003d a D d + b \u003d 0.00005 × 6276 + 0.3 \u003d 0.61 m 2 0 C / W.
The choice of translucent structures is carried out according to the value of the reduced resistance to heat transfer R o r, obtained as a result of certification tests or according to Appendix L of the Code of Rules. If the reduced heat transfer resistance of the selected translucent structure R o r, more or equal Rreg, then this design satisfies the requirements of the norms.
Conclusion: for a residential building in Omsk, we accept PVC-bound windows with double-glazed windows made of hard glass selective coating and filling with argon of the inter-glass space in which R about r \u003d 0.65 m 2 0 C / W more R reg \u003d 0.61 m 2 0 C / W.
LITERATURE
- SNiP 23-02-2003. Thermal protection of buildings.
- SP 23-101-2004. Thermal protection design.
- SNiP 23-01-99*. Building climatology.
- SNiP 31-01-2003. Residential multi-apartment buildings.
- SNiP 2.08.02-89 *. Public buildings and structures.
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:
- account by designers when planning the location of rooms, load-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.
Extremely uneconomical to heating system, which 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 load-bearing structures, which will ensure long-term operation without freezing and overheating of ceilings 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 building on the cardinal points, that is, taking into account the absorption of 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 condensate on the walls, as well as a negative impact 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- 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 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 of 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 "margin", so you have to install such heating equipment, which 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 of automated control for 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- 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 the construction is carried out using modern building materials, then the indicator 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
By using computer programs from the company "ZVSOFT" you can calculate all the materials spent on heating, as well as make a detailed floor plan of communications with the display of radiators, specific heat, 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 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 to design all engineering systems, including for 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.
Example of thermotechnical calculation of enclosing structures
1. Initial data
Technical task. In connection with the unsatisfactory heat and humidity regime of the building, it is necessary to insulate its walls and mansard roof. To this end, perform calculations of thermal resistance, heat resistance, air and vapor permeability of the building envelope with an assessment of the possibility of moisture condensation in the thickness of the fences. Determine the required thickness of the heat-insulating layer, the need to use wind and vapor barriers, the order of the layers in the structure. Develop a design solution that meets the requirements of SNiP 23-02-2003 "Thermal protection of buildings" for building envelopes. Perform calculations in accordance with the set of rules for the design and construction of SP 23-101-2004 "Design of thermal protection of buildings".
General characteristics of the building. A two-story residential building with an attic is located in the village. Sviritsa Leningrad region. The total area of external enclosing structures - 585.4 m 2; total wall area 342.5 m 2; the total area of windows is 51.2 m 2; roof area - 386 m 2; basement height - 2.4 m.
The structural scheme of the building includes bearing walls, reinforced concrete floors from multi-hollow panels, 220 mm thick and a concrete foundation. The outer walls are made of brickwork and plastered inside and out with a mortar layer of about 2 cm.
The roof of the building has a truss structure with a steel seam roof, made along the crate with a step of 250 mm. Insulation 100 mm thick is made of mineral wool boards laid between the rafters
The building is provided with stationary electric-thermal storage heating. The basement has a technical purpose.
climatic parameters. According to SNiP 23-02-2003 and GOST 30494-96, we take the estimated average temperature of the indoor air equal to
t int= 20 °С.
According to SNiP 23-01-99 we accept:
1) the estimated temperature of the outside air in the cold season for the conditions of the village. Sviritsa Leningrad region
t ext= -29 °С;
2) the duration of the heating period
z ht= 228 days;
3) the average outdoor temperature for the heating period
t ht\u003d -2.9 ° С.
Heat transfer coefficients. The values of the heat transfer coefficient of the inner surface of the fences are accepted: for walls, floors and smooth ceilings α int\u003d 8.7 W / (m 2 ºС).
The values of the heat transfer coefficient of the outer surface of the fences are accepted: for walls and coatings α ext=23; attic floors α ext\u003d 12 W / (m 2 ºС);
Normalized resistance to heat transfer. Degree-days of the heating period G d are determined by formula (1)
G d\u003d 5221 ° С day.
Because the value G d differs from table values, standard value R req determined by formula (2).
According to SNiP 23-02-2003 for the obtained degree-day value, the normalized resistance to heat transfer R req, m 2 ° С / W, is:
For external walls 3.23;
Coverings and ceilings over driveways 4.81;
Fencing over unheated undergrounds and basements 4.25;
windows and balcony doors 0,54.
2. Thermotechnical calculation of external walls
2.1. Resistance of external walls to heat transfer
Exterior walls are made of hollow ceramic bricks and have a thickness of 510 mm. The walls are plastered from the inside with lime-cement mortar 20 mm thick, from the outside - with cement mortar of the same thickness.
The characteristics of these materials - density γ 0, dry thermal conductivity coefficient 0 and vapor permeability coefficient μ - are taken from Table. Clause 9 of the application. In this case, in the calculations we use the coefficients of thermal conductivity of materials W for operating conditions B, (for wet operating conditions), which are obtained by formula (2.5). We have:
For lime-cement mortar
γ 0 \u003d 1700 kg / m 3,
W\u003d 0.52 (1 + 0.168 4) \u003d 0.87 W / (m ° C),
μ=0.098 mg/(m h Pa);
For brickwork from hollow ceramic bricks on cement-sand mortar
γ 0 \u003d 1400 kg / m 3,
W\u003d 0.41 (1 + 0.207 2) \u003d 0.58 W / (m ° C),
μ=0.16 mg/(m h Pa);
For cement mortar
γ 0 \u003d 1800 kg / m 3,
W\u003d 0.58 (1 + 0.151 4) \u003d 0.93 W / (m ° C),
μ=0.09 mg/(m h Pa).
The heat transfer resistance of a wall without insulation is
R o \u003d 1 / 8.7 + 0.02 / 0.87 + 0.51 / 0.58 + 0.02 / 0.93 + 1/23 \u003d 1.08 m 2 ° C / W.
In the presence of window openings that form the slopes of the wall, the coefficient of thermal uniformity of brick walls, 510 mm thick, is taken r = 0,74.
Then the reduced resistance to heat transfer of the walls of the building, determined by formula (2.7), is equal to
R r o \u003d 0.74 1.08 \u003d 0.80 m 2 ° C / W.
The obtained value is much lower than the standard value of heat transfer resistance, so a device is needed external thermal insulation and subsequent plastering with protective and decorative plaster mortar compositions with fiberglass reinforcement.
In order for the thermal insulation to dry out, the plaster layer covering it must be vapor-permeable, i.e. porous with low density. We choose a porous cement-perlite mortar having the following characteristics:
γ 0 \u003d 400 kg / m 3,
0 \u003d 0.09 W / (m ° C),
W\u003d 0.09 (1 + 0.067 10) \u003d 0.15 W / (m ° C),
\u003d 0.53 mg / (m h Pa).
The total resistance to heat transfer of the added layers of thermal insulation R t and plaster lining R w must be at least
R t+ R w \u003d 3.23 / 0.74-1.08 \u003d 3.28 m 2 ° C / W.
Preliminarily (with subsequent clarification), we accept the thickness of the plaster lining as 10 mm, then its resistance to heat transfer is equal to
R w \u003d 0.01 / 0.15 \u003d 0.067 m 2 ° C / W.
When used for thermal insulation of mineral wool boards manufactured by CJSC Mineralnaya Vata, Facade Butts brand 0 \u003d 145 kg / m 3, 0 \u003d 0.033, W \u003d 0.045 W / (m ° C) the thickness of the heat-insulating layer will be
δ=0.045 (3.28-0.067)=0.145 m.
Rockwool boards are available in thicknesses from 40 to 160 mm in 10 mm increments. We accept a standard thickness of thermal insulation of 150 mm. Thus, the slabs will be laid in one layer.
Checking compliance with energy saving requirements. The calculation scheme of the wall is shown in fig. 1. The characteristics of the layers of the wall and the total resistance of the wall to heat transfer, excluding vapor barrier, are given in Table. 2.1.
Table 2.1
Characterization of the layers of the wall andtotal resistance of the wall to heat transfer
|
layer material |
Density γ 0, kg / m 3 |
Thickness δ, m |
Design coefficient of thermal conductivity λ W, W/(m K) |
Estimated resistance to heat transfer R, m 2 ° С) / W |
|
|
Internal plaster (lime-cement mortar) |
|||||
|
Hollow ceramic brick masonry |
|||||
|
External plaster ( cement mortar) |
|||||
|
Mineral wool insulation FACADE BATTS |
|||||
|
Protective and decorative plaster (cement-perlite mortar) |
|||||
The heat transfer resistance of the walls of the building after insulation will be:
R o = 1/8.7+4.32+1/23=4.48 m 2 °C/W.
Taking into account the coefficient of thermal engineering uniformity of the outer walls ( r= 0.74) we get the reduced resistance to heat transfer
R o r\u003d 4.48 0.74 \u003d 3.32 m 2 ° C / W.
Received value R o r= 3.32 exceeds the standard R req= 3.23, since the actual thickness of the heat-insulating plates is greater than the calculated one. This situation meets the first requirement of SNiP 23-02-2003 for the thermal resistance of the wall - R o ≥ R req .
Verification of compliance with the requirements forsanitary and hygienic and comfortable conditions in the room. Estimated difference between the temperature of the indoor air and the temperature of the inner surface of the wall Δ t 0 is
Δ t 0 =n(t int – t ext)/(R o r ·α int)=1.0(20+29)/(3.32 8.7)=1.7 ºС.
According to SNiP 23-02-2003, for the outer walls of residential buildings, a temperature difference of not more than 4.0 ºС is permissible. Thus, the second condition (Δ t 0 ≤Δ t n) performed.
P
check the third condition ( τ
int >t grew), i.e. is it possible to condense moisture on the inner surface of the wall at the estimated outdoor temperature t ext\u003d -29 ° С. Inner surface temperature τ
int enclosing structure (without heat-conducting inclusion) is determined by the formula
τ int = t int –Δ t 0 \u003d 20–1.7 \u003d 18.3 ° С.
The elasticity of water vapor in the room e int is equal to
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 of external brick walls with effective insulation of 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
Consequently,
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 |
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.
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. One of them: heaters and gas silicate blocks. Thanks to these materials, for example, the thickness of brickwork can be reduced to 250 mm.
Now walls and ceilings are most often made of 2 or 3 layers, one layer of which is a material with good thermal insulation properties. And in order to determine the optimal thickness of this material, a thermal calculation is carried out and the dew point is determined.
How the calculation is made to determine the dew point, you can find on the next page. Here, the heat engineering calculation will be considered using an example.
Required regulatory documents
For the calculation, you will need two SNiPs, one joint venture, one GOST and one allowance:
- SNiP 23-02-2003 (SP 50.13330.2012). "Thermal protection of buildings". Updated edition from 2012.
- SNiP 23-01-99* (SP 131.13330.2012). "Construction climatology". Updated edition from 2012.
- SP 23-101-2004. "Design of thermal protection of buildings".
- GOST 30494-96 (replaced by GOST 30494-2011 since 2011). "Residential and public buildings. Indoor microclimate parameters".
- Benefit. E.G. Malyavin "Heat loss of the building. Reference guide".
Calculated parameters
In the process of performing a heat engineering calculation, the following are determined:
- thermal characteristics 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. Thermal characteristics materials
The values of the characteristics of the materials are summarized in the table.
Note (*): These characteristics can also be found from manufacturers of thermal insulation materials.
Calculation
4. Determining the thickness of the insulation
To calculate the thickness of the heat-insulating layer, it is necessary to determine the heat transfer resistance of the enclosing structure based on the requirements of sanitary standards and energy saving.
4.1. Determination of the norm of thermal protection according to the condition of energy saving
Determination of degree-days of the heating period according to clause 5.3 of SNiP 23-02-2003:
D d = ( t int - tht) z ht = (20 + 4.1)215 = 5182°С×day
Note: also degree-days have the designation - GSOP.
The normative value of the reduced resistance to heat transfer should be taken not less than the normalized values determined by SNIP 23-02-2003 (Table 4) depending on the degree-day of the construction area:
R req \u003d a × D d + b \u003d 0.00035 × 5182 + 1.4 \u003d 3.214m 2 × °С/W,
where: Dd - degree-day of the heating period in Nizhny Novgorod,
a and b - coefficients taken according to table 4 (if SNiP 23-02-2003) or according to table 3 (if SP 50.13330.2012) for the walls of a residential building (column 3).
4.1. Determination of the norm of thermal protection according to the condition of sanitation
In our case, it is considered as an example, since this indicator is calculated for industrial buildings with 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 = 0.09 / 0.96 = 0.094 m 2 × °С/W .
3rd layer (silicate brick): R 3 = 0.25 / 0.87 = 0.287 m 2 × °С/W .
4th layer (plaster): R 4 = 0.02 / 0.87 = 0.023 m 2 × °С/W .
Determination of the minimum allowable (required) thermal resistance thermal insulation material(formula 5.6 E.G. Malyavin "Heat loss of the building. Reference manual"):
where: R int = 1/α int = 1/8.7 - resistance to heat transfer on the inner surface;
R ext \u003d 1/α ext \u003d 1/23 - resistance to heat transfer on the outer surface, α ext is taken according to table 14 for external walls;
ΣR i = 0.094 + 0.287 + 0.023 - the sum of thermal resistances of all layers of the wall without a layer of insulation, determined taking into account the coefficients of thermal conductivity of materials taken in column A or B (columns 8 and 9 of Table D1 SP 23-101-2004) in accordance with the humidity conditions of the wall, m 2 ° С /W
The thickness of the insulation is (formula 5.7):
where: λ ut - coefficient of thermal conductivity of the insulation material, W / (m ° C).
Determination of the thermal resistance of the wall from the condition that the total thickness of the insulation will be 250 mm (formula 5.8):
where: ΣR t, i - the sum of thermal resistances of all layers of the fence, including the insulation layer, of the accepted structural thickness, m 2 ·°С / W.
From the result obtained, it can be concluded that
R 0 \u003d 3.503m 2 × °С/W> R tr0 = 3.214m 2 × °С/W→ therefore, the thickness of the insulation is selected right.
Influence of the air gap
In the case when in a three-layer masonry, mineral wool, glass wool or other slab insulation, it is necessary to install an air ventilated layer between the outer masonry and the insulation. The thickness of this layer should be at least 10 mm, and preferably 20-40 mm. It is necessary in order to drain the insulation, which gets wet from condensate.
This air layer is not a closed space, therefore, if it is present in the calculation, it is necessary to take into account the requirements of clause 9.1.2 of SP 23-101-2004, namely:
a) structural layers located between the air gap and the outer surface (in our case, this is a decorative brick (besser)) are not taken into account in the heat engineering calculation;
b) on the surface of the structure facing towards the layer ventilated by the outside air, the heat transfer coefficient α ext = 10.8 W/(m°C) should be taken.
Note: the influence of the air gap is taken into account, for example, in the heat engineering calculation of plastic double-glazed windows.