On the thermotechnical uniformity of a two-layer wall structure. On the thermal uniformity of a two-layer wall structure Thermal uniformity of enclosing structures
Description:
In a number of cases*, the specific consumption of thermal energy in old panel buildings and modern monolithic-frame houses with two-layer walls made of aerated concrete and facing bricks is practically the same. One of the reasons for this phenomenon is that double wall structures are often overestimated in terms of their thermal performance.
A. S. Gorshkov, cand. tech. Sci., Director of the Scientific and Educational Center "Monitoring and Rehabilitation of Natural Systems" FSAEI HE "St. Petersburg State Polytechnic University"
P. P. Rymkevich, cand. Phys.-Math. Sci., Professor, Department of Physics, FGKVOU VPO “Military Space Academy named after V.I. A. F. Mozhaisky»
N. I. Vatin, doctor of tech. Sciences, Professor, Director of the Institute of Civil Engineering, St. Petersburg State Polytechnic University
In a number of cases, * the specific consumption of thermal energy in old panel buildings and modern monolithic-frame houses with two-layer walls made of aerated concrete and facing bricks is practically the same. One of the reasons for this phenomenon is that double wall structures are often overestimated in terms of their thermal performance. Therefore, the calculation of the reduced resistance to heat transfer of a two-layer wall structure was carried out, which showed that its thermal characteristics do not meet not only the required, but also the minimum allowable regulatory requirements. At the design stage for this constructive solution usually lay the coefficient of thermal uniformity of 0.9, which for many cases is too high. In addition, designers use unreasonable values of the thermal conductivity of aerated concrete.
Currently, in the practice of designing and building buildings with a monolithic reinforced concrete frame and floor-by-floor support of the outer walls on monolithic or prefabricated-monolithic reinforced concrete floors, one of the most common options for filling the outer heat-shielding shell is a constructive solution of the wall, consisting of two layers (Fig. 1):
– an internal non-bearing layer made of aerated concrete blocks with a thickness of 300–400 mm, depending on the region of construction and its climatic parameters;
- an outer facing layer of face brick with a thickness of one or two bricks.
Description of the design of the wall fence
In the considered constructive solution, the inner layer of the wall fencing performs the function of thermal insulation, the outer one - the function of protection from external climatic influences, provides the required durability of the facades and forms the architectural appearance of the building. It is believed that this constructive solution satisfies the requirements of thermal protection for most regions of the Russian Federation.
In St. Petersburg, the traditional solution is wall fencing, in which the thickness of the aerated concrete layer is 375 mm (Fig. 1a).
Regulatory requirements
In SNiP 23-02-2003 "Thermal protection of buildings" (hereinafter - SNiP 23-02), three indicators of thermal protection are established for buildings:
a) individual elements of the building envelope;
b) sanitary and hygienic, including the temperature difference between the temperatures of the internal air and on the surface of the enclosing structures and the temperature on the internal surface above the dew point temperature;
c) specific consumption of thermal energy for heating the building, which makes it possible to vary the values of the heat-shielding properties of various types of enclosing structures of buildings, taking into account the space-planning decisions of the building and the choice of microclimate maintenance systems to achieve the normalized value of this indicator.
Reduced resistance to heat transfer R r 0 of enclosing structures should be taken not less than the normalized values 1 R req , determined by 2 depending on the degree-days of the heating period (hereinafter - GSOP) of the construction area.
The GSOP for residential buildings located on the territory of St. Petersburg is 3 4 796 °C days, and the normalized value of the reduced heat transfer resistance for the outer walls of residential buildings is 4 3.08 m 2 °C / W. At the same time, it is allowed to reduce the normalized value of the reduced resistance to heat transfer for the walls of residential and public buildings by 37% when meeting the requirements of SNIP 23-02 (clause 5.1).
Thus, in relation to the case under consideration, the minimum allowable value of the reduced resistance to heat transfer for the outer walls of residential buildings designed on the territory of St. Petersburg should not be lower than 6 R min = 1.94 m 2 °C / W.
Purpose and objectives of the study
Reduced resistance to heat transfer R r 0 for external walls should be calculated for the facade of the building or for one intermediate floor, taking into account the slopes of the openings, without taking into account their fillings 7 . Consider on specific example how this requirement is met in practice.
To do this, we will calculate the reduced resistance to heat transfer of the outer walls of the intermediate floor of a typical multi-apartment residential building with a structural monolithic-frame scheme and two-layer outer walls (Fig. 1) and compare the obtained value with the normalized R req and the minimum allowable R min values of the reduced resistance to heat transfer of the outer walls of a residential multi-apartment building.
Initial data for thermal engineering calculation
Construction area - St. Petersburg.
The purpose of the building is residential.
Design temperature: indoor air t c = 20 °С; outside air t n = -26 °С.
The wet zone is wet.
Humidity regime of the premises of the building is normal.
Operating conditions of enclosing structures - "B".
Thermal characteristics of materials used as part of wall fencing:
- cement-sand mortar γ o \u003d 1,800 kg / m 3, λ B \u003d 0.93 W / (m ° C);
– brickwork from an ordinary clay brick on a cement-sand mortar γ o \u003d 1,800 kg / m 3, λ B \u003d 0.80 W / (m ° C);
- masonry of unreinforced wall blocks made of autoclaved aerated concrete with a density of γ o = 400 kg / m 3, λ B = 0.14 W / (m ° C).
Border conditions:
Estimated heat transfer coefficient:
- the inner surface of the wall α int \u003d 8.7 W / (m 2 ° C);
- window blocks α int \u003d 8 W / (m 2 ° С);
- the outer surface of walls, windows α ext \u003d 23 W / (m 2 ° С).
Calculation schemes of fragments of external walls are presented in fig. 2.
Calculation results
The reduced resistance to heat transfer of the considered fragments of the heat-shielding shell of the building is calculated based on the calculation of temperature fields. The essence of the method lies in modeling the stationary process of heat transfer through the enclosing structures of buildings using computer programs 8 . The method is designed to assess the temperature regime and calculate the reduced heat transfer resistance of building envelopes or their fragments, taking into account the geometric shape, location and characteristics of structural and heat-insulating layers, ambient air temperatures, heat transfer coefficients of surfaces.
The value of the reduced resistance to heat transfer of the middle intermediate floorR r 0 is determined based on the calculation of the reduced resistance of a number of sections (fragments) R r 0,i, taking into account heat losses through the ends of floor slabs, slopes of window openings and balcony doors (see table), in particular the following fragments:
- a blank wall without openings, dimensions: in height - the height of the floor h= 3.0 m, in width - 1.2 m (Fig. 2a);
- walls with window openings, dimensions: in height - floor height h= 3.0 m, in width - the distance between the axes of window openings (Fig. 2b);
– walls with a balcony door, dimensions: in height – floor height h= 3.0 m, in width - the distance between the axes of the walls (Fig. 2c).
Reduced resistance to heat transfer of the outer walls of the middle intermediate floor of an apartment building R r 0, taking into account the areas of wall sections along the facades of the building, calculated by formula (1) (see Calculation formulas), is 1.81 m 2 °C / W.
Having calculated the conditional (without taking into account the influence of heat-conducting inclusions on the thermal uniformity of walls) resistance to heat transfer R 0 of the considered constructive solution (formula (2), Calculation formulas), we get 2.99 m 2 °C / W.
Hence the coefficient of heat engineering uniformity r, considered in the example of the outer wall of a typical intermediate floor, taking into account the slopes of the openings without taking into account their fillings, will be equal to 0.61 (formula (3), Calculation formulas).
What affects the coefficient of thermal inhomogeneity?
For a similar design solution, an even lower calculated value of the coefficient of thermal engineering uniformity was obtained r = 0,48.
Differences in the coefficients of heat engineering homogeneity can be due to differences in the design solutions used in the project, the quantitative and qualitative composition of heat-conducting inclusions. Also, the thermal inhomogeneity of the wall structure depends on the quality of the installation.
In particular, it was noted that according to the results of shooting 15 thermograms, the heat transfer resistance of a two-layer outer wall measured in natural conditions was 1.3–1.5 m 2 °C / W (with the conditional heat transfer resistance of a wall fence R 0 \u003d 3.92 m 2 ° C / W). It turns out that the actual coefficient of heat engineering uniformity may be even less than the calculated value and be according to r= (1.3÷1.5) / 3.92 = 0.33÷0.38.
As one of the possible reasons for the identified discrepancy, low-quality construction is noted, due to the receipt of construction site blocks of irregular shape. Indeed, the presence of cracks, breaks, potholes and other product defects can lead to overspending mortar, which acts as an additional heat-conducting inclusion that is not taken into account in the calculation.
It should be noted that the actual moisture content of aerated concrete products in the initial period of operation can significantly exceed the calculated one. In this regard, the thermal conductivity of aerated concrete products may turn out to be higher compared to the calculated values \u200b\u200baccepted in the project, since the thermal conductivity of the material depends on the mass content of moisture.
Based on the obtained calculations, we formulate the following conclusions:
- Reduced resistance to heat transfer R r 0 of a two-layer wall structure, consisting of an internal self-supporting layer of non-reinforced aerated concrete wall blocks of D400 density grade and an outer facing layer of facing ceramic bricks 120 mm thick, calculated based on the calculation of temperature fields for a typical intermediate floor of an apartment building, is 1.81 m 2 °C / W.
- The design of the considered wall fencing (Fig. 1) does not meet the regulatory requirements for thermal protection ( R req \u003d 3.08 m 2 ° C / W).
- The design of the wall enclosure (Fig. 1) does not meet the minimum allowable requirements for thermal protection ( R min \u003d 1.94 m 2 ° С / W).
- Thermal uniformity coefficient r the construction of the outer wall, made of masonry from aerated concrete blocks of the D400 density grade with a facing layer of face brick, does not exceed 0.61.
- The actual value of the coefficient of thermal engineering homogeneity of the design solution under consideration, taking into account the quality of the products delivered to the facility and the quality of their installation, may turn out to be significantly lower than the calculated value.
- To meet the regulatory requirements for the level of thermal protection of the outer walls of buildings as part of a wall enclosure (Fig. 1), one should either increase the thickness of aerated concrete blocks as part of a two-layer wall structure, or use an intermediate layer of heat-insulating materials with a calculated thermal conductivity of not more than 0.05 W/m °C The thermal insulation layer should be placed between the aerated concrete and the front (facing) layers.
- In all cases for efficient removal moisture from the composition of the wall fencing between the thermal insulation layer and the front brick, a ventilated gap should be provided, the effective section of which (thickness) should be determined by calculation.
Literature
- Krivoshein A. D., Fedorov S. V. On the issue of calculating the reduced resistance to heat transfer // Engineering and Construction Journal. 2010. No. 8.
- Krivoshein A. D., Fedorov S. V. User's manual for the TEMPER software package for calculating the temperature fields of building envelopes. Omsk: SibADI, 1997.
- Sokolov N. A., Gorshkov A. S. Thermal conductivity of building materials and products: the level of harmonization of Russian and European building standards // Construction Materials, equipment, technologies of the XXI century. 2014. No. 6 (185).
- Gagarin VG Thermophysical problems of modern wall enclosing structures of multi-storey buildings // Academia. Architecture and construction. 2009. No. 5.
- Nemova D.V., Spiridonova T.I., Kurazhova V.G. Unknown properties of the known material // Construction of unique buildings and structures. 2012. No. 1.
* Data on the actual energy consumption of residential buildings of different years of construction were collected and analyzed by the authors of the article. - Approx. ed.
1 In accordance with the requirements of SNiP 23-02 (clause 5.3).
2 According to SNiP 23-02, table 4.
3 According to the requirements of RMD 23-16–2012 “St. Petersburg. Recommendations for Ensuring Energy Efficiency of Residential and Public Buildings”, Table 3.
4 Ibid., Table 9.
5 According to the requirements of SNiP 23-02, clause 5.13.
6 See SNiP 23-02, formula (8).
7 According to the requirements of SNIP 23-02, clause 5.6.
8 In our case, the calculation was performed using the TEMPER 3D software package.
Figure H.1 - Schemes of heat-conducting inclusions in enclosing structures
H.1 CALCULATION OF THE THERMAL UNIFORMITY COEFFICIENT USING FORMULA (12)
OF THE PRESENT CODE OF RULES
Table H.1 - Definition of factor
Coefficient at (Figure H.1) |
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Note - The designations are taken from Figure H.1. |
Calculation example
Determine the reduced resistance to heat transfer of a panel with effective insulation (polystyrene foam) and steel cladding of an industrial building.
Initial data
The size of the panel is 6x2 m. Structural and thermal characteristics of the panel:
steel sheathing thickness 0.001 m, thermal conductivity coefficient ;
thickness of polystyrene foam insulation 0.2 m, thermal conductivity coefficient .
Flanging of sheet material along the extended sides of the panel leads to the formation of a heat-conducting inclusion of type IIb (Figure H.1), having a width of = 0.002 m.
Calculation procedure
Heat transfer resistance far from the inclusion and along the thermally conductive inclusion:
The value of the dimensionless parameter of the heat-conducting inclusion according to Table H.2
0.002 58/(0.2 0.04)=14.5.
Table H.2 - Definition of coefficient
#G0Scheme of thermally conductive connection according to figure H.1 |
The values of the coefficient at (according to Figure H.1 |
|||||||||
According to Table H.2, by interpolation, we determine the value
0,43+[(0,665-0,43)4,5]/10=0,536.
Coefficient , by formula (13)
Coefficient of heat engineering homogeneity of the panel according to the formula (12)
Reduced resistance to heat transfer according to the formula (11)
H.2 CALCULATION OF THE COEFFICIENT OF THERMAL TECHNICAL HOMOGENEITY USING FORMULA (14)
OF THE PRESENT CODE OF RULES
Calculation example
Determine the reduced resistance to heat transfer of a single-module three-layer reinforced concrete panel on flexible connections with a window opening of a large-panel residential building of the III-133 series.
Initial data
A 300 mm thick panel contains outer and inner reinforced concrete layers, which are interconnected by two hangers (in piers), a strut located in the lower zone of the window sill area, and spacers: 10 - at horizontal joints and 2 - in the area of the window slope (Figure N. 2).
1 - spacers; 2 - loop; 3 - pendants;
4 - concrete thickenings (= 75 mm of the inner reinforced concrete layer); 5 - brace
Figure H.2 - Structure of a three-layer panel on flexible ties
In #M12293 0 1200037434 4120950664 4294967273 80 2997211231 403162211 2325910542 403162211 2520 Table H.4#S shows design parameters panels.
In the zone of hangers and hinges, the inner concrete layer has thickenings that replace part of the insulation layer.
Calculation procedure
The design of the fence contains the following heat-conducting inclusions: horizontal and vertical joints, window slopes, thickenings of the inner reinforced concrete layer and flexible connections (suspension brackets, brace, spacers).
To determine the coefficient of influence of individual heat-conducting inclusions, we first calculate according to formula (7) thermal resistances individual sections panels:
in the zone of thickening of the inner reinforced concrete layer
along the horizontal line
along the vertical joint
thermal resistance of the panel away from heat-conducting inclusions
Conditional resistance to heat transfer away from heat-conducting inclusions
Since the panel has a vertical axis of symmetry, the determination of subsequent values is carried out for half of the panel.
Determine the area of half the panel without taking into account the window opening
Panel thickness =0.3 m.
Let us determine the area of zones of influence and the coefficient for each heat-conducting inclusion of the panel:
for horizontal joint
2,95/3,295=0,895.
According to table H.3 = 0.1. The area of the zone of influence according to the formula (15)
;
for vertical joint
Table H.3 — Determination of the influence coefficient
#G0Type of thermally conductive inclusion |
Influence coefficient |
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Without adjoining internal fences |
With adjoining internal fences |
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without ribs |
With ribs thickness, mm |
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Window slopes |
without ribs |
With ribs thickness, mm: |
|||
Flexible connections with a diameter, mm: |
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Notes |
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1 The table shows - thermal resistances, respectively, panels outside the heat-conducting inclusion, joint, thickening of the inner reinforced concrete layer, determined by formula (8); - distances, m, from the longitudinal axis of the window frame to its edge and to the inner surface of the panel. |
|||||
2 Intermediate values should be determined by interpolation. |
According to table H.3 = 0.375. The area of the zone of influence according to the formula (15)
;
for window slopes at = 0.065 m and = 0.18 m, according to table H.3 = 0.374. The area of the zone of influence of half of the window opening, taking into account the corner sections, is determined by the formula (16)
for concrete thickenings of the inner reinforced concrete layer in the suspension and hinge area at = 1.546 / 3.295 = 0.469 according to table M.3 * = 0.78. The total area of the zone of influence of the bulges of the suspension and loop is found by the formula (17)
for suspension (rod diameter 8 mm) according to table H.3 = 0.16, area of influence zone according to formula (17)
for strut (rod diameter 8 mm) according to table H.3 = 0.16, according to formula (17)
for spacers (rod diameter 4 mm) according to table H.3 = 0.05.
When determining the total area of the zone of influence of five spacers, it should be taken into account that the width of the zone of influence from the side of the joint is limited by the edge of the panel and is 0.09 m. According to the formula (18)
Calculate by formula (14)
The reduced resistance to heat transfer of the panel is determined by the formula (11)
Table H.4
#G0Layer material |
Layer thickness, mm |
|||||
Far from inclusions |
in the suspension and hinge area |
horizontal joint |
vertical joint |
|||
External reinforced concrete layer | ||||||
Thermal insulation layer - expanded polystyrene | ||||||
Mineral wool liners | ||||||
Internal reinforced concrete layer |
APPENDIX P
Technical article
Freezing of structures in winter and overheating in summer, the formation of condensate and, as a result, a reduction in their service life, high energy consumption of the building are the main results of errors made in thermal calculations. In modern construction, the level of thermal resistance is an important parameter of enclosing structures along with their bearing capacity. The requirements for creating a reliable, environmentally safe living environment with reasonable energy consumption are formed by the Scientific Research Institute of Building Physics subordinate to the Ministry of Construction of the Russian Federation. Russian Academy architecture and building sciences” (NIISF RAASN). Since the entry into force of the set of rules developed by him SP 50.13330.2012 “Thermal protection of buildings. Updated version of SNiP 23-02-2003” approach to the determination of the reduced resistance of enclosing structures has changed significantly. Now, instead of the usual tabular values of the coefficient of thermal uniformity of building envelopes, it is required to calculate each building envelope separately. What advantages does new technique calculation in practice?
As an example of a building envelope, consider the combined roofing of a residential apartment building. When carrying out the calculation in accordance with the methodology for determining the reduced resistance described in SNiP 23-02-2003, we will not find tabular values of uniformity for such types of structures. Therefore, it remains only to rely on your intuition and choose these values at random. Or, relying on the data of structures close in value, such as attic floors, the uniformity value of which is in the range from 0.5 to 0.9.
When solving the problem according to the standards described in Appendix E of SP 50.13330.2012, we can already accurately, on the basis of a specific geometry, determine the value of the coefficient of thermal engineering uniformity of the structure or fragment under consideration. For a combined roofing, we determine the flat, linear and point elements that make up the building envelope. We list the most common of them. The flat area includes the area of the roof along the smooth surface, the linear ones - the junction to different types parapets, roof exits, ventilation shafts, etc., and to the point ones - insulation and waterproofing fasteners. Next, you need to find the specific geometric indicator of each of the elements present on the roof. By defining: its area for flat elements, length for linear elements, and number of pieces for point elements. As a rule, for such types of structures, among the linear elements, the junctions to the parapet have the highest specific geometric index.
Then, it is necessary to calculate the specific heat losses passing through the element. To determine this parameter, you can use the ready-made tabular values given in SP 230.1325800.2015, or you can simulate the node in a specialized program for calculating thermal fields and determine the specific heat loss through the node yourself. The results obtained are entered in the table in the form E2 SP 50.13330.2012 and the reduced heat transfer resistance of the considered fragment of the building envelope is calculated according to the formula E1 SP 50.13330.2012.
Now, using an example, consider the combined roof of a conditional section of a residential apartment building. In the calculation of the reduced resistance, we will take two elements that have the largest geometric index: the roof area along the surface and the junction with an uninsulated parapet. The remaining elements are not taken into account in the calculation.
Initial data for calculation:
The roof surface area is 263 m 2 ;
The length of the junctions to the parapet is 101 m;
The conditional heat transfer resistance of the homogeneous part of the roof is 5.526 m 2 * 0 C / W;
Thermal resistance of the insulation layer on the wall 3 m 2 * 0 C / W;
Thermal conductivity of the base of the parapet 0.6 W / m 2 * 0 C;
Thermal resistance of the insulation layer on the coating slab 5 m 2 * 0 C / W;
There is no additional insulation of the parapet.
We will calculate according to the available parameters, the results will be entered in table 1 (a form similar to table E2). The values of specific heat losses through the parapet are taken on the basis of the data in Table D.42 of SP 230.1325800.2015.
Table 1
The reduced resistance for such a design will be equal to R pr \u003d 2.978 m 2 * 0 C / W. And the value of the coefficient of thermal uniformity r=0.54.
Example 1: Temperature fields of a parapet junction. Option 1.*
Let's make adjustments to the original data. We will reduce the thermal conductivity of the base to 0.2 W / m 2 * 0 C and add 500 mm high insulation to the parapet. The values of specific heat losses through the parapet are taken on the basis of the data in Table D.47 of SP 230.1325800.2015.
Let's correct table 1.
Now the reduced resistance for the same design will be equal to R pr \u003d 3.973 m 2 * 0 C / W. And the coefficient of thermal homogeneity r=0.72.
Example 2: Temperature fields of the junction to the parapet. Option 2.*
Thus, by making small changes in the design of the junction to the parapet and without changing the thickness of the main insulation, we get an increase in the value of the reduced resistance by 33% in relation to the initial value.
Based on the foregoing, we can conclude: the more detailed and rational, not only from the point of view of bearing capacity, but also from the point of view of heat engineering, all nodes are worked out, the less the building will lose heat through the building envelope, and the higher the efficiency of using insulation in such structures.
In TECHNONICOL, you can order a complete thermal calculation of the building, according to the methodology of SP 50.13330.2012, or the calculation of a specific unit to determine heat loss and meet sanitary and hygienic requirements.
Calculation of the coefficient of thermal uniformity of enclosing structures according to tabular values
- 1. Calculation of the coefficient of thermal homogeneity r according to the formula (2.7)
- Table B.1
- Table for determining the coefficient ki
-
Heat transfer circuit λm / λ Coefficient ki at α / δ 0,1 0,2 0,4 0,6 0,8 1 1,5 2 I 2 1,02 1,01 1,01 1,01 1 1 1 15 1,16 1,11 1,07 1,05 1,04 1,03 1,02 1,01 10 1,33 1,25 1,15 1,1 1,08 1,06 1,04 1,03 30 1,63 1,47 1,27 1,18 1,14 1,11 1,07 1,05 II 10 - 40 2,65 2,2 1,77 1,6 1,55 - - -III at c / δ 0,25 2 1,02 1,01 1,01 1,01 1,01 1,01 1,01 15 1,12 1,08 1,05 1,04 1,03 1,03 1,02 1,01 10 1,18 1,13 1,07 1,05 1,04 1,04 1,03 1,02 30 1,21 1,16 1,1 1,07 1,05 1,04 1,03 1,02 0,5 2 1,05 1,04 1,03 1,02 1,01 1,01 1,01 1,015 1,28 1,21 1,13 1,09 1,07 1,06 1,04 1,03 10 1,42 1,34 1,22 1,14 1,11 1,09 1,07 1,05 30 1,62 1,49 1,3 1,19 1,14 1,12 1,09 1,06 0,75 2 1,06 1,04 1,03 1,02 1,02 1,01 1,01 1,015 1,25 1,2 1,14 1,1 1,08 1,07 1,05 1,03 10 1,53 1,42 1,25 1,16 1,12 1,11 1,08 1,05 30 1,85 1,65 1,38 1,24 1,18 1,15 1,11 1,08 IV at c / δ 0,25 2 1,03 1,02 1,02 1,01 1,01 1,01 1 15 1,12 1,10 1,07 1,05 1,04 1,03 1,02 1,01 10 1,2 1,16 1,1 1,07 1,06 1,05 1,03 1,02 30 1,28 1,22 1,14 1,09 1,07 1,06 1,04 1,03 0,5 5 1,32 1,25 1,17 1,13 1,1 1,08 1,06 1,04 10 1,54 1,42 1,27 1,19 1,14 1,12 1,09 1,06 30 1,79 1,61 1,38 1,26 1,19 1,16 1,12 1,08 0,75 2 1,07 1,05 1,04 1,03 1,02 1,02 1,01 1,015 1,36 1,28 1,18 1,14 1,11 1,09 1,07 1,05 10 1,64 1,51 1,33 1,23 1,18 1,15 1,11 1,08 30 2,05 1,82 1,5 1,33 1,25 1,21 1,16 1,11 - Table B.2
- Table for determining the coefficient ψ
-
Scheme of thermally conductive connection Coefficient ψ at αλt / δisol λisol 0,25 0,5 1 2 5 10 20 50 150 I 0,024 0,041 0,066 0,093 0,121 0,137 0,147 0,155 0,19IIb - - - 0,09 0,231 0,43 0,665 1,254 2,491III at c / δ 0,25 0,016 0,02 0,023 0,026 0,028 0,029 0,03 0,03 0,0310,5 0,036 0,054 0,072 0,083 0,096 0,102 0,107 0,109 0,11 0,75 0,044 0,066 0,095 0,122 0,146 0,161 0,168 0,178 0,194 IV at c / δ 0,25 0,015 0,02 0,024 0,026 0,029 0,031 0,033 0,039 0,0480,5 0,037 0,056 0,076 0,09 0,103 0,12 0,128 0,136 0,15 0,75 0,041 0,067 0,01 0,13 0,16 0,176 0,188 0,205 0,22 - Calculation example
- Determine the reduced resistance to heat transfer of a panel with effective insulation (polystyrene foam) and steel cladding of a public building.
- A. Initial data
- Panel dimensions 6×2 m. Structural and thermal characteristics of the panel:
- steel sheathing thickness 0.001 m, thermal conductivity coefficient λ = 58 W/(m °С), polystyrene foam insulation thickness 0.2 m, thermal conductivity coefficient 0.04 W/(m °С).
- Flanging of sheet material along the extended sides of the panel leads to the formation of a heat-conducting inclusion of type IIb (Appendix 5* SNiP II-3-79* (ed. 1998)), having a width a = 0.002 m.
- B. Calculation procedure
- Heat transfer resistances away from the Rocon inclusion and across the thermally conductive Ro' connection:
- Rocon \u003d 1 / 8.7 + 2 (0.001 / 58) + 0.2 / 0.04 + 1 / 23 \u003d 5.16 m2 ° C / W;
- Ro′ \u003d 1 / 8.7 + (2 0.001 + 0.2) / 58 + 1 / 23 \u003d 0.162 m2 ° C.
- The value of the dimensionless parameter of the heat-conducting inclusion for the table. B.2
- aλt / δisolλisol = 0002 58 / (0.2 0.04) = 14.5
- According to the table B.2 by interpolation we determine the value ψ
- ψ = 0.43 + [(0.665 - 0.665) 4.5] / 10 = 0.536
- Coefficient ki according to formula (2.8)
- ki = 1 + 0.536 = 52.94
- Thermal homogeneity coefficient of the panel according to the formula (2.7)
- r = 1 / (0.002 6 52.94) = 0.593
- Reduced resistance to heat transfer according to the formula (2.6)
- Ror \u003d 0.593 5.16 \u003d 3.06 m2 ° C / W.
- 2. Calculation of the coefficient of thermal engineering uniformity r according to the formula (2.9)
- Table B.3
- Table for determining the influence factor fi
-
Type of heat-conducting inclusion Influence factor fi joints without adjoining internal fences with adjoining internal fences without ribs with ribs thickness, mm 10 20 Rcm/Rkcon: 1 or more - - 0,07 0,120,9 - 0,1 0,14 0,170,8 0,01 0,13 0,17 0,190,7 0,02 0,2 0,24 0,260,6 0,03 0,27 0,31 0,340,5 0,04 0,33 0,38 0,410,4 0,05 0,39 0,45 0,480,3 0,06 0,45 0,52 0,55Window slopes without ribs with ribs thick 10 mm 20 mmδF′ / δw′: 0,2 0,45 0,58 0,670,3 0,41 0,54 0,620,4 0,35 0,47 0,550,5 0,29 0,41 0,480,6 0,23 0,34 0,410,7 0,17 0,28 0,350,8 0,11 0,21 0,28Thickening of the inner reinforced concrete layer Ry/Rkcon: 0,9 0,02 - -0,8 0,12 - -0,7 0,28 - -0,6 0,51 - -0,5 0,78 - -Flexible connections with a diameter, mm: 4 0,05 - -6 0,1 - -8 0,16 - -10 0,21 - -12 0,25 - -14 0,33 - -16 0,43 - -18 0,54 - -20 0,67 - - - Calculation example
- Determine the reduced heat transfer resistance Ror of a single-module three-layer reinforced concrete panel on flexible connections with a window opening of a large-panel residential building of series III.
- A. Initial data
- A 300 mm thick panel contains outer and inner reinforced concrete layers, which are interconnected by two hangers (in piers), a brace located in the lower zone of the window sill area, and spacers: 10 - at horizontal joints and 2 - in the window slope area (Fig. B .one).
- Rice. B.1. Three-layer panel construction on flexible ties
- 1 - spacers; 2 - loop; 3 - pendants; 4 - concrete thickenings (δ = 75 mm of the inner reinforced concrete layer); 5 - brace
- In table. B.4 shows the design parameters of the panel.
- In the zone of hangers and hinges, the inner concrete layer has thickenings that replace part of the insulation layer.
- Table B.4
- B. Calculation procedure
- The design of the fence contains the following heat-conducting inclusions: horizontal and vertical joints, window slopes, thickenings of the inner reinforced concrete layer and flexible connections (suspension brackets, brace, spacers).
- To determine the coefficient of influence of individual heat-conducting inclusions, we first calculate the thermal resistances of individual sections of the panel using formula (2.2):
- in the zone of thickening of the inner reinforced concrete layer
- Ry = 0.175 / 2.04 + 0.06 / 0.042 + 0.065 / 2.04 = 1.546 m2 °C / W;
- along the horizontal line
- Rjng \u003d 0.1 / 2.04 + 0.135 / 0.047 + 0.065 / 2.04 \u003d 2.95 m2 ° C / W;
- along the vertical joint
- Rjnv \u003d 0.175 / 2.04 + 0.06 / 0.047 + 0.065 / 2.04 \u003d 1.394 m2 ° C / W;
- thermal resistance of the panel away from heat-conducting inclusions
- Rkcon = 0.1 / 2.04 + 0.135 / 0.042 + 0.065 / 2.04 = 3.295 m2 °C / W.
- Conditional resistance to heat transfer away from heat-conducting inclusions
- Rocon \u003d 1 / 8.7 + 3.295 + 1 / 23 \u003d 3.453 m2 ° C / W.
- Since the panel has a vertical axis of symmetry, we determine the following values for half of the panel:
- Determine the area of half the panel without taking into account the window opening
- Ao = 0.5 (2.8 2.7 - 1.48 1.51) = 2.66 m2.
- Panel thickness δw = 0.3 m.
- Let us determine the area of zones of influence Ai and the coefficient fi for each heat-conducting inclusion of the panel:
- for horizontal joint
- Rjng / Rkcon = 2.95 / 3.295 = 0.895
- According to the table B.3 fi = 0.1. The area of the zone of influence according to the formula (2.10)
- Ai = 0.3 2 1.25 = 0.75 m2;
- for vertical joint
- Rjnv / Rkcon = 1.394 / 3.295 = 0.423
- According to the table B.3 fi = 0.375. The area of the zone of influence according to the formula (2.10)
- Ai = 0.3 2.8 = 0.84 m2.
- for window slopes at δF′ = 0.065 m and δw′ = 0.18 m, according to Table B.3 fi = 0.374. The area of the zone of influence of half of the window opening, taking into account the corner sections, is determined by the formula (2.11)
- Ai = 0.5 = 1.069 m2;
- for concrete thickenings of the inner reinforced concrete layer in the suspension and hinge area at Ry′ / Rkcon = 1.546 / 3.295 = 0.469, according to Table. B.3 fi = 0.78. The total area of the zone of influence of the bulges of the suspension and loop is found by the formula (2.12)
- Ai = (0.6 + 2 0.3)(0.47 + 0.1) + (0.2 + 0.3 + 0.1)(0.42 + 0.3 + 0.075) = 1.161 m2 ;
- for suspension (rod diameter 8 mm) according to the table. D.3 fi = 0.16, the area of the call of influence according to the formula (2.12)
- Ai \u003d (0.13 + 0.3 + 0.14) (0.4 + 2 0.3) \u003d 0.57 m2;
- for strut (rod diameter 8 mm) according to the table. B.3 fi = 0.16, according to the formula (2.12)
- Ai = (0.13 + 0.3)(0.22 + 0.3 + 0.09) = 0.227 m2.
- for spacers (rod diameter 4 mm) according to the table. B.3 fi = 0.05.
- When determining the total area of the zone of influence of five spacers, it should be taken into account that the width of the zone of influence from the side of the joint is limited by the edge of the panel and is 0.09 m. According to the formula (2.13):
- Ai \u003d 5 (0.3 + 0.3) (0.3 + 0.09) \u003d 1.17 m2.
- Calculate r using formula (2.9)
- r = 1 / (1 + (0.84 0.375 + 0.75 0.1 + 1.069 0.374 + 1.161 0.78 + 0.57 0.16 + 0.227 0.16 + 1.17 0.05)) = 0.71
- The reduced resistance to heat transfer of the panel is determined by the formula (2.6)
- Ror = 0.71 3.453 = 2.45 m2 °C/W.
Note. Designations and schemes are adopted according to the appendix. 5* SNiP II-3-79* (ed. 1998)
Notes:
1. The table shows Rkcon, Rcm, Ry - thermal resistances, m2 °C / W, respectively, panels outside the heat-conducting inclusion, joint, thickening of the inner reinforced concrete layer, determined by formula (2.2); δF' and δw' - distances, m, from the longitudinal axis of the window frame to its edge and to the inner surface of the panel.
2. Intermediate values should be determined by interpolation.
Figure H.1 - Schemes of heat-conducting inclusions in enclosing structures
H.1 CALCULATION OF THE THERMAL UNIFORMITY COEFFICIENT USING FORMULA (12)
OF THE PRESENT CODE OF RULES
Table H.1 - Definition of factor
Coefficient at (Figure H.1) |
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Note - The designations are taken from Figure H.1. |
Calculation example
Determine the reduced resistance to heat transfer of a panel with effective insulation (polystyrene foam) and steel cladding of an industrial building.
Initial data
The size of the panel is 6x2 m. Structural and thermal characteristics of the panel:
steel sheathing thickness 0.001 m, thermal conductivity coefficient ;
thickness of polystyrene foam insulation 0.2 m, thermal conductivity coefficient .
Flanging of sheet material along the extended sides of the panel leads to the formation of a heat-conducting inclusion of type IIb (Figure H.1), having a width of = 0.002 m.
Calculation procedure
Heat transfer resistance far from the inclusion and along the thermally conductive inclusion:
The value of the dimensionless parameter of the heat-conducting inclusion according to Table H.2
0.002 58/(0.2 0.04)=14.5.
Table H.2 - Definition of coefficient
#G0Scheme of thermally conductive connection according to figure H.1 |
The values of the coefficient at (according to Figure H.1 |
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According to Table H.2, by interpolation, we determine the value
0,43+[(0,665-0,43)4,5]/10=0,536.
Coefficient , by formula (13)
Coefficient of heat engineering homogeneity of the panel according to the formula (12)
Reduced resistance to heat transfer according to the formula (11)
H.2 CALCULATION OF THE COEFFICIENT OF THERMAL TECHNICAL HOMOGENEITY USING FORMULA (14)
OF THE PRESENT CODE OF RULES
Calculation example
Determine the reduced resistance to heat transfer of a single-module three-layer reinforced concrete panel on flexible connections with a window opening of a large-panel residential building of the III-133 series.
Initial data
A 300 mm thick panel contains outer and inner reinforced concrete layers, which are interconnected by two hangers (in piers), a strut located in the lower zone of the window sill area, and spacers: 10 - at horizontal joints and 2 - in the area of the window slope (Figure N. 2).
1 - spacers; 2 - loop; 3 - pendants;
4 - concrete thickenings (= 75 mm of the inner reinforced concrete layer); 5 - brace
Figure H.2 - Structure of a three-layer panel on flexible ties
See #M12293 0 1200037434 4120950664 4294967273 80 2997211231 403162211 2325910542 403162211 2520 Table H.4#S shows the design panel parameters.
In the zone of hangers and hinges, the inner concrete layer has thickenings that replace part of the insulation layer.
Calculation procedure
The design of the fence contains the following heat-conducting inclusions: horizontal and vertical joints, window slopes, thickenings of the inner reinforced concrete layer and flexible connections (suspension brackets, brace, spacers).
To determine the coefficient of influence of individual heat-conducting inclusions, we first calculate the thermal resistances of individual sections of the panel using formula (7):
in the zone of thickening of the inner reinforced concrete layer
along the horizontal line
along the vertical joint
thermal resistance of the panel away from heat-conducting inclusions
Conditional resistance to heat transfer away from heat-conducting inclusions
Since the panel has a vertical axis of symmetry, the determination of subsequent values is carried out for half of the panel.
Determine the area of half the panel without taking into account the window opening
Panel thickness =0.3 m.
Let us determine the area of zones of influence and the coefficient for each heat-conducting inclusion of the panel:
for horizontal joint
2,95/3,295=0,895.
According to table H.3 = 0.1. The area of the zone of influence according to the formula (15)
for vertical joint
Table H.3 — Determination of the influence coefficient
#G0Type of thermally conductive inclusion |
Influence coefficient |
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Without adjoining internal fences |
With adjoining internal fences |
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without ribs |
With ribs thickness, mm |
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Window slopes |
without ribs |
With ribs thickness, mm: |
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Flexible connections with a diameter, mm: |
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Notes |
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1 The table shows - thermal resistances, respectively, panels outside the heat-conducting inclusion, joint, thickening of the inner reinforced concrete layer, determined by formula (8); - distances, m, from the longitudinal axis of the window frame to its edge and to the inner surface of the panel. |
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2 Intermediate values should be determined by interpolation. |
According to table H.3 = 0.375. The area of the zone of influence according to the formula (15)
;
for window slopes at = 0.065 m and = 0.18 m, according to table H.3 = 0.374. The area of the zone of influence of half of the window opening, taking into account the corner sections, is determined by the formula (16)
for concrete thickenings of the inner reinforced concrete layer in the suspension and hinge area at = 1.546 / 3.295 = 0.469 according to table M.3 * = 0.78. The total area of the zone of influence of the bulges of the suspension and loop is found by the formula (17)
for suspension (rod diameter 8 mm) according to table H.3 = 0.16, area of influence zone according to formula (17)
for strut (rod diameter 8 mm) according to table H.3 = 0.16, according to formula (17)
for spacers (rod diameter 4 mm) according to table H.3 = 0.05.
When determining the total area of the zone of influence of five spacers, it should be taken into account that the width of the zone of influence from the side of the joint is limited by the edge of the panel and is 0.09 m. According to the formula (18)
Calculate by formula (14)
The reduced resistance to heat transfer of the panel is determined by the formula (11)
Table H.4
#G0Layer material |
Layer thickness, mm |
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Far from inclusions |
in the suspension and hinge area |
horizontal joint |
vertical joint |
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External reinforced concrete layer | ||||||
Thermal insulation layer - expanded polystyrene | ||||||
Mineral wool liners | ||||||
Internal reinforced concrete layer |
APPENDIX P