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Heating and ventilation of residential buildings
Educational and methodical manual for practical exercises
By discipline
"Network engineering. Heat and ventilation"
(calculation examples)
Samara 2011
Compiled by: Dezhurova Natalya Yurievna
Nokhrina Elena Nikolaevna
UDC 628.81/83 07
Heating and ventilation of residential buildings: teaching aid to the control work and practical exercises in the discipline “Engineering networks. Heat and gas supply and ventilation / Comp.:
N.Yu. Dezhurova, E.N. Nokhrina; Samara State arch. - building. un-t. - Samara, 2011. - 80 p.
The methodology for conducting practical classes and performing tests on the course "Engineering networks and equipment of buildings" Heat and gas supply and ventilation is outlined. Given tutorial gives a wide range of options for constructive solutions for external walls, options for typical floor plans, reference data for calculations.
Designed for full-time and part-time students
specialty 270102.65 "Industrial and civil construction", and can also be used by students of specialty 270105.65 "Urban construction and economy".
1 Requirements for the design and content of the control
work (practical exercises) and initial data …………………..5
energy efficient buildings ……………………………………………………………………………11
3 Thermotechnical calculation of external enclosing structures ... .16
3.1 Thermal calculation of the outer wall (calculation example) ... ..20
(calculation example)……………………………………………………25
3.3 Thermal engineering calculation attic floor
(calculation example) …………………………………………………...26
4 Calculation of heat loss by the premises of the building …………………………....28
4.1 Calculation of heat losses in the premises of the building (calculation example) ... 34
5 System development central heating ………………………..44
6 Calculation heating appliances ……………………………………..46
6.1 Calculation example for heaters ……………………………………………………………………………………………………………………………………………………….
7 Constructive solutions for ventilation of a residential building ………………..55
7.1 Aerodynamic calculation of natural draft
ventilation ………………………………………………………...59
7.2 Channel calculation natural ventilation ……………………….62
Bibliographic list ……………………………………………….66
Annex A Map of humidity zones …………………….…………….67
Annex B Operating conditions of enclosing structures
depending on the humidity regime of rooms and humidity zones ………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………
Appendix B Thermal properties of materials …….. ..69
Appendix D Variants of typical floor sections …………………...70
Annex D Values of the water leakage coefficient in instrument assemblies with sectional and panel radiators ... ..75
Annex E Heat flow 1 m open vertical smooth metal pipes, painted oil paint, q, W/m ………………………………………….76
Annex G Table for the calculation of round steel air ducts with t in= 20 ºС …………………………………………..77
Annex 3 Correction factors for frictional pressure losses, taking into account the roughness of the material
air ducts ………………………………………….78
Application And Coefficients local resistance for various
air duct elements …………………………….79
1 Requirements for the design and content of the control
work (practical exercises) and initial data
The control work consists of a settlement and explanatory note and a graphic part.
All necessary initial data are taken according to table 1 in accordance with the last digit of the student's cipher.
The settlement and explanatory note contains the following sections:
1. Climate data
2. The choice of enclosing structures and their heat engineering
payment
3. Calculation of heat loss by the premises of the building
4. Development of a central heating scheme (placement of heating devices, risers, mains and control unit)
5. Calculation of heating devices
6. Constructive solution of the natural ventilation system
7. Aerodynamic calculation of the ventilation system.
An explanatory note is made on A4 sheets or squared notebooks.
The graphic part is made on graph paper, pasted into a notebook and contains:
1. Plan of a typical floor section M 1:100 (see appendix)
2. Basement plan M 1:100
3. Attic plan M 1:100
4. Axonometric diagram of the heating system M 1:100.
Basement and attic plans are drawn based on the plan
typical floor.
The control work provides for the calculation of a two-story residential building, calculations are made for one section. The heating system is single-pipe with upper wiring, dead-end.
The constructive solution for floors above an unheated basement and a warm attic should be taken by analogy with the calculation example.
The climatic characteristics of the construction area given in Table 1 are issued from SNiP 23-01-99 * Building climatology:
1) the average temperature of the coldest five-day period with a security of 0.92 (Table 1, column 5);
2) the average temperature of the heating period (Table 1
column 12);
3) the duration of the heating period (Table 1
column 11);
4) the maximum of the average wind speeds in points for January (table 1 column 19).
The thermophysical characteristics of the fencing materials are taken depending on the operating conditions of the structure, which are determined by the humidity regime of the room and the humidity zone of the construction site.
We accept the humidity regime of the living quarters normal, based on the set temperature +20 ºС and relative humidity of the internal air 55%.
According to the map, Appendix A and Appendix B determine the conditions
operation of building envelopes. Further, according to Appendix B, we accept the main thermophysical characteristics of the materials of the layers of the fence, namely the coefficients:
thermal conductivity, W / (m ºС);
heat absorption, W / (m 2 ºС);
vapor permeability, mg / (m h Pa).
Table 1
Initial data for execution control work
| Initial data | Numerical values depending on the last digit of the cipher | ||||||||||
| Variant number of the typical floor section plan (Appendix D) | |||||||||||
| Floor height (floor to floor) | 2,7 | 3,0 | 3,1 | 3,2 | 2,9 | 3,0 | 3,1 | 2,7 | 3,2 | 2,9 | |
| External wall design option (table 2) | |||||||||||
| City Options | Moscow | Saint Petersburg | Kaliningrad | Cheboksary | Nizhny Novgorod | Voronezh | Saratov | Volgograd | Orenburg | Penza | |
| , ºС | -28 | -26 | -19 | -32 | -31 | -26 | -27 | -25 | -31 | -29 | |
| , ºС | -3,1 | -1,8 | 1,1 | -4,9 | -4,1 | -3,1 | -4,3 | -2,4 | -6,3 | -4,5 | |
| , days | |||||||||||
| , m/s | 4,9 | 4,2 | 4,1 | 5,0 | 5,1 | 5,1 | 5,6 | 8,1 | 5,5 | 5,6 | |
| Orientation to the cardinal points | WITH | YU | Z | V | SW | NW | SE | SW | V | Z | |
| Floor thickness | 0,3 | 0,25 | 0,22 | 0,3 | 0,25 | 0,22 | 0,3 | 0,25 | 0,22 | 0,3 | |
| Kitchens with two-burner three-burner four-burner stove | + - - | - + - | - - + | + - - | - + - | - - + | + - - | - + - | + - - | - + - |
Windows size 1.8 x 1.5 (for living rooms); 1.5 x 1.5 (for the kitchen)
External door size 1.2 x 2.2
table 2
Variants of constructive solutions for external walls
 | Option 1 1 layer - lime-sand mortar; 2 layer - monolithic expanded clay concrete  |
 | Option 2 1 layer - lime-sand mortar; 2 layer - monolithic expanded clay concrete  ; 3 layer - cement-sand mortar; 4th layer - textured layer of the facade system |
 | Option 3 1 layer - lime-sand mortar; 2 layer - monolithic expanded clay concrete  3 layer - cement-sand mortar; 4th layer - textured layer of the facade system |
 | Option 4 1 layer - lime-sand mortar; 2nd layer - silicate brick masonry; 3 layer - monolithic expanded clay concrete  |
| | Option 5 1 layer - lime-sand mortar; 2nd layer - ceramic brick masonry; 3 layer - monolithic expanded clay concrete, ; 4 layer - cement-sand mortar; 5th layer - textured layer of the facade system |
 | Option 6  |
 | Option 7 1 layer - lime-sand mortar; 2 layer - monolithic expanded clay concrete, ; 3rd layer - ceramic brick masonry |
 | Option 8 1 layer - lime-sand mortar; 2 layer - monolithic expanded clay concrete, |
 | Option 9 1 layer - lime-sand mortar; 2 layer - monolithic expanded clay concrete, ; 3rd layer - silicate brick masonry |
 | Option 10 1 layer - lime-sand mortar; 2nd layer - silicate brick masonry; 3 layer - monolithic expanded clay concrete, ; 4 layer - ceramic brick masonry |
Table 3
The values of the coefficient of thermal engineering uniformity
| No. p / p | Type of construction of the outer wall | r |
| Single layer load-bearing exterior walls | 0,98 0,92 | |
| Single-layer self-supporting external walls in monolithic-frame buildings | 0,78 0,8 | |
| Double-layer external walls with internal insulation | 0.82 0,85 | |
| Two-layer external walls with non-ventilated facade systems of the LNPP type | 0,92 0,93 | |
| Double-layer exterior walls with ventilated façade | 0,76 0,8 | |
| Three-layer exterior walls with effective insulation | 0,84 0,86 |
2 Structural solutions for external walls
energy efficient buildings
Constructive solutions for external walls of energy-efficient buildings used in the construction of residential and public
structures can be divided into 3 groups (Fig. 1):
1) single-layer;
2) two-layer;
3) three-layer.
Single-layer external walls are made of cellular concrete blocks, which, as a rule, are designed as self-supporting with floor-by-floor support on floor elements, with mandatory protection from external atmospheric influences by applying plaster,
facings, etc. The transfer of mechanical forces in such structures is carried out through reinforced concrete columns.
Two-layer outer walls contain load-bearing and heat-insulating layers. In this case, the heater can be located as
outside as well as inside.
At the beginning of the energy saving program in the Samara region, internal insulation was mainly used. As thermal insulation material Styrofoam and URSA staple fiber boards were used. From the side of the room, the heaters were protected by drywall or plaster. For
to protect the heaters from moisture and moisture accumulation, a vapor barrier was installed in the form of a polyethylene film.
During the further operation of buildings, many defects were revealed related to the violation of air exchange in the premises, the appearance dark spots, mold and fungus on the interior surfaces of exterior walls. Therefore, at present, internal insulation is used only when installing supply and exhaust mechanical ventilation. As heaters, materials with low water absorption are used, for example, foam plastic and sprayed polyurethane foam.
Systems with external insulation have a number of significant
benefits. These include: high thermal uniformity, maintainability, the possibility of implementing architectural solutions of various shapes.
In construction practice, two options are used
facade systems: with an external plaster layer; with ventilated air gap.
In the first version of the facade systems as
Insulation boards are mainly used styrofoam boards.
The insulation is protected from external atmospheric influences by a base adhesive layer reinforced with fiberglass and a decorative layer.
Rice. 1. Types of external walls of energy efficient buildings:
a - single-layer, b - two-layer, c - three-layer;
1 - plaster; 2- cellular concrete;
3 – protective layer; 4 - outer wall;
5 - insulation; 6 - facade system;
7 - windproof membrane;
8 - ventilated air gap;
11 – facing brick; 12 - flexible connections;
13 - expanded clay concrete panel; 14 - textured layer.
In ventilated facades, only non-combustible insulation in the form of basalt fiber slabs is used. The insulation is protected from
exposure to atmospheric moisture facade plates, which are attached to the wall with brackets. An air gap is provided between the plates and the insulation.
When designing ventilated façade systems, the most favorable heat and moisture regime of the outer walls is created, since water vapor passing through the outer wall mixes with the outside air entering through the air gap and is released into the street through the exhaust ducts.
Three-layer walls, erected earlier, were used mainly in the form of well masonry. They were made of small-piece products located between the outer and inner layers of insulation. The coefficient of thermal engineering homogeneity of structures is relatively small ( r< 0,5) из-за наличия кирпичных перемычек. При реализации в России второго этапа энергосбережения достичь требуемых значений приведенного сопротивления теплопередаче с помощью
well masonry is not possible.
In construction practice, three-layer walls with the use of flexible ties, for the manufacture of which steel reinforcement is used, with the corresponding anti-corrosion properties of steel or protective coatings. Cellular concrete is used as the inner layer, and polystyrene foam, mineral plates and penoizol are used as heat-insulating materials. The facing layer is made of ceramic bricks.
Three-layer concrete walls in large-panel housing construction, they have been used for a long time, but with a lower value of the reduced
heat transfer resistance. To increase the thermal
homogeneity of panel structures must be used
flexible steel connections in the form of individual rods or their combinations. Expanded polystyrene is often used as an intermediate layer in such structures.
At present, three-layer
sandwich panels for construction shopping centers and industrial facilities.
As a middle layer in such structures,
effective heat-insulating materials - mineral wool, expanded polystyrene, polyurethane foam and penoizol. Three-layer enclosing structures are characterized by heterogeneity of materials in cross section, complex geometry and joints. For structural reasons, for the formation of bonds between the shells, it is necessary that stronger materials pass through or enter the thermal insulation, thereby violating the uniformity of the thermal insulation. In this case, the so-called cold bridges are formed. Typical examples of such cold bridges are framing ribs in three-layer panels with effective insulation of residential buildings, corner fastening with a wooden bar of three-layer panels with chipboard cladding and insulation, etc.
3 Thermal engineering calculation of external enclosing structures
The reduced heat transfer resistance of enclosing structures R 0 should be taken in accordance with the design assignment, but not less than the required values of R 0 tr, determined based on sanitary and hygienic conditions, according to formula (1), and energy saving conditions according to table 4.
1. We determine the required resistance to heat transfer of the fence, based on sanitary and hygienic and comfortable conditions:
(1)
where n- coefficient taken depending on the position of the outer surface of the building envelope in relation to the outside air, table 6;
Estimated winter temperature of the outside air, equal to the average temperature of the coldest five-day period with a security of 0.92;
Normalized temperature difference, ° С, table 5;
The heat transfer coefficient of the inner surface of the building envelope, taken according to Table. 7, W / (m 2 ºС).
2. We determine the required reduced resistance to heat transfer of the fence, based on the condition of energy saving.
Degree days of the heating period (GSOP) should be determined by the formula:
GSOP= , (2)
where the average temperature, ºС, and the duration of the heating period with an average daily air temperature of 8 ºС. The value of the required reduced resistance to heat transfer is determined from Table. 4
Table 4
Required reduced resistance to heat transfer
building envelopes
| Buildings and premises | Degree days of the heating period, °C day. | Reduced resistance to heat transfer of enclosing structures, (m 2 ° С) / W: | |||
| walls | coverings and ceilings over driveways | attic ceilings, over cold undergrounds and basements | windows and balcony doors | ||
| Residential, medical and preventive and children's institutions, school boarding schools. | 2,1 2,8 3,5 4,2 4,9 5,6 | 3,2 4,2 5,2 6,2 7,2 8,2 | 2,8 3,7 4,6 5,5 6,4 7,3 | 0,30 0,45 0,60 0,70 0,75 0,80 | |
| Public, except for the above, administrative and domestic, with the exception of premises with a damp or wet regime | 1,6 2,4 3,0 3,6 4,2 4,8 | 2,4 3,2 4,0 4,8 5,6 6,4 | 2,0 2,7 3,4 4,1 4,8 5,5 | 0,30 0,40 0,50 0,60 0,70 0,80 | |
| Production with dry and normal modes | 2,0 2,5 3,0 3,5 4,0 4,5 | 1,4 1,8 2,2 2,6 3,0 3,4 | 0,25 0,30 0,35 0,40 0,45 0,50 | ||
| Notes: 1. Intermediate values R 0 tr should be determined by interpolation. 2. Norms of resistance to heat transfer of translucent enclosing structures for premises industrial buildings with damp and wet conditions, with excess sensible heat from 23 W / m 3, as well as for premises of public, administrative and domestic buildings with wet or wet conditions, should be taken as for rooms with dry and normal conditions of industrial buildings. 3. The reduced heat transfer resistance of the blind part of balcony doors must be at least 1.5 times higher than the heat transfer resistance of the translucent part of these products. 4. In certain justified cases related to specific design solutions for filling window and other openings, it is allowed to use window and balcony door structures with a reduced heat transfer resistance of 5% lower than that specified in the table. |
The values of the reduced resistance to heat transfer of individual enclosing structures should be taken equal to at least
values determined by formula (3) for the walls of residential and public buildings, or by formula (4) - for other enclosing
designs:
(3)
(4)
where are the normalized heat transfer resistances that meet the requirements of the second stage of energy saving, (m 2 · ° С) / W.
3. Find the reduced resistance to heat transfer
building envelope according to the formula
, (5)
where R 0 arb.
r- coefficient of heat engineering uniformity, determined according to table 2.
We determine the value R 0 arb. for multilayer exterior wall
(m 2 ° С) / W, (6)
where R to- thermal resistance of the building envelope, (m 2 ·°С) / W;
is the heat transfer coefficient (for winter conditions) the outer surface of the enclosing structure, determined according to table 7, W / (m 2 ° C); 23 W / (m 2 ° C).
(m 2 ° С) / W, (7)
where R 1 , R 2 , …R n- thermal resistance of individual layers of the structure, (m 2 · ° С) / W.
Thermal resistance R, (m 2 ° C) / W, multilayer layer
enclosing structure should be determined by the formula
where is the layer thickness, m;
Estimated coefficient of thermal conductivity of the layer material,
W/(m °C) (Appendix B).
the value r pre-set depending on the design of the designed outer wall.
4. We compare the heat transfer resistance with the required values, based on comfortable conditions and energy saving conditions, choosing greater value.
There must be inequality
If it is fulfilled, then the design meets the thermal requirements. Otherwise, you need to increase the thickness of the insulation and repeat the calculation.
Based on actual heat transfer resistance R 0 arb. find
heat transfer coefficient of the enclosing structure K, W / (m 2 ºС), according to the formula
Thermal engineering calculation of the outer wall (calculation example)
Initial data
1. Construction area - Samara.
2. The average temperature of the coldest five-day period with a probability of 0.92 t n 5 \u003d -30 ° С.
3. Average temperature of the heating period = -5.2 °С.
4. The duration of the heating period is 203 days.
5. Air temperature inside the building t in=20 °С.
6. Relative air humidity =55%.
7. Humidity zone - dry (Appendix A).
8. Operating conditions of enclosing structures - A
(Appendix B).
Table 5 shows the composition of the fence, and Figure 2 shows the order of the layers in the structure.
Calculation procedure
1. We determine the required resistance to heat transfer of the outer wall, based on sanitary and comfortable
conditions:
where n- coefficient taken depending on the position
the outer surface of the building envelope in relation to the outside air; for exterior walls n = 1;
Design temperature of internal air, °C;
Estimated winter temperature of the outside air, equal to the average temperature of the coldest five-day period
security 0.92;
Normative temperature difference, °С, table 5, for external walls of residential buildings 4 °С;
The heat transfer coefficient of the inner surface of the building envelope, taken according to Table. 7, 8.7 W / (m 2 ºС).
Table 5
The composition of the fence
2. We determine the required reduced resistance to heat transfer of the outer wall, based on the condition of energy saving. Degree days of the heating period (GSOP) are determined by the formula
GSOP \u003d (20 + 5.2) 203 \u003d 5116 (ºС day);
where the average temperature, ºС, and the duration of the heating period with an average daily air temperature of 8 ºС
(m 2 ºС) / W.
Required reduced resistance to heat transfer
determined according to the table. 4 by interpolation method.
3. Of the two values \u200b\u200bof 1.43 (m 2 ºС) / W and 3.19 (m 2 ºС) / W
we take the largest value of 3.19 (m 2 ºС) / W.
4. Determine the required thickness of the insulation from the condition.
The reduced resistance to heat transfer of the enclosing structure is determined by the formula
where R 0 arb.– resistance to heat transfer of the surface of the outer wall without taking into account the influence of external corners, joints and ceilings, window slopes and heat-conducting inclusions, (m 2 ° C) / W;
r- coefficient of thermal uniformity, depending on the structure of the wall, determined according to table 2.
Accept for double layer curtain wall with
external insulation, see table. 3.
(m 2 ° C) / W
6. Determine the thickness of the insulation
M is the standard value of the insulation.
We accept the standard value.
7. Determine the reduced heat transfer resistance
enclosing structures, based on the standard thickness of the insulation
(m 2 ° C) / W
(m 2 ° C) / W
The condition must be met
3.38 > 3.19 (m 2 ° С) / W - the condition is met
8. According to the actual heat transfer resistance of the building envelope, we find the heat transfer coefficient of the outer wall
W / (m 2 ° С)
9. Wall thickness
Windows and balcony doors
According to table 4 and according to GSOP = 5116 ºС day we find for windows and balcony doors (m 2 °С) / W
W / (m 2 ° C).
External doors
In the building we accept external double doors with a vestibule
between them (m 2 ° C) / W.
Heat transfer coefficient of outer door
W / (m 2 ° C).
3.2 Thermotechnical calculation of the attic floor
(calculation example)
Table 6 shows the composition of the attic floor structure, and Figure 3 shows the order of the layers in the structure.
Table 6
Construction composition
| No. p / p | Name | Thickness, m | Density, kg / m 3 | Thermal conductivity coefficient, W / (m o C) |
| Reinforced concrete floor slab hollow | 0,22 | 1,294 | ||
| Grouting with cement-sand mortar | 0,01 | 0,76 | ||
| Waterproofing - one layer of EPP technoelast | 0,003 | 0,17 | ||
| Expanded clay concrete | 0,05 | 0,2 | ||
| Screed from cement-sand mortar | 0,03 | 0,76 |
Thermotechnical calculation of the overlap of a warm attic
For the residential building in question:
14 ºС; 20 ºС; -5.2 ºС; 203 days; - 30 ºС;
GSOP = 5116 ºС day.
We define
|
We determine the value of the required heat transfer resistance of the warm attic floor, according to.
Where 
4.76 0.12 \u003d 0.571 (m 2 ° C) / W.
where 12 W / (m 2 ºС) for attic floors, r= 1
1/8,7+0,22/1,294+0,01/0,76+
0,003/0,17+0,05/0,2+ 0,03/0,76+
1/12 \u003d 0.69 (m 2 o C) / W.
Heat transfer coefficient of a warm attic floor
W / (m 2 ° С)
Attic floor thickness
3.3 Thermal engineering calculation of the overlap over
unheated basement
Table 7 shows the composition of the fence. Figure 4 shows the order of the layers in the structure.
For floors above an unheated basement, the air temperature in the basement is assumed to be 2 ºС; 20 ºС; -5.2 ºС 203 days; GSOP = 5116 ºС day;
The required heat transfer resistance is determined from Table. 4th in GSOP
4.2 (m 2 ° C) / W.
According to where 
4.2 0.36 \u003d 1.512 (m 2 ° C) / W.
Table 7
Construction composition
We determine the reduced resistance of the structure:
where 6 W / (m 2 ºС) tab. 7, - for ceilings over an unheated basement, r= 1
1/8.7+0.003/0.38+0.03/0.76+0.05/0.044+0.22/1.294+1/6=1.635(m 2 o C)/W.
Heat transfer coefficient of the floor over an unheated basement
W / (m 2 ° С)
Ceiling thickness over unheated basement
4 Calculation of heat loss by the premises of the building
The calculation of heat loss by external fences is carried out for each room on the first and second floors for half of the building.
Heat losses of heated premises consist of main and additional. Heat loss by the premises of a building is defined as the sum of heat losses through individual building envelopes
(walls, windows, ceiling, floor above an unheated cellar) rounded up to 10 W. ; H - 16 ºС.
The lengths of the enclosing structures are taken according to the floor plan. In this case, the thickness of the outer walls must be drawn in accordance with the data of the heat engineering calculation. The height of the enclosing structures (walls, windows, doors) is taken according to the initial task data. When determining the height of the outer wall, the thickness of the floor structure or attic floor should be taken into account (see Fig. 5).
;
where the height of the outer wall, respectively, of the first and
second floors;
The thicknesses of the floors above the unheated basement and
attic (accepted from the heat engineering calculation);
The thickness of the intermediate floor.
|
|
Rice. 5. Determining the dimensions of enclosing structures when calculating the heat loss of a room (HC - external walls,
Pl - floor, Fri - ceiling, O - windows):
a - section of the building; b - building plan.
In addition to the main heat losses, it is necessary to take into account
heat loss for heating the infiltration air. Infiltration air enters the room at a temperature close to
outside air temperature. Therefore, in cold period year it must be heated to room temperature.
The heat consumption for heating the infiltration air is taken according to the formula
where specific consumption removed air, m 3 / h; for residential
buildings, 3 m 3 / h is taken per 1 m 2 of the floor area of \u200b\u200bthe living room and kitchen;
For the convenience of calculating heat losses, it is necessary to number all rooms of the building. The numbering should be done floor by floor, starting, for example, with corner rooms. The premises of the first floor are assigned numbers 101, 102, 103 ..., the second - 201, 202, 203 .... The first digit indicates on which floor the room in question is located. In the assignment, students are given a typical floor plan, so room 201 is located above room 101, and so on. Staircases are designated LK-1, LK-2.
The name of the enclosing structures is advisable
abbreviated as: outer wall - NS, double window - TO, balcony door- DB, inner wall - BC, ceiling - Fri, floor - Pl, outer door ND.
The orientation of the enclosing structures facing north - N, east - B, southwest - SW, northwest - NW, etc. is recorded in abbreviated form.
When calculating the area of the walls, it is more convenient not to subtract the area of the windows from them; thus, the heat loss through the walls is somewhat overestimated. When calculating the heat loss through the windows, the value of the heat transfer coefficient is taken equal to . The same is done if there are balcony doors in the outer wall.
Calculation of heat loss is carried out for the premises of the first floor, then - the second. If the room has a layout and orientation to the cardinal points similar to the previously calculated room, then the heat loss is not recalculated, and in the heat loss form opposite the room number is written: “The same as for No.”
(the number of a previously calculated similar room is indicated) and the final value of heat loss for this room.
The heat loss of the staircase is determined as a whole over its entire height, as for one room.
Heat losses through building fences between adjacent heated rooms, for example, through internal walls, should be taken into account only if the difference between the calculated temperatures of the internal air of these rooms is more than 3 ºС.
Table 8
Room heat loss
| room number | Room name and indoor temperature | Fence characteristic | Heat transfer coefficient k, W / (m 2o C) | Estimated temperature difference (t in - t n5) n | Additional heat loss | The amount of additional heat loss | Heat loss through fences Qo, W | Heat consumption for heating infiltration air Q inf, W | Household heat generation Q household, W | Room heat loss Q pom, W | ||||
| Name | orientation | dimensions a x b, m | surface area F, m 2 | for orientation | others | |||||||||
Now, in times of ever-rising energy prices, high-quality insulation has become one of the priorities in the construction of new and repair of already built houses. The cost of work associated with improving the energy efficiency of the house almost always pays off within a few years. The main thing in their implementation is not to make mistakes that will nullify all efforts in best case, at worst - they will also harm.
Modern market building materials just littered with all sorts of heaters. Unfortunately, the manufacturers, or rather, the sellers are doing everything so that we, ordinary developers, choose their material and give our money to them. And this leads to the fact that in various sources of information (especially on the Internet) there are many erroneous and misleading recommendations and advice. Get tangled up in them common man pretty easy.
To be fair, it must be said that modern heaters really quite effective. But in order to use their properties one hundred percent, firstly, the installation must be correct, corresponding to the manufacturer's instructions, and, secondly, the use of insulation must always be appropriate and expedient in each specific case. So how do you do the right thing effective insulation Houses? Let's try to understand this issue in more detail ...
home insulation mistakes
There are three main mistakes that developers most often make:
- incorrect selection of materials and their sequence for the "pie" of the building envelope (walls, floors, roofs ...);
- inappropriate, chosen "at random" thickness of the insulation layer;
- not correct installation with non-compliance with the technology for each specific type of insulation.
The consequences of these mistakes can be very sad. This is the deterioration of the microclimate in the house with an increase in humidity and constant fogging of windows in the cold season, and the appearance of condensate in places where it is not permissible, and the appearance of an unpleasantly smelling fungus with gradual decay of the interior decoration or building envelope.
The choice of insulation method
The most important rule to follow at all times is: insulate the house from the outside, not from the inside! The meaning of this important recommendation clearly shown in the following figure:
The blue-red line in the figure shows the change in temperature in the thickness of the "pie" of the wall. It clearly shows that if the insulation is made from the inside, in the cold season the wall will freeze through.
Here is an example of such a case, by the way, based on very real events. lives good man in a corner apartment of a multi-storey building panel house and in winter, especially in windy weather, it freezes. Then he decides to insulate the cold wall. And since his apartment is on the fifth floor, it’s impossible to think of anything better than to insulate from the inside. At the same time, on one Saturday afternoon, he watches a TV program about repairs and sees how, in a similar apartment, the walls are also insulated from the inside with the help of mineral wool mats.
And everything there seemed to be shown correctly and beautifully: they put up the frame, laid the insulation, closed it vapor barrier film and covered with drywall. But they just didn’t explain that they used mineral wool, not because it is the most suitable material for wall insulation from the inside, but because the sponsor of today's release is a major manufacturer of mineral wool insulation.
And so our good man decides to repeat it. It does everything the same as on TV, and the apartment immediately becomes noticeably warmer. Only his joy from this does not last long. After a while, he begins to feel that some foreign smell has appeared in the room and the air seems to have become heavier. And a few days later, dark damp spots began to appear on the drywall at the bottom of the wall. It's good that the wallpaper did not have time to paste. So what happened?
And what happened was that the panel wall, closed from internal heat by a layer of insulation, quickly froze. Water vapor, which is contained in the air and, due to the difference in partial pressures, always tends from the inside of a warm room to the outside, began to enter the insulation, despite the vapor barrier, through poorly glued or not glued joints at all, through holes from stapler brackets and drywall fastening screws. Upon contact of vapors with a frozen wall, condensation began to fall on it. The insulation began to dampen and accumulate more and more moisture, which led to an unpleasant musty smell and the appearance of a fungus. Also wet mineral wool quickly loses its heat-saving properties.
The question arises - what then should a person do in this situation? Well, for starters, you still need to try to find an opportunity to make insulation from the outside. Fortunately, now there are more and more organizations involved in such work, regardless of height. Of course, their prices will seem very high to many - 1000 ÷ 1500 rubles per 1 m² on a turnkey basis. But this is only at first glance. If we fully calculate all the costs for internal insulation (insulation, its lining, putties, primers, new painting or new wallpaper plus wages for employees), then in the end the difference with external insulation becomes irrelevant and of course it is better to prefer it.
Another thing is if it is not possible to obtain permission for external insulation (for example, the house has some architectural features). In this extreme case, if you have already decided to insulate the walls from the inside, use heaters with minimal (almost zero) vapor permeability, such as foam glass, extruded polystyrene foam.
Foam glass is a more environmentally friendly material, but unfortunately more expensive. So if 1 m³ of extruded polystyrene foam costs about 5,000 rubles, then 1 m³ of foam glass costs about 25,000 rubles, i.e. five times more expensive.
Technology details internal insulation walls will be discussed in a separate article. Now we note only the moment that when installing the insulation, it is necessary to exclude the violation of its integrity to the maximum. So, for example, it is better to glue EPPS to the wall and completely abandon the dowels (as in the figure), or reduce their number to a minimum. As a finish, the insulation is covered with gypsum plaster mixtures, or also pasted over with sheets of drywall without any frames and without any self-tapping screws.
How to determine the required thickness of insulation?
With the fact that it is better to insulate a house from the outside than from the inside, we have more or less figured it out. Now the next question is how much insulation should be laid in each case? It will depend on the following parameters:
- what are the climatic conditions in the region;
- what is the required microclimate in the room;
- what materials make up the "pie" of the building envelope.
A little about how to use it:
Calculation of insulation of the walls of the house
Let's say the "pie" of our wall consists of a layer of drywall - 10 mm ( interior decoration), gas silicate block D-600 - 300 mm, mineral wool insulation -? mm and siding.
We enter the initial data into the program in accordance with the following screenshot:
So point by point:
1) Perform the calculation according to:- we leave a dot in front of "SP 50.13330.2012 and SP 131.13330.2012", as we see these norms are more recent.
2) Locality:- choose "Moscow" or any other that is on the list and is closer to you.
3) Type of buildings and premises- install "Residential."
4) Type of enclosing structure- select "External walls with a ventilated facade." , as our walls are sheathed on the outside with siding.
5) Estimated average temperature and relative humidity of indoor air are determined automatically, we do not touch them.
6) Coefficient of thermal homogeneity "r"- its value is selected by clicking on the question mark. We are looking for what suits us in the tables that appear. If nothing fits, we accept the value “r” from the instructions of the Moscow State Expertise (indicated at the top of the page above the tables). For our example, we took the value r=0.85 for walls with window openings.
This coefficient is not available in most similar online programs for thermal calculation. Its introduction makes the calculation more accurate, since it characterizes the heterogeneity of wall materials. For example, when calculating brickwork this coefficient takes into account the presence of mortar joints, the thermal conductivity of which is much greater than that of the brick itself.
7) Calculation options:- check the boxes next to the items "Calculation of vapor permeability resistance" and "Calculation of dew point".
8) We enter into the table the materials that make up our “pie” of the wall. Please note - it is fundamentally important to make them in order from the outer layer to the inner.
Note: If the wall has an outer layer of material separated by a layer of ventilated air (in our example, this is siding), this layer is not included in the calculation. It is already taken into account when choosing the type of enclosing structure.
So we entered into the table the following materials- KNAUF mineral wool insulation, gas silicate with a density of 600 kg / m³ and lime-sand plaster. In this case, the values of the coefficients of thermal conductivity (λ) and vapor permeability (μ) automatically appear.
The thicknesses of the layers of gas silicate and plaster are known to us initially, we enter them into the table in millimeters. And we select the desired thickness of the insulation until the inscription “ R 0 pr >R 0 norms (... > ...) the design meets the requirements for heat transfer.«
In our example, the condition begins to be fulfilled when the thickness of the mineral wool is 88 mm. We round this value up to 100 mm, since it is this thickness that is commercially available.
Also under the table we see inscriptions that say that moisture accumulation in the heater is impossible and condensation is not possible. This indicates the correct choice of insulation scheme and the thickness of the insulation layer.
By the way, this calculation allows us to see what was said in the first part of this article, namely, why it is better not to insulate the walls from the inside. Let's swap the layers, i.e. put a heater inside the room. See what happens in the following screenshot:
It can be seen that although the design still meets the requirements for heat transfer, the vapor permeability conditions are no longer met and condensation is possible, as indicated under the material plate. The consequences of this have been discussed above.
Another advantage of this online program is that by clicking on the " Report» at the bottom of the page, you can get the entire thermotechnical calculation in the form of formulas and equations with the substitution of all values. Someone might be interested in this.
Calculation of attic floor insulation
An example of a thermal calculation of an attic floor is shown in the following screenshot:
From this it is clear that in this example the required thickness of mineral wool for attic insulation is at least 160 mm. Cover - by wooden beams, the “pie” is - insulation, pine boards 25 mm thick, fiberboard - 5 mm, air gap - 50 mm and drywall filing - 10 mm. The air gap is present in the calculation due to the presence of a frame for drywall.
Calculation of basement floor insulation
An example of a heat engineering calculation for a basement floor is shown in the following screenshot:
In this example, when the basement is a monolithic reinforced concrete 200 mm thick and the house has an unheated underground, the minimum required thickness of insulation with extruded polystyrene foam is about 120 mm.
Thus, the implementation of thermal engineering calculation allows you to correctly arrange the "pie" of the building envelope, choose required thickness each layer and finally perform effective insulation of the house. After that, the main thing is to make a high-quality and correct installation of insulation. Their choice is now very large and each has its own characteristics in working with them. This will certainly be discussed in other articles on our site devoted to the topic of home insulation.
We would love to see your comments on this topic!
A long time ago, buildings and structures were built without thinking about what heat-conducting qualities the enclosing structures have. In other words, the walls were simply made thick. And if you ever happened to be in old merchant houses, then you might have noticed that the outer walls of these houses are made of ceramic bricks, the thickness of which is about 1.5 meters. This thickness brick wall provided and still provides quite a comfortable stay of people in these houses even in the most severe frosts.
At present, everything has changed. And now it is not economically profitable to make the walls so thick. Therefore, materials have been invented that can reduce it. Some 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. 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 indoor air from the condition of no condensation on the inner surfaces of the outer 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.
Payment
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 calculated for industrial buildings with sensible heat excesses of more than 23 W / m 3 and buildings intended for seasonal operation(in autumn or spring), as well as buildings with an estimated internal air temperature of 12 ° C and below the reduced heat transfer resistance of enclosing structures (with the exception of translucent ones).
Determination of the normative (maximum allowable) resistance to heat transfer according to the condition of sanitation (formula 3 SNiP 23-02-2003):
where: n \u003d 1 - coefficient adopted according to table 6 for the outer wall;
t int = 20°C - value from the initial data;
t ext \u003d -31 ° С - value from the initial data;
Δt n \u003d 4 ° С - normalized temperature difference between the temperature of the indoor air and the temperature of the inner surface of the building envelope, is taken according to table 5 in this case for the outer walls of residential buildings;
α int \u003d 8.7 W / (m 2 × ° С) - heat transfer coefficient of the inner surface of the building envelope, taken according to table 7 for external walls.
4.3. Thermal protection rate
From the above calculations for the required heat transfer resistance, we choose R req from the condition of energy saving and denote it now R tr0 \u003d 3.214 m 2 × °С/W .
5. Determining the thickness of the insulation
For each layer of a given wall, it is necessary to calculate the thermal resistance using the formula:
where: δi - layer thickness, mm;
λ i - calculated coefficient of thermal conductivity of the layer material W/(m × °С).
1 layer (decorative brick): R 1 = 0.09 / 0.96 = 0.094 m 2 × °С/W .
3rd layer (silicate brick): R 3 = 0.25 / 0.87 = 0.287 m 2 × °С/W .
4th layer (plaster): R 4 = 0.02 / 0.87 = 0.023 m 2 × °С/W .
Determination of the minimum allowable (required) thermal resistance of a heat-insulating material (formula 5.6 by E.G. Malyavin "Heat loss of a building. Reference manual"):
where: R int = 1/α int = 1/8.7 - resistance to heat transfer on the inner surface;
R ext \u003d 1/α ext \u003d 1/23 - resistance to heat transfer on the outer surface, α ext is taken according to table 14 for external walls;
ΣR i = 0.094 + 0.287 + 0.023 - sum thermal resistance 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 ° C / 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 mineral wool, glass wool or other slab insulation is used as a heater in a three-layer masonry, 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 gap is enclosed 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.
In modern conditions, people are increasingly thinking about rational use resources. Electricity, water, materials. To save all this in the world came for a long time and everyone understands how to do it. But the main amount in the bills for payment is heating, and not everyone understands how to reduce the expense for this item.
What is thermal engineering calculation?
Thermal engineering calculation is performed in order to select the thickness and material of building envelopes and bring the building in line with thermal protection standards. The main regulatory document regulating the ability of a structure to resist heat transfer is SNiP 23-02-2003 "Thermal protection of buildings".
The main indicator of the enclosing surface in terms of thermal protection was the reduced resistance to heat transfer. This is a value that takes into account the heat-shielding characteristics of all layers of the structure, taking into account cold bridges.
A detailed and competent heat engineering calculation is quite laborious. When building private houses, the owners try to take into account the strength characteristics of materials, often forgetting about heat conservation. This can lead to rather disastrous consequences.
Why is the calculation performed?
Before starting construction, the customer can choose whether he will take into account the thermal characteristics or ensure only the strength and stability of the structures.
The cost of insulation will definitely increase the estimate for the construction of the building, but will reduce the cost of further operation. Individual houses are built for decades, perhaps they will serve the next generations. During this time, the cost of an effective insulation will pay off several times.
What does the owner get correct execution calculations:
- Savings on space heating. The heat losses of the building are reduced, respectively, the number of radiator sections will decrease when classical system heating and power of the underfloor heating system. Depending on the heating method, the owner's costs for electricity, gas or hot water become smaller;
- Savings on repairs. At proper insulation a comfortable microclimate is created in the room, condensation does not form on the walls, and microorganisms dangerous to humans do not appear. The presence of a fungus or mold on the surface requires repair, and a simple cosmetic one will not bring any results and the problem will arise again;
- Security for residents. Here, as in the previous paragraph, we are talking about dampness, mold and fungus, which can cause various diseases in people who are constantly in the room;
- respect for environment. There is a shortage of resources on the planet, so reducing the consumption of electricity or blue fuel has a positive effect on the ecological situation.
Normative documents for performing the calculation
The reduced resistance and its compliance with the normalized value - the main objective calculation. But for its implementation, you will need to know the thermal conductivity of the materials of the wall, roof or ceiling. Thermal conductivity is a value that characterizes the ability of a product to conduct heat through itself. The lower it is, the better.
During the calculation of heat engineering, they rely on the following documents:
- SP 50.13330.2012 "Thermal protection of buildings". The document was reissued on the basis of SNiP 23-02-2003. The main standard for calculation;
- SP 131.13330.2012 "Construction climatology". New edition of SNiP 23-01-99*. This document allows you to determine the climatic conditions of the settlement in which the object is located;
- SP 23-101-2004 "Design of thermal protection of buildings" in more detail than the first document in the list, reveals the topic;
- GOST 30494-96 (replaced by GOST 30494-2011 since 2011) Residential and public buildings;
- Manual for students of construction universities E.G. Malyavin “Heat loss of the building. Reference manual".
Thermal engineering calculation is not complicated. It can be performed by a person without special education according to the template. The main thing is to approach the issue very carefully.
An example of calculating a three-layer wall without an air gap
Let's take a closer look at an example of a heat engineering calculation. First you need to decide on the source data. As a rule, you choose the materials for the construction of the walls yourself. We will calculate the thickness of the insulation layer based on the materials of the wall.
Initial data
The data is individual for each construction object and depends on the location of the object.
1. Climate and microclimate
- Construction area: Vologda.
- Purpose of the object: residential.
- Relative air humidity for a room with a normal humidity regime is 55% (item 4.3. Table 1).
- The temperature inside the residential premises tint is set by regulatory documents (Table 1) and is equal to 20 degrees Celsius.
text is the estimated outside air temperature. It is set by the temperature of the coldest five days of the year. The value can be found in, table 1, column 5. For a given area, the value is -32ᵒС.
zht = 231 days - the number of days of the period when it is necessary additional heating premises, that is, the average daily temperature outside is less than 8ᵒС. The value is looked up in the same table as the previous one, but in column 11.
tht = -4.1ᵒС – average outside air temperature during the heating period. The value is in column 12.
2. Wall materials
All layers should be taken into account (even a layer of plaster, if any). This will allow the most accurate calculation of the design.
In this embodiment, consider a wall consisting of the following materials:
- a layer of plaster, 2 centimeters;
- an inner verst made of ordinary solid ceramic brick with a thickness of 38 centimeters;
- a layer of Rockwool mineral wool insulation, the thickness of which is selected by calculation;
- outer verst of front ceramic brick, 12 centimeters thick.
3. Thermal conductivity of adopted materials
All properties of materials must be presented in the passport from the manufacturer. Many companies represent full information about products on their websites. The characteristics of the selected materials are summarized in a table for convenience.
Calculation of the thickness of the insulation for the wall
1. Energy saving condition
Calculation of the value of degree-days of the heating period (GSOP) is carried out according to the formula:
Dd = (tint - tht) zht.
Everything letter designations, presented in the formula, are deciphered in the source data.
Dd \u003d (20-(-4.1)) * 231 \u003d 5567.1 ᵒС * day.
The normative resistance to heat transfer is found by the formula:
The coefficients a and b are taken according to table 4, column 3.
For initial data a=0.00045, b=1.9.
Rreq = 0.00045*5567.1+1.9=3.348 m2*ᵒС/W.
2. Calculation of the norm of thermal protection based on the conditions of sanitation
This indicator is not calculated for residential buildings and is given as an example. The calculation is carried out with an excess of sensible heat exceeding 23 W / m3, or the operation of the building in spring and autumn. Also, calculations are required at a design temperature of less than 12ᵒС indoors. Formula 3 is used:
The coefficient n is taken according to table 6 of the SP "Thermal protection of buildings", αint according to table 7, Δtn according to the fifth table.
Rreq = 1*(20+31)4*8.7 = 1.47 m2*ᵒС/W.
Of the two values obtained in the first and second paragraph, the largest is selected, and further calculation is carried out on it. In this case, Rreq = 3.348 m2*ᵒС/W.
3. Determination of the thickness of the insulation
The heat transfer resistance for each layer is obtained by the formula:
where δ is the layer thickness, λ is its thermal conductivity.
a) plaster R pcs \u003d 0.02 / 0.87 \u003d 0.023 m2 * ᵒС / W;
b) ordinary brick R row.brick. \u003d 0.38 / 0.48 \u003d 0.79 m2 * ᵒС / W;
c) facing brick Rut = 0.12 / 0.48 = 0.25 m2 * ᵒС / W.
The minimum heat transfer resistance of the entire structure is determined by the formula (, formula 5.6):
Rint = 1/αint = 1/8.7 = 0.115 m2*ᵒС/W;
Rext = 1/αext = 1/23 = 0.043 m2*ᵒС/W;
∑Ri = 0.023+0.79+0.25 = 1.063 m2*ᵒC/W, i.e. the sum of the numbers obtained in point 3;
R_tr ^ ut \u003d 3.348 - (0.115 + 0.043 + 1.063) \u003d 2.127 m2 * ᵒС / W.
The thickness of the insulation is determined by the formula (formula 5.7):
δ_tr^ut \u003d 0.038 * 2.127 \u003d 0.081 m.
The value found is the minimum. The insulation layer is taken not less than this value. In this calculation, we finally accept the thickness of the mineral wool insulation as 10 centimeters, so that you do not have to cut the purchased material.
To calculate the heat losses of a building, which are performed for the design of heating systems, it is necessary to find the actual value of the heat transfer resistance with the found thickness of the insulation.
Rо = Rint+Rext+∑Ri = 1/8.7 + 1/23 + 0.023 + 0.79 + 0.1/0.038 + 0.25 = 3.85 m2*ᵒС/W > 3.348 m2*ᵒС/W.
The condition is met.
Influence of the air gap on the heat-shielding characteristics
When constructing a wall protected slab insulation a ventilated layer is possible. It allows you to remove condensate from the material and prevent it from getting wet. The minimum thickness of the gap is 1 centimeter. This space is not closed and has direct communication with the outside air.
In the presence of an air-ventilated layer, only those layers that are up to it from the side are taken into account in the calculation. warm air. For example, a wall pie consists of plaster, internal masonry, insulation, an air gap and external masonry. Only plaster, internal masonry and insulation are taken into account. outer layer masonry goes after the ventilation gap, so it is not taken into account. In this case, the outer masonry performs only an aesthetic function and protects the insulation from external influences.
Important: when considering structures where the airspace is closed, it is taken into account in the calculation. For example, in the case of window fillings. The air between the panes plays the role of an effective insulation.
Teremok program
To perform the calculation using a personal computer, specialists often use the program for thermal calculation "Teremok". It exists online and as an application for operating systems.
The program performs calculations based on all necessary normative documents. Working with the application is extremely simple. It allows you to work in two modes:
- calculation of the required layer of insulation;
- verification of an already thought-out design.
The database contains all the necessary characteristics for the settlements of our country, you just need to select the one you need. It is also necessary to select the type of construction: external wall, mansard roof, ceiling over a cold basement or attic.
When you press the continue button, a new window appears that allows you to "assemble" the structure. Many materials are available in the program memory. They are divided into three groups for ease of search: structural, heat-insulating and heat-insulating-structural. You only need to set the layer thickness, the program will indicate the thermal conductivity itself.
Without necessary materials you can add them yourself, knowing the thermal conductivity.
Before making calculations, you must select the type of calculation above the plate with the wall structure. Depending on this, the program will give either the thickness of the insulation, or report on the compliance of the enclosing structure with the standards. After the calculations are completed, you can generate a report in text format.
"Teremok" is very convenient to use and even a person without a technical education is able to deal with it. For specialists, it significantly reduces the time for calculations and preparation of a report in electronic form.
The main advantage of the program is the fact that it is able to calculate the thickness of the insulation not only of the outer wall, but of any structure. Each of the calculations has its own characteristics, and it is quite difficult for a non-professional to understand all of them. To build a private house, it is enough to master this application and you don't have to go through all the complexity. Calculation and verification of all enclosing surfaces will take no more than 10 minutes.
Thermal engineering calculation online (calculator overview)
Thermal engineering calculation can be done on the Internet online. Not bad, as in my opinion, is the service: rascheta.net. Let's take a quick look at how to work with it.
Going to the website of the online calculator, the first step is to select the standards for which the calculation will be made. I choose the 2012 rulebook as it is a newer document.
Next, you need to specify the region in which the object will be built. If your city is not available, choose the nearest big city. After that, we indicate the type of buildings and premises. Most likely you will calculate a residential building, but you can choose public, administrative, industrial and others. And the last thing you need to choose is the type of enclosing structure (walls, ceilings, coatings).
Estimated average temperature, relative humidity and coefficient of thermotechnical homogeneity are left the same if you do not know how to change them.
In the calculation options, set all two checkboxes except the first one.
In the table, we indicate the wall cake starting from the outside - we select the material and its thickness. On this, in fact, the whole calculation is completed. Below the table is the result of the calculation. If any of the conditions is not met, we change the thickness of the material or the material itself until the data complies with regulatory documents.
If you want to see the calculation algorithm, then click on the "Report" button at the bottom of the site page.
