Peeling skin can be caused by a variety of factors. To deal with ...
Heating and ventilation of residential buildings
Study guide for practical exercises
By discipline
« Network engineering... Heat and ventilation"
(calculation examples)
Samara 2011
Compiled by: Dezhurova Natalia Yurievna
Nokhrina Elena Nikolaevna
UDC 628.81 / 83 07
Heating and ventilation of residential buildings: teaching aid for test work and practical training in the discipline "Engineering networks. Heat and gas supply and ventilation / Comp .:
N.Yu. Dezhurova, E.N. Nokhrina; Samara State arch. - builds. un-t. - Samara, 2011 .-- 80 p.
The method of conducting practical exercises and performing control works on the course "Engineering networks and equipment of buildings" is described. Heat and gas supply and ventilation. This tutorial gives a wide range of options for constructive solutions of external walls, options for plans for typical floors, provides 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 the specialty 270105.65 "Urban construction and economy".
1 Requirements for the design and content of the control
works (practical exercises) and initial data ………………… ..5
energy efficient buildings …………………………………………… 11
3 Thermal calculation of external enclosing structures ... .16
3.1 Thermal calculation outer wall(calculation example)… ..20
(calculation example) …………………………………………………… 25
3.3 Heat engineering calculation attic floor
(calculation example) ………………………………………………… ... 26
4 Calculation of heat loss by the premises of the building ………………………… .... 28
4.1 Calculation of heat loss in the premises of a building (example of calculation) ... 34
5 System design central heating ………………………..44
6 Calculation of heating devices …………………………………… ..46
6.1 An example of calculating heating devices ……………………… 50
7 Constructive solutions for ventilation of a residential building ……………… ..55
7.1 Aerodynamic calculation of natural exhaust
ventilation ……………………………………………………… ... 59
7.2 Calculation of natural ventilation channels ……………………… .62
Bibliography …………………………………………… .66
Appendix A Map of moisture zones ……………………. …………… .67
Appendix B Operating conditions of enclosing structures
depending on the humidity conditions of the premises and humidity zones …………………………………………………………………………………………………………………… 68
Appendix B Thermophysical characteristics of materials …… .. ..69
Appendix D Section Variants typical floor …………………...70
Appendix D Values of the coefficient of water inflow in instrument assemblies with sectional and panel radiators ... ..75
Appendix E Heat flow of 1 m of open-laid vertical smooth metal pipes colored oil paint, q, W / m ……………………………………… .76
Appendix G Table for calculating round steel ducts at t in= 20 ºС ………………………………………… ..77
Appendix 3 Correction factors for friction pressure loss, taking into account the roughness of the material
air ducts ………………………………………… .78
Appendix I Coefficients of local resistances for various
air duct elements …………………………… .79
1 Requirements for the design and content of the control
work (practical training) and initial data
The test consists of a settlement and explanatory note and a graphic part.
All the 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. Climatic data
2. The choice of enclosing structures and their thermal 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, highways and a 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 a squared notebook.
The graphic part is made on graph paper, glued into a notebook and contains:
1. Section plan of a typical floor M 1: 100 (see appendix)
2. Basement plan M 1: 100
3. Plan of the attic 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 test provides for the calculation of a two-story residential building, calculations are made for one section. Heating system - one-pipe with top wiring, dead-end.
The constructive solution of floors over 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 written out from SNiP 23-01-99 * Construction climatology:
1) average temperature the coldest five-day period with a security of 0.92, (Table 1, column 5);
2) the average temperature of the heating period (tab. 1
column 12);
3) the duration of the heating period (tab. 1
column 11);
4) the maximum of the average wind speeds in terms of points for January (talb. 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.
The humidity regime of the dwelling is accepted normal, based on the set temperature +20 ºС and the relative humidity of the indoor air 55%.
On the map, Appendix A and Appendix B determine the conditions
operation of enclosing structures. Further, according to Appendix B, we take the main thermophysical characteristics of the materials of the fencing layers, namely the coefficients:
thermal conductivity, W / (m · ºС);
heat assimilation, W / (m 2 · ºС);
vapor permeability, mg / (m · h · Pa).
Table 1
Initial data for execution test work
| Initial data | Numerical values depending on the last digit of the cipher | ||||||||||
| The number of the variant of the plan of the section of the typical floor (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 Parameters | Moscow | St. 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 | NS | Z | V | SV | SZ | SE | SW | V | Z | |
| Interfloor slab thickness | 0,3 | 0,25 | 0,22 | 0,3 | 0,25 | 0,22 | 0,3 | 0,25 | 0,22 | 0,3 | |
| Kitchens with two-plate, three-plate, four-plate | + - - | - + - | - - + | + - - | - + - | - - + | + - - | - + - | + - - | - + - |
Window size 1.8 x 1.5 (for living rooms); 1.5 x 1.5 (for the kitchen)
Outer door size 1.2 x 2.2
table 2
Variants of constructive solutions for external walls
 | Option 1 1 layer - lime-sand mortar; 2nd layer - monolithic expanded clay concrete  |
 | Option 2 1 layer - lime-sand mortar; 2nd layer - monolithic expanded clay concrete  ; 3rd layer - cement-sand mortar; 4th layer - textured layer of the facade system |
 | Option 3 1 layer - lime-sand mortar; 2nd layer - monolithic expanded clay concrete  3rd layer - cement-sand mortar; 4th layer - textured layer of the facade system |
 | Option 4 1 layer - lime-sand mortar; 2nd layer - silicate brick masonry; 3rd layer - monolithic expanded clay concrete  |
| | Option 5 1 layer – lime-sand mortar; 2nd layer - ceramic brick masonry; 3rd layer - monolithic expanded clay concrete, ; 4th layer - cement-sand mortar; 5 layer - textured layer of the facade system |
 | Option 6  |
 | Option 7 1 layer - lime-sand mortar; 2nd layer - monolithic expanded clay concrete, ; 3 layer - ceramic brick masonry |
 | Option 8 1 layer - lime-sand mortar; 2nd layer - monolithic expanded clay concrete, |
 | Option 9 1 layer - lime-sand mortar; 2nd layer - monolithic expanded clay concrete, ; 3rd layer - silicate brick masonry |
 | Option 10 1 layer - lime-sand mortar; 2nd layer - silicate brick masonry; 3rd layer - monolithic expanded clay concrete, ; 4th layer - ceramic brickwork |
Table 3
Values of the coefficient of heat engineering uniformity
| P / p No. | Exterior wall structure type | 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 | |
| Double-layer external walls with non-ventilated facade systems such as LNPP | 0,92 0,93 | |
| Double-layer exterior walls with ventilated façade | 0,76 0,8 | |
| Three-layer exterior walls using effective insulation | 0,84 0,86 |
2 Structural solutions of external walls
energy efficient buildings
Constructive solutions for external walls of energy-efficient buildings used in the construction of residential and public buildings
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 support on floor elements, with compulsory protection from external atmospheric influences by applying plaster,
cladding, etc. The transfer of mechanical forces in such structures is carried out through reinforced concrete columns.
Double-layer exterior walls contain load-bearing and heat-insulating layers. In this case, the insulation can be located as
outside and inside.
At the beginning of the implementation of the energy saving program in the Samara region, internal insulation was mainly used. As thermal insulation material polystyrene foam and URSA staple fiberglass boards were used. From the side of the room, the insulation was protected with plasterboard or plaster. For
a vapor barrier in the form of a polyethylene film was installed to protect heaters from moisture and moisture accumulation.
During the further operation of the buildings, many defects were revealed associated with impaired air exchange in the premises, the appearance of dark spots, mold and fungi on the inner surfaces of the outer walls. Therefore, at present, internal insulation is used only when installing supply and exhaust mechanical ventilation. Materials with low water absorption are used as heaters, for example, penoplex and sprayed polyurethane foam.
Systems with external insulation have a number of essential
advantages. These include: high heat engineering uniformity, maintainability, the possibility of implementation architectural solutions of various shapes.
In construction practice, two options are used
facade systems: with external plastering layer; with ventilated air gap.
With the first version of the facade systems as
heaters are mainly used polystyrene foam plates.
Insulation from external weathering is protected by a base adhesive layer, reinforced with fiberglass mesh 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 by facade plates, which are attached to the wall with brackets. An air gap is provided between the plates and the insulation.
When designing ventilated facade systems, the most favorable thermal and humidity conditions of the outer walls are created, since water vapor passing through the outer wall is mixed with the outside air entering through the air gap and thrown out into the street through the exhaust ducts.
Previously erected three-layer walls were used mainly in the form of well masonry. They were made from 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 brick lintels... When implementing the second stage of energy saving in Russia, achieve the required values of the reduced heat transfer resistance using
well masonry is not possible.
In construction practice wide application found three-layer walls using flexible ties, for the manufacture of which steel reinforcement is used, with the corresponding anti-corrosion properties of steel or protective coatings... Aerated concrete is used as an inner layer, and polystyrene foam, mineral plates and foam insulation are used for thermal insulation 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 given
resistance to heat transfer. To improve the heat engineering
homogeneity panel structures nessesary to use
flexible steel ties in the form of individual rods or their combinations. Expanded polystyrene is often used as an intermediate layer in such structures.
Currently, three-layer
sandwich panels for construction shopping centers and industrial facilities.
As a middle layer in such structures, they are used
effective heat-insulating materials - mineral wool, expanded polystyrene, polyurethane foam and penoizol. Three-layer enclosing structures are distinguished 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 the stronger materials pass through the thermal insulation or enter it, thereby disrupting the homogeneity 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 mount wooden beams three-layer panels with chipboard cladding and insulation, etc.
3 Thermal calculation of external enclosing structures
The reduced resistance to heat transfer of the 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 enclosing structure 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 provision of 0.92;
Standardized temperature difference, ° С, table 5;
The heat transfer coefficient of the inner surface of the enclosing structure, taken according to table. 7, W / (m 2 · ºС).
2. Determine the required reduced resistance to heat transfer of the fence, based on the condition of energy saving.
Degree days of the heating period (GSSD) should be determined by the formula:
GSOP =, (2)
where is 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 envelope
| Buildings and premises | Degree days of the heating period, ° С | 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 prophylactic and children's institutions, 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 household, with the exception of rooms with a wet or wet mode | 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 | |
| Dry and normal production | 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. The norms of resistance to heat transfer of translucent enclosing structures for premises of industrial buildings with humid and wet modes, with an excess of sensible heat from 23 W / m 3, as well as for premises of public, administrative and residential buildings with wet or wet modes should be taken as for premises with dry and normal modes of industrial buildings. 3. The reduced resistance to heat transfer of the blind part of balcony doors should be at least 1.5 times higher than the resistance to heat transfer of the translucent part of these products. 4. In certain justified cases related to specific constructive solutions filling window and other openings, it is allowed to use constructions of windows and balcony doors with a reduced heat transfer resistance 5% lower than that set 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 walls of residential and public buildings, or by formula (4) - for the rest of the enclosing
designs:
(3)
(4)
where are the normalized heat transfer resistances corresponding to the requirements of the second stage of energy saving, (m 2 ° С) / W.
3. Find the reduced resistance to heat transfer
the enclosing structure according to the formula
, (5)
where R 0 conv.
r- coefficient of thermal homogeneity, determined according to table 2.
Determine the value R 0 conv for a multi-layer outer wall
(m 2 ° С) / W, (6)
where R to- thermal resistance of the enclosing structure, (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 · ° С); 23 W / (m 2 ° C).
(m 2 ° C) / 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 ° С) / W, layer of multilayer
the enclosing structure should be determined by the formula
where is the layer thickness, m;
The calculated coefficient of thermal conductivity of the layer material,
W / (m ° C) (Appendix B).
The quantity r preset depending on the design of the projected outer wall.
4. We compare the resistance to heat transfer with the required values, based on comfortable conditions and energy saving conditions, choosing a higher value.
Inequality must be respected
If it is fulfilled, then the design meets the thermal engineering requirements. Otherwise, you need to increase the thickness of the insulation and repeat the calculation.
By actual resistance to heat transfer R 0 conv find
heat transfer coefficient of the building envelope K, W / (m 2 ºС), according to the formula
Thermal calculation of the outer wall (calculation example)
Initial data
1. Construction area - Samara.
2. The average temperature of the coldest five-day period of availability is 0.92 t n 5 = -30 ° C.
3. Average temperature of the heating season = -5.2 ° С.
4. The duration of the heating period is 203 days.
5. Air temperature inside the building t in= 20 ° C.
6. Relative humidity = 55%.
7. Moisture 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. Determine the required resistance to heat transfer of the outer wall, based on sanitary and hygienic and comfortable
conditions:
where n- coefficient taken depending on the position
the outer surface of the enclosing structure in relation to the outside air; for exterior walls n = 1;
Design temperature of internal air, ° С;
Estimated winter outdoor temperature equal to the average temperature of the coldest five-day period
security 0.92;
Standard temperature difference, ° С, table 5, for external walls of residential buildings 4 ° С;
The heat transfer coefficient of the inner surface of the enclosing structure, taken according to table. 7, 8.7 W / (m 2 · ºС).
Table 5
Fencing composition
2. 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 = = (20 + 5.2) · 203 = 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
we determine according to the table. 4 by interpolation.
3. Of the two values 1.43 (m 2 · ºС) / W and 3.19 (m 2 · ºС) / W
accept greatest value 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 conv.- resistance to heat transfer of the surface of the outer wall without taking into account the influence of outer corners, joints and ceilings, window slopes and heat-conducting inclusions, (m 2 · ° С) / W;
r- coefficient of heat engineering uniformity, depending on the structure of the wall, determined according to table 2.
Acceptable for double layer outer 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 ° C) / W - the condition is met
8. According to the actual heat transfer resistance of the enclosing structure, we find the heat transfer coefficient of the outer wall
W / (m 2 ° C)
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.
Outside door heat transfer coefficient
W / (m 2 ° C).
3.2 Thermal 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
Structure composition
| P / p No. | Name | Thickness, m | Density, kg / m 3 | Thermal conductivity coefficient, W / (m о С) |
| Reinforced concrete slab hollow ceilings | 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 |
Thermal calculation of overlap 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 resistance to heat transfer of the overlap of the warm attic, according to.
Where 
4.76 0.12 = 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 = 0.69 (m 2 about C) / W.
Heat transfer coefficient of the ceiling of a warm attic
W / (m 2 ° C)
Attic floor thickness
3.3 Thermal calculation of overlapping
unheated basement
Table 7 shows the composition of the fence. Figure 4 shows the order of the layers in the structure.
For ceilings above an unheated basement, the air temperature in the basement is taken as 2 ºС; 20 ºС; -5.2 ºС for 203 days; GSOP = 5116 ºС · day;
The required heat transfer resistance is determined from table. 4th largest GSN
4.2 (m 2 ° C) / W.
According to where 
4.2 0.36 = 1.512 (m 2 ° C) / W.
Table 7
Structure composition
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 о С) / W.
Heat transfer coefficient of a floor over an unheated basement
W / (m 2 ° C)
Slab thickness over unheated basement
4 Calculation of heat loss by the premises of the building
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 the heated premises consist of main and additional ones. Heat losses in the premises of a building are defined as the sum of heat losses through individual building envelopes
(walls, windows, ceiling, floor above an unheated basement) rounded 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 should be drawn in accordance with the data heat engineering calculation... The height of the enclosing structures (walls, windows, doors) is taken according to the initial data of the assignment. When determining the height of the outer wall, the thickness of the floor or attic structure should be taken into account (see Fig. 5).
;
where is the height of the outer wall, respectively, of the first and
second floors;
The thicknesses of the ceilings above the unheated basement and
attic (taken from the heat engineering calculation);
The thickness of the floor slab.
|
|
Rice. 5. Determination of the dimensions of the enclosing structures when calculating the heat loss of the room (NS - 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 infiltration air. Infiltration air enters the room with a temperature close to
outdoor temperature. Therefore, in the cold season, it must be heated to room temperature.
The heat consumption for heating the infiltration air is taken according to the formula
where is the specific consumption of the removed air, m 3 / h; for residential
buildings are assumed to be 3 m 3 / h per 1 m 2 of floor area of living quarters and kitchens;
For the convenience of calculating heat loss, it is necessary to number all the premises of the building. Numbering should be done floor by floor, starting, for example, from corner rooms... The rooms on the first floor are assigned numbers 101, 102, 103 ..., on the second - 201, 202, 203 .... The first number indicates on which floor the room in question is located. In the assignment, students are given a plan for a typical floor, so room 201 is located above room 101, etc. Stairwells are designated LK-1, LK-2.
The name of the enclosing structures is advisable
abbreviated as: outer wall - NS, double window - DO, balcony door- DB, inner wall - Sun, ceiling - Fri, floor - Pl, outer door ND.
The orientation of the enclosing structures facing north - N, east - E, south-west - SW, north-west - 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 same heat loss through the windows, the value of the heat transfer coefficient is taken to be. Do the same if there are balcony doors in the outer wall.
Calculation of heat loss is carried out for the premises of the first floor, then for 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 re-calculated, and in the heat loss form opposite the room number it is written: "Same as for No."
(the number of a previously calculated similar room is indicated) and the total value of heat loss for this room.
Heat loss staircase determined as a whole over its entire height, as for one room.
Heat loss through building fences between adjacent heated rooms, for example, through interior walls, should be taken into account only when the difference in the design temperatures of the internal air of these rooms is more than 3 ºС.
Table 8
Heat loss of premises
| Room No. | Room name and internal temperature | Fence characteristic | Heat transfer coefficient k, W / (m 2о С) | Calculated temperature difference (t in - t n5) n | Additional heat losses | The sum of additional heat losses | Heat loss through fences Q o, W | Heat consumption for heating infiltration air Q inf, W | Household heat dissipation Q life, W | Heat loss of the room Q pom, W | ||||
| Name | orientation | dimensions a x b, m | surface area F, m 2 | orientation | others | |||||||||
Now, at a time of constantly rising energy prices, high-quality insulation has become one of the primary tasks in the construction of new and repair of already built houses. The work costs associated with improving the energy efficiency of a home almost always pay off within a few years. The main thing during their implementation is not to make mistakes that will nullify all efforts at best, and at worst - they will also harm.
Modern market building materials just littered with all kinds of insulation. Unfortunately, manufacturers, or more precisely, 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 information (especially on the Internet), there are many erroneous and misleading recommendations and advice. Get entangled in them common man pretty easy.
In fairness, it must be said that modern heaters are really quite effective. But in order to use their properties one hundred percent, firstly, the correct installation must be carried out in accordance with 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 and effective insulation at home? Let's try to deal with this issue in more detail ...
Errors when insulating a house
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 to the standards, 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 dire. This is the deterioration of the microclimate in the house with an increase in humidity and constant fogging of the windows in the cold season, and the appearance of condensation in places where this is not permissible, and the appearance of an unpleasant-smelling fungus with gradual decay of the interior decoration or enclosing structures.
Choosing a method of insulation
The most important rule, which is always better to follow, is - insulate the house from the outside, not from the inside! The meaning of this important recommendation is clearly demonstrated in the following figure:
The blue-red line in the figure shows the change in temperature in the thickness of the "cake" of the wall. It clearly shows that if the insulation is made from the inside, the wall will freeze through during the cold season.
For example, such a case, by the way based on quite real events. Lives good man in a corner apartment of a multi-storey 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, there is nothing better than insulating it from the inside. At the same time, one Saturday afternoon, he watches a TV program about repairs and sees how the walls in a similar apartment 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 sheathed with plasterboard. But they just did not 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 their today's release is a large manufacturer of mineral wool insulation.
And so our good man decides to repeat it. He does everything the same way 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 kind of foreign smell has appeared in the room and the air has become as if 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 I didn't have time to stick the wallpaper. So what happened?
What happened is that panel wall, closed from internal heat with a layer of insulation, quickly froze over. Water vapors that are contained in the air and, due to the difference in partial pressures, always tend to the outside from the inside of a warm room, began to fall into the insulation, despite the vapor barrier made, through poorly glued or not glued joints at all, through holes from stapler brackets and drywall fixing screws. When the vapors came into contact with the frozen wall, condensation began to fall out on it. The insulation began to get damp and accumulate more and more moisture, which led to an unpleasant musty smell and the appearance of fungus. In addition, wet mineral wool quickly loses its heat-saving properties.
The question arises - what then should a person do in this situation? Well, first you need to still try to find an opportunity to make insulation outside. Fortunately, now more and more organizations appear that are engaged in such work, regardless of height. Of course, their prices will seem very high to many - 1000 ÷ 1500 rubles for 1m² turnkey. But this is only at first glance. If we calculate in full all the costs for internal insulation (insulation, its cladding, putties, primers, new painting or new wallpaper plus employees' salaries), then in the end the difference with external insulation becomes not fundamental and of course it is better to prefer it.
It is another matter if it is not possible to obtain a permit 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 insulation with minimal (almost zero) vapor permeability, such as foam glass, extruded polystyrene foam.
Foam glass is more environmentally friendly material but unfortunately also more expensive. So if 1 m³ of extruded polystyrene foam costs about 5000 rubles, then 1 m³ of foam glass - about 25000 rubles, i.e. five times more expensive.
Technology details internal insulation walls will be discussed in a separate article. Now we will note only the moment that during the installation of the insulation it is necessary to exclude the violation of its integrity to the maximum. So, for example, it is better to glue the EPSP to the wall and abandon the dowels altogether (as in the figure), or reduce their number to a minimum. As a finish, the insulation is covered with plaster plaster mixes, or also pasted over with sheets of drywall without any frames and without any 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 more or less figured out. Now the next question is - how much insulation do you need to lay in each specific case? It will depend on the following parameters:
- what are the climatic conditions in the region;
- what is the required indoor climate;
- what materials make up the "pie" of the enclosing structure.
A little about how to use it:
Calculation of the insulation of the walls of the house
Let's say the "cake" 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) Calculate according to:- we leave the dot opposite "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 closer to you.
3) Type of buildings and premises- install "Residential."
4) Type of enclosing structure- choose "External walls with a ventilated facade." as our walls are sheathed with siding on the outside.
5) Estimated average temperature and relative humidity of indoor air are determined automatically, we do not touch them.
6) Thermal homogeneity coefficient "r"- select its value by clicking on the question mark. We are looking for what suits us in the tables that appear. If nothing fits, we take 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 absent in most of such online programs for heat engineering calculation. Its introduction makes the calculation more accurate, since it characterizes the heterogeneity of the wall materials. For example, when calculating brickwork this coefficient takes into account the presence of mortar joints, the thermal conductivity of which is much higher than that of the brick itself.
7) Calculation options:- put a tick in front of the items "Calculation of resistance to vapor permeation" and "Calculation of the dew point".
8) We enter in the table the materials that make up our "cake" of the wall. Please note that it is fundamentally important to make them in order from the outer layer to the inner one.
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 has already been taken into account when choosing the type of enclosing structure.
So, we have added the following materials to the table - 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) appear automatically.
The thicknesses of the layers of gas silicate and plaster are known to us initially, we enter them in the table in millimeters. And we select the required 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 starts to be fulfilled when the thickness of the mineral wool is equal to 88 mm. We round this value up to 100 mm, since this is the thickness that is commercially available.
Also, under the table, we see inscriptions saying that moisture accumulation in the insulation is impossible and no condensation possible... This indicates a correctly selected 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. we will put the insulation inside the room. What happens in this case, see 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 were discussed above.
Another advantage of this online program is that by clicking on the " Report»At the bottom of the page, you can get all the heat engineering calculation in the form of formulas and equations with the substitution of all values. Someone may be interested in this.
Calculation of attic floor insulation
An example of a thermal engineering calculation for an attic floor is shown in the following screenshot:
From this it can be seen that in this example, the required thickness of mineral wool for insulating an attic is at least 160 mm. Overlap - by wooden beams, "Pie" is made up - insulation, pine boards 25 mm thick, fiberboard - 5 mm, air gap - 50 mm and plasterboard filing - 10 mm. The air gap is present in the calculation due to the presence of a frame for drywall.
Calculation of basement insulation
An example of a heat engineering calculation for a basement floor is shown in the following screenshot:
In this example, when basement floor is monolithic reinforced concrete thickness 200 mm and the house has an unheated underground, the minimum required thickness of insulation with extruded polystyrene foam is about 120 mm.
Thus, performing a heat engineering calculation allows you to correctly compose the "pie" of the enclosing structure, choose required thickness of each layer and, in the end, perform effective insulation of the house. After that, the main thing is to make a high-quality and correct installation of insulation. The choice of them is now very large and in working with each has its own peculiarities. This will certainly be discussed in other articles on our site dedicated to the topic of home insulation.
We will be glad to see your comments on this topic!
The heat in the house directly depends on many factors, including the thickness of the insulation. The thicker it is, the better your home will be protected from cold and freezing, and the less you will pay for heating.
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To understand the best way to insulate a house, you need to take into account several factors:
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How much insulation and how thick do you need for your home?
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In order for the house to be warm in the most severe frosts, it is necessary to choose the right thermal insulation system - for this, the thermal engineering calculation of the outer wall is performed. The result of the calculations shows how effective the real or projected method of insulation is.
How to make a heat engineering calculation of an outer wall
First, you should prepare the initial data. On design parameter influenced by the following factors:
- the climatic region in which the house is located;
- the purpose of the premises is a residential building, an industrial building, a hospital;
- the operating mode of the building is seasonal or year-round;
- the presence of door and window openings in the structure;
- indoor humidity, the difference between indoor and outdoor temperatures;
- number of floors, overlap features.
After collecting and recording the initial information, the thermal conductivity coefficients of the building materials from which the wall is made are determined. The degree of heat absorption and heat transfer depends on how humid the climate is. In this regard, moisture maps are used to calculate the coefficients, compiled for Russian Federation... After that, all the numerical values necessary for the calculation are entered into the corresponding formulas. 
Thermal calculation of an external wall, an example for a foam concrete wall
As an example, the heat-shielding properties of a wall lined with foam blocks, insulated with expanded polystyrene with a density of 24 kg / m3 and plastered on both sides with a lime-sand mortar are calculated. Calculations and selection of tabular data are based on building regulations.
Initial data: construction area - Moscow; relative humidity - 55%, average temperature in the house tв = 20О С. The thickness of each layer is set: δ1, δ4 = 0.01m (plaster), δ2 = 0.2m (foam concrete), δ3 = 0.065m (expanded polystyrene "SP Radoslaw" ).
The purpose of the heat engineering calculation of the outer wall is to determine the required (Rtr) and actual (Rf) resistance to heat transfer.
Payment
- According to table 1 of SP 53.13330.2012 under the given conditions, the humidity mode is assumed to be normal. The required value of Rtr is found by the formula:
Rtr = a GSOP + b,
where a, b are taken according to table 3 of SP 50.13330.2012. For a residential building and an outer wall a = 0.00035; b = 1.4.
GSOP - degree-day of the heating period, they are found by the formula (5.2) SP 50.13330.2012:
GSOP = (tv-tot) zfrom,
where tв = 20О С; tfrom - the average temperature of the outside air during the heating period, according to table 1 SP131.13330.2012tot = -2.2ОС; zfrom = 205 days. (duration of the heating season according to the same table).
Substituting the tabular values, they find: GSOP = 4551О С * day; Rtr = 2.99 m2 * С / W - According to table 2 SP50.13330.2012 for normal humidity, the thermal conductivity coefficients of each layer of the "pie" are selected: λB1 = 0.81W / (m ° C), λB2 = 0.26W / (m ° C), λB3 = 0.041W / (m ° C), λB4 = 0.81W / (m ° C).
According to the formula E.6 SP 50.13330.2012, the conditional resistance to heat transfer is determined:
R0condition = 1 / αint + δn / λn + 1 / αext.
where αext = 23 W / (m2 ° С) from clause 1 of table 6 SP 50.13330.2012 for external walls.
Substituting the numbers, you get R0con = 2.54m2 ° C / W. Refine it using the coefficient r = 0.9, depending on the homogeneity of structures, the presence of ribs, reinforcement, cold bridges:
Rf = 2.54 0.9 = 2.29m2 ° C / W.
The obtained result shows that the actual thermal resistance is less than the required one, therefore, it is necessary to reconsider the structure of the wall.
Thermal calculation of the outer wall, the program simplifies the calculations
Uncomplicated computer services speed up computational processes and the search for the desired coefficients. It is worth familiarizing yourself with the most popular programs. 
- "TeReMok". The initial data are entered: the type of building (residential), the internal temperature is 20 °, the humidity mode is normal, the area of residence is Moscow. In the next window, the calculated value of the standard heat transfer resistance opens - 3.13 m2 * оС / W.
Based on the calculated coefficient, the thermal engineering calculation of the outer wall of foam blocks (600 kg / m3), insulated with extruded polystyrene "Flurmat 200" (25 kg / m3) and plastered with cement-lime mortar, is carried out. Choose from the menu necessary materials, putting down their thickness (foam block - 200 mm, plaster - 20 mm), leaving the cell with the thickness of the insulation empty.
By pressing the "Calculation" button, the desired thickness of the heat insulator layer is obtained - 63 mm. The convenience of the program does not eliminate its drawback: it does not take into account the different thermal conductivity of the masonry material and mortar. Thanks to the author can be said at this address http://dmitriy.chiginskiy.ru/teremok/ - The second program is offered by the site http://rascheta.net/. Its difference from the previous service is that all thicknesses are set independently. The coefficient of heat engineering homogeneity r is introduced into the calculation. It is selected from the table: for aerated concrete blocks with wire reinforcement in horizontal seams, r = 0.9.
After filling in the fields, the program gives a report on what is the actual thermal resistance of the selected structure, whether it meets climatic conditions... In addition, a sequence of calculations is provided with formulas, normative sources and intermediate values.
When building a house or carrying out thermal insulation work, it is important to assess the effectiveness of the outer wall insulation: a heat engineering calculation performed independently or with the help of a specialist allows you to do this quickly and accurately.
A long time ago, buildings and structures were built without thinking about the heat-conducting qualities of the enclosing structures. In other words, the walls were made just 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 of the brick wall has ensured and still ensures quite comfortable stay of people in these houses, even in the most severe frosts.
Nowadays, everything has changed. And now it is not economically viable 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 the brickwork can be reduced to 250 mm.
Now walls and ceilings are most often made in 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 heat engineering calculation is carried out and the dew point is determined.
You can find out how the dew point calculation is performed on the next page. Here, the heat engineering calculation will be considered using an example.
Required regulatory documents
The calculation will require two SNiPs, one joint venture, one GOST and one manual:
- SNiP 23-02-2003 (SP 50.13330.2012). "Thermal protection of buildings". Updated edition from 2012.
- SNiP 23-01-99 * (SP 131.13330.2012). "Building climatology". Updated edition from 2012.
- SP 23-101-2004. "Design of thermal protection of buildings".
- GOST 30494-96 (replaced by GOST 30494-2011 since 2011). "Residential and public buildings. Parameters of indoor microclimate".
- Benefit. E.G. Malyavin "Heat loss of a building. Reference book".
Calculated parameters
In the process of performing a heat engineering calculation, the following is determined:
- thermotechnical characteristics of building materials of enclosing structures;
- reduced heat transfer resistance;
- compliance of this reduced resistance with the standard value.
Example. Heat 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 is t int = 20 ° С (GOST 30494-96 table 1).
Estimated outdoor temperature t ext, determined by the temperature of the coldest five-day period with a security of 0.92 = -31 ° C (SNiP 23-01-99 Table 1, column 5);
The duration of the heating period with an average daily outside air temperature of 8 ° C is z ht = 215 days (SNiP 23-01-99 table 1, column 11);
The average temperature of the outside air for the heating period is t ht = -4.1 ° С (SNiP 23-01-99 table 1, column 12).
2. Wall construction
The wall consists of the following layers:
- Decorative brick (besser) 90 mm thick;
- insulation (mineral wool slab), in the figure its thickness is indicated by an "X", since it will be found during the calculation;
- silicate brick 250 mm thick;
- plaster (complex solution), an additional layer to obtain a more objective picture, since its influence is minimal, but it is there.
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. Determination of the thickness of the insulation
To calculate the thickness of the insulating layer, it is necessary to determine the heat transfer resistance of the enclosing structure based on the requirements of sanitary standards and energy saving.
4.1. Determination of the rate of thermal protection by the condition of energy saving
Determination of the degree-day of the heating period according to clause 5.3 of SNiP 23-02-2003:
D d = ( t int - t ht) z ht = (20 + 4.1) 215 = 5182 ° С × day
Note: also degree-days are designated - GSOP.
The standard value of the reduced resistance to heat transfer should be taken not less than the standardized values determined according to SNIP 23-02-2003 (Table 4), depending on the degree-day of the construction area:
R req = a × D d + b = 0.00035 × 5182 + 1.4 = 3.214m 2 × ° C / W,
where: Dd - degree-day of the heating period in Nizhny Novgorod,
a and b are 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 rate of thermal protection according to the condition of sanitation
In our case, it is considered as an example, since this indicator is calculated for industrial buildings with an excess of sensible heat of more than 23 W / m 3 and buildings intended for seasonal operation (in autumn or spring), as well as buildings with an estimated internal air temperature of 12 ° С and below is the resistance to heat transfer of enclosing structures (excluding translucent ones).
Determination of the normative (maximum permissible) resistance to heat transfer according to the condition of sanitation (formula 3 SNiP 23-02-2003):
where: n = 1 - coefficient adopted according to table 6 for the outer wall;
t int = 20 ° С - value from the initial data;
t ext = -31 ° С - value from the initial data;
Δt n = 4 ° С is the normalized temperature difference between the temperature of the internal air and the temperature of the internal surface of the enclosing structure, taken according to table 5 in this case for exterior walls of residential buildings;
α int = 8.7 W / (m 2 × ° С) is the heat transfer coefficient of the inner surface of the enclosing structure, 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 = 3.214m 2 × ° C / W .
5. Determination of 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 is the 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 × ° C / W .
3rd layer (silicate brick): R 3 = 0.25 / 0.87 = 0.287 m 2 × ° C / W .
4th layer (plaster): R 4 = 0.02 / 0.87 = 0.023 m 2 × ° C / 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 = 1 / α ext = 1/23 - resistance to heat transfer on the outer surface, α ext is taken according to table 14 for external walls;
ΣR i = 0.094 + 0.287 + 0.023 - the sum of thermal resistances of all layers of the wall without a layer of insulation, determined taking into account the coefficients of thermal conductivity of materials adopted in column A or B (columns 8 and 9 of table D1 SP 23-101-2004) in accordance with the humidity conditions of the wall, m 2 ° С / W
The thickness of the insulation is (formula 5.7):
where: λ ut - thermal conductivity coefficient of the insulation material, W / (m · ° С).
Determination of the thermal resistance of the wall from the condition that the total thickness of the insulation will be 250 mm (formula 5.8):
where: ΣR t, i is the sum of thermal resistances of all layers of the fence, including the insulation layer, of the accepted structural thickness, m 2 ° C / W.
From the result obtained, we can conclude that
R 0 = 3.503m 2 × ° C / W> R tr0 = 3.214m 2 × ° C / 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, you need a ventilated air gap between external masonry and insulation. The thickness of this layer should be at least 10 mm, and preferably 20-40 mm. It is necessary in order to drain the insulation that gets wet from condensation.
This air gap is not a closed space, therefore, if it is present in the calculation, it is necessary to take into account the requirements of clause 9.1.2 of SP 23-101-2004, namely:
a) the layers of the structure located between the air gap and the outer surface (in our case, this is a decorative brick (besser)), are not taken into account in the heat engineering calculation;
b) on the surface of the structure facing the layer ventilated with 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 thermal engineering calculation of plastic insulating glass units.
