Thermal engineering calculation for a concrete wall. Calculation of the thermal conductivity of the wall

17.06.2019

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: a teaching aid for tests 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. This tutorial provides a wide range of options for constructive solutions for external walls, options for typical floor plans, and 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 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 Development of a central heating system ………………………..44

6 Calculation of heating devices ……………………………………..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 Section Options typical floor …………………...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

Appendix I Coefficients of local resistances 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. Structural solution natural ventilation systems

7. Aerodynamic calculation of the ventilation system.

Explanatory note performed 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	W	V	SW	NW	SE	SW	V	W
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; 2 layer - masonry from ceramic brick; 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 exterior walls 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 implementation of the energy saving program in the Samara region, it was mainly used internal insulation. Expanded polystyrene and URSA staple fiberglass slabs were used as heat-insulating material. 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 associated with a violation of 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. 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 facade systems, the most favorable heat and moisture regime of the outer walls is created, since the 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 are widely used, for the manufacture of which it is used steel reinforcement, with appropriate 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 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 thermal insulation 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 residential buildings, corner fastening with wooden beams 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 outer surface building envelope in relation to 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. The norms of resistance to heat transfer of translucent enclosing structures for premises of 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 damp or wet conditions should be taken as for premises 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 standard thickness insulation

(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 slab ceilings 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
	Cement-sand mortar screed	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

Rice. 1.8.1

to cover the warm attic of a residential building according to the table. 4 \u003d 4.76 (m 2 ° C) / W.

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 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 the specific consumption of the 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. Numbering should be done floor by floor, starting, for example, from the 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

It is required to determine the thickness of the insulation in a three-layer brick outer wall in a residential building located in Omsk. Wall structure: inner layer - brickwork of ordinary clay bricks with a thickness of 250 mm and a density of 1800 kg / m 3, the outer layer - brickwork of facing bricks with a thickness of 120 mm and a density of 1800 kg / m 3; located between the outer and inner layers effective insulation from expanded polystyrene with a density of 40 kg / m 3; the outer and inner layers are interconnected by fiberglass flexible ties with a diameter of 8 mm, located at a step of 0.6 m.

1. Initial data

The purpose of the building is a residential building

Construction area - Omsk

Estimated indoor air temperature t int= plus 20 0 С

Estimated outdoor temperature text= minus 37 0 C

Estimated indoor air humidity - 55%

2. Determination of the normalized resistance to heat transfer

It is determined according to table 4 depending on the degree-days of the heating period. Degree-days of the heating period, D d , °С×day, determined by formula 1, based on the average outdoor temperature and the duration of the heating period.

According to SNiP 23-01-99 * we determine that in Omsk the average outdoor temperature of the heating period is equal to: t ht \u003d -8.4 0 С, duration of the heating period z ht = 221 days The degree-day value of the heating period is:

D d = (t int - tht) z ht \u003d (20 + 8.4) × 221 \u003d 6276 0 C day.

According to Table. 4. normalized resistance to heat transfer Rreg exterior walls for residential buildings corresponding to the value D d = 6276 0 С day equals Rreg \u003d a D d + b \u003d 0.00035 × 6276 + 1.4 \u003d 3.60 m 2 0 C / W.

3. The choice of a constructive solution for the outer wall

The constructive solution of the outer wall was proposed in the task and is a three-layer fence with an inner layer of brickwork 250 mm thick, with an outer layer of brickwork 120 mm thick, between the outer and inner layers there is a polystyrene foam insulation. The outer and inner layers are interconnected by flexible fiberglass ties with a diameter of 8 mm, located at 0.6 m increments.

4. Determining the thickness of the insulation

The thickness of the insulation is determined by formula 7:

d ut \u003d (R reg ./r - 1 / a int - d kk / l kk - 1 / a ext) × l ut

where Rreg. – normalized resistance to heat transfer, m 2 0 C / W; r- coefficient of heat engineering uniformity; a int is the heat transfer coefficient of the inner surface, W / (m 2 × ° C); a ext is the heat transfer coefficient of the outer surface, W / (m 2 × ° C); d kk- the thickness of the brickwork, m; l kk- the calculated coefficient of thermal conductivity of brickwork, W/(m×°С); l ut- the calculated coefficient of thermal conductivity of the insulation, W/(m×°С).

The normalized resistance to heat transfer is determined: R reg \u003d 3.60 m 2 0 C / W.

The thermal uniformity coefficient for a brick three-layer wall with fiberglass flexible ties is about r=0.995, and may not be taken into account in the calculations (for information - if steel flexible connections are used, then the coefficient of thermal engineering uniformity can reach 0.6-0.7).

The heat transfer coefficient of the inner surface is determined from Table. 7 a int \u003d 8.7 W / (m 2 × ° C).

The heat transfer coefficient of the outer surface is taken according to table 8 a e xt \u003d 23 W / (m 2 × ° C).

The total thickness of the brickwork is 370 mm or 0.37 m.

The design coefficients of thermal conductivity of the materials used are determined depending on the operating conditions (A or B). Operating conditions are determined in the following sequence:

According to the table 1 determine the humidity regime of the premises: since the estimated temperature of the indoor air is +20 0 С, the calculated humidity is 55%, the humidity regime of the premises is normal;

According to Appendix B (map of the Russian Federation), we determine that the city of Omsk is located in a dry zone;

According to the table 2 , depending on the humidity zone and the humidity regime of the premises, we determine that the operating conditions of the enclosing structures are A.

App. D determine the coefficients of thermal conductivity for operating conditions A: for expanded polystyrene GOST 15588-86 with a density of 40 kg / m 3 l ut \u003d 0.041 W / (m × ° С); for brickwork from ordinary clay bricks on a cement-sand mortar with a density of 1800 kg / m 3 l kk \u003d 0.7 W / (m × ° С).

Let us substitute all the determined values into formula 7 and calculate the minimum thickness of the polystyrene foam insulation:

d ut \u003d (3.60 - 1 / 8.7 - 0.37 / 0.7 - 1/23) × 0.041 \u003d 0.1194 m

We round the resulting value up to the nearest 0.01 m: d ut = 0.12 m. We perform a verification calculation according to formula 5:

R 0 \u003d (1 / a i + d kk / l kk + d ut / l ut + 1 / a e)

R 0 \u003d (1 / 8.7 + 0.37 / 0.7 + 0.12 / 0.041 + 1/23) \u003d 3.61 m 2 0 C / W

5. Limitation of temperature and moisture condensation on the inner surface of the building envelope

Δt o, °С, between the temperature of the internal air and the temperature of the internal surface of the enclosing structure should not exceed the normalized values Δtn, °С, established in table 5, and defined as follows

Δt o = n(t int – text)/(R 0 a int) \u003d 1 (20 + 37) / (3.61 x 8.7) \u003d 1.8 0 C i.e. less than Δt n , = 4.0 0 C, determined from table 5.

Conclusion: t thickness of foam polystyrene insulation in a three-layer brick wall is 120 mm. At the same time, the heat transfer resistance of the outer wall R 0 \u003d 3.61 m 2 0 C / W, which is greater than the normalized resistance to heat transfer Rreg. \u003d 3.60 m 2 0 C / W on the 0.01m 2 0 C/W. Estimated temperature difference Δt o, °С, between the temperature of the internal air and the temperature of the internal surface of the enclosing structure does not exceed the standard value Δtn,.

Example of thermotechnical calculation of translucent enclosing structures

Translucent enclosing structures (windows) are selected according to the following method.

Rated resistance to heat transfer Rreg determined according to table 4 of SNiP 23-02-2003 (column 6) depending on the degree-days of the heating period D d. However, the type of building and D d are taken as in the previous example of the heat engineering calculation of opaque enclosing structures. In our case D d = 6276 0 From days, then for the window of an apartment building Rreg \u003d a D d + b \u003d 0.00005 × 6276 + 0.3 \u003d 0.61 m 2 0 C / W.

The choice of translucent structures is carried out according to the value of the reduced resistance to heat transfer R o r, obtained as a result of certification tests or according to Appendix L of the Code of Rules. If the reduced heat transfer resistance of the selected translucent structure R o r, more or equal Rreg, then this design satisfies the requirements of the norms.

Conclusion: for a residential building in the city of Omsk, we accept windows in PVC binding with double-glazed windows made of glass with a hard selective coating and filling the inter-glass space with argon R about r \u003d 0.65 m 2 0 C / W more R reg \u003d 0.61 m 2 0 C / W.

LITERATURE

SNiP 23-02-2003. Thermal protection of buildings.
SP 23-101-2004. Thermal protection design.
SNiP 23-01-99*. Building climatology.
SNiP 31-01-2003. Residential multi-apartment buildings.
SNiP 2.08.02-89 *. Public buildings and structures.

To keep the house warm in the most very coldy, it is necessary to choose the right thermal insulation system - for this, a thermal engineering calculation of the outer wall is performed. The result of the calculations shows how effective the actual or projected method of insulation is.

How to make a thermal calculation of the outer wall

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

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

After collecting and recording the initial information, the thermal conductivity coefficients are determined building materials from which the wall is made. The degree of heat absorption and heat transfer depends on how damp the climate is. In this regard, to calculate the coefficients, moisture maps compiled for Russian Federation. After that, all the numerical values \u200b\u200bnecessary for the calculation are entered into the appropriate formulas.

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

As an example, the heat-shielding properties of a wall made of foam blocks, insulated with expanded polystyrene with a density of 24 kg / m3 and plastered on both sides with a lime-sand mortar are calculated. Calculations and selection of tabular data are carried out on the basis of building regulations.Initial data: construction area - Moscow; relative humidity - 55%, average temperature in the house tv = 20 ° C. The thickness of each layer is set: δ1, δ4 = 0.01m (plaster), δ2 = 0.2m (foam concrete), δ3 = 0.065m (expanded polystyrene "SP Radoslav" ).
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 given conditions, the humidity regime is assumed to be normal. The required value of Rtr is found by the formula:
Rtr=a GSOP+b,
where a, b are taken according to Table 3 of SP 50.13330.2012. For a residential building and an outer wall, a = 0.00035; b = 1.4.
GSOP - degree-days of the heating period, they are found according to the formula (5.2) SP 50.13330.2012:
GSOP=(tin-tot)zot,
where tv \u003d 20O C; tot is the average outdoor temperature during the heating season, according to Table 1 SP131.13330.2012 tot = -2.2°C; zot = 205 days (duration heating season according to the same table).
Substituting the tabular values, they find: GSOP = 4551O C * day; Rtr \u003d 2.99 m2 * C / 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 of SP 50.13330.2012, the conditional resistance to heat transfer is determined:
R0cond=1/αint+δn/λn+1/αext.
where αext \u003d 23 W / (m2 ° С) from clause 1 of Table 6 of SP 50.13330.2012 for external walls.
Substituting the numbers, get R0usl = 2.54 m2 ° C / W. It is refined using the coefficient r = 0.9, which depends on the homogeneity of structures, the presence of ribs, reinforcement, cold bridges:
Rf=2.54 0.9=2.29m2 °C/W.

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

Thermotechnical calculation of the outer wall, the program simplifies calculations

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

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

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

In modern conditions, people are increasingly thinking about the rational use of 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 the building envelope 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 with a classical heating system and the system power warm floors. Depending on the heating method, the owner's costs for electricity, gas or hot water become less;
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 well 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 is the main goal of the 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 in the period when additional space heating is needed, 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 provide complete product information 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.

All 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.

For calculations of heat losses of the building, which are performed for design heating systems, it is necessary to find the actual value of the resistance to heat transfer 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 located before it from the side of warm air are taken into account in the calculation. 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. The outer layer of masonry goes after the ventilation gap, therefore 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 makes calculations based on all the necessary regulatory 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 choose 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 do not have to delve into all the difficulties. 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.

By going to the site online calculator, the first step is to choose the standards by 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).

We leave the calculated average temperature, relative humidity and thermal uniformity coefficient 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.

When and determining the need additional insulation at home, it is important to know the heat loss of its structures, in particular. An online wall thermal conductivity calculator will help you make calculations quickly and accurately.

In contact with

Why do you need a calculation

The thermal conductivity of this element of the building is the property of the structure to conduct heat through a unit of its area with a temperature difference between inside and outside the room of 1 deg. WITH.

The heat engineering calculation of enclosing structures performed by the service mentioned above is necessary for the following purposes:

to select heating equipment and the type of system that allows not only to compensate for heat loss, but also to create a comfortable temperature inside residential premises;
to determine the need for additional insulation of the building;
when designing and constructing a new building to select a wall material that provides the least heat loss in certain climatic conditions;
to create a comfortable temperature indoors not only during the heating period, but also in summer in hot weather.

Attention! When performing independent heat engineering calculations of wall structures, they use the methods and data described in such regulatory documents as SNiP II 03 79 "Construction Heat Engineering" and SNiP 23-02-2003 "Thermal Protection of Buildings".

What does thermal conductivity depend on?

Heat transfer depends on factors such as:

The material from which the building is built various materials differ in their ability to conduct heat. Yes, concrete different kinds bricks contribute to a large loss of heat. On the contrary, galvanized logs, beams, foam and gas blocks, with a smaller thickness, have lower thermal conductivity, which ensures the preservation of heat inside the room and much lower costs for insulation and heating of the building.
Wall thickness - the larger this value, the less heat transfer occurs through its thickness.
Humidity of the material - the greater the moisture content of the raw material from which the structure is erected, the more it conducts heat and the faster it collapses.
The presence of air pores in the material - air-filled pores prevent accelerated heat loss. If these pores are filled with moisture, heat loss increases.
The presence of additional insulation - lined with a layer of insulation outside or inside the wall in terms of heat loss, have values \u200b\u200bmany times less than non-insulated ones.

In construction, along with the thermal conductivity of walls, such a characteristic as thermal resistance (R) has become widespread. It is calculated taking into account the following indicators:

coefficient of thermal conductivity of the wall material (λ) (W/m×0С);
construction thickness (h), (m);
the presence of a heater;
moisture content of the material (%).

The lower the thermal resistance value, the more the wall is subject to heat loss.

The thermotechnical calculation of enclosing structures according to this characteristic is performed according to the following formula:

R=h/λ; (m2×0С/W)

Example of thermal resistance calculation:

Initial data:

the load-bearing wall is made of dry pine timber 30 cm (0.3 m) thick;
thermal conductivity coefficient is 0.09 W/m×0С;
result calculation.

Thus, the thermal resistance of such a wall will be:

R=0.3/0.09=3.3 m2×0С/W

The values obtained as a result of the calculation are compared with the normative ones in accordance with SNiP II 03 79. At the same time, such an indicator as the degree-day of the period during which the heating season continues is taken into account.

If the obtained value is equal to or greater than the standard value, then the material and thickness of the wall structures are selected correctly. Otherwise, the building should be insulated to achieve the standard value.

In the presence of a heater, its thermal resistance is calculated separately and summarized with the same value of the main wall material. Also, if the material of the wall structure has high humidity, apply the appropriate coefficient of thermal conductivity.

For a more accurate calculation of the thermal resistance of this design, similar values \u200b\u200bof windows and doors facing the street are added to the result obtained.

Valid values

When performing a heat engineering calculation of the outer wall, the region in which the house will be located is also taken into account:

For southern regions with warm winters and small temperature differences, walls of small thickness can be erected from materials with an average degree of thermal conductivity - ceramic and clay fired single and double, and of high density. The thickness of the walls for such regions can be no more than 20 cm.
At the same time, for the northern regions, it is more expedient and cost-effective to build enclosing wall structures of medium and large thickness from materials with high thermal resistance - logs, gas and foam concrete of medium density. For such conditions, wall structures up to 50–60 cm thick are erected.
For regions with temperate climate and alternating temperature regime in winter they are suitable with high and medium thermal resistance - gas and foam concrete, timber, medium diameter. Under such conditions, the thickness of wall enclosing structures, taking into account heaters, is no more than 40–45 cm.

Important! The thermal resistance of wall structures is most accurately calculated by the heat loss calculator, which takes into account the region where the house is located.

Heat transfer of various materials

One of the main factors affecting the thermal conductivity of the wall is the building material from which it is built. This dependence is explained by its structure. So, materials with a low density have the lowest thermal conductivity, in which the particles are arranged quite loosely and there is a large number of pores and voids filled with air. These include various types of wood, light porous concrete - foam, gas, slag concrete, as well as hollow silicate bricks.

Materials with high thermal conductivity and low thermal resistance include various types of heavy concrete, monolithic silicate brick. This feature is explained by the fact that the particles in them are located very close to each other, without voids and pores. This contributes to faster heat transfer in the thickness of the wall and a large heat loss.

Table. Thermal conductivity coefficients of building materials (SNiP II 03 79)

Calculation of a sandwich structure

The thermotechnical calculation of the outer wall, consisting of several layers, is carried out as follows:

according to the formula described above, the value of thermal resistance of each of the layers of the "wall cake" is calculated;
the values of this characteristic of all layers are added together, obtaining the total thermal resistance of the wall multilayer structure.

Based on this technique, it is possible to calculate the thickness. To do this, it is necessary to multiply the thermal resistance missing to the norm by the thermal conductivity coefficient of the insulation - as a result, the thickness of the insulation layer will be obtained.

With the help of the TeReMOK program, the thermotechnical calculation is performed automatically. In order for the wall thermal conductivity calculator to perform calculations, it is necessary to enter the following initial data into it:

type of building - residential, industrial;
wall material;
construction thickness;
region;
required temperature and humidity inside the building;
presence, type and thickness of insulation.

Useful video: how to independently calculate the heat loss in the house

Thus, the thermotechnical calculation of enclosing structures is very important both for a house under construction and for a building that has already been built for a long time. In the first case, the correct heat calculation will save on heating, in the second case, it will help to choose the insulation that is optimal in thickness and composition.