Thermal engineering calculation of an external brick wall. How to make a heat engineering calculation of the outer walls of a low-rise building? Thermotechnical calculation of the outer wall with verification

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

1. Initial data

The purpose of the building is a residential building

Construction area - Omsk

Estimated indoor air temperature t int= plus 20 0 C

Estimated outdoor temperature text= minus 37 0 С

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. Choice constructive solution outer wall

The constructive solution of the outer wall was proposed in the assignment and is a three-layer fence with an inner layer of brickwork 250 mm thick, an outer layer of brickwork 120 mm thick, and an expanded polystyrene insulation is located between the outer and inner layers. 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 The thickness of the expanded polystyrene insulation in a three-layer brick wall is 120 mm. At the same time, the heat transfer resistance of the outer wall R 0 \u003d 3.61 m 2 0 C / W, which is greater than the normalized resistance to heat transfer Rreg. \u003d 3.60 m 2 0 C / W on 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.

Initial data

Place of construction - Omsk

z ht = 221 days

t ht = -8.4ºС.

t ext = -37ºС.

t int = + 20ºС;

air humidity: = 55%;

Operating conditions of enclosing structures - B. Heat transfer coefficient of the inner surface of the fence A i nt \u003d 8.7 W / m 2 ° С.

a ext \u003d 23 W / m 2 ° C.

The necessary data on the structural layers of the wall for thermal calculation are summarized in the table.

1. Determination of degree-days of the heating period according to the formula (2) SP 23-101-2004:

D d \u003d (t int - t ht) z th \u003d (20–(8.4)) 221 \u003d 6276.40

2. The normalized value of the heat transfer resistance of the outer walls according to the formula (1) SP 23-101-2004:

R reg \u003d a D d + b \u003d 0.00035 6276.40+ 1.4 \u003d 3.6 m 2 ° C / W.

3. Reduced resistance to heat transfer R 0 r external brick walls with effective insulation residential buildings is calculated by the formula

R 0 r = R 0 arb r,

where R 0 conv - heat transfer resistance of brick walls, conditionally determined by formulas (9) and (11) without taking into account heat-conducting inclusions, m 2 ·°С / W;

R 0 r - reduced resistance to heat transfer, taking into account the coefficient of thermal uniformity r, which for walls is 0.74.

The calculation is carried out from the condition of equality

hence,

R 0 conditional \u003d 3.6 / 0.74 \u003d 4.86 m 2 ° C / W

R 0 conv \u003d R si + R k + R se

R k \u003d R reg - (R si + R se) \u003d 3.6- (1 / 8.7 + 1/23) \u003d 3.45 m 2 ° C / W

4. Thermal resistance of outer brick wall layered structure can be represented as the sum of the thermal resistances of the individual layers, i.e.

R to \u003d R 1 + R 2 + R ut + R 4

5. Determine the thermal resistance of the insulation:

R ut \u003d R k + (R 1 + R 2 + R 4) \u003d 3.45– (0.037 + 0.79) \u003d 2.62 m 2 ° С / W.

6. Find the thickness of the insulation:

\u003d R ut \u003d 0.032 2.62 \u003d 0.08 m.

We accept the thickness of the insulation 100 mm.

The final wall thickness will be (510+100) = 610 mm.

We perform a check taking into account the accepted thickness of the insulation:

R 0 r \u003d r (R si + R 1 + R 2 + R ut + R 4 + R se) \u003d 0.74 (1 / 8.7 + 0.037 + 0.79 + 0.10 / 0.032 + 1/23 ) \u003d 4.1m 2 ° C / W.

Condition R 0 r \u003d 4.1> \u003d 3.6m 2 ° C / W is performed.

Checking compliance with sanitary and hygienic requirements

building thermal protection

1. Check the condition :

∆t = (t int- t ext)/ R 0r a int \u003d (20-(37)) / 4.1 8.7 \u003d 1.60 ºС

According to Table. 5SP 23-101-2004 ∆ t n = 4 °C, therefore, the condition ∆ t = 1,60< ∆t n = 4 ºС is fulfilled.

2. Check the condition :

] = 20 – =

20 - 1.60 = 18.40ºС

3. According to Appendix Sp 23-101–2004 for indoor air temperature t int = 20 ºС and relative humidity = 55% dew point temperature t d = 10.7ºС, therefore, the condition τsi = 18.40> t d= performed.

Conclusion. The enclosing structure satisfies regulatory requirements thermal protection of the building.

4.2 Thermotechnical calculation of attic roofing.

Initial data

Determine the thickness of the attic floor insulation, consisting of insulation δ = 200 mm, vapor barrier, prof. sheet

Attic floor:

Combined coverage:

Place of construction - Omsk

The length of the heating period z ht = 221 days.

Average design temperature of the heating period t ht = -8.4ºС.

The temperature of the cold five-day t ext = -37ºС.

The calculation was made for a five-story residential building:

indoor air temperature t int = + 20ºС;

air humidity: = 55%;

the humidity regime of the room is normal.

Operating conditions of enclosing structures - B.

Heat transfer coefficient of the inner surface of the fence A i nt \u003d 8.7 W / m 2 ° С.

Heat transfer coefficient of the outer surface of the fence a ext \u003d 12 W / m 2 ° C.

Name of material Y 0 , kg / m³ δ , m λ , mR, m 2 ° С / W

1. Determination of degree-days of the heating period according to the formula (2) SP 23-101-2004:

D d \u003d (t int - t ht) z th \u003d (20 -8.4) 221 \u003d 6276.4 ° C day

2. Rationing the value of the resistance to heat transfer of the attic floor according to the formula (1) SP 23-101-2004:

R reg \u003d a D d + b, where a and b are selected according to table 4 of SP 23-101-2004

R reg \u003d a D d + b \u003d 0.00045 6276.4+ 1.9 \u003d 4.72 m² ºС / W

3. Thermal engineering calculation is carried out from the condition that the total thermal resistance R 0 is equal to the normalized R reg , i.e.

4. From formula (8) SP 23-100-2004 we determine the thermal resistance of the building envelope R k (m² ºС / W)

R k \u003d R reg - (R si + R se)

Rreg = 4.72m² ºС / W

R si \u003d 1 / α int \u003d 1 / 8.7 \u003d 0.115 m² ºС / W

R se \u003d 1 / α ext \u003d 1/12 \u003d 0.083 m² ºС / W

R k \u003d 4.72– (0.115 + 0.083) \u003d 4.52 m² ºС / W

5. The thermal resistance of the building envelope (attic floor) can be represented as the sum of the thermal resistances of individual layers:

R k \u003d R cb + R pi + R tss + R ut → R ut \u003d R c + (R cb + R pi + R cs) \u003d R c - (d / λ) \u003d 4.52 - 0.29 \u003d 4 .23

6. Using the formula (6) SP 23-101-2004, we determine the thickness of the insulating layer:

d ut = R ut λ ut = 4.23 0.032= 0.14 m

7. We accept the thickness of the insulating layer 150mm.

8. We consider the total thermal resistance R 0:

R 0 \u003d 1 / 8.7 + 0.005 / 0.17 + 0.15 / 0.032 + 1 / 12 \u003d 0.115 + 4.69 + 0.083 \u003d 4.89m² ºС / W

R 0 ≥ R reg 4.89 ≥ 4.72 satisfies the requirement

Condition check

1. Check the fulfillment of the condition ∆t 0 ≤ ∆t n

The value of ∆t 0 is determined by the formula (4) SNiP 23-02-2003:

∆t 0 = n (t int - t ext) / R 0 a int 6

∆t 0 \u003d 1 (20 + 37) / 4.89 8.7 \u003d 1.34ºС

According to Table. (5) SP 23-101-2004 ∆t n = 3 ºС, therefore, the condition ∆t 0 ≤ ∆t n is fulfilled.

2. Check the fulfillment of the condition τ >t d

Value τ we calculate according to the formula (25) SP 23-101-2004

tsi = t int– [n(t int–text)]/(R o a int)

τ \u003d 20- 1 (20 + 26) / 4.89 8.7 \u003d 18.66 ºС

3. According to Appendix R SP 23-01-2004 for indoor air temperature t int = +20 ºС and relative humidity φ = 55% dew point temperature t d = 10.7 ºС, therefore, the condition τ >t d is executed.

Conclusion: attic floor meets regulatory requirements.

Example of thermotechnical calculation of enclosing structures

1. Initial data

Technical task. In connection with the unsatisfactory heat and humidity regime of the building, it is necessary to insulate its walls and mansard roof. To this end, perform calculations of thermal resistance, heat resistance, air and vapor permeability of the building envelope with an assessment of the possibility of moisture condensation in the thickness of the fences. Determine the required thickness of the heat-insulating layer, the need to use wind and vapor barriers, the order of the layers in the structure. Develop a design solution that meets the requirements of SNiP 23-02-2003 "Thermal protection of buildings" for building envelopes. Perform calculations in accordance with the set of rules for the design and construction of SP 23-101-2004 "Design of thermal protection of buildings".

General characteristics of the building. A two-story residential building with an attic is located in the village. Sviritsa Leningrad region. The total area of external enclosing structures - 585.4 m 2; total wall area 342.5 m 2; the total area of windows is 51.2 m 2; roof area - 386 m 2; basement height - 2.4 m.

The structural scheme of the building includes load-bearing walls, reinforced concrete floors from multi-hollow panels, 220 mm thick and a concrete foundation. The outer walls are made of brickwork and plastered inside and out with a mortar layer of about 2 cm.

The roof of the building has a truss structure with a steel seam roof, made along the crate with a step of 250 mm. Insulation 100 mm thick is made of mineral wool boards laid between the rafters

The building is provided with stationary electric-thermal storage heating. The basement has a technical purpose.

climatic parameters. According to SNiP 23-02-2003 and GOST 30494-96, we take the estimated average temperature of the indoor air equal to

t int= 20 °С.

According to SNiP 23-01-99 we accept:

1) the estimated temperature of the outside air in the cold season for the conditions of the village. Sviritsa Leningrad region

t ext= -29 °С;

2) the duration of the heating period

z ht= 228 days;

3) average temperature outdoor air during the heating period

t ht\u003d -2.9 ° С.

Heat transfer coefficients. The values of the heat transfer coefficient of the inner surface of the fences are accepted: for walls, floors and smooth ceilings α int\u003d 8.7 W / (m 2 ºС).

The values of the heat transfer coefficient of the outer surface of the fences are accepted: for walls and coatings α ext=23; attic floors α ext\u003d 12 W / (m 2 ºС);

Normalized resistance to heat transfer. Degree-days of the heating period G d are determined by formula (1)

G d\u003d 5221 ° С day.

Because the value G d differs from table values, standard value R req determined by formula (2).

According to SNiP 23-02-2003 for the obtained degree-day value, the normalized resistance to heat transfer R req, m 2 ° С / W, is:

For external walls 3.23;

Coverings and ceilings over driveways 4.81;

Fencing over unheated undergrounds and basements 4.25;

windows and balcony doors 0,54.

2. Thermotechnical calculation of external walls

2.1. Resistance of external walls to heat transfer

Exterior walls are made of hollow ceramic bricks and have a thickness of 510 mm. The walls are plastered from the inside with lime-cement mortar 20 mm thick, from the outside - with cement mortar of the same thickness.

The characteristics of these materials - density γ 0, dry thermal conductivity coefficient  0 and vapor permeability coefficient μ - are taken from Table. Clause 9 of the application. In this case, in the calculations we use the coefficients of thermal conductivity of materials  W for operating conditions B, (for wet operating conditions), which are obtained by formula (2.5). We have:

For lime-cement mortar

γ 0 \u003d 1700 kg / m 3,

 W\u003d 0.52 (1 + 0.168 4) \u003d 0.87 W / (m ° C),

μ=0.098 mg/(m h Pa);

For brickwork from hollow ceramic bricks on cement-sand mortar

γ 0 \u003d 1400 kg / m 3,

 W\u003d 0.41 (1 + 0.207 2) \u003d 0.58 W / (m ° C),

μ=0.16 mg/(m h Pa);

For cement mortar

γ 0 \u003d 1800 kg / m 3,

 W\u003d 0.58 (1 + 0.151 4) \u003d 0.93 W / (m ° C),

μ=0.09 mg/(m h Pa).

The heat transfer resistance of a wall without insulation is

R o \u003d 1 / 8.7 + 0.02 / 0.87 + 0.51 / 0.58 + 0.02 / 0.93 + 1/23 \u003d 1.08 m 2 ° C / W.

In the presence of window openings that form the slopes of the wall, the coefficient of thermal uniformity of brick walls, 510 mm thick, is taken r = 0,74.

Then the reduced resistance to heat transfer of the walls of the building, determined by formula (2.7), is equal to

R r o \u003d 0.74 1.08 \u003d 0.80 m 2 ° C / W.

The obtained value is much lower than the normative value of heat transfer resistance, therefore, it is necessary to install external thermal insulation and subsequent plastering with protective and decorative plaster compositions with fiberglass reinforcement.

In order for the thermal insulation to dry out, the plaster layer covering it must be vapor-permeable, i.e. porous with low density. We choose a porous cement-perlite mortar having the following characteristics:

γ 0 \u003d 400 kg / m 3,

 0 \u003d 0.09 W / (m ° C),

 W\u003d 0.09 (1 + 0.067 10) \u003d 0.15 W / (m ° C),

 \u003d 0.53 mg / (m h Pa).

The total resistance to heat transfer of the added layers of thermal insulation R t and plaster lining R w must be at least

R t+ R w \u003d 3.23 / 0.74-1.08 \u003d 3.28 m 2 ° C / W.

Preliminarily (with subsequent clarification), we accept the thickness of the plaster lining as 10 mm, then its resistance to heat transfer is equal to

R w \u003d 0.01 / 0.15 \u003d 0.067 m 2 ° C / W.

When used for thermal insulation of mineral wool boards manufactured by CJSC Mineralnaya Vata, Facade Butts brand  0 \u003d 145 kg / m 3,  0 \u003d 0.033,  W \u003d 0.045 W / (m ° C) the thickness of the heat-insulating layer will be

δ=0.045 (3.28-0.067)=0.145 m.

Rockwool boards are available in thicknesses from 40 to 160 mm in 10 mm increments. We accept a standard thickness of thermal insulation of 150 mm. Thus, the slabs will be laid in one layer.

Checking compliance with energy saving requirements. The calculation scheme of the wall is shown in fig. 1. The characteristics of the layers of the wall and the total resistance of the wall to heat transfer, excluding vapor barrier, are given in Table. 2.1.

Table 2.1

Characterization of the layers of the wall andtotal resistance of the wall to heat transfer

layer material	Density γ 0, kg / m 3	Thickness δ, m	Design coefficient of thermal conductivity λ W, W/(m K)	Estimated resistance to heat transfer R, m 2 ° С) / W
layer material	Density γ 0, kg / m 3	Thickness δ, m	Design coefficient of thermal conductivity λ W, W/(m K)	Estimated resistance to heat transfer R, m 2 ° С) / W
Internal plaster (lime-cement mortar)
Hollow ceramic brick masonry
External plaster ( cement mortar)
Mineral wool insulation FACADE BATTS
Protective and decorative plaster (cement-perlite mortar)

The heat transfer resistance of the walls of the building after insulation will be:

R o = 1/8.7+4.32+1/23=4.48 m 2 °C/W.

Taking into account the coefficient of thermal engineering uniformity of the outer walls ( r= 0.74) we get the reduced resistance to heat transfer

R o r\u003d 4.48 0.74 \u003d 3.32 m 2 ° C / W.

Received value R o r= 3.32 exceeds the standard R req= 3.23, since the actual thickness of the heat-insulating plates is greater than the calculated one. This situation meets the first requirement of SNiP 23-02-2003 for the thermal resistance of the wall - R o ≥ R req .

Verification of compliance with the requirements forsanitary and hygienic and comfortable conditions in the room. Estimated difference between the temperature of the indoor air and the temperature of the inner surface of the wall Δ t 0 is

Δ t 0 =n(t int – t ext)/(R o r ·α int)=1.0(20+29)/(3.32 8.7)=1.7 ºС.

According to SNiP 23-02-2003, for the outer walls of residential buildings, a temperature difference of not more than 4.0 ºС is permissible. Thus, the second condition (Δ t 0 ≤Δ t n) done.

P
check the third condition ( τ int >t grew), i.e. is it possible to condense moisture on the inner surface of the wall at the estimated outdoor temperature t ext\u003d -29 ° С. Inner surface temperature τ int enclosing structure (without heat-conducting inclusion) is determined by the formula

τ int = t int –Δ t 0 \u003d 20–1.7 \u003d 18.3 ° С.

The elasticity of water vapor in the room e int is equal to

In the climatic conditions of the northern geographical latitudes, for builders and architects, a correctly made thermal calculation of the building is extremely important. The obtained indicators will provide the necessary information for the design, including the materials used for construction, additional insulation, ceilings and even finishing.

In general, heat calculation affects several procedures:

consideration by designers when planning the location of rooms, bearing walls and fences;
creation of a project for a heating system and ventilation facilities;
selection of building materials;
analysis of the operating conditions of the building.

All this is connected by single values obtained as a result of settlement operations. In this article, we will tell you how to make a thermal calculation of the outer wall of a building, as well as give examples of the use of this technology.

Tasks of the procedure

A number of goals are relevant only for residential buildings or, on the contrary, industrial premises, but most of the problems to be solved are suitable for all buildings:

Preservation of comfortable climatic conditions inside the rooms. The term "comfort" includes both the heating system and the natural conditions for heating the surface of walls, roofs, and the use of all heat sources. The same concept includes the air conditioning system. Without proper ventilation, especially in production, the premises will be unsuitable for work.
Saving electricity and other resources for heating. The following values take place here:
- specific heat capacity of the materials and skins used;
- climate outside the building;
- heating power.

It is extremely uneconomical to install a heating system that simply will not be used to the right extent, but will be difficult to install and expensive to maintain. The same rule can be attributed to expensive building materials.

Thermotechnical calculation - what is it

The heat calculation allows you to set the optimal (two boundaries - minimum and maximum) thickness of the walls of enclosing and supporting structures, which will ensure long-term operation without freezing and overheating of floors and partitions. In other words, this procedure allows you to calculate the real or assumed, if it is carried out at the design stage, the thermal load of the building, which will be considered the norm.

The analysis is based on the following data:

the design of the room - the presence of partitions, heat-reflecting elements, ceiling height, etc .;
features of the climatic regime in a given area - maximum and minimum temperature limits, difference and rapidity of temperature changes;
the location of the structure on the cardinal points, that is, taking into account absorption solar heat, at what time of the day is the maximum susceptibility of heat from the sun;
mechanical effects and physical properties of the building object;
indicators of air humidity, the presence or absence of protection of walls from moisture penetration, the presence of sealants, including sealing impregnations;
the work of natural or artificial ventilation, the presence of a "greenhouse effect", vapor permeability and much more.

At the same time, the assessment of these indicators must comply with a number of standards - the level of resistance to heat transfer, air permeability, etc. Let us consider them in more detail.

Requirements for the heat engineering calculation of the premises and related documentation

State inspection bodies that manage the organization and regulation of construction, as well as checking the implementation of safety regulations, compiled SNiP No. 23-02-2003, which details the norms for carrying out measures for the thermal protection of buildings.

The document proposes engineering solutions that will provide the most economical consumption heat energy that is spent on heating premises (residential or industrial, municipal) during the heating period. These guidelines and requirements have been developed with regard to ventilation, air conversion, and the location of heat entry points.

SNiP is a bill at the federal level. Regional documentation is presented in the form of TSN - territorial building codes.

Not all buildings fall within the jurisdiction of these vaults. In particular, those buildings that are heated irregularly or are completely constructed without heating are not checked according to these requirements. Mandatory heat calculation is for the following buildings:

residential - private and apartment buildings;
public, municipal - offices, schools, hospitals, kindergartens, etc.;
industrial - factories, concerns, elevators;
agricultural - any heated buildings for agricultural purposes;
storage - barns, warehouses.

The text of the document contains the norms for all those components that are included in the thermal analysis.

Design requirements:

Thermal insulation. This is not only the preservation of heat in the cold season and the prevention of hypothermia, freezing, but also protection against overheating in the summer. Isolation, therefore, must be mutual - prevention of influences from outside and return of energy from within.
The permissible value of the temperature difference between the atmosphere inside the building and the thermal regime of the interior of the building envelope. This will lead to the accumulation of condensation on the walls, as well as to negative influence on the health of people in the room.
Heat resistance, that is, temperature stability, preventing sudden changes in the heated air.
Breathability. Balance is important here. On the one hand, it is impossible to allow the building to cool down due to active heat transfer, on the other hand, it is important to prevent the appearance of the "greenhouse effect". It happens when synthetic, "non-breathing" insulation is used.
Absence of dampness. High humidity is not only a reason for the appearance of mold, but also an indicator due to which serious losses of heat energy occur.

How to make a heat engineering calculation of the walls of a house - the main parameters

Before proceeding with the direct heat calculation, you need to collect detailed information about the building. The report will include responses to the following items:

The purpose of the building is residential, industrial or public premises, a specific purpose.
Geographic latitude of the area where the object is located or will be located.
Climatic features of the area.
The direction of the walls to the cardinal points.
Dimensions input structures And window frames- their height, width, permeability, type of windows - wooden, plastic, etc.
The power of heating equipment, the layout of pipes, batteries.
The average number of residents or visitors, workers, if these are industrial premises that are inside the walls at a time.
Building materials from which floors, ceilings and any other elements are made.
Presence or absence of supply hot water, the type of system that is responsible for this.
Features of ventilation, both natural (windows) and artificial - ventilation shafts, air conditioning.
The configuration of the entire building - the number of floors, the total and individual area of \u200b\u200bthe premises, the location of the rooms.

When these data are collected, the engineer can proceed to the calculation.

We offer you three methods that are most often used by specialists. You can also use the combined method, when the facts are taken from all three possibilities.

Variants of thermal calculation of enclosing structures

Here are three indicators that will be taken as the main one:

building area from the inside;
volume outside;
specialized coefficients of thermal conductivity of materials.

Heat calculation by area

Not the most economical, but the most frequent, especially in Russia, method. It involves primitive calculations based on the area indicator. This does not take into account the climate, band, minimum and maximum temperature values, humidity, etc.

Also, the main sources of heat loss are not taken into account, such as:

Ventilation system - 30-40%.
Roof slopes - 10-25%.
Windows and doors - 15-25%.
Walls - 20-30%.
Floor on the ground - 5-10%.

These inaccuracies are due to the neglect of the majority important elements lead to the fact that the heat calculation itself can have a strong error in both directions. Usually, engineers leave a "reserve", so you have to install such heating equipment that is not fully activated or threatens severe overheating. It is not uncommon for a heating and air conditioning system to be installed at the same time, since they cannot correctly calculate heat losses and heat gains.

Use "aggregated" indicators. Cons of this approach:

expensive heating equipment and materials;
uncomfortable indoor climate;
additional installation automated control over temperature regime;
possible freezing of the walls in winter.

Q=S*100W (150W)

Q is the amount of heat required for a comfortable climate in the entire building;
W S - heated area of the room, m.

The value of 100-150 watts is a specific indicator of the amount of thermal energy required to heat 1 m.

If you choose this method, then heed the following tips:

If the height of the walls (to the ceiling) is not more than three meters, and the number of windows and doors per surface is 1 or 2, then multiply the result by 100 watts. Usually all residential buildings, both private and multi-family, use this value.
If the design contains two window openings or a balcony, a loggia, then the figure increases to 120-130 watts.
For industrial and warehouse premises, a factor of 150 W is more often taken.
When choosing heaters (radiators), if they are located near the window, it is worth adding their projected power by 20-30%.

Thermal calculation of enclosing structures according to the volume of the building

Usually this method is used for those buildings where high ceilings are more than 3 meters. That is industrial facilities. The downside of this method is that the air conversion is not taken into account, that is, the fact that the top is always warmer than the bottom.

Q=V*41W (34W)

V is the external volume of the building in cubic meters;
41 W is the specific amount of heat required to heat one cubic meter of a building. If the construction is carried out using modern building materials, then the indicator is 34 watts.

Glass in windows:
- double package - 1;
- binding - 1.25.
Insulation materials:
- new modern developments - 0.85;
- standard brickwork in two layers - 1;
- small wall thickness - 1.30.
Air temperature in winter:
- -10 – 0,7;
- -15 – 0,9;
- -20 – 1,1;
- -25 – 1,3.
Percentage of windows compared to the total surface:
- 10% – 0,8;
- 20% – 0,9;
- 30% – 1;
- 40% – 1,1;
- 50% – 1,2.

All these errors can and should be taken into account, however, they are rarely used in real construction.

An example of a thermotechnical calculation of the external enclosing structures of a building by analyzing the used insulation

If you are building a residential building or a cottage on your own, then we strongly recommend that you think through everything to the smallest detail in order to ultimately save money and create an optimal climate inside, ensuring long-term operation of the facility.

To do this, you need to solve two problems:

make the correct heat calculation;
install a heating system.

Example data:

corner living room;
one window - 8.12 square meters;
region - Moscow region;
wall thickness - 200 mm;
area according to external parameters - 3000 * 3000.

It is necessary to find out how much power is needed to heat 1 square meter of the room. The result will be Qsp = 70 W. If the insulation (wall thickness) is less, then the values \u200b\u200bare also lower. Compare:

100 mm - Qsp \u003d 103 W.
150 mm - Qsp \u003d 81 W.

This indicator will be taken into account when laying heating.

Heating system design software

With the help of computer programs from the ZVSOFT company, you can calculate all the materials spent on heating, as well as make a detailed floor plan communications with the display of radiators, specific heat capacity, energy consumption, nodes.

The firm offers basic CAD for design work any complexity. In it, you can not only design a heating system, but also create detailed diagram for the construction of the whole house. This can be realized due to the large functionality, the number of tools, as well as work in two- and three-dimensional space.

You can install an add-on to the base software. This program is designed for the design of all engineering systems, including heating. With the help of easy line tracing and the plan layering function, you can design several communications on one drawing - water supply, electricity, etc.

Before building a house, make a thermal calculation. This will help you not to make a mistake with the choice of equipment and the purchase of building materials and insulation.

Thermal engineering calculation allows you to determine the minimum thickness of building envelopes so that there are no cases of overheating or freezing during the operation of the building.

Enclosing structural elements of heated public and residential buildings, with the exception of the requirements of stability and strength, durability and fire resistance, economy and architectural design, must primarily meet thermal engineering standards. Enclosing elements are selected depending on the design solution, climatological characteristics of the building area, physical properties, humidity and temperature conditions in the building, as well as in accordance with the requirements of resistance to heat transfer, air permeability and vapor permeability.

What is the meaning of calculation?

If, during the calculation of the cost of a future building, only strength characteristics are taken into account, then, naturally, the cost will be less. However, this is a visible savings: subsequently, much more money will be spent on heating the room.
Properly selected materials will create an optimal microclimate in the room.
When planning a heating system, a heat engineering calculation is also necessary. In order for the system to be cost-effective and efficient, it is necessary to have an understanding of the real possibilities of the building.

Thermal requirements

It is important that the external structures comply with the following thermal requirements:

They had sufficient heat-shielding properties. In other words, it should not be allowed summer time overheating of the premises, and in winter - excessive heat loss.
The air temperature difference between the internal elements of the fences and the premises should not be higher than the standard value. Otherwise, excessive cooling of the human body by heat radiation on these surfaces and moisture condensation of the internal air flow on the enclosing structures may occur.
In the event of a change in heat flow, temperature fluctuations inside the room should be minimal. This property is called heat resistance.
It is important that the air tightness of the fences does not cause strong cooling of the premises and does not worsen the heat-shielding properties of the structures.
Fences must have a normal humidity regime. Since waterlogging of fences increases heat loss, causes dampness in the room, and reduces the durability of structures.

In order for the structures to meet the above requirements, they perform a thermal calculation, and also calculate the heat resistance, vapor permeability, air permeability and moisture transfer according to the requirements of regulatory documentation.

Thermotechnical qualities

From the thermal characteristics of the external structural elements of buildings depends:

Moisture regime of structural elements.
The temperature of internal structures, which ensures that there is no condensation on them.
Constant humidity and temperature in the premises, both in the cold and in the warm season.
The amount of heat lost by a building winter period time.

So, based on all of the above, the heat engineering calculation of structures is considered an important stage in the process of designing buildings and structures, both civil and industrial. Designing begins with the choice of structures - their thickness and sequence of layers.

Tasks of thermotechnical calculation

So, the heat engineering calculation of enclosing structural elements is carried out in order to:

Compliance of structures with modern requirements for thermal protection of buildings and structures.
Ensuring a comfortable microclimate in the interior.
Ensuring optimal thermal protection of fences.

Basic parameters for calculation

To determine the heat consumption for heating, as well as to make a heat engineering calculation of the building, it is necessary to take into account many parameters that depend on the following characteristics:

Purpose and type of building.
Geographic location of the building.
The orientation of the walls to the cardinal points.
Dimensions of structures (volume, area, number of storeys).
Type and size of windows and doors.
Characteristics of the heating system.
The number of people in the building at the same time.
The material of the walls, floor and ceiling of the last floor.
The presence of a hot water system.
Type of ventilation systems.
Other design features buildings.

Thermal engineering calculation: program

To date, many programs have been developed that allow you to make this calculation. As a rule, the calculation is carried out on the basis of the methodology set out in the regulatory and technical documentation.

These programs allow you to calculate the following:

Thermal resistance.
Heat loss through structures (ceiling, floor, door and window openings, and walls).
The amount of heat required to heat the infiltrating air.
Selection of sectional (bimetallic, cast iron, aluminum) radiators.
Selection of panel steel radiators.

Thermotechnical calculation: calculation example for external walls

For the calculation, it is necessary to determine the following main parameters:

t in \u003d 20 ° C is the temperature of the air flow inside the building, which is taken to calculate the fences according to the minimum values of the most optimal temperature relevant building and structure. It is accepted in accordance with GOST 30494-96.

According to the requirements of GOST 30494-96, the humidity in the room should be 60%, as a result, a normal humidity regime will be provided in the room.
In accordance with Appendix B of SNiPa 23-02-2003, the humidity zone is dry, which means that the operating conditions of the fences are A.
t n \u003d -34 ° C is the temperature of the outdoor air flow in the winter period, which is taken according to SNiP based on the coldest five-day period, which has a security of 0.92.
Z ot.per = 220 days - this is the duration of the heating period, which is taken according to SNiP, while the average daily temperature environment≤ 8°C.
T from.per. = -5.9 °C is the ambient temperature (average) during the heating season, which is accepted according to SNiP, at a daily ambient temperature ≤ 8 °C.

Initial data

In this case, the thermotechnical calculation of the wall will be carried out in order to determine the optimal thickness of the panels and the heat-insulating material for them. Sandwich panels will be used as external walls (TU 5284-001-48263176-2003).

Comfortable conditions

Consider how the thermal engineering calculation of the outer wall is performed. First you need to calculate the required heat transfer resistance, focusing on comfortable and sanitary conditions:

R 0 tr \u003d (n × (t in - t n)) : (Δt n × α in), where

n = 1 is a factor that depends on the position of the external structural elements in relation to the outside air. It should be taken according to SNiP 23-02-2003 from Table 6.

Δt n \u003d 4.5 ° C is the normalized temperature difference between the internal surface of the structure and internal air. Accepted according to SNiP data from table 5.

α in \u003d 8.7 W / m 2 ° C is the heat transfer of internal enclosing structures. Data are taken from table 5, according to SNiP.

We substitute the data in the formula and get:

R 0 tr \u003d (1 × (20 - (-34)) : (4.5 × 8.7) \u003d 1.379 m 2 ° C / W.

Energy Saving Conditions

When performing a thermal engineering calculation of the wall, based on the conditions of energy saving, it is necessary to calculate the required heat transfer resistance of the structures. It is determined by GSOP (heating degree-day, °C) using the following formula:

GSOP = (t in - t from.per.) × Z from.per, where

t in is the temperature of the air flow inside the building, °C.

Z from.per. and t from.per. is the duration (days) and temperature (°C) of the period with an average daily air temperature ≤ 8 °C.

Thus:

GSOP = (20 - (-5.9)) × 220 = 5698.

Based on the conditions of energy saving, we determine R 0 tr by interpolation according to SNiP from table 4:

R 0 tr \u003d 2.4 + (3.0 - 2.4) × (5698 - 4000)) / (6000 - 4000)) \u003d 2.909 (m 2 ° C / W)

R 0 = 1/ α in + R 1 + 1/ α n, where

d is the thickness of the thermal insulation, m.

l = 0.042 W/m°C is the thermal conductivity of the mineral wool board.

α n \u003d 23 W / m 2 ° C is the heat transfer of external structural elements, taken according to SNiP.

R 0 \u003d 1 / 8.7 + d / 0.042 + 1/23 \u003d 0.158 + d / 0.042.

Insulation thickness

Thickness thermal insulation material is determined based on the fact that R 0 \u003d R 0 tr, while R 0 tr is taken under the conditions of energy saving, thus:

2.909 = 0.158 + d/0.042, whence d = 0.116 m.

We select the brand of sandwich panels from the catalog with optimum thickness thermal insulation material: DP 120, while the total thickness of the panel should be 120 mm. The heat engineering calculation of the building as a whole is carried out in a similar way.

The need to perform the calculation

Designed on the basis of a competently executed heat engineering calculation, building envelopes can reduce heating costs, the cost of which is regularly increasing. In addition, heat conservation is considered an important environmental task, because it is directly related to a decrease in fuel consumption, which leads to a decrease in the impact of negative factors on the environment.

In addition, it is worth remembering that improperly performed thermal insulation can lead to waterlogging of structures, which will result in the formation of mold on the surface of the walls. The formation of mold, in turn, will lead to spoilage interior decoration(peeling of wallpaper and paint, destruction of the plaster layer). In particularly advanced cases, radical intervention may be necessary.

Often construction companies tend to use in their activities modern technologies and materials. Only a specialist can understand the need to use one or another material, both separately and in combination with others. It is the heat engineering calculation that will help determine the most optimal solutions, which will ensure the durability of structural elements and minimal financial costs.