An example of the heat engineering calculation of enclosing structures
1. Source data
Technical task. In connection with the unsatisfactory heat and humidity regime of the building, it is necessary to insulate its walls and dersighted roof. To this end, perform calculations of thermal resistance, heat-resistant, air and vapor permeability of the building structures with an assessment of the possibility of moisture condensation in the thickness of the fences. To establish the necessary thickness of the heat insulating layer, the need to use wind and vaporizolation, the order of layers in the design. Develop a project solution that meets the requirements of SNiP 23-02-2003 " Heavy protection buildings "to enclosing structures. Calculations to fulfill in accordance with the arrangement of the Rules for Design and Construction of the SP 23-101-2004 "Design of heat protection of buildings".
General characteristic of the building. A two-story residential building with an attic is located in the village. Svits Leningrad region. The total area of \u200b\u200bexternal enclosing structures is 585.4 m 2; The total area of \u200b\u200bthe walls is 342.5 m 2; The total area of \u200b\u200bwindows is 51.2 m 2; Roof area - 386 m 2; The height of the basement is 2.4 m.
Constructive building scheme includes bearing walls, reinforced concrete floors from multi-retained panels, 220 mm thick and concrete foundation. The outer walls are made of brick masonry and plastered from the inside and outside the mortar with a layer of about 2 cm.
The covering of the building has a rafter design with steel folded roof, made on the crate in 250 mm increments. The insulation of 100 mm thick is made of mineral wool plates laid between rafyles
The building provides stationary electric heat accumulation heating. The basement has a technical purpose.
Climatic parameters. According to SNiP 23-02-2003 and GOST 30494-96, the estimated average temperature of the internal air is taken equal to
t. int. \u003d 20 ° C.
According to SNiP 23-01-99, we accept:
1) the calculated outdoor air temperature in cold period years for the conditions of the village. Sweets of the Leningrad region
t. eXT. \u003d -29 ° C;
2) the duration of the heating period
z. hT \u003d 228 days;
3) the average outdoor temperature for the heating period
t. hT \u003d -2.9 ° C.
The heat transfer coefficients.The values \u200b\u200bof the coefficient of heat transfer coefficient of the fences are accepted: for walls, floors and smooth ceilings α int. \u003d 8.7 W / (m 2 · ºС).
The values \u200b\u200bof the heat transfer coefficient of the external fences are accepted: for walls and coatings α eXT. \u003d 23; Blacksmiths of attic α. eXT. \u003d 12 W / (m 2 · ºС);
The normalized heat transfer resistance. Degree-day of the heating period G. d. are determined by formula (1)
G. d. \u003d 5221 ° С · day.
Since value G. d. differs from table values regulatory value R. rEQ. Determine by formula (2).
According to SNiP 23-02-2003 for the obtained value of the degree, the normalized heat transfer resistance R. rEQ. , m 2 · ° C / W is:
For external walls 3.23;
Coatings and overlaps over drives 4.81;
Fences over unheated underground and basements of 4.25;
Windows I. balcony doors 0,54.
2. Heat engineering calculation of exterior walls
2.1. Resistance to the outer walls heat transfer
Exterior walls made of hollow ceramic bricks and have a thickness of 510 mm. Walls are plastered from within a lime-cement solution with a thickness of 20 mm, outside the cement solution of the same thickness.
The characteristics of these materials are the density Γ 0, the thermal conductivity coefficient in the dry state 0 and the parry permeability coefficient μ - we accept the table. P.9 applications. At the same time, in the calculations we use the coefficients of thermal conductivity of materials W. For operating conditions b, (for wet operating conditions), which we obtain according to 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 \u003d 0.098 mg / (m · h · par);
For brick masonry From the hollow ceramic brick on cement-sandy solution
γ 0 \u003d 1400 kg / m 3,
W. \u003d 0.41 (1 + 0.207 · 2) \u003d 0.58 W / (m · С),
μ \u003d 0.16 mg / (m · h · par);
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 \u003d 0.09 mg / (m · h · par).
The heat transfer resistance of the wall without insulation is equal
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 forming the slope, the coefficient of thermal uniformity of brick walls, 510 mm thick, we accept r. = 0,74.
Then the resistance of the heat transfer of the walls of the building, determined by formula (2.7), is equal
R. r. O \u003d 0.74 · 1.08 \u003d 0.80 m 2 · ° C / W.
The resulting value is much lower than the regulatory value of heat transfer resistance, so the device is necessary outdoor thermal insulation And the subsequent plastering by the protective and decorative compositions of the plastering solution with the reinforcement of the glass wall.
For the possibility of drying the heat insulation, its plaster layer must be vapor permeable, i.e. porous with low density. Select the picked cement-perlite solution, which has 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 · par).
Total heat transfer resistance added heat insulation layers R. t and plastering R. W should be at least
R. T +. R. Ш \u003d 3,23 / 0.74-1.08 \u003d 3.28 m 2 · ° C / W.
Previously (followed by clarification) We accept the thickness of the plastering of 10 mm, then the resistance of its heat transfer is equal to
R. Ш \u003d 0.01 / 0.15 \u003d 0.067 m 2 · ° C / W.
When used for thermal insulation of mineral wool plates of production CJSC "Mineral Wat" of the brand of the facade of Batts 0 \u003d 145 kg / m 3, 0 \u003d 0.033, W. \u003d 0.045 W / (m · ° С) The thickness of the heat insulating layer will be
δ \u003d 0.045 · (3.28-0.067) \u003d 0.145 m.
Rockwool plates are produced with a thickness of 40 to 160 mm in 10 mm increments. We accept the standard thickness of thermal insulation 150 mm. Thus, laying the plates will be made in one layer.
Checking the fulfillment of requirements for energy saving.The calculated wall circuit is presented in Fig. 1. Characteristics of the layers of the wall and the overall resistance of the heat transfer wall without taking into account vaporizolation are shown in Table. 2.1.
Table 2.1
Characteristics of the layers of the wall and General heat transfer wall resistance
|
Layer material |
Density γ 0, kg / m 3 |
Thickness δ, m |
The calculated coefficient of thermal conductivity λ W. , W / (M K) |
The calculated resistance of heat transfer R., m 2 · ° С) / W |
|
|
Internal plaster (lime-cement solution) |
|||||
|
Masonry from hollow ceramic brick |
|||||
|
Exterior plaster ( cement mortar) |
|||||
|
Mineral wool insulation facade batts |
|||||
|
Stucco Protective-Decorative (cement-perlite solution) |
|||||
The resistance of the heat transfer walls of the building after insulation will be:
R. O. = 1 / 8.7 + 4.32 + 1/2 \u003d 4.48 m 2 · ° C / W.
Taking into account the coefficient of heat engineering homogeneity of the outer walls ( r. \u003d 0.74) We obtain the reduced heat transfer resistance
R. O. r. \u003d 4.48 · 0.74 \u003d 3.32 m 2 · ° C / W.
Received R. O. r. \u003d 3.32 exceeds the normative R. rEQ. \u003d 3.23, since the actual thickness of thermal insulation plates is greater than the calculated one. This provision meets the first requirement of SNiP 23-02-2003 to the thermal resistance of the wall - R. O ≥ R. rEQ. .
Checking the fulfillment of requirements forsanitary and hygienic and comfortable conditions indoors.Estimated differential between the temperature of the inner 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.) \u003d 1.0 (20 + 29) / (3.32 · 8.7) \u003d 1.7 ºС.
According to SNiP 23-02-2003 for the outer walls of residential buildings, we will admit a temperature drop of not more than 4.0 ºС. Thus, the second condition (δ t. 0 ≤Δ t. n.) done.
P
rubber third condition ( τ
int. >t. Ros), i.e. Is the condensation of moisture on the inner surface of the wall at the calculated temperature of the outer air t. eXT. \u003d -29 ° C. Temperature of the inner surface τ
int. Fighting design (without heat-conducting inclusion) determine the formula
τ int. = t. int. –Δ t. 0 \u003d 20-1.7 \u003d 18.3 ° C.
Elasticity of water vapor indoors e. int. equal
The heat engineering calculation allows you to determine the minimum thickness of the enclosing structures so that there are no cases of overheating or freezing during the operation of the structure.
Fencing structural elements of heated public and residential buildings, with the exception of the requirements of stability and strength, durability and fire resistance, efficiency and architectural design, must be responsible primarily to the heat engineering standards. Choose enclosing elements depending on the constructive solution, the climatological characteristics of the building area, physical properties, wet-temperature regime in the building, as well as in accordance with the requirements of heat transfer resistance, air absorption and vapor permeation.
It is important that external structures correspond to the following heat engineering requirements:
In order for the designs to comply with the above requirements, the heat engineering calculation is performed, and also calculate heat resistance, vapor permeability, breathability and intensifying the requirements of regulatory documentation.
From heat engineering outdoor constructive elements Buildings depends:
So, based on the above, the heat engineering calculation of the structures is considered an important stage in the process of designing buildings and structures, both civil and industrial. Design begins with the choice of structures - their thickness and sequences of the layers.
So, the heat engineering calculation of the enclosing structural elements is carried out in order to:
To determine the heat consumption for heating, as well as produce a heat engineering of the building, it is necessary to take into account the set of parameters depending on the following characteristics:
To date, many programs have been developed that allow this calculation. As a rule, the calculation is carried out on the basis of the technique set out in the regulatory and technical documentation.
Programs data allow us to calculate the following:
To calculate, it is necessary to define the following main parameters:
In this case, the heat engineering calculation of the wall will be carried out in order to determine the optimal thickness of the panels and the thermal insulation material for them. Sandwich panels will be used as outer walls (TU 5284-001-48263176-2003).
Consider how the heat engineering is performed outdoor Wall. To begin, it is necessary to calculate the required heat transfer resistance, focusing on comfortable and sanitary and hygienic conditions:
R 0 TR \u003d (N × (T B - T H)): (ΔT H × α c), where
n \u003d 1 is a coefficient that depends on the position of the outer structural elements relative to the outer air. It should be taken according to the SNIVA data 23-02-2003 from table 6.
ΔT H \u003d 4.5 ° C is a normalized temperature difference in the inner surface of the design and internal air. Accepted according to SNIPA data from Table 5.
α B \u003d 8.7 W / m 2 ° C is heat transfer of internal enclosing structures. Data is taken from Table 5, SNUP.
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.
Performing the heat engineering of the wall, based on the power saving conditions, it is necessary to calculate the required resistance of heat transfer structures. It is determined by HSOP (degree of heating period, ° C) by the following formula:
HSOP \u003d (T B - T. Per.) × Z. Perfect, where
t B is the temperature of the air flow inside the building, ° C.
Z OT.PER. and t from Per. - This duration (day) and temperature (° C) of a period having an average daily air temperature ≤ 8 ° C.
In this way:
HSOP \u003d (20 - (-5.9)) × 220 \u003d 5698.
Based on the conditions of energy saving, we determine the R 0 TP method of interpolation on low-rise 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 \u003d 1 / α B + R 1 + 1 / α n, where
d is the thickness of thermal insulation, m.
l \u003d 0.042 W / m ° C is the thermal conductivity of the mineral wool plate.
α H \u003d 23 W / m 2 ° C is the heat transfer of external structural elements adopted by SNU.
R 0 \u003d 1 / 8.7 + d / 0.042 + 1/2 23 \u003d 0.158 + d / 0.042.
Thickness heat insulating material It is determined based on the fact that R 0 \u003d R 0 TR, while R 0 TP is taken under the conditions of energy saving, so:
2.909 \u003d 0.158 + d / 0.042, from where d \u003d 0.116 m.
We select the brand of sandwich panels according to the catalog with the optimal thickness of the insulating material: DP 120, while the total thickness of the panel should be 120 mm. Similarly, the heat engineering calculation of the building as a whole is produced.
Designed on the basis heat engineering calculation, performed competently, enclosing structures make it possible to reduce the cost of heating, the cost of which increases regularly. In addition, heat saving 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 environmental factors.
In addition, it is worth remembering that incorrectly made thermal insulation is able to reduce the designs, which will result in the formation of mold on the surface of the walls. The formation of mold, in turn, will lead to damage interior decoration (peeling wallpaper and paint, destruction of the plastering layer). In particularly launched cases, radical intervention may be needed.
Often construction companies in their activities seek to use modern technologies and materials. Only a specialist to understand the need to use a particular material, both separately and in conjunction with others. It is the heat engineering that will help determine the most optimal solutions that will ensure the durability of the structural elements and the minimum financial costs.
Initial data
Construction site - Omsk
z. HT \u003d 221 days
t. HT \u003d -8.4ºС.
t. Ext \u003d -37ºС.
t. int \u003d + 20ºС;
air humidity: \u003d 55%;
Conditions of operation of enclosing structures - B. The heat transfer coefficient of the inner surface of the fence but I NT \u003d 8.7 W / m 2 ° C.
a. Ext \u003d 23 W / m 2 · ° C.
The necessary data on the structural layers of the wall for the heat engineering calculation is reduced to the table.
1. Determination of the degree of the heating period by formula (2) of the SP 23-101-2004:
D 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 by the formula (1) of the 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. The reduced heat transfer resistance R. 0 R external brick walls with an effective insulation of residential buildings is calculated by the formula
R 0 R \u003d R 0 SELL R,
where R 0 SEL is the resistance of the heat transfer of brick walls, conditionally determined by formulas (9) and (11) without taking into account the heat-conducting inclusions, m 2 · ° C / W;
R 0 R is the reduced resistance to heat transfer taking into account the coefficient of heat engineering homogeneity r.which for the walls is 0.74.
Calculation is carried out from the condition of equality
hence,
R 0 SEL \u003d 3.6 / 0.74 \u003d 4.86м 2 · ° C / W
R 0 SEL \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. The thermal resistance of the outer brick wall of the layered structure can be represented as the sum of the thermal resistances of individual layers, i.e.
R K \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 · ° C / W.
6. We find the thickness of the insulation:
| R. |
We accept the thickness of the insulation 100 mm.
The final wall thickness will be equal to (510 + 100) \u003d 610 mm.
We produce checks taking into account the accepted solar thickness:
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 4.1 m 2 · ° C / W.
Condition R. 0 r \u003d 4,1\u003e \u003d 3.6 m 2 · ° C / W is performed.
Checking the performance of sanitary and hygienic requirements
thermal protection of the building
1. Check the condition 
:
∆t. = (t. int - t. EXT) / R. 0 R. a. int \u003d (20- (37)) / 4.1 · 8.7 \u003d 1.60 ºС
According to the table. 5SP 23-101-2004 Δ. t. N \u003d 4 ° C, therefore, condition Δ t. = 1,60< ∆t. N \u003d 4 ºС is performed.
2. Check the condition 
:
] = 20 – =
20 - 1,60 \u003d 18.40ºС
3. According to Appendix SP 23-101-2004 for internal air temperature t. INT \u003d 20 ºС and relative humidity \u003d 55% dew point temperature t. D \u003d 10.7ºС, therefore, the condition τsi \u003d 18.40\u003e t. d \u003d 
performed.
Output. Fencing design satisfies regulatory requirements Thermal protection of the building.
4.2 Heat engineering calculation of horseproofing.
Initial data
Determine the thickness of the insulation of the attic overlap consisting of the insulation δ \u003d 200 mm, vaporizolation, prof. Sheet
Attic overlap:
Combined coating:
Construction site - Omsk
The duration of the heating period z. HT \u003d 221 days.
The average estimated temperature of the heating period t. HT \u003d -8.4ºС.
Temperature of cold five days t. Ext \u003d -37ºС.
Calculation is manufactured for a five-story residential building:
the temperature of the inner air t. int \u003d + 20ºС;
air humidity: \u003d 55%;
the humidity mode of the room is normal.
Conditions of operation of enclosing structures - B.
The heat transfer coefficient of the inner surface of the fence but I NT \u003d 8.7 W / m 2 ° C.
The 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 · ° C / W
1. Determination of the degree of the heating period by formula (2) of the SP 23-101-2004:
D d d \u003d (t int - t ht) z th \u003d (20 -8.4) · 221 \u003d 6276,4º dumplings
2. rationing The value of the impedance of the ingredial overlap by the formula (1) of the SP 23-101-2004:
R reg \u003d a · d d + b, where a and b - select on table 4 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. The heat engineering is carried out from the condition of equality of the general thermal resistance R 0 normalized R REG, i.e.
4. From formula (8) SP 23-100-2004 Determine the thermal resistance of the enclosing structure R k (m² · ºС / W)
R k \u003d r REG - (R Si + R SE)
R reg \u003d 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 4.52m² · ºС / W
5. The thermal resistance of the enclosing structure (attic overlap) can be represented as the sum of the thermal resistances of individual layers:
R K \u003d R zhb + r Pi + R CS + R UT → R ut \u003d R K + (R zhb + r Pi + R CA) \u003d R to - (d / λ) \u003d 4.52 - 0.29 \u003d 4 23.
6. We use the formula (6) of the SP 23-101-2004, we define the thickness of the insulation layer:
d ut \u003d R ut · λ ut \u003d 4.23 · 0.032 \u003d 0.14 m
7. Take the thickness of the insulation 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
Verification check
1. Check the performance of the condition Δt 0 ≤ Δt n
The value of Δt 0 is determined by the formula (4) SNiP 23-02-2003:
ΔT 0 \u003d N · (T int - T EXT) / R 0 · A int where, N is a coefficient, taking into account the dependence of the position of the outer surface to the outer air in the table. 6.
ΔT 0 \u003d 1 (20 + 37) / 4.89 · 8.7 \u003d 1.34ºС
According to the table. (5) SP 23-101-2004Δt n \u003d 3 ºС, therefore, the condition Δt 0 ≤ Δt n is performed.
2. Check the execution of the condition τ \u003e T D.
Value τ. calculate by formula (25) SP 23-101-2004
t SI = t int– [n.(t int–t Ext)]/(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 the temperature of the inner air T int \u003d +20 ºС and the relative humidity φ \u003d 55% the temperature of the dew point T d d \u003d 10.7 ºС, therefore, the condition τ \u003e T D is performed.
Output: attic overlap Satisfies regulatory requirements.
It is required to determine the soap-thickness thickness in a three-layer brick outer wall in a residential building located in Omsk. Wall design: inner layer - brickwork from ordinary clay brick 250 mm thick and a density of 1800 kg / m 3, the outer layer is a brickwork made 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 of expanded polystyrene density with a density of 40 kg / m 3; The outer and inner layers are combined between fiberglass flexible bonds with a diameter of 8 mm, located in 0.6 m increments.
1. Source data
Appointment of the building - residential building
Construction area - Omsk
Calculated indoor air temperature t int \u003d plus 20 0 s
Calculated outdoor air temperature t Ext \u003d minus 37 0 s
Calculated internal air humidity - 55%
2. Determination of the normalized heat transfer resistance
Depends on Table 4, depending on the degree-day of the heating period. Degree-day of the heating period, D d, ° С × day Determine by formula 1, based on the average temperature of the outer air and the duration of the heating period.
Snip 23-01-99 * Determine that in Omsk average temperature Outdoor air of the heating period is: t HT \u003d -8.4 0 C, duration of the heating period z ht \u003d 221 days. The magnitude of the degree-day of the heating period is:
D D. = (t int - t HT) z ht \u003d (20 + 8.4) × 221 \u003d 6276 0 s day.
According to the table. 4. Normated heat transfer resistance R REG exterior walls for residential buildings corresponding to value D d \u003d 6276 0 s dayequally R reg \u003d a d d + b \u003d 0.00035 × 6276 + 1,4 \u003d 3.60 m 2 0 C / W.
3. Choosing a constructive solution of the outer wall
The design solution of the outer wall is proposed in the task and is a three-layer fencing with an inner layer of brick masonry with a thickness of 250 mm, the outer layer of brickwork with a thickness of 120 mm, between the outer and the inner layer is the insulation of expanded polystyrene. The outer and inner layer are connected to the flexible bonds of fiberglass with a diameter of 8 mm, located in 0.6 m increments.
4. Determination of the thickness of the insulation
The thickness of the insulation is determined by the formula 7:
d ut \u003d (r reg ./r - 1 / a int - D KK / L KK - 1 / a ext) × l
where R REG. - the normalized heat transfer resistance, m 2 0 C / W; R. - the coefficient of thermal uniformity; a int - the heat transfer coefficient of the inner surface, W / (m 2 × ° C); A Ext - the heat transfer coefficient of the outer surface, W / (m 2 × ° C); D QK - brick masonry thickness, m.; l QK - the estimated thermal conductivity coefficient of brick masonry, W / (m × ° С); l Ut. - the estimated thermal conductivity coefficient of the insulation, W / (m × ° С).
The normalized heat transfer resistance is determined: R reg \u003d 3.60 m 2 0 C / W.
The coefficient of thermal homogeneity for the brick three-layer wall with fiberglass flexible ties is about r \u003d 0.995and in the calculations may not be taken into account (for information - if steel flexible bonds applied, the thermal homogeneity coefficient can reach 0.6-0.7).
The heat transfer coefficient of the inner surface is determined by table. 7. a int \u003d 8.7 w / (m 2 × ° C).
The heat transfer coefficient of the outer surface is accepted on Table 8 a E Xt \u003d 23 W / (m 2 × ° C).
The total thickness of the brick masonry is 370 mm or 0.37 m.
The calculated thermal conductivity coefficients of the materials used are determined depending on the operating conditions (A or B). Operating conditions are determined in the following sequence:
Table. 1 Determine the humidity mode of the premises: since the calculated temperature of the inner air is +20 0 s, the calculated humidity is 55%, the humidity mode of the premises is normal;
By Appendix B (Map of the Russian Federation), we define that Omsk is located in the dry zone;
Table. 2, depending on the moisture content zone and humidity regime, determine that the conditions of operation of the enclosing structures - BUT.
By arrival D Determine the coefficients of thermal conductivity for the operating conditions A: for expanded polystyrene GOST 15588-86 with a density of 40 kg / m 3 l ut \u003d 0.041 W / (M × ° C); For brick masonry from clay ordinary brick on cement-sandy solution with a density of 1800 kg / m 3 l QC \u003d 0.7 W / (M × ° C).
We substitute all certain values \u200b\u200bin formula 7 and calculate the minimum thickness of the insulation of expanded polystyrene:
d ut \u003d (3.60 - 1 / 8.7 - 0.37 / 0.7 - 1/23) × 0.041 \u003d 0.1194 m
Rounded the resulting value to the large side up to 0.01 m: d ut \u003d 0.12 m.We carry out the 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. Restriction of temperature and condensation of moisture on the inner surface of the enclosing construction
Δt O., ° C, between the temperature of the inner air and the temperature of the inner surface of the enclusive design should not exceed the normalized values Δt N., ° С installed in Table 5 and defined as follows
Δt o \u003d n (t int – t Ext)/( R 0 a int) \u003d 1 (20 + 37) / (3.61 x 8.7) \u003d 1.8 0 s i.e. less than Δt n, \u003d 4.0 0 s, defined in table 5.
Conclusion: T.oil insulation of polystyrene foam in three-layer brick wall It is 120 mm. At the same time, the resistance of the outer wall heat transfer R 0 \u003d 3.61 m 2 0 C / Wthat more normalized heat transfer resistance R REG. \u003d 3.60 m 2 0 C / Won the 0.01m 2 0 C / W.Settlement temperature difference Δt O., ° С, between the temperature of the inner air and the temperature of the inner surface of the enclosing structure does not exceed the regulatory value Δt n,.
An example of a heat engineering calculation of translucent enclosing structures
Translucent enclosing structures (windows) are selected according to the following procedure.
Normated heat transfer resistance R REG Defined on table 4 Snip 23-02-2003 (column 6) depending on the degree of day of the heating period D D.. At the same time the type of building and D D. Adopted as in the previous example of the heat engineering calculation of light-proximal enclosing structures. In our case D D. = 6276 0 s day, Then for the window of a residential building R reg \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 by the meaning of the resistance of heat transfer R O R.Received as a result of certification tests or an application of the rules of the rules. If the reduced heat transfer resistance of the selected translucent design R O R., more or equal R REGThis design satisfies the requirements of the norms.
Output:for a residential building in Omsk We accept windows in PVC bindings with double-chamber glass windows made of glass with a solid selective coating and filling in argon of the interconnect space R o r \u003d 0.65 m 2 0 s / wmore R reg \u003d 0.61 m 2 0 C / W.
LITERATURE
If you are going to build
A small brick cottage, then you will certainly have questions: "What
Thicks should be a wall? "," Does the insulation need? "," From which side to put
insulation? " etc. etc.
In this article we will try in
It is sorting out and answer all your questions.
Heat engineering
Fencing design is needed, first of all, in order to find out what
Thicks should be your outer wall.
First, you need to decide how much
Floors will be in your building and, depending on this, calculates
Fencing structures on the bearing capacity (not in this article).
On this calculation we define
The number of bricks in the masonry of your building.
For example, it turned out 2 clay
bricks without voids, brick length 250 mm,
the thickness of the solution is 10 mm, the total is 510 mm (brick density 0.67
In the future, we will be useful). Outdoor surface you decided to cover
facing tiles, thickness 1 cm (when buying it is necessary to know it
Density), and the inner surface of ordinary plaster, thickness of layer 1.5
See also Do not forget to learn its density. In the amount of 535mm.
In order for the building is not
collapsed this is of course enough, but unfortunately in most cities
Winter is cold and therefore such walls will freeze. And not
The walls were frozen, needed a layer of insulation.
The thickness of the insulation layer is calculated
in the following way:
1. On the Internet you need to download SNiP
II 3-79 * -
"Construction heat engineering" and SNiP 23-01-99 - "Construction climatology".
2. Open SNIP building
Climatology and find your city in Table 1 *, and we look at the intersection
The column "The air temperature is the coldest five days, ° C, secure
0.98 "and rows with your city. For the city of Penza, for example t n \u003d -32 about S.
3. Calculated internal air temperature
Take
t B \u003d 20 o C.
The heat transfer coefficient for inner wallsa. B \u003d 8.7W / m 2 · ˚С
Heat transfer coefficient for exterior walls in winter conditionsa. H \u003d 23W / m 2 · ˚С
Regulatory temperature difference between the temperature of the internal
air and temperature of the inner surface of the enclosing structuresΔt N \u003d 4 O S.
4. Next
Determine the required heat transfer resistance by formula # G0 (1a) from construction heat engineering
HSOP \u003d (T B - T OT.PER.) Z OTP , HSOP \u003d (20 + 4.5) · 207 \u003d 507,15 (for the city
Penza).
By formula (1), we calculate:
(where sigma is directly thick
Material, and lambda density. Itook as insulation
Polyurethanistanovapanel with density 0.025)
We accept the thickness of the insulation of 0.054 m.
From here the thickness of the wall will be:
d. = d. 1 + d. 2 + d. 3 + d. 4 =
0,01+0,51+0,054+0,015=0,589
m.
The repair season came up. Head broke: how to make good repair For less money. There is no thoughts about the loan. Support only on available ...
Instead of postponing the main repair from year to year, you can get ready for him to survive it in moderation ...
For the beginning you need to remove everything that remains from the old company that worked there. We break the artificial partition. After that, I'm moving away everything ...
