Air volume reduction factor in a building. Calculation of solar radiation in winter. Thermal engineering calculations of enclosing structures

(determining the thickness of the attic insulation layer

floors and coverings)
A. Initial data

Humidity zone is normal.

z ht = 229 days.

Average design temperature of the heating period t ht = –5.9 ºС.

Cold five-day temperature t ext = –35 °С.

t int = + 21 °С.

Relative humidity air: = 55%.

Estimated air temperature in the attic t int g = +15 С.

Heat transfer coefficient inner surface attic floor
= 8.7 W/m 2 ·С.

Heat transfer coefficient of the outer surface of the attic floor
= 12 W/m 2 °C.

Heat transfer coefficient of the inner surface of the coating of a warm attic
= 9.9 W/m 2 °C.

Heat transfer coefficient of the outer surface of the covering of a warm attic
= 23 W/m 2 °C.
Building type – 9-storey residential building. Kitchens in apartments are equipped gas stoves. Height attic space– 2.0 m. Coverage area (roof) A g. c = 367.0 m 2, warm attic floors A g. f = 367.0 m 2, external walls of the attic A g. w = 108.2 m2.

The warm attic contains the upper distribution of pipes for heating and water supply systems. Design temperatures of the heating system – 95 °C, hot water supply– 60 °C.

The diameter of the heating pipes is 50 mm with a length of 55 m, hot water supply pipes are 25 mm with a length of 30 m.
Attic floor:

Rice. 6 Calculation scheme

The attic floor consists of the structural layers shown in the table.

№	Name of material (structures)	, kg/m 3	δ, m	,W/(m °C)	R, m 2 °C/W
1	Rigid mineral wool slabs with bitumen binders (GOST 4640)	200	X	0,08	X
2	Vapor barrier – Rubitex 1 layer (GOST 30547)	600	0,005	0,17	0,0294
3	Reinforced concrete hollow core slabs PC (GOST 9561 - 91)		0,22		0,142

Combined coverage:

Rice. 7 Calculation scheme

The combined covering above the warm attic consists of the structural layers shown in the table.

B. Calculation procedure
Determination of the degree-day of the heating period using formula (2) SNiP 23-02–2003:
D d = ( t int – t ht) z ht = (21 + 5.9) 229 = 6160.1.
The normalized value of the heat transfer resistance of the coating of a residential building according to formula (1) SNiP 23-02–2003:

R req = a· D d+ b=0.0005·6160.1 + 2.2 = 5.28 m 2 ·С/W;
Using formula (29) SP 23-101–2004, we determine the required heat transfer resistance of the floor of a warm attic
, m 2 °C /W:

,
Where
– standardized resistance to heat transfer of the coating;

n– coefficient determined by formula (30) SP 230101–2004,
(21 – 15)/(21 + 35) = 0,107.
Based on the found values
And n define
:
= 5.28·0.107 = 0.56 m2·С/W.

Required coating resistance over a warm attic R 0 g. c is set using formula (32) SP 23-101–2004:
R 0 g.c = ( – t ext)/(0.28 G ven With(t ven – ) + ( t int – )/ R 0 g.f +
+ (
)/A g.f – ( – t ext) A g.w/ R 0 g.w ,
Where G ven – reduced (per 1 m2 of attic) air flow in the ventilation system, determined from the table. 6 SP 23-101–2004 and equal to 19.5 kg/(m 2 h);

c – specific heat air, equal to 1 kJ/(kg °C);

t ven – temperature of the air leaving ventilation ducts, °С, taken equal to t int + 1.5;

q pi – linear density heat flow through the thermal insulation surface per 1 m of pipeline length, taken to be 25 for heating pipes, and 12 W/m for hot water supply pipes (Table 12 SP 23-101–2004).

The given heat inputs from pipelines of heating and hot water supply systems are:
()/A g.f = (25·55 + 12·30)/367 = 4.71 W/m2;
a g. w – reduced area of ​​the outer walls of the attic m 2 / m 2, determined by formula (33) SP 23-101–2004,

= 108,2/367 = 0,295;

– normalized resistance to heat transfer of the external walls of a warm attic, determined through the degree-day of the heating period at the internal air temperature in the attic = +15 ºС.

– t ht)· z ht = (15 + 5.9)229 = 4786.1 °C day,
m 2 °C/W
We substitute the found values ​​into the formula and determine the required heat transfer resistance of the coating above the warm attic:
(15 + 35)/(0.28 19.2(22.5 – 15) + (21 – 15)/0.56 + 4.71 –
– (15 + 35) 0.295/3.08 = 50/50.94 = 0.98 m 2 °C/W

We determine the thickness of the insulation in the attic floor when R 0 g. f = 0.56 m 2 °C/W:

= (R 0 g. f – 1/– R reinforced concrete – R rub – 1/) ut =
= (0.56 – 1/8.7 – 0.142 –0.029 – 1/12)0.08 = 0.0153 m,
we take the insulation thickness = 40 mm, since minimum thickness mineral wool slabs 40 mm (GOST 10140), then the actual heat transfer resistance will be

R 0 g. f fact. = 1/8.7 + 0.04/0.08 + 0.029 + 0.142 + 1/12 = 0.869 m 2 °C/W.
We determine the amount of insulation in the coating when R 0 g. c = = 0.98 m 2 °C/W:
= (R 0 g. c – 1/ – R reinforced concrete – R rub - R c.p.r – R t – 1/) ut =
= (0.98 – 1/9.9 – 0.017 – 0.029 – 0.022 – 0.035 – 1/23) 0.13 = 0.0953 m,
we accept the thickness of the insulation ( aerated concrete slab) 100 mm, then the actual value of heat transfer resistance attic covering will be almost equal to the calculated one.
B. Checking compliance with sanitary and hygienic requirements

thermal protection of the building
I. Checking the fulfillment of the condition
for the attic floor:

= (21 – 15)/(0.869·8.7) = 0.79 °C,
According to table. 5 SNiP 23-02–2003 ∆ t n = 3 °С, therefore, the condition ∆ t g = 0.79 °C t n =3 °C is satisfied.
We check the external enclosing structures of the attic to ensure that condensation does not form on their internal surfaces, i.e. to fulfill the condition
:

– for covering above a warm attic, taking
W/m 2 °С,
15 – [(15 + 35)/(0.98 9.9] =
= 15 – 4.12 = 10.85 °C;
– for the external walls of a warm attic, taking
W/m 2 °С,
15 – [(15 + 35)]/(3.08 8.7) =
= 15 – 1.49 = 13.5 °C.
II. Calculating the dew point temperature t d , °C, in the attic:

– calculate the moisture content of the outside air, g/m 3, at the design temperature t ext:

=
– the same, air from a warm attic, taking the increment in moisture content ∆ f for houses with gas stoves equal to 4.0 g/m3:
g/m 3 ;
– determine the partial pressure of water vapor in the air in a warm attic:

According to Appendix 8 by value E= e g find the dew point temperature t d = 3.05 °C.

The obtained dew point temperature values ​​are compared with the corresponding values
And
:
=13,5 > t d = 3.05 °C; = 10.88 > t d = 3.05 °C.
The dew point temperature is significantly lower than the corresponding temperatures on the internal surfaces of external fences, therefore, condensation will not form on the internal surfaces of the coating and on the walls of the attic.

Conclusion. Horizontal and vertical fences for a warm attic satisfy regulatory requirements thermal protection of the building.

Example5
Calculation of specific heat energy consumption for heating a 9-story single-section residential building (tower type)
Dimensions typical floor A 9-story residential building is shown in the figure.

Fig. 8 Typical floor plan of a 9-story one-section residential building

A. Initial data
Place of construction - Perm.

Climatic region – IV.

Humidity zone is normal.

The humidity level of the room is normal.

Operating conditions of enclosing structures – B.

Duration of the heating season z ht = 229 days.

Average temperature of the heating season t ht = –5.9 °С.

Indoor air temperature t int = +21 °С.

Cold five-day outdoor air temperature t ext = = –35 °С .

The building is equipped with a “warm” attic and a technical basement.

Internal air temperature of the technical basement = = +2 °С

Height of the building from the floor level of the first floor to the top of the exhaust shaft H= 29.7 m.

Floor height – 2.8 m.

The maximum of the average wind speeds by rumba for January v= 5.2 m/s.
B. Calculation procedure
1. Determination of the areas of enclosing structures.

The determination of the areas of enclosing structures is based on the plan of a typical floor of a 9-story building and the initial data of section A.

Total floor area of ​​the building
A h = (42.5 + 42.5 + 42.5 + 57.38) 9 = 1663.9 m2.
Living area of ​​apartments and kitchens
A l = (27,76 + 27,76 + 27,76 + 42,54 + 7,12 + 7,12 +
+ 7,12 + 7,12)9 = 1388.7 m2.
Floor area above the technical basement A b .с, attic floor A g. f and coverings above the attic A g. c
A b .с = A g. f = A g. c = 16·16.2 = 259.2 m2.
The total area of ​​window fillings and balcony doors A F with their number on the floor:

– window fillings 1.5 m wide – 6 pcs.,

– window fillings 1.2 m wide – 8 pcs.,

– balcony doors 0.75 m wide – 4 pcs.

Window height – 1.2 m; the height of the balcony doors is 2.2 m.
A F = [(1.5 6+1.2 8) 1.2+(0.75 4 2.2)] 9 = 260.3 m2.
Square entrance doors into the stairwell with a width of 1.0 and 1.5 m and a height of 2.05 m
A ed = (1.5 + 1.0) 2.05 = 5.12 m2.
Area of ​​window fillings in the staircase with a window width of 1.2 m and a height of 0.9 m

= (1.2·0.9)·8 = 8.64 m2.
The total area of ​​external doors of apartments with a width of 0.9 m, a height of 2.05 m and a number of 4 pcs per floor.
A ed = (0.9 2.05 4) 9 = 66.42 m2.
The total area of ​​the external walls of the building, taking into account window and door openings

= (16 + 16 + 16.2 + 16.2) 2.8 9 = 1622.88 m2.
The total area of ​​the external walls of the building without windows and doorways

A W = 1622.88 – (260.28 + 8.64 + 5.12) = 1348.84 m2.
The total area of ​​the internal surfaces of external enclosing structures, including the attic floor and the floor above the technical basement,

= (16 + 16 + 16.2 + 16.2) 2.8 9 + 259.2 + 259.2 = 2141.3 m2.
Heated volume of the building

V n = 16·16.2·2.8·9 = 6531.84 m3.
2. Determination of the degree-day of the heating period.

Degree days are determined by formula (2) SNiP 23-02–2003 for the following enclosing structures:

– external walls and attic floors:

D d 1 = (21 + 5.9) 229 = 6160.1 °C day,
– covering and external walls of a warm “attic”:
D d 2 = (15 + 5.9) 229 = 4786.1 °C day,
– ceilings above the technical basement:
D d 3 = (2 + 5.9) 229 = 1809.1 °C day.
3. Determination of the required heat transfer resistance of enclosing structures.

The required heat transfer resistance of enclosing structures is determined from the table. 4 SNiP 23-02–2003 depending on the degree-day values ​​of the heating period:

– for external walls of a building
= 0.00035 6160.1 + 1.4 = 3.56 m 2 °C/W;
– for attic flooring
= n· = 0.107(0.0005 6160.1 + 2.2) = 0.49 m2,
n =
=
= 0,107;
– for external walls of the attic
= 0.00035 4786.1 + 1.4 = 3.07 m 2 °C/W,
– for covering above the attic

=
=
= 0.87 m 2 °C/W;
– for covering over a technical basement

= n b. c R reg = 0.34(0.00045 1809.1 + 1.9) = 0.92 m 2 °C/W,

n b. c =
=
= 0,34;
– for window fillings and balcony doors with triple glazing in wooden frames (Appendix L SP 23-101–2004)

= 0.55 m 2 °C/W.
4. Determination of thermal energy consumption for heating the building.

To determine the consumption of thermal energy for heating a building during the heating period, it is necessary to establish:

– total heat loss of the building through external fences Q h, MJ;

– domestic heat gains Q int, MJ;

– heat gain through windows and balcony doors from solar radiation, MJ.

When determining the total heat loss of a building Q h , MJ, two coefficients need to be calculated:

– reduced heat transfer coefficient through the external building envelope
, W/(m 2 °C);
L v = 3 A l= 3 1388.7 = 4166.1 m 3 / h,
Where A l– area of ​​living quarters and kitchens, m2;

– determined average air exchange rate of the building during the heating period n a, h –1, according to formula (D.8) SNiP 23-02–2003:
n a =
= 0.75 h –1.
We accept a coefficient for reducing the air volume in the building, taking into account the presence of internal fences, B v = 0.85; specific heat capacity of air c= 1 kJ/kg °С, and the coefficient taking into account the influence of counter heat flow in translucent structures k = 0,7:

=
= 0.45 W/(m 2 °C).
The value of the overall heat transfer coefficient of the building K m, W/(m 2 °C), determined by formula (D.4) SNiP 23-02–2003:
K m = 0.59 + 0.45 = 1.04 W/(m 2 °C).
We calculate the total heat loss of the building during the heating period Q h, MJ, according to formula (D.3) SNiP 23-02–2003:
Q h = 0.0864 1.04 6160.1 2141.28 = 1185245.3 MJ.
Household heat gains during the heating season Q int , MJ, determined by formula (G.11) SNiP 23-02–2003, taking the value of specific household heat release q int equal to 17 W/m2:
Q int = 0.0864·17·229·1132.4 = 380888.62 MJ.
Heat input into the building from solar radiation during the heating period Q s , MJ, determined by formula (G.11) SNiP 23-02–2003, taking into account the values ​​of the coefficients taking into account the shading of light openings by opaque filling elements τ F = 0.5 and the relative penetration of solar radiation for light-transmitting window fillings k F = 0.46.

The average value of solar radiation during the heating period vertical surfaces I avg, W/m 2, we accept according to Appendix (D) SP 23-101–2004 for geographical latitude location of Perm (56° N):

I av = 201 W/m2,
Q s = 0.5 0.76(100.44 201 + 100.44 201 +
+ 29.7·201 + 29.7·201) = 19880.18 MJ.
Thermal energy consumption for heating the building during the heating period , MJ, determined by formula (D.2) SNiP 23-02–2003, taking numerical value the following coefficients:

– coefficient of reduction of heat input due to thermal inertia of enclosing structures = 0,8;

– coefficient taking into account the additional heat consumption of the heating system associated with the discreteness of the nominal heat flow of the product range heating devices for tower buildings = 1,11.
= ·1.11 = 1024940.2 MJ.
We establish the specific thermal energy consumption of the building
, kJ/(m 2 °C day), according to formula (D.1) SNiP 23-02–2003:
=
= 25.47 kJ/(m 2 °C day).
According to the data in Table. 9 SNiP 23-02–2003, the standardized specific heat energy consumption for heating a 9-story residential building is 25 kJ/(m 2 °C day), which is 1.02% lower than the calculated specific heat energy consumption = 25.47 kJ /(m 2 °C day), therefore, during the thermal engineering design of enclosing structures, it is necessary to take this difference into account.

Description:

In accordance with the latest SNiP " Thermal protection buildings”, the “Energy Efficiency” section is mandatory for any project. The main purpose of the section is to prove that the specific heat consumption for heating and ventilation of the building is below the standard value.

Calculation of solar radiation in winter time

Flux of total solar radiation arriving during the heating period on horizontal and vertical surfaces under actual cloudy conditions, kWh/m2 (MJ/m2)

Flux of total solar radiation arriving for each month of the heating period on horizontal and vertical surfaces under actual cloudy conditions, kWh/m2 (MJ/m2)

As a result of the work done, data were obtained on the intensity of total (direct and diffuse) solar radiation falling on differently oriented vertical surfaces for 18 Russian cities. This data can be used in real design.

Literature

1. SNiP 23–02–2003 “Thermal protection of buildings.” – M.: Gosstroy of Russia, FSUE TsPP, 2004.

2. Scientific and applied reference book on the climate of the USSR. Parts 1–6. Vol. 1–34. – St. Petersburg. : Gidrometeoizdat, 1989–1998.

3. SP 23–101–2004 “Design of thermal protection of buildings.” – M.: Federal State Unitary Enterprise TsPP, 2004.

4. MGSN 2.01–99 “Energy saving in buildings. Standards for thermal protection and heat and water power supply.” – M.: State Unitary Enterprise “NIAC”, 1999.

5. SNiP 23–01–99* “Building climatology”. – M.: Gosstroy of Russia, State Unitary Enterprise TsPP, 2003.

6. Construction climatology: Reference manual for SNiP. – M.: Stroyizdat, 1990.

MINISTRY OF EDUCATION AND SCIENCE OF THE RUSSIAN FEDERATION

Federal state budget educational institution higher professional education

"State University - educational, research and production complex"

Institute of Architecture and Construction

Department: " City building and farming"

Discipline: “Structural Physics”

COURSE WORK

"Thermal protection of buildings"

Completed by student: Arkharova K.Yu.

• Introduction
• Assignment form
• 1 . Climate certificate
• 2 . Thermal calculation
• 2.1 Thermal engineering calculation of enclosing structures
• 2.2 Calculation of enclosing structures of “warm” basements
• 2.3 Thermal calculation of windows
• 3 . Calculation of specific heat energy consumption for heating during the heating period
• 4 . Heat absorption of floor surfaces
• 5 . Protection of the building envelope from waterlogging
• Conclusion
• List of sources and literature used
• Appendix A

Introduction

Thermal protection is a set of energy saving measures and technologies that allows increasing the thermal insulation of buildings for various purposes, reduce heat loss in rooms.

The task of ensuring the necessary thermal technical qualities of external enclosing structures is solved by giving them the required heat resistance and heat transfer resistance.

The heat transfer resistance must be high enough to ensure hygienically acceptable temperature conditions on the surface of the structure facing the room during the coldest period of the year. The thermal stability of structures is assessed by their ability to maintain a relative constant temperature in the premises during periodic fluctuations in the temperature of the air surrounding the structures and the flow of heat passing through them. The degree of thermal stability of a structure as a whole is largely determined by the physical properties of the material from which the outer layer of the structure is made, which can withstand sudden temperature fluctuations.

In this course work Thermal engineering calculation of the building envelope will be performed individual house, the construction area of ​​which is Arkhangelsk.

Assignment form

1 Construction area:

Arkhangelsk.

2 Wall structure (name construction material, insulation, thickness, density):

1st layer - polystyrene concrete modified with slag-Portland cement (=200 kg/m3; ?=0.07 W/(m*K); ?=0.36 m)

2nd layer - extruded polystyrene foam (=32 kg/m3; ?=0.031 W/(m*K); ?=0.22 m)

3rd layer - perlite concrete (=600 kg/m3; ?=0.23 W/(m*K); ?=0.32 m

3 Material of heat-conducting inclusion:

perlibeton (=600 kg/m3; ?=0.23 W/(m*K); ?=0.38 m

4 Floor design:

1st layer - linoleum (=1800 kg/m 3; s=8.56 W/(m 2 °C); ?=0.38 W/(m 2 °C); ?=0.0008 m

2nd layer - cement-sand screed (=1800 kg/m 3; s=11.09 W/(m 2 °C); ?=0.93 W/(m 2 °C); ?=0.01 m)

3rd layer - polystyrene foam boards (=25 kg/m 3; s=0.38 W/(m 2 °C); ?=0.44 W/(m 2 °C); ?=0.11 m )

4th layer - foam concrete slab (=400 kg/m 3; s=2.42 W/(m 2 °C); ?=0.15 W/(m 2 °C); ?=0.22 m )

1 . Climate certificate

Development area - Arkhangelsk.

Climatic region - II A.

Humidity zone - wet.

Indoor air humidity? = 55%;

estimated room temperature = 21°C.

The humidity level of the room is normal.

Operating conditions - B.

Climatic parameters:

Estimated outside air temperature (Outside air temperature of the coldest five-day period (probability 0.92)

Duration of the heating period (with an average daily outside air temperature of 8°C) - = 250 days;

The average temperature of the heating period (with an average daily outside air temperature? 8°C) - = - 4.5 °C.

enclosing heat absorption heating

2 . Thermal calculation

2 .1 Thermal engineering calculation of enclosing structures

Calculation of degree-days of the heating period

GSOP = (t in - t from) z from, (1.1)

where is the estimated room temperature, °C;

Estimated outside air temperature, °C;

Duration of the heating season, days

GSOP =(+21+4.5) 250=6125°Сday

We calculate the required heat transfer resistance using formula (1.2)

where, a and b are coefficients, the values ​​of which should be taken according to Table 3 of SP 50.13330.2012 “Thermal protection of buildings” for the corresponding groups of buildings.

We accept: a = 0.00035 ; b=1.4

0.00035 6125 +1.4=3.54m 2 °C/W.

Design outer wall

a) We cut the structure with a plane parallel to the direction of heat flow (Fig. 1):

Figure 1 - External wall design

Table 1 - Parameters of external wall materials

Heat transfer resistance R a is determined by formula (1.3):

where A i is the area of ​​the i-th section, m 2 ;

R i - heat transfer resistance of the i-th section, ;

A is the sum of the areas of all plots, m2.

We determine the heat transfer resistance for homogeneous areas using formula (1.4):

Where, ? - layer thickness, m;

Thermal conductivity coefficient, W/(mK)

We calculate the heat transfer resistance for non-uniform areas using formula (1.5):

R= R 1 +R 2 +R 3 +…+R n +R VP, (1.5)

where, R 1 , R 2 , R 3 ...R n is the heat transfer resistance of individual layers of the structure, ;

R VP - resistance to heat transfer of the air layer, .

We find R a using formula (1.3):

b) We cut the structure with a plane perpendicular to the direction of heat flow (Fig. 2):

Figure 2 - External wall design

Heat transfer resistance R b is determined by formula (1.5)

R b = R 1 +R 2 +R 3 +…+R n +R vp, (1.5)

We will determine the air permeation resistance for homogeneous areas using formula (1.4).

We determine the air permeation resistance for non-uniform areas using formula (1.3):

We find Rb using formula (1.5):

R b =5.14+3.09+1.4= 9.63.

The conditional resistance to heat transfer of the outer wall is determined by formula (1.6):

where, R a is the heat transfer resistance of the enclosing structure, cut parallel to the heat flow;

R b - heat transfer resistance of the enclosing structure, cut perpendicular to the heat flow, .

The reduced resistance to heat transfer of the outer wall is determined by formula (1.7):

Heat transfer resistance on the outer surface is determined by formula (1.9)

where, heat transfer coefficient of the inner surface of the enclosing structure = 8.7;

where, is the heat transfer coefficient of the outer surface of the enclosing structure, = 23;

The calculated temperature difference between the temperature of the internal air and the temperature of the internal surface of the enclosing structure is determined by formula (1.10):

where n is a coefficient that takes into account the dependence of the position of the outer surface of the enclosing structures in relation to the outside air, we take n=1;

estimated room temperature, °C;

design temperature of outside air during the cold season, °C;

heat transfer coefficient of the internal surface of enclosing structures, W/(m 2 °C).

The temperature of the inner surface of the enclosing structure is determined by formula (1.11):

2 . 2 Calculation of enclosing structures of “warm” basements

Required heat transfer resistance of the part basement wall, located above the ground level, we take equal to the reduced resistance to heat transfer of the outer wall:

The reduced heat transfer resistance of the enclosing structures of the buried part of the basement, located below ground level.

The height of the recessed part of the basement is 2m; basement width - 3.8m

According to table 13 SP 23-101-2004 “Design of thermal protection of buildings” we accept:

We calculate the required heat transfer resistance of the basement floor above the “warm” basement using formula (1.12)

where, the required heat transfer resistance of the basement floor is found from Table 3 of SP 50.13330.2012 “Thermal protection of buildings”.

where, air temperature in the basement, °C;

the same as in formula (1.10);

the same as in formula (1.10)

Let’s take it equal to 21.35 °C:

We determine the air temperature in the basement using formula (1.14):

where, the same as in formula (1.10);

Linear heat flux density; ;

Air volume in the basement, ;

Length of pipeline of i-th diameter, m; ;

Air exchange rate in the basement; ;

Air density in the basement;

c - specific heat capacity of air;;

Basement area, ;

The area of ​​the floor and walls of the basement in contact with the ground;

The area of ​​the external walls of the basement above ground level, .

2 . 3 Thermal calculation of windows

We calculate the degree-day of the heating period using formula (1.1)

GSOP =(+21+4.5) 250=6125°Sd.

The reduced heat transfer resistance is determined according to Table 3 of SP 50.13330.2012 “Thermal protection of buildings” by interpolation method:

We select windows based on the found heat transfer resistance R0:

Regular glass and single-chamber double-glazed windows in separate frames made of glass with a hard selective coating - .

Conclusion: The reduced heat transfer resistance, temperature difference and temperature of the internal surface of the enclosing structure comply with the required standards. Consequently, the designed structure of the outer wall and the thickness of the insulation are selected correctly.

Due to the fact that we took the wall structure as the enclosing structure in the recessed part of the basement, we received an unacceptable resistance to heat transfer of the basement floor, which affects the temperature difference between the temperature of the internal air and the temperature of the inner surface of the enclosing structure.

3 . Calculation of specific heat energy consumption for heating during the heating period

The estimated specific consumption of thermal energy for heating buildings during the heating period is determined by formula (2.1):

where, thermal energy consumption for heating the building during the heating period, J;

Sum of floor areas of apartments or usable area premises of the building, with the exception of technical floors and garages, m 2

Thermal energy consumption for heating the building during the heating period is calculated using formula (2.2):

where, the total heat loss of the building through the external enclosing structures, J;

Household heat input during the heating period, J;

Heat gain through windows and skylights from solar radiation during the heating season, J;

Heat gain reduction coefficient due to thermal inertia of enclosing structures, recommended value = 0.8;

A coefficient that takes into account the additional heat consumption of the heating system associated with the discreteness of the nominal heat flow of the range of heating devices, their additional heat losses through the behind-the-radiator sections of the fences, increased air temperature in corner rooms, heat losses of pipelines passing through unheated premises, for buildings with heated basements = 1.07;

The total heat loss of the building, J, during the heating period is determined by formula (2.3):

where, is the overall heat transfer coefficient of the building, W/(m 2 °C), determined by formula (2.4);

Total area of ​​enclosing structures, m 2 ;

where, is the reduced heat transfer coefficient through the external building envelope, W/(m 2 °C);

Conditional heat transfer coefficient of a building, taking into account heat loss due to infiltration and ventilation, W/(m 2 °C).

The reduced heat transfer coefficient through the external building envelope is determined by formula (2.5):

where, area, m 2 and reduced heat transfer resistance, m 2 °C/W, of external walls (except for openings);

The same, filling light openings (windows, stained glass, lanterns);

The same for external doors and gates;

the same, combined coverings (including over bay windows);

the same, attic floors;

the same, basement floors;

Same, .

0.306 W/(m 2 °C);

The conditional heat transfer coefficient of the building, taking into account heat loss due to infiltration and ventilation, W/(m 2 °C), is determined by formula (2.6):

where, is the coefficient of reduction in air volume in the building, taking into account the presence of internal enclosing structures. We accept sv = 0.85;

Volume of heated premises;

The coefficient for taking into account the influence of oncoming heat flow in translucent structures, equal to 1 for windows and balcony doors with separate sashes;

Average density supply air for the heating period, kg/m 3, determined by formula (2.7);

Average ratio air exchange of the building during the heating period, h 1

The average air exchange rate of a building during the heating period is calculated from the total air exchange due to ventilation and infiltration using formula (2.8):

where, is the amount of supply air into the building with unorganized inflow or the standardized value with mechanical ventilation, m 3 / h, equal for residential buildings intended for citizens taking into account the social norm (with an estimated occupancy of an apartment of 20 m 2 of total area or less per person) - 3 A; 3 A = 603.93 m 2;

Living area; =201.31m2;

Number of operating hours of mechanical ventilation during a week, h; ;

Number of hours of infiltration recording during the week, h;=168;

The amount of air infiltrated into the building through the enclosing structures, kg/h;

The amount of air infiltrating into the staircase of a residential building through leaks in the filling of the openings will be determined by formula (2.9):

where, - respectively for the staircase, the total area of ​​windows and balcony doors and external entrance doors, m 2;

accordingly, for the staircase, the required air permeability resistance of windows and balcony doors and external entrance doors, m 2 °C/W;

Accordingly, for the staircase, the calculated difference in pressure of external and internal air for windows and balcony doors and external entrance doors, Pa, determined by formula (2.10):

where, n, in - specific gravity respectively external and internal air, N/m 3, determined by formula (2.11):

Maximum of average wind speeds by direction for January (SP 131.13330.2012 “Building climatology”); =3.4 m/s.

3463/(273 + t), (2.11)

n = 3463/(273 -33) = 14.32 N/m 3 ;

in = 3463/(273+21) = 11.78 N/m 3 ;

From here we find:

We find the average air exchange rate of the building during the heating period using the data obtained:

0.06041 h 1 .

Based on the data obtained, we calculate using formula (2.6):

0.020 W/(m 2 °C).

Using the data obtained in formulas (2.5) and (2.6), we find the overall heat transfer coefficient of the building:

0.306+0.020= 0.326 W/(m 2 °C).

We calculate the total heat loss of the building using formula (2.3):

0.08640.326317.78=J.

Household heat input during the heating period, J, is determined by formula (2.12):

where, the amount of household heat generation per 1 m 2 of residential premises or the estimated area of ​​a public building, W/m 2, is accepted;

area of ​​residential premises; =201.31m2;

Heat gain through windows and skylights from solar radiation during the heating period, J, for four facades of buildings oriented in four directions, will be determined by formula (2.13):

where, are coefficients taking into account the darkening of the light opening by opaque elements; for a single-chamber double-glazed window made of ordinary glass with a hard selective coating - 0.8;

Relative penetration coefficient of solar radiation for light-transmitting fillings; for a single-chamber double-glazed window made of ordinary glass with a hard selective coating - 0.57;

The area of ​​light openings of the building facades, respectively oriented in four directions, m 2 ;

The average value of solar radiation on vertical surfaces during the heating period under actual cloudy conditions, respectively oriented along the four facades of the building, J/(m2, determined according to table 9.1 SP 131.13330.2012 “Building climatology”;

Heating season:

January, February, March, April, May, September, October, November, December.

We take the latitude of 64°N for the city of Arkhangelsk.

C: A 1 =2.25m2; I 1 =(31+49)/9=8.89 J/(m2;

I 2 =(138+157+192+155+138+162+170+151+192)/9=161.67J/(m2;

B: A 3 =8.58; I 3 =(11+35+78+135+153+96+49+22+12)/9=66 J/(m 2 ;

Z: A 4 =8.58; I 4 =(11+35+78+135+153+96+49+22+12)/9=66 J/(m2.

Using the data obtained from calculating formulas (2.3), (2.12) and (2.13), we find the consumption of thermal energy for heating the building using formula (2.2):

Using formula (2.1), we calculate the specific consumption of thermal energy for heating:

KJ/(m 2 °C day).

Conclusion: the specific consumption of thermal energy for heating a building does not correspond to the standardized consumption determined according to SP 50.13330.2012 “Thermal protection of buildings” and equal to 38.7 kJ/(m 2 °C day).

4 . Heat absorption of floor surfaces

Thermal inertia of floor structure layers

Figure 3 - Floor diagram

Table 2 - Parameters of floor materials

Let us calculate the thermal inertia of the layers of the floor structure using formula (3.1):

where, s is the heat absorption coefficient, W/(m 2 °C);

Thermal resistance determined by formula (1.3)

Calculated indicator of heat absorption of the floor surface.

The first 3 layers of the floor structure have a total thermal inertia but the thermal inertia of 4 layers.

Therefore, we will determine the heat absorption rate of the floor surface sequentially by calculating the heat absorption rate of the surfaces of the layers of the structure, starting from the 3rd to the 1st:

for the 3rd layer according to formula (3.2)

for the i-th layer (i=1,2) according to formula (3.3)

W/(m 2 °C);

W/(m 2 °C);

W/(m 2 °C);

The heat absorption rate of the floor surface is assumed to be equal to the heat absorption rate of the surface of the first layer:

W/(m 2 °C);

The normalized value of the heat absorption index is determined according to SP 50.13330.2012 “Thermal protection of buildings”:

12 W/(m 2 °C);

Conclusion: the calculated heat absorption rate of the floor surface corresponds to the standardized value.

5 . Protection of the building envelope from waterlogging

Climatic parameters:

Table 3 - Average monthly temperatures and water vapor pressure of outdoor air

Average partial pressure of water vapor of outdoor air over an annual period

Figure 4 - External wall design

Table 4 - Parameters of external wall materials

We find the vapor permeability resistance of the layers of the structure using the formula:

where, is the layer thickness, m;

Vapor permeability coefficient, mg/(mchPa)

We determine the resistance to vapor permeation of the layers of the structure from the outer and inner surfaces to the plane of possible condensation (the plane of possible condensation coincides with outer surface insulation):

The heat transfer resistance of the wall layers from the inner surface to the plane of possible condensation is determined by formula (4.2):

where, is the resistance to heat transfer on the inner surface, determined by formula (1.8)

Length of seasons and average monthly temperatures:

winter (January, February, March, December):

summer (May, June, July, August, September):

spring, autumn (April, October, November):

where, the reduced resistance to heat transfer of the outer wall, ;

calculated room temperature, .

We find the corresponding value of water vapor pressure:

We find the average value of water vapor pressure per year using formula (4.4):

where E 1, E 2, E 3 are the values ​​of water vapor pressure by season, Pa;

duration of seasons, months

The partial vapor pressure of the internal air is determined by formula (4.5):

where, partial pressure of saturated water vapor, Pa, at the temperature of the indoor air in the room; for 21: 2488 Pa;

relative humidity of indoor air, %

We find the required resistance to vapor permeation using formula (4.6):

where, the average partial pressure of water vapor of the outside air over the annual period, Pa; accept = 6.4 hPa

From the condition of inadmissibility of moisture accumulation in the enclosing structure over the annual period of operation, we check the condition:

We find the water vapor pressure of the outside air for a period with negative average monthly temperatures:

We find the average outside air temperature for a period with negative average monthly temperatures:

We determine the temperature value in the plane of possible condensation using formula (4.3):

This temperature corresponds to

We determine the required resistance to vapor permeation using formula (4.7):

where, the duration of the period of moisture accumulation, days, taken equal to the period with negative average monthly temperatures; take =176 days;

density of the wetted layer material, kg/m 3 ;

thickness of the wetted layer, m;

maximum permissible increase in humidity in the material of the wetted layer, % by weight, during the period of moisture accumulation, taken according to table 10 SP 50.13330.2012 “Thermal protection of buildings”; accept for expanded polystyrene = 25%;

coefficient determined by formula (4.8):

where, the average partial pressure of water vapor of the outside air for the period with negative average monthly temperatures, Pa;

the same as in formula (4.7)

From here we calculate using formula (4.7):

From the condition of limiting moisture in the enclosing structure for a period with negative average monthly outdoor temperatures, we check the condition:

Conclusion: due to the fulfillment of the condition of limiting the amount of moisture in the enclosing structure during the period of moisture accumulation, an additional vapor barrier device is not required.

Conclusion

The thermal properties of external building enclosures depend on: favorable microclimate buildings, that is, ensuring the temperature and humidity in the room is not lower than regulatory requirements; the amount of heat lost by the building in winter; the temperature of the inner surface of the fence, which guarantees against the formation of condensation on it; the humidity regime of the fencing design, which affects its heat-protective qualities and durability.

The task of ensuring the necessary thermal technical qualities of external enclosing structures is solved by giving them the required heat resistance and heat transfer resistance. The permissible permeability of structures is limited by a given resistance to air permeation. The normal moisture state of structures is achieved by reducing the initial moisture content of the material and installing moisture insulation, and in layered structures, in addition, by the appropriate arrangement of structural layers made of materials with different properties.

During the course project Calculations were carried out related to the thermal protection of buildings, which were carried out in accordance with the codes of practice.

List sources used and literature

1. SP 50.13330.2012. Thermal protection of buildings (Updated edition of SNiP 23-02-2003) [Text] /Ministry of Regional Development of Russia. - M.: 2012. - 96 p.

2. SP 131.13330.2012. Construction climatology (Updated version of SNiP 23-01-99*) [Text] / Ministry of Regional Development of Russia. - M.: 2012. - 109 p.

3. Kupriyanov V.N. Design of thermal protection of enclosing structures: Textbook [Text]. - Kazan: KGASU, 2011. - 161 p..

4. SP 23-101-2004 Design of thermal protection of buildings [Text]. - M.: Federal State Unitary Enterprise TsPP, 2004.

5. T.I. Abasheva. Album technical solutions to increase the thermal protection of buildings, insulate structural units during overhaul housing stock [Text]/ T.I. Abasheva, L.V. Bulgakov. N.M. Vavulo et al. M.: 1996. - 46 pages.

Appendix A

Energy passport of the building

general information

Design conditions

Functional purpose, type and constructive solution building

Geometric and thermal energy indicators

Odds

Comprehensive indicators

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Thermal engineering calculation of technical underground

Thermal calculations enclosing structures

The areas of external enclosing structures, the heated area and volume of the building, necessary for calculating the energy passport, and thermal characteristics building envelopes are determined in accordance with the adopted design decisions in accordance with the recommendations of SNiP 23-02 and TSN 23 - 329 - 2002.

The heat transfer resistance of enclosing structures is determined depending on the number and materials of layers, as well as physical properties building materials according to the recommendations of SNiP 23-02 and TSN 23 - 329 - 2002.

1.2.1 External walls of the building

There are three types of external walls in a residential building.

First type - brickwork with floor support 120 mm thick, insulated with polystyrene concrete 280 mm thick, with a facing layer of sand-lime brick. Second type - reinforced concrete panel 200 mm, insulated with 280 mm thick polystyrene concrete, with a facing layer of sand-lime brick. The third type, see Fig. 1. Thermal engineering calculations are given for two types of walls, respectively.

1). Composition of layers of the outer wall of the building: protective covering- cement-lime mortar 30 mm thick, λ = 0.84 W/(m× o C). The outer layer is 120 mm - made of sand-lime brick M 100 with frost resistance grade F 50, λ = 0.76 W/(m× o C); filling 280 mm – insulation – polystyrene concrete D200, GOST R 51263-99, λ = 0.075 W/(m× o C); inner layer 120 mm - from sand-lime brick, M 100, λ = 0.76 W/(m× o C). Internal walls plastered with lime-sand mortar M 75, 15 mm thick, λ = 0.84 W/(m× o C).

Rw= 1/8.7+0.030/0.84+0.120/0.76+0.280/0.075+0.120/0.76+0.015/0.84+1/23 = 4.26 m 2 × o C/W.

Heat transfer resistance of building walls, with facade area
Aw= 4989.6 m2, equal to: 4.26 m 2 × o C/W.

Thermal uniformity coefficient of external walls r, determined by formula 12 SP 23-101:

a i– width of the heat-conducting inclusion, a i = 0.120 m;

L i– length of the heat-conducting inclusion, L i= 197.6 m (building perimeter);

k i – coefficient depending on the heat-conducting inclusion, determined according to adj. N SP 23-101:

k i = 1.01 for heat-conducting connection at ratios λm/λ= 2.3 and a/b= 0,23.

Then the reduced heat transfer resistance of the building walls is equal to: 0.83 × 4.26 = 3.54 m 2 × o C/W.

2). Composition of the layers of the outer wall of the building: protective coating - cement-lime mortar M 75, 30 mm thick, λ = 0.84 W/(m× o C). The outer layer is 120 mm - made of sand-lime brick M 100 with frost resistance grade F 50, λ = 0.76 W/(m× o C); filling 280 mm – insulation – polystyrene concrete D200, GOST R 51263-99, λ = 0.075 W/(m× o C); inner layer 200 mm – reinforced concrete Wall panel, λ= 2.04 W/(m× o C).

The heat transfer resistance of the wall is equal to:

Rw= 1/8,7+0,030/0,84+0,120/0,76+0,280/0,075+
+0.20/2.04+1/23 = 4.2 m 2 × o C/W.

Since the walls of the building have a homogeneous multilayer structure, the coefficient of thermal uniformity of the external walls is accepted r= 0,7.

Then the reduced heat transfer resistance of the building walls is equal to: 0.7 × 4.2 = 2.9 m 2 × o C/W.

Type of building - ordinary section of a 9-story residential building with a lower distribution of pipes for heating and hot water supply systems.

A b= 342 m2.

technical floor area underground - 342 m2.

Area of ​​external walls above ground level A b, w= 60.5 m2.

Design temperatures of the lower heating system are 95 °C, hot water supply 60 °C. Length of heating system pipelines with bottom wiring 80 m. The length of hot water supply pipelines was 30 m. Gas distribution pipes in technical. There is no underground, so the frequency of air exchange in those. underground I= 0.5 h -1 .

t int= 20 °C.

Basement area (above technical underground) - 1024.95 m2.

The width of the basement is 17.6 m. The height of the outer wall is technical. underground, buried in the ground - 1.6 m. Total length l cross section technical fencing underground, buried in the ground,

l= 17.6 + 2×1.6 = 20.8 m.

Air temperature in the premises of the first floor t int= 20 °C.

Resistance to heat transfer of external walls. underground spaces above ground level are accepted in accordance with SP 23-101 clause 9.3.2. equal to the heat transfer resistance of the external walls R o b . w= 3.03 m 2 ×°C/W.

Reduced resistance to heat transfer of enclosing structures of the buried part of the technical area. underground areas will be determined in accordance with SP 23-101 clause 9.3.3. as for non-insulated floors on the ground in the case where the floor and wall materials have calculated thermal conductivity coefficients λ≥ 1.2 W/(m o C). Reduced resistance to heat transfer of technical fences. underground, buried in the ground was determined according to table 13 SP 23-101 and amounted to R o rs= 4.52 m 2 ×°C/W.

The basement walls consist of: wall block, 600 mm thick, λ = 2.04 W/(m× o C).

Let's determine the air temperature in those. underground t int b

For the calculation we use the data from Table 12 [SP 23-101]. At air temperature in those. underground 2 °C the heat flux density from the pipelines will increase compared to the values ​​​​given in Table 12 by the value of the coefficient obtained from equation 34 [SP 23-101]: for heating system pipelines - by the coefficient [(95 - 2)/( 95 - 18)] 1.283 = 1.41; for hot water supply pipelines - [(60 - 2)/(60 - 18) 1.283 = 1.51. Then we calculate the temperature value t int b from the heat balance equation at a designated underground temperature of 2 °C

t int b= (20×342/1.55 ​​+ (1.41 25 80 + 1.51 14.9 30) - 0.28×823×0.5×1.2×26 - 26×430/4.52 - 26×60.5/3.03)/

/(342/1.55 ​​+ 0.28×823×0.5×1.2 + 430/4.52 +60.5/3.03) = 1316/473 = 2.78 °C.

The heat flow through the basement floor was

q b . c= (20 – 2.78)/1.55 ​​= 11.1 W/m2.

Thus, in those underground, thermal protection equivalent to the standards is provided not only by barriers (walls and floors), but also by heat from the pipelines of heating and hot water supply systems.

1.2.3 Overlapping over technical. underground

The fence has an area Af= 1024.95 m2.

Structurally, the overlap is made as follows.

2.04 W/(m× o C). Cement-sand screed 20 mm thick, λ =
0.84 W/(m× o C). Insulation extruded polystyrene foam "Rufmat", ρ o= 32 kg/m 3, λ = 0.029 W/(m× o C), thickness 60 mm according to GOST 16381. Air gap, λ = 0.005 W/(m× o C), thickness 10 mm. Boards for flooring, λ = 0.18 W/(m× o C), 20 mm thick according to GOST 8242.

Rf= 1/8,7+0,22/2,04+0,020/0,84+0,060/0,029+

0.010/0.005+0.020/0.180+1/17 = 4.35 m 2 × o C/W.

According to clause 9.3.4 SP 23-101, we will determine the value of the required heat transfer resistance of the basement floor above the technical underground Rс according to the formula

R o = nR req,

Where n- coefficient determined at the accepted minimum temperature underground air t int b= 2°C.

n = (t int - t int b)/(t int - t ext) = (20 - 2)/(20 + 26) = 0,39.

Then R with= 0.39 × 4.35 = 1.74 m 2 × ° C / W.

Let's check whether the thermal protection of the ceiling above the technical underground meets the requirement of the standard differential D tn= 2 °C for the floor of the first floor.

Using formula (3) SNiP 23 - 02, we determine the minimum permissible heat transfer resistance

R o min =(20 - 2)/(2×8.7) = 1.03 m 2 ×°C/W< R c = 1.74 m 2 ×°C/W.

1.2.4 Attic floor

Floor area A c= 1024.95 m2.

Reinforced concrete floor slab, thickness 220 mm, λ =
2.04 W/(m× o C). Insulation of mini-slabs JSC " Mineral wool», r =140-
175 kg/m 3, λ = 0.046 W/(m× o C), 200 mm thick according to GOST 4640. The coating on top has cement-sand screed 40 mm thick, λ = 0.84 W/(m× o C).

Then the heat transfer resistance is equal to:

Rc= 1/8.7+0.22/2.04+0.200/0.046+0.04/0.84+1/23=4.66 m 2 × o C/W.

1.2.5 Attic covering

Reinforced concrete floor slab, thickness 220 mm, λ =
2.04 W/(m× o C). Expanded clay gravel insulation, r=600 kg/m 3, λ =
0.190 W/(m× o C), thickness 150 mm according to GOST 9757; Mineral slab of Mineral Wool JSC, 140-175 kg/m3, λ = 0.046 W/(m×oC), 120 mm thick according to GOST 4640. The coating on top has a cement-sand screed 40 mm thick, λ = 0.84 W/ (m×o C).

Then the heat transfer resistance is equal to:

Rc= 1/8.7+0.22/2.04+0.150/0.190+0.12/0.046+0.04/0.84+1/17=3.37 m 2 × o C/W.

1.2.6 Windows

Modern translucent designs of heat-protective windows use double-glazed windows, and for making window frames and sashes, mainly PVC profiles or combinations thereof. When manufacturing double-glazed windows using float glass, the windows provide a calculated reduced heat transfer resistance of no more than 0.56 m 2 × o C/W, which meets the regulatory requirements for their certification.

Square window openings A F= 1002.24 m2.

Window heat transfer resistance is accepted R F= 0.56 m 2 × o C/W.

1.2.7 Reduced heat transfer coefficient

The reduced heat transfer coefficient through the external building envelope, W/(m 2 ×°C), is determined by formula 3.10 [TSN 23 - 329 - 2002] taking into account the structures adopted in the project:

1.13(4989.6 / 2.9+1002.24 / 0.56+1024.95 / 4.66+1024.95 / 4.35) / 8056.9 = 0.54 W/(m 2 × °C).

1.2.8 Conditional heat transfer coefficient

The conditional heat transfer coefficient of a building, taking into account heat loss due to infiltration and ventilation, W/(m 2 ×°C), is determined by formula G.6 [SNiP 23 - 02] taking into account the designs adopted in the project:

Where With– specific heat capacity of air equal to 1 kJ/(kg×°C);

β ν – coefficient of air volume reduction in the building, taking into account the presence of internal enclosing structures, equal to β ν = 0,85.

0.28×1×0.472×0.85×25026.57×1.305×0.9/8056.9 = 0.41 W/(m 2 ×°C).

The average air exchange rate of a building during the heating period is calculated from the total air exchange due to ventilation and infiltration using the formula

n a= [(3×1714.32) × 168/168+(95×0.9×

×168)/(168×1.305)] / (0.85×12984) = 0.479 h -1 .

– the amount of infiltrated air, kg/h, entering the building through the enclosing structures during the day of the heating period, is determined by formula G.9 [SNiP 23-02-2003]:

19.68/0.53×(35.981/10) 2/3 + (2.1×1.31)/0.53×(56.55/10) 1/2 = 95 kg/h.

– respectively, for the staircase, the calculated difference in pressure of external and internal air for windows and balcony doors and external entrance doors is determined by formula 13 [SNiP 23-02-2003] for windows and balcony doors with the value 0.55 replaced by 0, 28 and with the calculation of specific gravity according to formula 14 [SNiP 23-02-2003] at the corresponding air temperature, Pa.

∆р e d= 0.55× Η ×( γ ext -γ int) + 0.03× γ ext×ν 2 .

Where Η = 30.4 m – building height;

– specific gravity of external and internal air, respectively, N/m 3 .

γ ext = 3463/(273-26) = 14.02 N/m 3,

γ int = 3463/(273+21) = 11.78 N/m 3 .

∆р F= 0.28×30.4×(14.02-11.78)+0.03×14.02×5.9 2 = 35.98 Pa.

∆р ed= 0.55×30.4×(14.02-11.78)+0.03×14.02×5.9 2 = 56.55 Pa.

– average density of supply air during the heating period, kg/m3, ,

353/ = 1.31 kg/m3.

Vh= 25026.57 m3.

1.2.9 Overall heat transfer coefficient

The conditional heat transfer coefficient of a building, taking into account heat loss due to infiltration and ventilation, W/(m 2 ×°C), is determined by formula G.6 [SNiP 23-02-2003] taking into account the designs adopted in the project:

0.54 + 0.41 = 0.95 W/(m 2 ×°C).

1.2.10 Comparison of standardized and reduced heat transfer resistances

The results of the calculations are compared in table. 2 standardized and reduced heat transfer resistances.

Table 2 - Standardized Rreg and given R r o heat transfer resistance of building enclosures

1.2.11 Protection against waterlogging of enclosing structures

The temperature of the inner surface of the enclosing structures must be greater than the dew point temperature td=11.6 o C (3 o C for windows).

Temperature of the internal surface of enclosing structures τ int, is calculated using formula Ya.2.6 [SP 23-101]:

τ int = t int-(t int-t ext)/(R r× α int),

for building walls:

τ int=20-(20+26)/(3.37×8.7)=19.4 o C > td=11.6 o C;

for covering technical floor:

τ int=2-(2+26)/(4.35×8.7)=1.3 o C<td=1.5 o C, (φ=75%);

for windows:

τ int=20-(20+26)/(0.56×8.0)=9.9 o C > td=3 o C.

The temperature of condensation on the internal surface of the structure was determined by I-d humid air diagram.

The temperatures of internal structural surfaces satisfy the conditions for preventing moisture condensation, with the exception of the technical floor ceiling structures.

1.2.12 Space-planning characteristics of the building

The space-planning characteristics of the building are established in accordance with SNiP 23-02.

Glazing coefficient of building facades f:

f = A F /A W + F = 1002,24 / 5992 = 0,17

Building compactness indicator, 1/m:

8056.9 / 25026.57 = 0.32 m -1.

1.3.3 Thermal energy consumption for heating the building

Thermal energy consumption for heating the building during the heating period Q h y, MJ, determined by formula G.2 [SNiP 23 - 02]:

0.8 – coefficient of heat gain reduction due to thermal inertia of enclosing structures (recommended);

1.11 – coefficient that takes into account the additional heat consumption of the heating system associated with the discreteness of the nominal heat flow of the range of heating devices, their additional heat loss through the behind-the-radiator sections of the fences, increased air temperature in corner rooms, heat loss of pipelines passing through unheated rooms.

General heat loss of the building Q h, MJ, for the heating period are determined by formula G.3 [SNiP 23 - 02]:

Q h= 0.0864×0.95×4858.5×8056.9 = 3212976 MJ.

Household heat gains during the heating season Q int, MJ, are determined by formula G.10 [SNiP 23 - 02]:

Where q int= 10 W/m2 – the amount of household heat generation per 1 m2 of residential area or the estimated area of ​​a public building.

Q int= 0.0864×10×205×3940= 697853 MJ.

Heat gain through windows from solar radiation during the heating season Q s, MJ, are determined by formula 3.10 [TSN 23 - 329 - 2002]:

Q s =τ F ×k F ×(A F 1 ×I 1 +A F 2 ×I 2 +A F 3 ×I 3 +A F 4 ×I 4)+τ scy× k scy ×A scy ×I hor ,

Q s = 0.76×0.78×(425.25×587+25.15×1339+486×1176+66×1176)= 552756 MJ.

Q h y= ×1.11 = 2,566917 MJ.

1.3.4 Estimated specific heat energy consumption

The estimated specific consumption of thermal energy for heating a building during the heating period, kJ/(m 2 × o C × day), is determined by the formula
D.1:

10 3 × 2 566917 /(7258 × 4858.5) = 72.8 kJ/(m 2 × o S×day)

According to table. 3.6 b [TSN 23 – 329 – 2002] normalized specific heat energy consumption for heating a nine-story residential building is 80 kJ/(m 2 × o S×day) or 29 kJ/(m 3 × o S×day).

CONCLUSION

In the project of a 9-story residential building, special techniques were used to increase the energy efficiency of the building, such as:

¾ a constructive solution has been applied that allows not only to implement fast construction object, but also to use various structural elements in the external enclosing structure - insulation materials And architectural forms at the request of the customer and taking into account the existing capabilities of the construction industry of the region,

¾ the project includes thermal insulation of heating and hot water supply pipelines,

¾ modern ones are used thermal insulation materials, in particular, polystyrene concrete D200, GOST R 51263-99,

¾ in modern translucent designs of heat-insulating windows, double-glazed windows are used, and for the manufacture of window frames and sashes, mainly PVC profiles or combinations thereof. When manufacturing double-glazed windows using float glass, the windows provide a calculated reduced heat transfer resistance of 0.56 W/(m×oC).

The energy efficiency of the designed residential building is determined by the following main criteria:

¾ specific consumption of thermal energy for heating during the heating period q h des,kJ/(m 2 ×°C×day) [kJ/(m 3 ×°C×day)];

¾ indicator of building compactness k e,1m;

¾ glazing coefficient of the building facade f.

As a result of the calculations, the following conclusions can be drawn:

1. The enclosing structures of a 9-story residential building comply with the requirements of SNiP 23-02 for energy efficiency.

2. The building is designed to support optimal temperatures and air humidity while ensuring the lowest energy consumption costs.

3. Calculated building compactness index k e= 0.32 is equal to the normative one.

4. The glazing coefficient of the building façade f=0.17 is close to the standard value f=0.18.

5. The degree of reduction in thermal energy consumption for heating the building from normative value amounted to minus 9%. This value parameter matches normal class of thermal energy efficiency of the building according to Table 3 SNiP 02/23/2003 Thermal protection of buildings.

ENERGY PASSPORT OF THE BUILDING