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The coefficient of reduction of air volume in the building. Building Energy Performance Calculations

Thermal engineering calculation of the technical underground

Thermal engineering calculations of enclosing structures

The areas of external enclosing structures, the heated area and volume of the building required for the calculation of the energy passport, and the thermal performance of the enclosing structures of the building 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 the layers, as well as the physical properties of 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 exterior walls in a residential building.

The first type is brickwork with floor support 120 mm thick, insulated with polystyrene concrete 280 mm thick, with a facing layer of silicate brick. The second type is a 200 mm reinforced concrete panel, insulated with polystyrene concrete 280 mm thick, with a facing layer of silicate brick. The third type, see Fig.1. Thermal engineering calculation is given for two types of walls, respectively.

one). The composition of the layers of the outer wall of the building: protective coating - cement-lime mortar 30 mm thick, λ = 0.84 W / (m × o C). The outer layer of 120 mm is made of silicate brick M 100 with a 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 С); inner layer 120 mm - from silicate brick, M 100, λ = 0.76 W / (m × o C). The internal walls are plastered with lime-sand mortar M 75, 15 mm thick, λ=0.84 W/(m×o C).

Rw\u003d 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 \u003d 4.26 m 2 × o C / W.

Resistance to heat transfer of the walls of the building, with the area of the facades
A w\u003d 4989.6 m 2, equal to: 4.26 m 2 × about C / W.

Coefficient of thermal engineering uniformity of external walls r, determined by formula 12 SP 23-101:

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

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

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

k i = 1.01 for thermally conductive inclusion at ratios λm /λ= 2.3 and a/b= 0,23.

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

2). The composition of the layers of the outer wall of the building: protective coating - cement-lime mortar M 75 with a thickness of 30 mm, λ = 0.84 W / (m × o C). The outer layer of 120 mm is made of silicate brick M 100 with a 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:

Rw= 1/8,7+0,030/0,84+0,120/0,76+0,280/0,075+
+0, 20 / 2.04 + 1/23 \u003d 4.2 m 2 × o C / W.

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

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

Building type - an ordinary section of a 9-storey residential building with a lower piping of heating and hot water systems.

A b\u003d 342 m 2.

floor area of underground - 342 m 2.

External wall area above ground level A b , w\u003d 60.5 m 2.

Estimated temperature of the heating system of the lower wiring is 95 °С, hot water supply is 60 °С. The length of the pipelines of the heating system with the lower wiring is 80 m. The length of the hot water supply pipelines was 30 m. there is no underground, so the rate of air exchange in those. underground I= 0.5 h -1 .

t int= 20 °С.

Ground floor area (above technical underground) - 1024.95 m2.

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

l\u003d 17.6 + 2 × 1.6 \u003d 20.8 m.

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

Resistance to heat transfer of the outer walls of those. undergrounds above ground level are accepted in accordance with SP 23-101 clause 9.3.2. equal to the heat transfer resistance of the outer walls R.o.b. w\u003d 3.03 m 2 × ° C / W.

The reduced resistance to heat transfer of the enclosing structures of the buried part of those. undergrounds will be determined in accordance with SP 23-101 clause 9.3.3. as for non-insulated floors on the ground in the case when the floor and wall materials have design coefficients of thermal conductivity λ≥ 1.2 W / (m o C). Reduced resistance to heat transfer of fences of those. undergrounds buried in the ground is determined according to table 13 of SP 23-101 and amounted to R o rs\u003d 4.52 m 2 × ° C / W.

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

Determine the air temperature in those. underground t int b

For the calculation, we use the data in Table 12 [SP 23-101]. At the air temperature in those underground 2 °С, the heat flux density from 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 pipelines of the heating system - by the coefficient [(95 - 2)/( 95 - 18)] 1.283 = 1.41; for hot water 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) \u003d 1316/473 \u003d 2.78 ° С.

The heat flux through the basement was

q b . c\u003d (20 - 2.78) / 1.55 \u003d 11.1 W / m 2.

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

1.2.3 Overlapping over those. underground

The fence has an area A f\u003d 1024.95 m 2.

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\u003d 32 kg / m 3, λ \u003d 0.029 W / (m × o C), 60 mm thick according to GOST 16381. Air gap, λ \u003d 0.005 W / (m × o C), 10 mm thick. 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 \u003d 4.35 m 2 × o C / W.

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

R o = nR req,

where n- coefficient determined at the accepted minimum air temperature in the underground t int b= 2°С.

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

Then R with\u003d 0.39 × 4.35 \u003d 1.74 m 2 × ° C / W.

Let's check whether the thermal protection of the ceiling above the technical underground satisfies the requirement of the standard difference D t n= 2 °C for the floor of the first floor.

According to the formula (3) SNiP 23 - 02, we determine the minimum allowable resistance to heat transfer

R o min =(20 - 2) / (2 × 8.7) \u003d 1.03 m 2 × ° C / W< R c = 1.74 m 2 × ° C / W.

1.2.4 Attic floor

Cover area A c\u003d 1024.95 m 2.

Reinforced concrete floor slab, 220 mm thick, λ =
2.04 W / (m × o C). Insulation minplita CJSC "Mineral wool", r =140-
175 kg / m 3, λ \u003d 0.046 W / (m × o C), 200 mm thick according to GOST 4640. From above, the coating has a cement-sand screed 40 mm thick, λ = 0.84 W / (m × o C).

Then the heat transfer resistance is:

Rc\u003d 1 / 8.7 + 0.22 / 2.04 + 0.200 / 0.046 + 0.04 / 0.84 + 1/23 \u003d 4.66 m 2 × o C / W.

1.2.5 Attic roofing

Reinforced concrete floor slab, 220 mm thick, λ =
2.04 W / (m × o C). Expanded clay gravel insulation, r\u003d 600 kg / m 3, λ \u003d
0.190 W / (m × o C), 150 mm thick according to GOST 9757; min-slab of CJSC "Mineralnaya vata", 140-175 kg/m3, λ = 0.046 W/(m×оС), 120 mm thick according to GOST 4640. The top coating has a cement-sand screed 40 mm thick, λ = 0.84 W/ (m × o C).

Then the heat transfer resistance is:

Rc\u003d 1 / 8.7 + 0.22 / 2.04 + 0.150 / 0.190 + 0.12 / 0.046 + 0.04 / 0.84 + 1/17 \u003d 3.37 m 2 × o C / W.

1.2.6 Windows

In modern translucent designs of heat-shielding windows, double-glazed windows are used, and for the manufacture of window frames and sashes, mainly PVC profiles or their combinations. In the manufacture of double-glazed windows using float glass, the windows provide a calculated reduced heat transfer resistance of not more than 0.56 m 2 × o C / W., which meets the regulatory requirements for their certification.

Area of window openings A F\u003d 1002.24 m 2.

Heat transfer window accept R F\u003d 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 × ° С), 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 \u003d 0.54 W / (m 2 × °C).

1.2.8 Conditional heat transfer coefficient

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

where with– specific heat capacity of air, equal to 1 kJ/(kg×°С);

β ν - coefficient of reduction of air volume 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 building air exchange rate for the heating period is calculated from the total air exchange due to ventilation and infiltration according to 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 infiltrating air, kg/h, entering the building through the building envelope during the day of the heating period, is determined by formula D.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 pressure difference between the outside and inside 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 replacement of 0.55 by 0 in it, 28 and with the calculation of the specific gravity according to the formula 14 [SNiP 23-02-2003] at the corresponding air temperature, Pa.

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

where Η \u003d 30.4 m - the height of the building;

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

γ ext \u003d 3463 / (273-26) \u003d 14.02 N / m 3,

γint \u003d 3463 / (273 + 21) \u003d 11.78 N / m 3.

∆p 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.

- the average density of the supply air for the heating period, kg / m 3, ,

353 / \u003d 1.31 kg / m 3.

V h\u003d 25026.57 m 3.

1.2.9 Overall heat transfer coefficient

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

0.54 + 0.41 \u003d 0.95 W / (m 2 × ° C).

1.2.10 Comparison of standardized and reduced heat transfer resistances

As a result of the calculations are compared in table. 2 normalized and reduced heat transfer resistances.

Table 2 - Normalized Rreg and given R r o resistance to heat transfer of building fences

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 t d\u003d 11.6 ° C (3 ° C - for windows).

The temperature of the inner surface of the enclosing structures τ int, is calculated by the formula Ya.2.6 [SP 23-101]:

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

for building walls:

τ int\u003d 20-(20 + 26) / (3.37 × 8.7) \u003d 19.4 o C\u003e t d\u003d 11.6 about C;

to cover the technical floor:

τ int\u003d 2-(2 + 26) / (4.35 × 8.7) \u003d 1.3 o C<t d\u003d 1.5 about C, (φ \u003d 75%);

for windows:

τ int\u003d 20-(20 + 26) / (0.56 × 8.0) \u003d 9.9 about C\u003e t d\u003d 3 about C.

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

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

1.2.12 Space-planning characteristics of the building

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

Building façade glazing coefficient f:

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

Building compactness index, 1/m:

8056.9 / 25026.57 \u003d 0.32 m -1.

1.3.3 Thermal energy consumption for heating the building

Consumption of thermal energy for heating the building during the heating period Q h y, MJ, determined by formula D.2 [SNiP 23 - 02]:

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

1.11 - coefficient taking 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 radiator sections of the fences, the increased air temperature in the corner rooms, the heat losses of pipelines passing through unheated rooms.

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

Qh= 0.0864×0.95×4858.5×8056.9 = 3212976 MJ.

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

where q int\u003d 10 W / m 2 - the amount of household heat emissions per 1 m 2 of the area of \u200b\u200bresidential premises or the estimated area of a public building.

Q int= 0.0864×10×205×3940= 697853 MJ.

Heat gains through windows from solar radiation during the heating period Qs, 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 consumption

The estimated specific consumption of thermal energy for heating the 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 C × day)

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

CONCLUSION

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

¾ a constructive solution was applied that allows not only to carry out the rapid construction of the facility, but also to use various structural and insulating materials and architectural forms in the external enclosing structure at the request of the customer and taking into account the existing capabilities of the construction industry of the region,

¾ in the project, thermal insulation of heating and hot water pipelines is carried out,

¾ modern heat-insulating materials were used, in particular, polystyrene concrete D200, GOST R 51263-99,

¾ in modern translucent designs of heat-shielding windows, double-glazed windows are used, and for the manufacture of window frames and sashes, mainly PVC profiles or their combinations. In the manufacture of 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 building compactness index k e,1m;

¾ coefficient of glazing of the facade of the building f.

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

1. Enclosing structures of a 9-storey residential building comply with the requirements of SNiP 23-02 for energy efficiency.

2. The building is designed to maintain optimum temperature and humidity while ensuring the lowest energy consumption.

3. Calculated indicator of building compactness k e= 0.32 is equal to the standard.

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

5. The degree of reduction in the consumption of thermal energy for heating the building from the standard value was minus 9%. This parameter value corresponds to normal class of heat and power efficiency of the building according to Table 3 of SNiP 23-02-2003 Thermal protection of buildings.

ENERGY PASSPORT OF THE BUILDING

(determination of the thickness of the attic insulation layer

coverings and coverings)
A. Initial data

The humidity zone is normal.

z ht = 229 days.

Average design temperature of the heating period t ht \u003d -5.9 ºС.

The temperature of the cold five-day t ext \u003d -35 ° С.

t int \u003d + 21 ° С.

Relative humidity: = 55%.

Estimated air temperature in the attic t int g \u003d +15 С.

Heat transfer coefficient of the inner surface of the attic floor
\u003d 8.7 W / m 2 С.

Heat transfer coefficient of the outer surface of the attic floor
\u003d 12 W / m 2 · ° С.

Heat transfer coefficient of the inner surface of the warm attic coating
\u003d 9.9 W / m 2 · ° С.

The heat transfer coefficient of the outer surface of the warm attic coating
\u003d 23 W / m 2 · ° С.
Building type - 9-storey residential building. The kitchens in the apartments are equipped with gas stoves. The height of the attic space is 2.0 m. Covering areas (roofs) BUT g. c \u003d 367.0 m 2, warm attic floors BUT g. f \u003d 367.0 m 2, outer walls of the attic BUT g. w \u003d 108.2 m 2.

In a warm attic there is an upper wiring of pipes for heating and water supply systems. Estimated temperatures of the heating system - 95 °С, hot water supply - 60 °С.

The diameter of heating pipes is 50 mm with a length of 55 m, hot water 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.

№	Material name (designs)	, kg / m 3	δ, m	,W/(m °С)	R, m 2 ° С / W
1	Rigid mineral wool slabs on bituminous 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 coating over the warm attic consists of the structural layers shown in the table.

№	Material name (designs)	, kg / m 3	δ, m	,W/(m °С)	R, m 2 ° С / W
1	Technoelast	600	0,006	0,17	0,035
2	Cement-sand mortar	1800	0,02	0,93	0,022
3	Aerated concrete slabs	300	X	0,13	X
4	Ruberoid	600	0,005	0,17	0,029
5	reinforced concrete slab	2500	0,035	2,04	0,017

B. Calculation procedure
Determination of degree-days of the heating period according to the 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 resistance to heat transfer of the coating of a residential building according to the formula (1) SNiP 23-02-2003:

R req= a· D d+ b\u003d 0.0005 6160.1 + 2.2 \u003d 5.28 m 2 C / W;
According to the formula (29) SP 23-101–2004, we determine the required heat transfer resistance of the warm attic floor
, m 2 ° С / W:

,
where
- normalized resistance to heat transfer of the coating;

n- coefficient determined by the formula (30) SP 230101-2004,
(21 – 15)/(21 + 35) = 0,107.
According to the found values
and n determine
:
\u003d 5.28 0.107 \u003d 0.56 m 2 С / W.

Required coating resistance over a warm attic R 0g. c is determined by 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 +
+ (
)/BUT g.f - ( – t ext) a g.w/ R 0 g.w
where G ven - reduced (related to 1 m 2 of the attic) air flow in the ventilation system, determined according to table. 6 SP 23-101-2004 and equal to 19.5 kg / (m 2 h);

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

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

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

The reduced heat gains from pipelines of heating and hot water supply systems are:
()/BUT g.f \u003d (25 55 + 12 30) / 367 \u003d 4.71 W / m 2;
a g. w - reduced area of the outer walls of the attic m 2 / m 2, determined by the formula (33) SP 23-101-2004,

= 108,2/367 = 0,295;

- normalized resistance to heat transfer of the outer walls of a warm attic, determined through a degree-day of the heating period at an internal air temperature in the attic room = +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 over the warm attic:
(15 + 35) / (0.28 19.2 (22.5 - 15) + (21 - 15) / 0.56 + 4.71 -
- (15 + 35) 0.295 / 3.08 \u003d 50 / 50.94 \u003d 0.98 m 2 ° C / W

We determine the thickness of the insulation in the attic floor at R 0g. f \u003d 0.56 m 2 ° C / W:

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

R 0g. f fact. \u003d 1 / 8.7 + 0.04 / 0.08 + 0.029 + 0.142 + 1/12 \u003d 0.869 m 2 ° C / W.
Determine the amount of insulation in the coating at R 0g. c \u003d \u003d 0.98 m 2 ° C / W:
= (R 0g. c – 1/ – R f.b - R rub - R c.p.r - R t – 1/) ut =
\u003d (0.98 - 1 / 9.9 - 0.017 - 0.029 - 0.022 - 0.035 - 1/23) 0.13 \u003d 0.0953 m,
we accept the thickness of the insulation (aerated concrete slab) 100 mm, then the actual value of the resistance to heat transfer of the attic coating will be almost equal to the calculated one.
B. Checking compliance with sanitary and hygienic requirements

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

\u003d (21 - 15) / (0.869 8.7) \u003d 0.79 ° С,
According to Table. 5 SNiP 23-02–2003 ∆ t n = 3 °C, therefore, the condition ∆ t g = 0.79 °С t n =3 °С is fulfilled.
We check the outer enclosing structures of the attic for the conditions of non-condensation on their inner surfaces, i.e. to fulfill the condition
:

- for covering over a warm attic, taking
W / m 2 ° С,
15 - [(15 + 35)/(0.98 9.9] =
\u003d 15 - 4.12 \u003d 10.85 ° С;
- for the outer walls of a warm attic, taking
W / m 2 ° С,
15 - [(15 + 35)]/(3.08 8.7) =
\u003d 15 - 1.49 \u003d 13.5 ° С.
II. Calculate the dew point temperature t d, °С, in the attic:

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

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

By application 8 by value E= e g find the dew point temperature t d = 3.05 °С.

The obtained values of the dew point temperature are compared with the corresponding values
and
:
=13,5 > t d = 3.05 °С; = 10.88 > t d = 3.05 °С.
The dew point temperature is much lower than the corresponding temperatures on the inner surfaces of the outer fences, therefore, condensate will not fall on the inner surfaces of the coating and on the walls of the attic.

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

Example5
Calculation of the specific consumption of thermal energy for heating a 9-storey one-section residential building (tower type)
The dimensions of a typical floor of a 9-storey residential building are given in the figure.

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

A. Initial data
Place of construction - Perm.

Climatic region - IV.

The humidity zone is normal.

The humidity regime of the room is normal.

Operating conditions of enclosing structures - B.

The length of the heating period z ht = 229 days.

Average temperature of the heating period t ht \u003d -5.9 ° С.

Indoor air temperature t int \u003d +21 ° С.

The temperature of the cold five-day outdoor air t ext = = -35 °С.

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

The temperature of the internal air of the technical basement = = +2 °C

The 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 rhumb wind speeds for January v\u003d 5.2 m / s.
B. Calculation procedure
1. Determination of the areas of enclosing structures.

The determination of the area of enclosing structures is based on the plan of a typical floor of a 9-storey building and the initial data of Section A.

Total floor area of the building
BUT h \u003d (42.5 + 42.5 + 42.5 + 57.38) 9 \u003d 1663.9 m 2.
Living area of apartments and kitchens
BUT l = (27,76 + 27,76 + 27,76 + 42,54 + 7,12 + 7,12 +
+ 7,12 + 7,12)9 \u003d 1388.7 m 2.
Floor area above technical basement BUT b .c, attic floor BUT g. f and coverings over the attic BUT g. c
BUT b .c = BUT g. f= BUT g. c \u003d 16 16.2 \u003d 259.2 m 2.
Total area of window fillings and balcony doors BUT 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.

Windows height - 1.2 m; the height of the balcony doors is 2.2 m.
BUT F \u003d [(1.5 6 + 1.2 8) 1.2 + (0.75 4 2.2)] 9 \u003d 260.3 m 2.
The area of the entrance doors to the staircase with their width of 1.0 and 1.5 m and height of 2.05 m
BUT ed \u003d (1.5 + 1.0) 2.05 \u003d 5.12 m 2.
The area of the window fillings of the staircase with a window width of 1.2 m and a height of 0.9 m

\u003d (1.2 0.9) 8 \u003d 8.64 m 2.
The total area of the outer doors of the apartments with their width of 0.9 m, height of 2.05 m and the number of 4 pcs per floor.
BUT ed \u003d (0.9 2.05 4) 9 \u003d 66.42 m 2.
The total area of the outer walls of the building, taking into account window and door openings

\u003d (16 + 16 + 16.2 + 16.2) 2.8 9 \u003d 1622.88 m 2.
The total area of the outer walls of the building without window and door openings

BUT W \u003d 1622.88 - (260.28 + 8.64 + 5.12) \u003d 1348.84 m 2.
The total area of the internal surfaces of the external enclosing structures, including the attic floor and the floor above the technical basement,

\u003d (16 + 16 + 16.2 + 16.2) 2.8 9 + 259.2 + 259.2 \u003d 2141.3 m 2.
Heated volume of the building

V n \u003d 16 16.2 2.8 9 \u003d 6531.84 m 3.
2. Determination of degree-days of the heating period.

Degree days are determined by the formula (2) SNiP 23-02-2003 for the following building envelopes:

- external walls and attic floor:

D d 1 \u003d (21 + 5.9) 229 \u003d 6160.1 ° C day,
- coatings and external walls of a warm "attic":
D d 2 \u003d (15 + 5.9) 229 \u003d 4786.1 ° C day,
- floors above the technical basement:
D d 3 \u003d (2 + 5.9) 229 \u003d 1809.1 ° C day.
3. Determination of the required resistance to heat transfer of enclosing structures.

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

- for the outer walls of the building
\u003d 0.00035 6160.1 + 1.4 \u003d 3.56 m 2 ° C / W;
- for attic flooring
= n· \u003d 0.107 (0.0005 6160.1 + 2.2) \u003d 0.49 m 2,
n =
=
= 0,107;
- for the outer walls of the attic
\u003d 0.00035 4786.1 + 1.4 \u003d 3.07 m 2 ° C / W,
- for covering over the attic

=
=
\u003d 0.87 m 2 ° C / W;
– for overlapping over a technical basement

= n b. c R reg \u003d 0.34 (0.00045 1809.1 + 1.9) \u003d 0.92 m 2 ° C / W,

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

\u003d 0.55 m 2 ° C / W.
4. Determination of the consumption of thermal energy for heating the building.

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

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

- household heat inputs Q int , MJ;

- heat gains through windows and balcony doors from solar radiation, MJ.

When determining the total heat loss of a building Q h , MJ, it is necessary to calculate two coefficients:

- the reduced coefficient of heat transfer through the external building envelope
, W / (m 2 ° С);
L v = 3 A l\u003d 3 1388.7 \u003d 4166.1 m 3 / h,
where A l- the area of \u200b\u200bliving premises and kitchens, m 2;

- the determined average rate of air exchange of the building for the heating period n a , h –1 , according to formula (D.8) SNiP 23-02–2003:
n a =
= 0.75 h -1.
We accept the coefficient for reducing the volume of air 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 for taking into account the influence of the oncoming heat flow in translucent structures k = 0,7:

=
\u003d 0.45 W / (m 2 ° C).
The value of the building's total heat transfer coefficient K m, W / (m 2 ° С), determined by the formula (D.4) SNiP 23-02–2003:
K m \u003d 0.59 + 0.45 \u003d 1.04 W / (m 2 ° C).
We calculate the total heat loss of the building for 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 inputs during the heating season Q int , MJ, determined by the formula (D.11) SNiP 23-02-2003, assuming the value of specific household heat emissions q int equal to 17 W / m 2:
Q int = 0.0864 17 229 1132.4 = 380888.62 MJ.
Heat input to the building from solar radiation during the heating period Q s , MJ, determined by the formula (G.11) SNiP 23-02-2003, taking the values of the coefficients that take 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 for the heating period on vertical surfaces I cf, W / m 2, we accept according to Appendix (D) SP 23-101–2004 for the geographical latitude of the location of Perm (56 ° N):

I av \u003d 201 W / m 2,
Q s = 0.5 0.76(100.44 201 + 100.44 201 +
+ 29.7 201 + 29.7 201) = 19880.18 MJ.
Consumption of thermal energy for heating the building during the heating period , MJ, is determined by the formula (D.2) of SNiP 23-02-2003, taking the numerical value of the following coefficients:

- coefficient of heat gain reduction 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 flux of the range of heating devices for tower-type buildings = 1,11.
= 1.11 = 1024940.2 MJ.
We set the specific consumption of thermal energy of the building
, kJ / (m 2 °C day), according to the formula (D.1) SNiP 23-02–2003:
=
\u003d 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-storey 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 ·°С·day), therefore, in the heat engineering design of enclosing structures, this difference must be taken into account.

MINISTRY OF EDUCATION AND SCIENCE OF THE RUSSIAN FEDERATION

Federal State Budgetary Educational Institution of Higher Professional Education

"State University - educational-scientific-industrial complex"

Institute of Architecture and Construction

Department: "Urban construction and economy"

Discipline: "Construction physics"

COURSE WORK

"Thermal protection of buildings"

Completed by student: Arkharova K.Yu.

Introduction
Task form
1 . Climate reference
2 . Thermal engineering 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 the specific consumption of thermal energy for heating during the heating period
4 . Heat absorption of the floor surface
5 . Protection of the enclosing structure from waterlogging
Conclusion
List of used sources and literature
Annex A

Introduction

Thermal protection is a set of measures and technologies for energy saving, which makes it possible to increase the thermal insulation of buildings for various purposes, to reduce heat losses in premises.

The task of providing the necessary thermal properties of external enclosing structures is solved by giving them the required heat resistance and resistance to heat transfer.

The resistance to heat transfer 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 heat resistance of structures is assessed by their ability to maintain a relatively constant temperature in the premises with periodic fluctuations in the temperature of the air environment adjacent to the structures and the flow of heat passing through them. The degree of heat resistance of the 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 perceives sharp temperature fluctuations.

In this course work, a thermal calculation of the enclosing structure of a residential individual house will be performed, the construction area of \u200b\u200bwhich is the city of Arkhangelsk.

Task form

1 Construction area:

Arkhangelsk.

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

1st layer - polystyrene concrete modified on Portland slag cement (= 200 kg / m 3; ? = 0.07 W / (m * K); ? = 0.36 m)

2nd layer - extruded polystyrene foam (= 32 kg / m 3; ? = 0.031 W / (m * K); ? = 0.22 m)

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

3 Thermally conductive inclusion material:

pearl concrete (= 600 kg / m 3; ? = 0.23 W / (m * K); ? = 0.38 m

4 Floor construction:

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 - expanded polystyrene plates (= 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 reference

Building area - Arkhangelsk.

Climatic region - II A.

Humidity zone - wet.

Humidity in the room? = 55%;

design temperature in the room = 21°С.

The humidity regime of the room is normal.

Operating conditions - B.

Climatic parameters:

Estimated outdoor temperature (Outdoor temperature of the coldest five-day period (security 0.92)

The duration of the heating period (with an average daily outdoor temperature? 8 ° C) - \u003d 250 days;

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

enclosing heat absorption heating

2 . Thermal engineering 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, - design temperature in the room, ° С;

Estimated outdoor temperature, °С;

Duration of the heating period, days

GSOP \u003d (+ 21 + 4.5) 250 \u003d 6125 ° C day

The required resistance to heat transfer is calculated by the 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 respective groups of buildings.

We accept: a = 0.00035; b=1.4

0.00035 6125 +1.4=3.54m 2 °C/W.

Outer wall construction

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

Figure 1 - Construction of the outer wall

Table 1 - Material parameters of the outer wall

The heat transfer resistance R and is determined by the formula (1.3):

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

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

A is the sum of the areas of all plots, m 2.

The heat transfer resistance for homogeneous sections is determined by the formula (1.4):

where, ? - layer thickness, m;

Thermal conductivity coefficient, W/(mK)

We calculate the heat transfer resistance for inhomogeneous sections using formula (1.5):

R \u003d R 1 + R 2 + R 3 + ... + R n + R VP, (1.5)

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

R vp - resistance to heat transfer of the air gap, .

We find R and according to the formula (1.3):

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

Figure 2 - Construction of the outer wall

The heat transfer resistance R b is determined by the formula (1.5)

R b \u003d R 1 + R 2 + R 3 + ... + R n + R VP, (1.5)

The resistance to air penetration for homogeneous sections is determined by the formula (1.4).

The resistance to air penetration for inhomogeneous areas is determined by the formula (1.3):

We find R b according to the formula (1.5):

R b \u003d 5.14 + 3.09 + 1.4 \u003d 9.63.

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

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

R b - resistance to heat transfer of the building envelope, cut perpendicular to the heat flow,.

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

Resistance to heat transfer on the outer surface, is determined by the formula (1.9)

where, heat transfer coefficient of the inner surface of the building envelope, = 8.7;

where, is the heat transfer coefficient of the outer surface of the building envelope, = 23;

The calculated temperature difference between the temperature of the internal air and the temperature of the inner surface of the enclosing structure is determined by the 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;

design temperature in the room, °С;

estimated outdoor air temperature during the cold season, °С;

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

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

2 . 2 Calculation of enclosing structures of "warm" basements

The required heat transfer resistance of the part of the basement wall located above the planning mark of the soil is taken equal to the reduced heat transfer resistance of the outer wall:

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

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

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

The required resistance to heat transfer of the basement over the "warm" basement is calculated by the formula (1.12)

where, the required resistance to heat transfer of the basement floor, we find according to table 3 of SP 50.13330.2012 "Thermal protection of buildings".

where, air temperature in the basement, °С;

the same as in formula (1.10);

the same as in formula (1.10)

Let's take equal to 21.35 ° С:

The air temperature in the basement is determined by the formula (1.14):

where, the same as in formula (1.10);

Linear heat flux density,; ;

The volume of air in the basement, ;

The length of the pipeline of the i-th diameter, m; ;

The rate of air exchange in the basement; ;

The density of the air 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 outer walls of the basement above ground level,.

2 . 3 Thermal calculation of windows

The degree-day of the heating period is calculated by the formula (1.1)

GSOP \u003d (+ 21 + 4.5) 250 \u003d 6125 ° C day.

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

We select windows based on the found resistance to heat transfer R 0:

Ordinary glass and a single-chamber double-glazed window in separate covers from glass with a hard selective coating -.

Conclusion: The reduced resistance to heat transfer, the temperature difference and the temperature of the inner surface of the enclosing structure correspond to the required standards. Consequently, the designed design of the outer wall and the thickness of the insulation are chosen correctly.

Due to the fact that we took the structure of the walls for the enclosing structures in the deep 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 the specific consumption of thermal energy for heating during the heating period

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

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

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

The consumption of thermal energy for heating the building during the heating period is calculated by the formula (2.2):

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

Household heat inputs during the heating period, J;

Heat gains through windows and lanterns from solar radiation during the heating period, J;

Coefficient of heat input reduction due to thermal inertia of enclosing structures, recommended value = 0.8;

The coefficient taking 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 radiator sections of the fences, the increased air temperature in the corner rooms, the heat losses of pipelines passing through unheated rooms, for buildings with heated basements = 1, 07;

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

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

The total area of enclosing structures, m 2;

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

The conditional heat transfer coefficient of the building, taking into account heat losses due to infiltration and ventilation, W / (m 2 ° С).

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

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

The same, fillings of light apertures (windows, stained-glass windows, lanterns);

The same, external doors and gates;

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

the same, attic floors;

the same, basement ceilings;

too, .

0.306 W / (m 2 ° C);

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

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

The volume of heated rooms;

Coefficient of taking into account the influence of a counter heat flow in translucent structures, equal to windows and balcony doors with separate bindings 1;

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

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

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

where, is the amount of supply air into the building with unorganized inflow or the normalized value with mechanical ventilation, m 3 / h, equal to residential buildings intended for citizens, taking into account the social norm (with an estimated occupancy of the apartment of 20 m 2 of total area or less per person) - 3 A; 3 A \u003d 603.93m 2;

The area of residential premises; \u003d 201.31m 2;

The number of hours of mechanical ventilation during the week, h; ;

The number of hours of accounting for infiltration during the week, h;=168;

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

The amount of air infiltrating into the stairwell of a residential building through the gaps in the filling of openings is determined by the formula (2.9):

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

respectively, for the stairwell, the required resistance to air penetration of windows and balcony doors and entrance external doors, m 2 ·°С / W;

Accordingly, for the staircase, the calculated pressure difference between the outside and inside air for windows and balcony doors and entrance external doors, Pa, determined by the formula (2.10):

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

The maximum of the average wind speeds in points for January (SP 131.13330.2012 "Construction climatology"); =3.4 m/s.

3463/(273 + t), (2.11)

n \u003d 3463 / (273 -33) \u003d 14.32 N / m 3;

c \u003d 3463 / (273 + 21) \u003d 11.78 N / m 3;

From here we find:

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

0.06041 h 1 .

Based on the data obtained, we calculate according to the 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 \u003d 0.326 W / (m 2 ° C).

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

0.08640.326317.78=J.

Household heat inputs during the heating period, J, are determined by the formula (2.12):

where, the value of household heat emissions per 1 m 2 of the area of \u200b\u200bresidential premises or the estimated area of a public building, W / m 2, is accepted;

area of residential premises; \u003d 201.31m 2;

Heat gains through windows and lanterns from solar radiation during the heating period, J, for four facades of buildings oriented in four directions, we determine by the formula (2.13):

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

Coefficient of relative penetration 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 facades of the building, respectively oriented in four directions, m 2;

The average value of solar radiation for the heating period on vertical surfaces under actual cloudiness conditions, respectively oriented along the four facades of the building, J / (m 2), is determined according to table 9.1 of SP 131.13330.2012 "Construction climatology";

Heating season:

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

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

C: A 1 \u003d 2.25 m 2; I 1 \u003d (31 + 49) / 9 \u003d 8.89 J / (m 2;

I 2 \u003d (138 + 157 + 192 + 155 + 138 + 162 + 170 + 151 + 192) / 9 \u003d 161.67 J / (m 2;

B: A 3 \u003d 8.58; I 3 \u003d (11 + 35 + 78 + 135 + 153 + 96 + 49 + 22 + 12) / 9 \u003d 66 J / (m 2;

W: A 4 \u003d 8.58; I 4 \u003d (11 + 35 + 78 + 135 + 153 + 96 + 49 + 22 + 12) / 9 \u003d 66 J / (m 2.

Using the data obtained in the calculation of formulas (2.3), (2.12) and (2.13) we find the consumption of thermal energy for heating the building according to the formula (2.2):

According to 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 the building does not correspond to the normalized 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 the floor surface

Thermal inertia of floor construction layers

Figure 3 - Floor plan

Table 2 - Parameters of floor materials

The thermal inertia of the layers of the floor structure is calculated by the formula (3.1):

where, s is the coefficient of heat absorption, 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 index of the floor surface sequentially by calculating the heat absorption indexes of the surfaces of the layers of the structure, starting from the 3rd to the 1st:

for the 3rd layer according to the formula (3.2)

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

W / (m 2 ° C);

The index of heat absorption of the floor surface is taken equal to the index of heat absorption 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 indicator of heat absorption of the floor surface corresponds to the normalized value.

5 . Protection of the enclosing structure from waterlogging

Climatic parameters:

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

The average partial pressure of water vapor in the outdoor air for the annual period

Figure 4 - Construction of the outer wall

Table 4 - Parameters of outer wall materials

The resistance to vapor permeability of the layers of the structure is found by the formula:

where, - layer thickness, m;

Vapor permeability coefficient, mg/(mchPa)

We determine the resistance to vapor permeability 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 the outer surface of the insulation):

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

where, is the resistance to heat transfer on the inner surface, is determined by the 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, reduced resistance to heat transfer of the outer wall, ;

calculated room temperature, .

We find the corresponding value of water vapor elasticity:

We find the average value of water vapor elasticity for a year using the formula (4.4):

where, E 1 , E 2 , E 3 - values of water vapor elasticity by season, Pa;

duration of seasons, months

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

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

relative humidity of internal air, %

The required vapor permeability resistance is found by the formula (4.6):

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

From the condition of the inadmissibility of moisture accumulation in the building envelope for the annual period of operation, we check the condition:

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

We find the average outdoor temperature for the period with negative average monthly temperatures:

The temperature value in the plane of possible condensation is determined by the formula (4.3):

This temperature corresponds

The required vapor permeability resistance is determined by the formula (4.7):

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

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

wetted layer thickness, m;

maximum allowable moisture increment in the material of the moistened layer, % by weight, for the period of moisture accumulation, taken according to Table 10 of SP 50.13330.2012 "Thermal protection of buildings"; accept for expanded polystyrene \u003d 25%;

coefficient determined by formula (4.8):

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

the same as in formula (4.7)

From here we consider according to the formula (4.7):

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

Conclusion: in connection with the fulfillment of the condition for limiting the amount of moisture in the building envelope during the period of moisture accumulation, an additional vapor barrier device is not required.

Conclusion

From the heat engineering qualities of the external fences of buildings depend: a favorable microclimate of buildings, that is, ensuring the temperature and humidity of the air in the room are 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 condensate on it; humidity regime of the constructive solution of the fence, affecting its heat-shielding qualities and durability.

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

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

List used sources and literature

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

2. SP 131.13330.2012. Building 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: Tutorial [Text]. - Kazan: KGASU, 2011. - 161 p.

4. SP 23-101-2004 Design of thermal protection of buildings [Text]. - M. : FSUE TsPP, 2004.

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

Annex A

Energy passport of the building

general information

Design conditions

Name of design parameters	Parameter designation	unit of measurement	Estimated value
Estimated indoor air temperature
Estimated outdoor temperature
Estimated temperature of a warm attic
Estimated temperature of the technical underground
The length of the heating period
Average outdoor temperature during the heating period
Degree-days of the heating period

Functional purpose, type and constructive solution of the building

Geometric and thermal power indicators

	Indicator		Estimated (design) value of the indicator

Geometric indicators
	The total area of the building's external enclosing structures
	Including:

	windows and balcony doors
	stained glass windows

	entrance doors and gates
	coatings (combined)
	attic floors (cold attic)
	floors of warm attics
	ceilings over technical undergrounds

	ceilings above driveways and under bay windows
	floor on the ground
	Apartment area
	Useful area (public buildings)
	Residential area
	Estimated area (public buildings)
	Heated volume
	Building façade glazing factor
	Building compactness index
Thermal power indicators
Thermal performance
	Reduced resistance to heat transfer of external fences:	M 2 °C / W

	windows and balcony doors
	stained glass windows

	entrance doors and gates
	coatings (combined)
	attic floors (cold attics)
	floors of warm attics (including coating)
	ceilings over technical undergrounds
	ceilings over unheated basements or undergrounds
	ceilings above driveways and under bay windows
	floor on the ground
	Reduced building heat transfer coefficient	W / (m 2 ° С)
	The rate of air exchange of the building during the heating period
	Building air exchange rate during testing (at 50 Pa)
	Conditional heat transfer coefficient of the building, taking into account heat losses due to infiltration and ventilation	W / (m 2 ° С)
	Overall building heat transfer coefficient	W / (m 2 ° С)
Energy indicators
	Total heat loss through the building envelope during the heating period
	Specific household heat emissions in the building
	Household heat gains in the building during the heating period
	Heat input to the building from solar radiation during the heating period
	The need for thermal energy for heating the building during the heating period

Odds

Indicator	Indicator designation and unit of measure	Standard value of the indicator	The actual value of the indicator
Calculated coefficient of energy efficiency of the building district heating system from the heat source
Estimated coefficient of energy efficiency of apartment and autonomous heat supply systems of a building from a heat source

Coefficient for taking into account the counter heat flow
Accounting coefficient for additional heat consumption

Comprehensive indicators

Calculation of solar radiation in winter

The flux of total solar radiation coming during the heating period to horizontal and vertical surfaces under actual cloudiness conditions, kW h / m 2 (MJ / m 2)

The flux of total solar radiation coming for each month of the heating period to horizontal and vertical surfaces under actual cloudiness conditions, kW h / m 2 (MJ / m 2)

As a result of the work done, data were obtained on the intensity of the total (direct and scattered) solar radiation incident 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. Ch. 1–6. Issue. 1–34. - St. Petersburg. : Gidrometeoizdat, 1989–1998.

3. SP 23-101-2004 "Design of thermal protection of buildings". - M. : FSUE TsPP, 2004.

4. MGSN 2.01–99 “Energy saving in buildings. Standards for thermal protection and heat and water supply”. - M. : GUP "NIATs", 1999.

5. SNiP 23-01-99* "Construction climatology". - M .: Gosstroy of Russia, State Unitary Enterprise TsPP, 2003.

6. Building climatology: A reference guide to SNiP. - M .: Stroyizdat, 1990.

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