Heat transfer resistance of doors and gates. Entrance metal doors with thermal break

The required total heat transfer resistance for external doors (except balcony doors) must be at least 0.6
for the walls of buildings and structures, determined at the estimated winter temperature of the outside air, equal to the average temperature of the coldest five-day period with a probability of 0.92.

We accept the actual total heat transfer resistance of external doors
=
, then the actual heat transfer resistance of external doors is
, (m 2 ·С)/W,

, (18)

where t in, t n, n, Δt n, α in – the same as in equation (1).

The heat transfer coefficient of external doors k door, W/(m 2 ·С), is calculated using the equation:

.

Example 6. Thermal engineering calculation of external fences

Initial data.

Residential building, t = 20С .

Values ​​of thermal characteristics and coefficients tхп(0.92) = -29С (Appendix A);

α in = 8.7 W/(m 2 ·С) (Table 8); Δt n = 4С (Table 6).

Calculation procedure.

Determine the actual heat transfer resistance outer door
according to equation (18):

(m 2 ·С)/W.

The heat transfer coefficient of the external door k dv is determined by the formula:

W/(m 2 ·С).

2 Calculation of the heat resistance of external fences during the warm period

External fencing is checked for heat resistance in areas with an average monthly air temperature in July of 21°C and above. It has been established that fluctuations in the outside air temperature A t n, C, occur cyclically, obey the sinusoidal law (Figure 6) and, in turn, cause fluctuations in the actual temperature by inner surface fencing
, which also flow harmoniously according to the law of a sinusoid (Figure 7).

Thermal resistance is the property of a fence to maintain a relative constant temperature on the inner surface τ in, С, with fluctuations in external thermal influences
, С, and provide comfortable conditions indoors. As you move away from the outer surface, the amplitude of temperature fluctuations in the thickness of the fence, A τ , С, decreases, mainly in the thickness of the layer closest to the outside air. This layer with a thickness of δ pk, m, is called a layer of sharp temperature fluctuations A τ, С.

Figure 6 – Fluctuations in heat flows and temperatures on the surface of the fence

Figure 7 – Attenuation of temperature fluctuations in the fence

Thermal resistance testing is carried out for horizontal (covering) and vertical (wall) fences. First, the permissible (required) amplitude of temperature fluctuations of the internal surface is established
external fencing taking into account sanitary and hygienic requirements in the expression:

, (19)

where t nl is the average monthly outdoor temperature for July (summer month), С, .

These fluctuations occur due to fluctuations in the design temperatures of the outside air
,С, determined by the formula:

where A t n is the maximum amplitude of daily fluctuations in the outside air for July, С, ;

ρ – solar radiation absorption coefficient by the outer surface material (Table 14);

I max, I av - respectively the maximum and average values ​​of total solar radiation (direct and diffuse), W/m 3, accepted:

a) for external walls - as for vertical surfaces of western orientation;

b) for coatings - as for a horizontal surface;

α n - heat transfer coefficient of the outer surface of the fence under summer conditions, W/(m 2 ·С), equal to

where υ is the maximum of the average wind speeds for July, but not less than 1 m/s.

Table 14 – Solar radiation absorption coefficient ρ

The magnitude of actual vibrations on the inner plane
,С, will depend on the properties of the material, characterized by the values ​​of D, S, R, Y, α n and contributing to the attenuation of the amplitude of temperature fluctuations in the thickness of the fence A t. Attenuation coefficient determined by the formula:

where D is the thermal inertia of the enclosing structure, determined by the formula ΣD i = ΣR i ·S i ;

e = 2.718 – base of natural logarithm;

S 1 , S 2 , …, S n – calculated coefficients of heat absorption of the material of individual layers of the fence (Appendix A, table A.3) or table 4;

α n – heat transfer coefficient of the outer surface of the fence, W/(m 2 ·С), is determined by formula (21);

Y 1, Y 2,…, Y n is the coefficient of heat absorption of the material on the outer surface of the individual layers of the fence, determined by formulas (23 ÷ 26).

,

where δi is the thickness of individual layers of the enclosing structure, m;

λ i – thermal conductivity coefficient of individual layers of the enclosing structure, W/(m·С) (Appendix A, Table A.2).

The heat absorption coefficient of the outer surface Y, W/(m 2 ·С), of an individual layer depends on the value of its thermal inertia and is determined in the calculation, starting from the first layer from the inner surface of the room to the outer one.

If the first layer has D i ≥1, then the heat absorption coefficient of the outer surface of the layer Y 1 should be taken

Y 1 = S 1 . (23)

If the first layer has D i< 1, то коэффициент теплоусвоения наружной поверхности слоя следует определить расчетом для всех слоев ограждающей конструкции, начиная с первого слоя:

for the first layer
; (24)

for the second layer
; (25)

for nth layer
, (26)

where R 1 , R 2 ,…, R n – thermal resistance of the 1st, 2nd and nth layers of the fence, (m 2 ·С)/W, determined by the formula
;

α in – heat transfer coefficient of the inner surface of the fence, W/(m 2 ·С) (Table 8);

Based on known values And
determine the actual amplitude of temperature fluctuations of the internal surface of the enclosing structure
,C,

. (27)

The enclosing structure will meet the heat resistance requirements if the condition is met

(28)

In this case, the enclosing structure provides comfortable room conditions, protecting against the effects of external heat fluctuations. If
, then the enclosing structure is not heat-resistant, then it is necessary to use a material with a high heat absorption coefficient S, W/(m 2 ·С) for the outer layers (closer to the outside air).

Example 7. Calculation of the heat resistance of an external fence

Initial data.

Enclosing structure consisting of three layers: plaster made of cement-sand mortar with a volumetric mass γ 1 = 1800 kg/m 3, thickness δ 1 = 0.04 m, λ 1 = 0.76 W/(m·С); insulation layer made of ordinary clay brick γ 2 = 1800 kg/m 3, thickness δ 2 = 0.510 m, λ 2 = 0.76 W/(mС); facing sand-lime brickγ 3 = 1800 kg/m 3, thickness δ 3 = 0.125 m, λ 3 = 0.76 W/(m·С).

Construction area - Penza.

Estimated internal air temperature tв = 18 С .

The humidity level of the room is normal.

Operating condition – A.

Calculated values ​​of thermal characteristics and coefficients in the formulas:

t nl = 19.8С;

R 1 = 0.04/0.76 = 0.05 (m 2 °C)/W;

R 2 = 0.51/0.7 = 0.73 (m 2 °C)/W;

R 3 = 0.125/0.76 = 0.16 (m 2 °C)/W;

S 1 = 9.60 W/(m 2 °C); S 2 = 9.20 W/(m 2 °C);

S 3 = 9.77 W/(m 2 °C); (Appendix A, Table A.2);

V = 3.9 m/s;

A t n = 18.4 С;

I max = 607 W/m 2 , , I av = 174 W/m 2 ;

ρ= 0.6 (Table 14);

D = R i · S i = 0.05·9.6+0.73·9.20+0.16·9.77 = 8.75;

α in = 8.7 W/(m 2 °C) (Table 8),

Calculation procedure.

1. Determine the permissible amplitude of temperature fluctuations of the internal surface
external fencing according to equation (19):

2. Calculate the estimated amplitude of fluctuations in outside air temperature
according to formula (20):

where α n is determined by equation (21):

W/(m 2 ·С).

3. Depending on the thermal inertia of the enclosing structure D i = R i ·S i = 0.05 · 9.6 = 0.48<1, находим коэффициент теплоусвоения наружной поверхности для каждого слоя по формулам  (24 – 26):

W/(m 2 °C).

W/(m 2 °C).

W/(m 2 °C).

4. We determine the attenuation coefficient of the calculated amplitude of fluctuations of the external air V in the thickness of the fence using formula (22):

5. We calculate the actual amplitude of temperature fluctuations of the internal surface of the enclosing structure
, С.

If the condition, formula (28), is met, the structure meets the requirements of heat resistance.

The general scheme of the design procedure for thermal protection of buildings required in accordance with Scheme 1 is presented in Figure 2.1.

Where R req , R min – normalized and minimum value of heat transfer resistance, m 2 ×°C/W;

, – standard calculated specific heat energy consumption for heating buildings during the heating period, kJ/(m 2 °C day) or kJ/(m °C day).

method “b” method “a”

Change project

NO

YES

Where R int , Rext - heat transfer resistance on the inner and outer surfaces of the fence, (m 2 K)/W;

R to- thermal resistance of the layers of the enclosing structure, (m 2 × K)/W;

R pr– reduced thermal resistance of a non-uniform structure (structure with heat-conducting inclusions), (m 2 K)/W;

a int, a ext – heat transfer coefficients on the inner and outer surfaces of the fence, W/(m 2 K), are taken according to the table. 7 and table. 8;

d i– thickness of the layer of the enclosing structure, m;

l i– coefficient of thermal conductivity of the layer material, W/(m 2 K).

Since the thermal conductivity of materials largely depends on their humidity, their operating conditions are determined. According to Appendix “B”, the humidity zone is established on the territory of the country, then according to Table. 2, depending on the humidity regime of the room and the humidity zone, the operating conditions of the enclosing structure A or B are determined. If the humidity regime of the room is not specified, then it is allowed to accept it as normal. Then, according to Appendix “D”, depending on the established operating conditions (A or B), the thermal conductivity coefficient of the material is determined (see Appendix “E”).

If the fence includes structures with inhomogeneous inclusions (floor panels with air gaps, large blocks with heat-conducting inclusions, etc.), then the calculation of such structures is carried out using special methods. These methods are presented in appendices “M”, “N”, “P”. In the course project, such structures are the floor panels of the first floor and the ceiling of the last floor; their reduced thermal resistance is determined as follows.

A). By planes parallel to the heat flow, the panel is divided into sections of homogeneous and heterogeneous composition (Fig. 2.2, A). Areas of the same composition and size are assigned the same number. The total resistance of the floor panel will be equal to the average resistance. Due to their size, the sections have a unequal effect on the overall resistance of the structure. Therefore, the thermal resistance of the panel is calculated taking into account the areas occupied by the sections in the horizontal plane, using the formula:

Where l reinforced concrete – coefficient of thermal conductivity of reinforced concrete, taken depending on operating conditions A or B;

R a . g.─ thermal resistance of a closed air layer, taken according to table. 7 at a positive air temperature in the interlayer, (m 2 K)/W.

But the obtained thermal resistance of the floor panel does not coincide with the data of the laboratory experiment, so the second part of the calculation is performed.

B). By planes perpendicular to the direction of heat flow, the structure is also divided into homogeneous and inhomogeneous layers, which are usually denoted by capital letters of the Russian alphabet (Fig. 2.2, b). The total thermal resistance of the panel in this case is:

where is the thermal resistance of layers “A”, (m 2 K)/W;

RB– thermal resistance of layer “B”, (m 2 K)/W.

When calculating R B it is necessary to take into account the varying degrees of influence of areas on the thermal resistance of the layer due to their sizes:

The calculations can be averaged as follows: the calculations in both cases do not coincide with the laboratory experiment data, which are closer to the value R 2 .

The calculation of the floor panel must be done twice: for the case when the heat flow is directed from bottom to top (ceiling) and from top to bottom (floor).

The heat transfer resistance of external doors can be taken according to table. 2.3, windows and balcony doors - according to table. 2.2 of this manual

Using table A11, we determine the thermal resistance of external and internal doors: R nd = 0.21 (m 2 0 C)/W, therefore we accept double external doors; R ind1 = 0.34 (m 2 0 C)/W, R ind2 = 0.27 (m 2 0 C)/W.

Then, using formula (6), we determine the heat transfer coefficient of external and internal doors:

W/m 2 o C

W/m 2 o C

2 Calculation of heat losses

Heat losses are conventionally divided into basic and additional.

Heat losses through internal enclosing structures between rooms are calculated if the temperature difference on both sides is >3 0 C.

The main heat losses of premises, W, are determined by the formula:

where F is the estimated area of ​​the fence, m2.

Heat losses, according to formula (9), are rounded to 10 W. The temperature t in corner rooms is taken to be 2 0 C higher than the standard one. We calculate heat losses for external walls (NS) and internal walls (BC), partitions (PR), ceilings above the basement (PL), triple windows (TO), double external doors (DD), internal doors (ID), attic floors(PT).

When calculating heat losses through the floors above the basement, the temperature of the coldest five-day period with a probability of 0.92 is taken as the outside air temperature tn.

Additional heat losses include heat losses that depend on the orientation of the premises in relation to the cardinal directions, from wind blowing, from the design of external doors, etc.

The addition for the orientation of enclosing structures to the cardinal points is taken in the amount of 10% of the main heat losses if the fence is facing east (E), north (N), northeast (NE) and northwest (NW) and 5% - if west (W) and southeast (SE). Addition for heating the cold air rushing in through the external doors at a building height N, m, we take 0.27 N from the main heat losses outer wall.

Heat consumption for heating the supply ventilation air, W, is determined by the formula:

where L p – flow rate supply air, m 3 / h, for living rooms we take 3 m 3 / h per 1 m 2 of living space and kitchen area;

 n – density of outside air equal to 1.43 kg/m3;

c – specific heat capacity equal to 1 kJ/(kg 0 C).

Household heat emissions complement the heat output of heating devices and are calculated using the formula:

, (11)

where F p is the floor area of ​​the heated room, m 2.

The total (total) heat loss of a building Q floor is defined as the sum of heat losses from all rooms, including staircases.

Then we calculate the specific thermal characteristic of the building, W/(m 3 0 C), using the formula:

, (13)

where  is a coefficient taking into account the influence of local climatic conditions(for Belarus
);

V building – volume of the building, taken according to external measurements, m 3.

Room 101 – kitchen; t in =17+2 0 C.

We calculate heat loss through the outer wall with a northwest orientation (C):

external wall area F= 12.3 m2;

temperature difference t= 41 0 C;

coefficient taking into account the position of the outer surface of the enclosing structure in relation to the outside air, n=1;

heat transfer coefficient taking into account window openings k = 1.5 W/(m 2 0 C).

The main heat losses of the premises, W, are determined by formula (9):

Additional heat loss for orientation is 10% of Q main and is equal to:

W

Heat consumption for heating the supply ventilation air, W, is determined by formula (10):

Household heat emissions were determined using formula (11):

The heat consumption for heating the supply ventilation air Q veins and household heat emissions Q household remain the same.

For triple glazing: F = 1.99 m 2, t = 44 0 C, n = 1, heat transfer coefficient K = 1.82 W/m 2 0 C, it follows that the main heat loss of the window Q main = 175 W, and additional Q ext = 15.9 W. Heat loss of the outer wall (B) Q main = 474.4 W, and additional Q add = 47.7 W. Floor heat loss is: Q pl. =149 W.

We sum up the obtained values ​​of Q i and find the total heat loss for this room: Q = 1710 W. Similarly, we find heat loss for other rooms. The calculation results are entered into Table 2.1.

Continuation of Table 2.1

Continuation of Table 2.1

Continuation of Table 2.1

After the calculation, it is necessary to calculate the specific thermal characteristics of the building:

,

where α-coefficient, taking into account the influence of local climatic conditions (for Belarus - α≈1.06);

V building – volume of the building, taken according to external measurements, m 3

We compare the resulting specific thermal characteristic using the formula:

,

where H is the height of the building being calculated.

If the calculated value of the thermal characteristic deviates from the standard value by more than 20%, it is necessary to find out the reasons for this deviation.

,

Because <then we accept that our calculations are correct.

1.4 Heat transfer resistance of external doors and gates

For external doors, the required heat transfer resistance R o tr must be at least 0.6 R o tr of the walls of buildings and structures, determined by formulas (1) and (2).

0.6R o tr =0.6*0.57=0.3 m²·ºС/W.

Based on the accepted designs of external and internal doors according to Table A.12, their thermal resistances are accepted.

External wooden doors and double gates 0.43 m²·ºС/W.

Single internal doors 0.34 m²·ºС/W

1.5 Heat transfer resistance of light opening fillings

For the selected type of glazing, according to Appendix A, the value of thermal resistance to heat transfer of light openings is determined.

In this case, the heat transfer resistance of the fillings of external light openings R approx must be no less than the standard heat transfer resistance

determined according to table 5.1, and not less than the required resistance

R= 0.39, determined according to table 5.6

Heat transfer resistance of the fillings of light openings, based on the difference in the calculated temperatures of internal air t in (table A.3) and external air t n and using table A.10 (t n is the temperature of the coldest five-day period).

Rt= t in -(- t n)=18-(-29)=47 m²·ºС/W

R ok = 0.55 -

for triple glazing in wooden split-pair sashes.

When the ratio of the glazing area to the filling area of ​​the light opening in wooden frames is equal to 0.6 - 0.74, the specified value of R ok should be increased by 10%

R=0.55∙1.1=0.605 m 2 Cº/W.

1.6 Heat transfer resistance of internal walls and partitions

Section 13. - tee for passage 1 pc. z = 1.2; - outlet 2 pcs. z = 0.8; Section 14 - branch 1 pc. z = 0.8; - valve 1 pc. z = 4.5; The local resistance coefficients of the remaining sections of the heating system of a residential building and garage are determined similarly. 1.4.4. General provisions for designing a garage heating system. System...

Thermal protection of buildings. SNiP 3.05.01-85* Internal sanitary systems. GOST 30494-96 Residential and public buildings. Room microclimate parameters. GOST 21.205-93 SPDS. Symbols for elements of sanitary systems. 2. Determination of the thermal power of the heating system The building envelope is represented by external walls, the ceiling above the upper floor...

... ; m3; W/m3 ∙ °С. The condition must be met. The standard value is taken from Table 4 depending on. The value of the standardized specific thermal characteristics for a civil building (tourist base). Since 0.16< 0,35, следовательно, условие выполняется. 3 РАСЧЕТ ПОВЕРХНОСТИ НАГРЕВАТЕЛЬНЫХ ПРИБОРОВ Для поддержания в помещении требуемой температуры необходимо, ...

Designer. Internal sanitary and technical installations: at 3 o’clock – Ch 1 Heating; edited by I. G. Staroverov, Yu. I. Schiller. – M: Stoyizdat, 1990 – 344 p. 8. Lavrentieva V. M., Bocharnikova O. V. Heating and ventilation of a residential building: MU. – Novosibirsk: NGASU, 2005. – 40 p. 9. Eremkin A.I., Koroleva T.I. Thermal regime of buildings: Textbook. – M.: ASV Publishing House, 2000. – 369 p. ...