Coursework on the subject: "Architecture"

External Wall Heat Engineering Calculation

It is required to determine the thickness of the insulation in a three-layer brick outer wall in a residential building located in Volgograd. Wall construction: inner layer - brickwork from ordinary clay bricks 250 mm thick and 1800 kg/m3 density, outer layer - brickwork from facing bricks 120 mm thick and 1800 kg/m3 density; between the outer and inner layers there is an effective insulation of polystyrene foam with a density of 40 kg/m3; outer and inner layers are connected to each other by glass-plastic flexible bonds with diameter of 8 mm, located with pitch of 0.6 m.

1. Source Data

- Purpose of the building - residential building

- Construction district - Volgograd

- Design internal air temperature tint = plus 20 0С

- Design ambient air temperature text = minus 35 0С

- Design internal air humidity - 55%

2. Determination of normalized heat transfer resistance

It is determined according to Table 4 [1] depending on the degree of heating period. Degrees of the heating period, Dd, ° C × day, are determined by formula 1, based on the average temperature of the outside air and the duration of the heating period.

According to SNiP 230199 * [3], we determine that in Volgograd the average outside temperature of the heating period is: tht = 2.4 0С, the duration of the heating period zht = 177 days. The value of the degree of the heating period is:

Dd = (tint - tht) zht = (20 + 2.4) × 177 = 3964.8 0C day.

According to Table 4. [1] normalized heat transfer resistance Rreg of external walls for residential buildings corresponding to the value Dd = 3964.8 0С day is Rreg = a Dd + b = 0,00035 × 3964.8 + 1.4 = 2.79 m2 0С/W .

3. Selecting an Exterior Wall Design Solution

The structural solution of the outer wall is proposed in the specification and is a three-layer fence with an internal layer of brickwork with a thickness of 250 mm, an external layer of brickwork with a thickness of 120 mm, between the external and internal layer there is an insulation from polystyrene foam. The outer and inner layers are connected to each other by flexible connections made of fiberglass with a diameter of 8 mm, located with a pitch of 0.6 m.

4. Determination of insulation thickness

Thickness of insulation is determined by formula 7:

dut = (Rreg ./r - 1/aint - dkk/lkk - 1/aext) × lut

where is Rreg. - normalized heat transfer resistance, m2 0С/W; r - coefficient of thermal uniformity; aint - heat transfer coefficient of the inner surface, W/( m2 × ° С); atext - heat transfer coefficient of the outer surface, W/( m2 × ° С); dkk - thickness of brickwork, m; lkk - design coefficient of thermal conductivity of brickwork, W/( m × ° С); lut - design thermal conductivity coefficient of the insulation, W/( m × ° С).

Normalized heat transfer resistance is defined as Rreg = 2.79 m2 0C/W.

The coefficient of thermal uniformity for a brick three-layer wall with fiberglass flexible bonds is about r = 0.995, and may not be taken into account in the calculations (for information, if steel flexible bonds are used, then the coefficient of thermal uniformity can reach 0.60.7).

The heat transfer coefficient of the inner surface is determined according to Table 7 [1] aint = 8.7 W/( m2 × ° C).

The heat transfer coefficient of the outer surface is taken according to Table 8 [2] aext = 23 W/( m2 × ° С).

The total thickness of the brickwork is 370 mm or 0.37 m.

The design thermal conductivity factors of the materials used are determined depending on the operating conditions (A or B). Operating conditions are defined in the following sequence:

as per Table 1 [1] we determine the humidity mode of the rooms: since the design temperature of the internal air is + 20 0С, the design humidity is 55%, the humidity mode of the rooms is normal ;

according to Appendix B (map of the Russian Federation) [1] we determine that Volgograd is located in the dry zone;

according to Table 2 [1], depending on the humidity zone and humidity mode of the premises, we determine that the operating conditions of the enclosing structures are A.

According to Appendix D [2], we determine thermal conductivity coefficients for operating conditions A: for polystyrene foam GOST 1558886 with a density of 40 kg/m3 lut = 0.041 W/( m × ° С); for brickwork made of clay ordinary brick on cement sand mortar with a density of 1800 kg/m3 lkk = 0.7 W/( m × ° C).

We substitute all certain values ​ ​ into formula 7 and calculate the minimum thickness of the insulation from polystyrene foam:

dut = (2.79 - 1/8.7 - 0.37/0.7 - 1/23) × 0.041 = 0.0862 m

We round the obtained value upwards to the accuracy of 0.01 m: dut = 0.09 m. We perform a check calculation according to the formula 5:

R0 = (1/ai + dkk/lkk + dut/lut + 1/ae)

R0 = (1/8.7 + 0.37/0.7 + 0.09/0.041 + 1/23) = 2.88 m2 0C/W

5. Limitation of temperature and moisture condensation on the inner surface of the enclosing structure

The design temperature difference Δto, ° C between the internal air temperature and the internal surface temperature of the enclosing structure shall not exceed the rated values Δtn, ° C set in Table 5 [1] and shall be determined as follows:

Δto = n (tint - text )/( R0 aint) = 1 (20 + 35 )/( 2.88 x 8.7) = 2.2 0C i.e. less than Δtn, = 4.0 0 C determined from table 5 [1].

Conclusion: the thickness of the polystyrene foam insulation in the three-layer brick wall is 90 mm. In this case, the heat transfer resistance of the outer wall R0 = 2.88 m2 0C/W, which is more than the normalized heat transfer resistance Rreg. = 2.79 m2 0C/W by 0.09m2 0C/W. Design temperature difference Δto, ° С, between the internal air temperature and the internal surface temperature of the enclosing structure does not exceed the standard value Δtn,.

Heat engineering calculation of translucent enclosing structures

Translucent enclosing structures (windows) are selected according to the following procedure.

Rated heat transfer resistance Rreg is determined according to Table 4 of SNiP 23022003 [1] (column 6) depending on the degree of heating period Dd. At the same time, the building type and Dd are accepted as in the previous example of heat engineering calculation of light-transparent enclosing structures. In our case, Dd = 3964.8 0C day, then for the window of a residential building Rreg = a Dd + b = 0,00005 × 3964.8 + 0.3 = 0,498 m2 0C/W.

The choice of translucent structures is made according to the value of the reduced heat transfer resistance Ror obtained as a result of certification tests or according to Annex L of the Code [2]. If the reduced heat transfer resistance of the selected translucent structure Ror is greater than or equal to Rreg, then this design meets the requirements of the standards.

Conclusion: for a residential building in Volgograd, we accept windows in PVC bindings with two-chamber double-glazed windows in a single binding made of ordinary glass (with an inter-glass distance of 8 mm) in which Ror = 0.5 m2 0C/W is greater than Rreg = 0.498 m2 0C/W.

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