Walls

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 walls

U= 5.5 W/m K U= 2.7 W/m2 K U= 1.8 W/m2 K

R-l (ft2 °F h/Btu) R-2 R-3

The following conditions are assumed in calculating the design R-values:

1. Equilibrium or steady-state heat transfer, disregarding effects of heat storage

2. Surrounding surfaces at ambient air temperature

Example 5.4

Calculate the U-factor of the 38 mm by 90 mm stud wall. The studs are at 400 mm on center.

9

WallsThere is 90-mm mineral fiber batt insulation (R = 2.30 K m /W) in the stud space. The inside finish is 13-mm gypsum wallboard; the outside is finished with rigid foam insulating sheathing (R = 0.70 K m /W) and 13-mm by 200-mm wood bevel lapped siding. The insulated cavity occupies approximately 75% of the transmission area and the stud 25%.

1. Outside surface

2. Wood bevel lapped siding

3. Sheathing

4. Mineral fiber batt insulation

5. Wood stud

6. Gypsum wallboard

7. Inside surface

1 2 3 4 5

Solution:

6 7

Element

R R (Insulated cavity) (Studs)

1. Outside surface, 3.4 m/s wind

0.03

0.03

2. Wood bevel lapped siding

0.14

0.14

3. Rigid foam insulating sheathing

0.70

0.70

4. Mineral fiber batt insulation

2.30

5. Wood stud, 38 mm by 90 mm

0.63

6. Gypsum wallboard, 13 mm

0.10

0.10

7. Inside surface, still air

0.12

0.12

R, = 3.39

R2= 1.72

O

• Thermal bridges

Thermal conductivity of metals is a thousand times higher than that of insulation material. The less loss through the metal conduction is considerable.

Example 5.5

• 2 *

The thermal conductance of a 200 mm thick masonry wall is 0.4 W/m K. If there is a (J) =20 mm aluminum bar through the wall and thermal conductivity is 220 W/m K, compare the heat loss through 1 m of the masonry wall and the aluminum bar.

Solution

Posted in Fundamentals of Heating. Ventilating, and Air-Conditioning


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