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

There 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