Masonry Magazine June 1975 Page. 13
high heat transfer and the other a low heat transfer, or the paths involve large percentages of the total wall with small variations in the transfer coefficients for the paths.
Thermal bridges built into a wall may increase heat transfer substantially above the calculated amount if the bridge is ignored. Thermal bridges occur in several types of walls. Three examples of these are shown. Different methods are used in calculating the Uave for metallic and non-metallic bridges. Examples of both are shown.
Calculate U, at path B as follows:
Material
R
Outside air surface
0.17
4-in. face brick
0.44
1-in. air space
= 0.97
½-in. sheathing (fiberboard)
= 1.32
3½-in. batt insulation
= 11.00
½-in. drywall
= 0.45
Inside air surface
0.68
Rt
= 15.03
U1/Rbt 1/15.03 0.067
Calculate Uave as follows:
Uavga(U)+b(U)
= 0.094(0.119) + 0.906(0.067)= 0.072
If the thermal bridge at the studs were ignored, the U value would be the same as U, or 0.067.
This calculation reveals that, if the thermal bridge formed by the stud is considered, the Uave exceeds the U of the wall having the insulation (path B) by approximately 7½ percent. It is common practice for wood frame walls to calculate the U values for the insulation path by the series method and then multiply this value by 1.08 to obtain the Uavg for the wall.
Most masonry walls have parallel paths of heat flow which result from bonding the separate wythes together. This may be by masonry bonders or metal ties. However, for conventional constructions, the effect of the bonders is not significant, because of the relatively small area of the metal ties per square foot of wall, and the slight differences in conductivity or conductance of masonry units.
However, if masonry-bonded cavity walls with insulation in the cavity or walls with a large amount of headers are being considered, the parallel path method of calculation should be used. This is illustrated by the calculated U values of the brick and brick cavity wall, shown in Fig. 2.
The brick veneer-frame wall shown in Fig. I has thermal bridges which occur at the wood studs. The average U value of this wall is calculated by the parallel path method, by first calculating the U values in series of the two paths involved. Using the heat transmission coefficients for the various materials found in Table 1, the calculation is as follows:
Consider the path at the wood stud as path A and the path at the insulation as path B. Consider an area of wall 16 in. wide and 12 in. high; total area 1.33 x 1.0 = 1.33 sq ft. The area of path A 12 x 1.5/144 0.125 sq ft. Thus, a = 0.125/1.33 0.094, and b = (1.33 0.125)/1.33 0.906, both given as square feet per square foot of wall.
Calculate Un at path A as follows:
Material
R
Outside air surface
0.17
4-in. face brick
= 0.44
1-in. air space
0.97
½-in sheathing (fiberboard)
1.32
3½-in. wood stud
4.38
½-in. drywall
0.45
Inside air surface
0.68
Rat
8.41
Ua1/Rat 1/8.41 = 0.119