Question 11.10: The R Value of a Typical Wall Goal Calculate the R value of ...
The R Value of a Typical Wall
Goal Calculate the R value of a wall consisting of several layers of insulating material.
Problem Calculate the total R value for a wall constructed as shown in Figure 11.8a. Starting outside the house (to the left in the figure) and moving inward, the wall consists of 4.0 \mathrm{in}. brick, 0.50 \mathrm{in}. sheathing, an air space 3.5 \mathrm{in}. thick, and 0.50 in. drywall.
Strategy Add all the R values together, remembering the stagnant air layers inside and outside the house.
Refer to Table 11.4, and sum. All quantities are in units of \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} \cdot \mathrm{h} / \mathrm{Btu}.
\begin{aligned} & R_{\text {total }}=R_{\text {outside air layer }}+R_{\text {brick }}+R_{\text {sheath }}+R_{\text {air space }} \\ & \quad+R_{\text {drywall }}+R_{\text {inside air layer }}=(0.17+4.00+1.32+1.01 \\ & \quad+0.45+0.17) \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} \cdot \mathrm{h} / \mathrm{Btu} \\ & R_{\text {total }}=7.12 \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} \cdot \mathrm{h} / \mathrm{Btu} \end{aligned}
TABLE 11.4
R Values for Some Common Building Materials
| Material | R value (ft2 · °F · h/Btu) |
| Hardwood siding (1.0 in. thick) | 0.91 |
| Wood shingles (lapped) | 0.87 |
| Brick (4.0 in. thick) | 4.00 |
| Concrete block (filled cores) | 1.93 |
| Styrofoam (1.0 in. thick) | 5.0 |
| Fiber glass batting (3.5 in. thick) | 10.90 |
| Fiber glass batting (6.0 in. thick) | 18.80 |
| Fiber glass board (1.0 in. thick) | 4.35 |
| Cellulose fiber (1.0 in. thick) | 3.70 |
| Flat glass (0.125 in. thick) | 0.89 |
| Insulating glass (0.25-in. space) | 1.54 |
| Vertical air space (3.5 in. thick) | 1.01 |
| Stagnant layer of air | 0.17 |
| Dry wall (0.50 in. thick) | 0.45 |
| Sheathing (0.50 in. thick) | 1.32 |