Question

Determine the winter \(R\)-value and the \(U\)-factor of a masonry cavity wall that is built around 4-in-thick concrete blocks made of lightweight aggregate. The outside is finished with 4 -in face brick with \(\frac{1}{2}\)-in cement mortar between the bricks and concrete blocks. The inside finish consists of \(\frac{1}{2}\)-in gypsum wallboard separated from the concrete block by \(\frac{3}{4}\)-in-thick (1-in by 3 -in nominal) vertical furring whose center- to-center distance is \(16 \mathrm{in}\). Neither side of the \(\frac{3}{4}\) in- thick airspace between the concrete block and the gypsum board is coated with any reflective film. When determining the \(R\)-value of the airspace, the temperature difference across it can be taken to be \(30^{\circ} \mathrm{F}\) with a mean air temperature of \(50^{\circ} \mathrm{F}\). The airspace constitutes 80 percent of the heat transmission area, while the vertical furring and similar structures constitute 20 percent.

Short answer

Answer: The winter R-value of the masonry cavity wall is 47.70, and the U-factor is 0.021 Btu / (hr-sq ft-F).

Step-by-step solution

Determine the R-values of the individual layers

To determine the R-values of each layer in the wall assembly, we can use the following formula: R = thickness (in) / thermal conductivity (Btu-in / (hr-sq ft-F)) The thermal conductivity values for the materials are as follows: Face Brick: 0.15 Btu-in / (hr-sq ft-F) Cement Mortar: 1.40 Btu-in / (hr-sq ft-F) Lightweight Aggregate Concrete Blocks: 0.28 Btu-in / (hr-sq ft-F) Airspace: Using the given ΔT (30°F) and mean temperature (50°F), we need to first determine the effective emittance using tables (typical value: 0.82) and then use the convective and radiation formula for airspaces to find R: R = 1 / [hr × (ε × 0.17136)] Gypsum Wallboard: 0.45 Btu-in / (hr-sq ft-F)

Calculate the R-values for each layer in the wall

Face Brick R-value: R_brick = 4 / 0.15 = 26.67 Cement Mortar R-value: R_mortar = 0.5 / 1.40 = 0.36 Lightweight Aggregate Concrete Blocks R-value: R_aggregate = 4 / 0.28 = 14.29 Airspace R-value: Using the effective emittance value, we find R_air = 1 / (0.82 × 0.17136) = 7.15 Gypsum Wallboard R-value: R_gypsum = 0.5 / 0.45 = 1.11

Combine the R-values of the layers

We know that the airspace contributes to 80% of the wall's heat transmission area, and the remaining area is made up of vertical furring and other structures. We find the R-values for these two areas separately: R_wall_air = R_brick + R_mortar + R_aggregate + R_air + R_gypsum = 26.67 + 0.36 + 14.29 + 7.15 + 1.11 = 49.58 R_wall_other = R_brick + R_mortar + R_aggregate + R_gypsum = 26.67 + 0.36 + 14.29 + 1.11 = 42.43 Now we take the weighted average of the R-values, considering the 80% contribution of the airspace: R_wall = 0.8 × R_wall_air + 0.2 × R_wall_other = 0.8 × 49.58 + 0.2 × 42.43 = 47.70

Calculate the U-factor

The U-factor is the inverse of the R-value: U_factor = 1 / R_wall = 1 / 47.70 = 0.021 Btu / (hr-sq ft-F) The winter R-value of the masonry cavity wall is 47.70, and the U-factor is 0.021 Btu / (hr-sq ft-F).