Question

A wall is constructed of two layers of \(0.6\)-in-thick sheetrock $\left(k=0.10 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)$, which is a plasterboard made of two layers of heavy paper separated by a layer of gypsum, placed 7 in apart. The space between the sheetrocks is filled with fiberglass insulation $\left(k=0.020 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)$. Determine (a) the thermal resistance of the wall and (b) its \(R\)-value of insulation in English units.

Short answer

Based on the provided solution: (a) The total thermal resistance of the wall is 30.17 ft²·°F/Btu·h. (b) The R-value of insulation in English units for this wall is (30.17 ft²·°F/Btu·h)/A, where A is the area of the wall.

Step-by-step solution

Thermal Resistance of Sheetrock Layers

To find the thermal resistance of one sheetrock layer, we can use the formula \(R_\text{sheetrock} = \frac{L_{\text{sheetrock}}}{k_\text{sheetrock}}\). The thickness of one sheetrock layer is given as 0.6 inches and the thermal conductivity is 0.10 Btu/h·ft·°F. We have: \(R_\text{sheetrock} = \frac{0.6\text{ in}}{0.10 \,\text{Btu} / \text{h} \cdot \text{ft} \cdot { }^{\circ}\text{F}}\) Next, we need to convert the thickness from inches to feet: \(0.6 \text{ in} \times \frac{1 \text{ ft}}{12 \text{ in}} = 0.05 \text{ ft}\) Now we can find the thermal resistance of one sheetrock layer: \(R_\text{sheetrock} = \frac{0.05 \text{ ft}}{0.10 \,\text{Btu} / \text{h} \cdot \text{ft} \cdot { }^{\circ}\text{F}} = 0.5 \text{ ft}^2 \cdot { }^{\circ}\text{F} / \text{Btu}·\text{h}\) Since there are two sheetrock layers, we'll multiply by 2 to get the total sheetrock thermal resistance: \(R_\text{total-sheets} = 2 \times 0.5 = 1.0 \text{ ft}^2 \cdot { }^{\circ}\text{F} / \text{Btu}·\text{h}\)

Thermal Resistance of Fiberglass Insulation

To calculate the thermal resistance of the fiberglass insulation, we use the same formula but with the given values for the insulation's thickness and thermal conductivity: \(R_\text{fiberglass} = \frac{L_{\text{fiberglass}}}{k_\text{fiberglass}}\) The space between the sheetrocks filled with fiberglass insulation is 7 inches, and its thermal conductivity is 0.020 Btu/h·ft·°F. Convert the thickness to feet: \(7 \text{ in} \times \frac{1 \text{ ft}}{12 \text{ in}} = \frac{7}{12} \text{ ft}\) The thermal resistance of the fiberglass insulation is: \(R_\text{fiberglass} = \frac{7/12 \text{ ft}}{0.020 \,\text{Btu} / \text{h} \cdot \text{ft} \cdot { }^{\circ}\text{F}} = 29.17 \text{ ft}^2 \cdot { }^{\circ}\text{F} / \text{Btu}·\text{h}\)

Total Thermal Resistance of the Wall

Now that we have the thermal resistances of the sheetrock layers and the fiberglass insulation, we can find the total thermal resistance of the wall by adding them together: \(R_\text{total} = R_\text{total-sheets} + R_\text{fiberglass} = 1.0 + 29.17 = 30.17 \text{ ft}^2 \cdot { }^{\circ}\text{F} / \text{Btu}·\text{h}\) (a) The total thermal resistance of the wall is \(30.17 \text{ ft}^2 \cdot{ }^{\circ}\text{F} / \text{Btu}·\text{h}\).

R-Value of Insulation

Now, we can calculate the R-value of the insulation, which is simply the ratio of the total thermal resistance to the area of the wall. Since we are not given a specific area in this problem, we can use the general relationship that \(R\)-value = thermal resistance/area: \((b)\,\, R\text{-value} = \frac{R_\text{total}}{A} = \frac{30.17 \text{ ft}^2 \cdot { }^{\circ}\text{F} / \text{Btu} · \text{h}}{A}\) The \(R\)-value of insulation in English units for this wall is \(\frac{30.17 \text{ ft}^2 \cdot { }^{\circ}\text{F} / \text{Btu}·\text{h}}{A}\).