A wall is constructed of two layers of \(0.7\)-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.

We need to find the total thermal resistance caused by the two sheetrock layers. Given the thickness of each sheetrock layer is 0.7 inches and the thermal conductivity is \(0.10 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). Since there are two layers, we will calculate the resistance for one layer and then multiply by 2 to get the total resistance of the sheetrock layers. Let's first convert inches to feet to be consistent with the units: \(0.7 \text{ inch} = 0.7/12 = 0.0583 \text{ ft}\) Now we can calculate the thermal resistance of one layer: \(R_\text{sheetrock} = \frac{L}{k} = \frac{0.0583\text{ ft}}{0.10 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}}\) \(R_\text{sheetrock} = 0.583 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\) Total thermal resistance for both sheetrock layers: \(2\times R_\text{sheetrock} = 2\times 0.583 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu} = 1.166 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\)

Now we will find the thermal resistance caused by the fiberglass insulation. The thickness of the insulation is 7 inches and the thermal conductivity is \(0.020 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). Let's first convert inches to feet: \(7\text{ inches} = 7/12 = 0.583\text{ ft}\) Now we can calculate the thermal resistance of the fiberglass insulation: \(R_\text{fiberglass} = \frac{L}{k} = \frac{0.583\text{ ft}}{0.020 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}}\) \(R_\text{fiberglass} = 29.15 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\)

Now we will find the total thermal resistance of the wall by summing the thermal resistance of the sheetrock layers and the fiberglass insulation: \(R_\text{total} = R_\text{sheetrock\ layers} + R_\text{fiberglass} = 1.166 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu} + 29.15 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\) \(R_\text{total} = 30.316 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\) Answer (a): The total thermal resistance of the wall is \(30.316 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\).

Since the R-value is just the total thermal resistance, then: Answer (b): The R-value of insulation for the wall is \(30.316 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\).