Question

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.

Short answer

Answer: The total thermal resistance of the wall is 30.316 hours · °F / Btu, and the R-value of insulation for the wall is also 30.316 hours · °F / Btu.

Step-by-step solution

Calculate Thermal Resistance for Sheetrock Layers

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: \(2\times R_\text{sheetrock} = 2\times 0.583 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu} = 1.166 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\)

Calculate Thermal Resistance for Fiberglass Insulation

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: \(R_\text{fiberglass} = \frac{L}{k} = \frac{0.583\text{ ft}}{0.020 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}}\) \(R_\text{fiberglass} = 29.15 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\)

Calculate Total Thermal Resistance

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 \(30.316 \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\).

Calculate R-value of Insulation

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

Key concepts

Heat Transfer

Heat transfer is an essential concept when studying how thermal energy moves from one place to another. It describes how heat flows through various materials and layers in environments like buildings or walls.
Heat always travels from a warmer area to a cooler one. The rate of this transfer depends on the temperature difference between the two areas and the specific materials involved.
Consider a wall made from different layers like sheetrock and fiberglass. Heat will move through these materials at different rates. This transfer affects how well the wall insulates a building.
There are three main types of heat transfer:

• Conduction: This is the direct transfer of heat through a solid material. It happens when molecules in a substance collide and pass on their kinetic energy.
• Convection: This involves the movement of heat via a fluid (liquid or gas) moving across a surface.
• Radiation: The transfer of energy via electromagnetic waves, which doesn't need a medium to travel through.

In the case of the wall, conduction is the primary method of heat transfer as heat moves through the solid sheetrock and fiberglass insulation materials.

Thermal Conductivity

Thermal conductivity is a measure of how well a material can conduct heat. In other words, it tells us how easily heat can flow through a material.
Materials with high thermal conductivity, like metals, easily allow heat to pass through them. Conversely, materials with low thermal conductivity, like fiberglass or sheetrock, resist the flow of heat, making them good insulators.
The units used to measure thermal conductivity are typically \(\mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). This unit shows us how much heat, in British Thermal Units (Btu), passes through a foot of the material per hour, given a temperature difference of one degree Fahrenheit.

• Sheetrock used in the wall has a thermal conductivity of \(0.10 \, \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\).
• Fiberglass insulation has a lower thermal conductivity of \(0.020 \, \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\).

The lower the thermal conductivity, the better the material is at insulating, making it ideal for keeping buildings warmer in winter and cooler in summer.

R-value

The term 'R-value' is commonly used in construction and engineering to describe a material's resistance to heat flow. It is a measure of a material’s insulating ability.
The higher the R-value, the better the material is at insulating. It's crucial in determining how effective a wall or a roof is at preventing heat from escaping or entering a space.
Calculating the R-value involves adding up the resistances of all layers in a wall or ceiling setup. For instance, in our wall example:

• The R-value of a single layer of the sheetrock is calculated by dividing the thickness of the layer by its thermal conductivity.
• The same approach is applied to the fiberglass insulation, which tends to have a much higher resistance compared to sheetrock.

Combining these values gives the total R-value of the wall. For our example, the R-value of the wall is \(30.316 \, \mathrm{h} \cdot{ }^{\circ} \mathrm{F} / \mathrm{Btu}\), as it includes both the insulating fiberglass and the double layer of sheetrock.