Question

The thermal conductivity of fiberglass batting, which is 4.0 in thick, is $8.0\cdot {10}^{-6}\mathrm{BTU}/\left({\mathrm{ft}}^{\circ }\mathrm{F}\mathrm{s}\right).$ What is the $R$ value (in ${\mathrm{ft}}^2{}^{\circ }\mathrm{F}\mathrm{h}/\mathrm{BTU})?$

Short answer

Answer: The R-value of the fiberglass batting is $1.8\cdot {10}^5{\mathrm{ft}}^2{}^{\circ }\mathrm{F}\mathrm{h}/\mathrm{BTU})$.

Step-by-step solution

Write down the given data

The thermal conductivity (k) = $8.0\cdot {10}^{-6}\mathrm{BTU}/({\mathrm{ft}}^{\circ }\mathrm{F}\mathrm{s})$ and the thickness (d) = 4.0 in

Convert the thickness from inches to feet

We have the thickness in inches, but we need it in feet to match the units of the thermal conductivity. There are 12 inches in a foot, so: d = 4.0 in * (1 ft / 12 in) = $\frac{1}{3}\mathrm{ft}$

Calculate the R-value

The R-value is calculated as the ratio of the thickness (d) to the thermal conductivity (k): R-value = d / k = $\frac{(\frac{1}{3}\mathrm{ft})}{(8.0\cdot {10}^{-6}\mathrm{BTU}/({\mathrm{ft}}^{\circ }\mathrm{F}\mathrm{s}))}$

Simplify and convert the units

Simplify the expression and convert the units from seconds to hours: R-value = $\frac{\frac{1}{3}\mathrm{ft}}{8.0\cdot {10}^{-6}\mathrm{BTU}/({\mathrm{ft}}^{\circ }\mathrm{F}\mathrm{s})}\cdot \frac{3600\mathrm{s}}{1\mathrm{h}}=\frac{1}{2\cdot {10}^{-5}}\cdot \frac{3600}{1}{\mathrm{ft}}^2{}^{\circ }\mathrm{F}\mathrm{h}/\mathrm{BTU})$ R-value = $1.8\cdot {10}^5{\mathrm{ft}}^2{}^{\circ }\mathrm{F}\mathrm{h}/\mathrm{BTU})$ The R-value of the fiberglass batting is $1.8\cdot {10}^5{\mathrm{ft}}^2{}^{\circ }\mathrm{F}\mathrm{h}/\mathrm{BTU})$.

Key concepts

Thermal Conductivity

Understanding thermal conductivity is essential when examining materials that insulate or conduct heat. It refers to a material's ability to transfer heat through its mass. Measured in units like BTU/(ft²°F h), it quantifies the rate at which heat passes through a material.

When a material has high thermal conductivity, it means heat will pass through it quickly—like metal pans on a stove dispersing heat fast. Conversely, materials with low thermal conductivity, such as fiberglass used in insulation, do not transfer heat well, making them ideal for keeping energy within a space.

Calculations typically involve the thermal conductivity constant 'k', and understanding this property is crucial for engineers and architects when selecting materials to ensure proper insulation and energy efficiency in buildings.

Insulation Materials

Insulation materials play a pivotal role in energy conservation in buildings. These materials are chosen based on their ability to resist heat flow, which is quantified by the R-value.

Materials like fiberglass, cellulose, and foam have low thermal conductivities, making them excellent insulators. The thickness of these materials also affects their insulating abilities - thicker layers usually mean better insulation. The R-value is a cumulative indicator of an insulation material's effectiveness and factors in thickness and thermal conductivity.

Choosing the right insulation material is not only about thermal efficiency but also involves considerations related to moisture permeability, fire resistance, and durability.

Unit Conversion

In the context of R-value calculations, unit conversion is a necessary step to ensure compatibility of measurements. As seen in the problem solution, converting the thickness from inches to feet was important because the thermal conductivity was given in units of BTU/(ft²°F h).

Unit conversion frequently involves multiplying or dividing by conversion factors. For example, converting seconds to hours requires multiplying by 3600 seconds per hour, as in the exercise. Understanding unit conversion can help avoid costly mistakes in calculations and allows professionals in various fields to communicate effectively using standard units of measure.

Students should practice unit conversion to gain fluency and avoid errors in scientific and engineering contexts.