Question

Determine the thermal conductivity of a piece of building material 0.25 in0.25 \mathrm{~in}. thick that has an RR value of 1.6 ft2°F/Btu/h1.6 \mathrm{~ft}^2{ }{\degree} \mathrm{F} / \mathrm{Btu} / \mathrm{h}.

Solution

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Answered 2 years ago
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Given\textbf{Given}

L=0.25 in=0.02083 ftL=0.25 \text{ in}= 0.02083 \text{ ft}

R=1.6 ft2 h BtuR=1.6 \text{ } \dfrac{ \text{ft}^2 \text { h }^\circ \text{F }}{\text{Btu}}

The R value is defined as

R=LKR= \dfrac{L}{K}

where L is the thickness of the material and K is the thermal conductivity of the material

Reorganizing the formula we express K:

K=LRK= \dfrac{L}{R}

By including the given data we get:

K=0.020831.6K=0.013  Btu ftF h\begin{align*} K&=\dfrac{0.02083}{1.6}\\ &\boxed{K= 0.013 \text{ } \dfrac{\text{ Btu}}{\text{ ft} ^\circ \text{F h} } } \end{align*}

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