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Consider a compound slab consisting of two different materials having equal lengths, thicknesses and thermal conductivities $K$ and $2 K$ respectively. The equivalent thermal conductivity of the slab is
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Verified Answer
The correct answer is:
$\frac{4}{3} K$
Equivalent thermal conductivity of the compound slab,
$\begin{aligned}
K_{\mathrm{eq}} & =\frac{l_1+l_2}{\frac{l_1}{K_1}+\frac{l_2}{K_2}}=\frac{l+l}{\frac{l}{K}+\frac{l}{2 K}} \\
& =\frac{2 l}{\frac{3 l}{2 K}}=\frac{4}{3} K
\end{aligned}$
$\begin{aligned}
K_{\mathrm{eq}} & =\frac{l_1+l_2}{\frac{l_1}{K_1}+\frac{l_2}{K_2}}=\frac{l+l}{\frac{l}{K}+\frac{l}{2 K}} \\
& =\frac{2 l}{\frac{3 l}{2 K}}=\frac{4}{3} K
\end{aligned}$
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