Question

A composite slab is prepared by pasting two plates of thicknesses $L_{1}$ and $L_{2}$ and thermal conductivities $K_{1}$ and $K_{2}$. The slabs have equal cross-sectional area. Find the equivalent conductivity of the slab.

• A. $K_{e q}=\frac{L_{1}+L_{2}}{\frac{L_{1}}{K_{1}}+\frac{L_{2}}{K_{2}}}$
• B. $K_{e q}=\frac{L_{1}+L_{2}}{K_{1} K_{2}}$
• C. $K_{e q}=\frac{L_{1}}{K_{1}+K_{2}}+\frac{L_{2}}{K_{1}+K_{2}}$
• D. $K_{e q}=\frac{L_{1} L_{2}}{L_{1}+L_{2}}$

Answer 1

Answer: (A)

$R_{e q} =R_{1}+R_{2} $
$ \Rightarrow \frac{L_{1}+L_{2}}{K_{e q} A} =\frac{L_{1}}{K_{1} A}+\frac{L_{2}}{K_{2} A} $
$\Rightarrow K_{e q} =\frac{L_{1}+L_{2}}{\frac{L_{1}}{K_{1}}+\frac{L_{2}}{K_{2}}} $