Question

A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities $K_1$ and $K_2$. The equivalent thermal conductivity of the slab is :

• A. $\frac{K_1+K_2}{K_1{K}_2}$
• B. $\frac{2K_1{K}_2}{K_1+K_2}$
• C. $\frac{K_1+K_2}{2}$
• D. $K_1+K_2$

Answer 1

Answer: (C)

At thermal equilibrium
$H=H_1+H_2$
Given that two slab are in parallel and have same dimensions.
$\frac{K\times A\times (T_2-T_1)}{L}=\frac{K_1\times \frac{A}{2}\times (T_2-T_1)}{L}+\frac{K_2\times \frac{A}{2}\times (T_2-T_1)}{L}$,
where $K$ is the effective thermal conductivity, $T_2-T_1$ is the temperature difference between the ends of the slab, $A$ is area of the slab and $L$ is the thickness of the each layer of the slab
So, $\frac{K\times A}{L}=\frac{K_1\times \frac{A}{2}}{L}+\frac{K_2\times \frac{A}{2}}{L}$
or, $K=\frac{K_1+K_2}{2}$