Question

Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities K and 2K respectively. Find the equivalent thermal conductivity of the slab.

Answer 1

Let the length be $L$, area $A$, the temperature of the sink $T_1$, the temperature of the source $T_2$ and the temperature of the junction $T$.
For the first conductor, the thermal rate is given by
$\frac{H_1}{t}=KA\frac{T-T_1}{L}$ . . . . (1)
For the second conductor,
$\frac{H_2}{t}=2KA\frac{T_2-T}{L}$ . . . . (2)
The two slabs are connected in series so the heat current flowing through them will be same. So,
$KA\frac{T-T_1}{L}=2KA\frac{T_2-T}{L}$
$\Rightarrow (T-T_1)=2(T_2-T)$
$\Rightarrow T=\frac{1}{3}(2T_2+T_1)$ . . . . (3)
Let the equivalent coefficient of thermal conductivity be K'
Adding 1 and 2
$\frac{H_1+H_2}{t}=KA\frac{T-T_1}{L}+2KA\frac{T_2-T}{L}$
$\Rightarrow \frac{H}{t}=\frac{[KA\left\{\frac{1}{3}\right(2T_2+T_1)-T_1\}+2KA\{T_2-\frac{1}{3}(2T_2+T_1\left)\right\}]}{L}$
$\Rightarrow \frac{H}{t}=KA\frac{4}{3}\frac{(T_2-T_1)}{L}$
$\Rightarrow \frac{H}{t}=\left(\frac{4}{3}K\right)\frac{(T_2-T_1)}{L}$
$\Rightarrow \frac{H}{t}=K^{\prime }\frac{T_2-T_1}{L}$
So, equivalent thermal conductivity, $K^{\prime }=\frac{4}{3}K$