Question

A bi-metallic slab is made (refer figure) by fusing two material. The components have same thickness and lengths, but are of thermal conductivities $K_1$ and $K_2$. If the heat were to conduct across the faces $(T_1 > T_2)$, then effective thermal conductivity of the composite slab is

• A. $2(K_1 + K_2)$
• B. $\frac{K_1 \times K_2}{K_1 +K_2}$
• C. $2 \frac{K_1 \times K_2}{K_1 +K_2}$
• D. $\frac{K_1 \times K_2}{2}$

Answer 1

Answer: (C)

$2 \frac{K_1 \times K_2}{K_1 +K_2}$

Answer 2

Under steady condition, heat current through metallic slab of conductivity $K_1$ is same that in other slab of conductivity $K_2$
$\frac{dQ}{dt} = \frac{dQ_{1}}{dt} = \frac{dQ_{2}}{dt}$
or $\frac{K_{1 }A_{1} \left(T_{1 } -T\right)}{L_{1}} = \frac{K_{2}A_{2} \left(T -T_{2}\right)}{L_{2}}$
$A_{1 } =A_{2} =A , L_{1} =L_{2} =L$
So, $K_{1}=\left(T_{1} -T\right)= K_{2}\left(T - T_{2}\right)$
or , $K_{1}T_{1} +K_{2}T_{2} =T\left(K_{1} + K_{2}\right)$
\or , $T= \frac{K_{1}T_{1} +K_{2} T_{2}}{ K_{1} + K_{2}}$
Also, $\frac{dQ}{dt} = \frac{K_{1}A \left(T_{1} -T\right)}{L_{1} }$
$\, \, \, \, = \frac{K_{1}A}{L} \left(T_{1} - \frac{K_{1}T_{1} +K_{2} T_{2}}{K_{1} +K_{2}}\right)$
$\, \, \, \, \, = \frac{A\left(T_{1} -T_{2}\right)}{L\left(\frac{1}{K_{1}} + \frac{1}{K_{2}}\right)} \, \, \, \, \, \,$ ...(i)
Let equivalent thermal conductivity of bi-metallic slab be Kand length of bi-metalic slab = $L_1 + L_2 = 2L.$
So, $\frac{dQ}{dt} = \frac{K A\left(T_{1} -T_{2}\right)}{2L} \, \, \, \, \,$ ....(ii)
From eqn (i) and (ii)
$\frac{K}{2} = \frac{K_{1}K_{2}}{K_{1} +K_{2}}$ or,$K= \frac{2K_{1} K_{2}}{K_{1 } + k_{2}}$