A composite slab consists of two materials having coefficient of thermal conductivity $K$ and $2…
A composite slab consists of two materials having coefficient of thermal conductivity $K$ and $2 \mathrm{~K}$, thickness $\mathrm{x}$ and $4 \mathrm{x}$ respectively. The temperature of the two outer surfaces of a composite slab are $\mathrm{T}_2$ and $\mathrm{T}_1\left(\mathrm{~T}_2>\mathrm{T}_1\right)$. The rate of heat transfer through the slab in a steady state is $\left[\frac{\mathrm{A}\left(\mathrm{T}_2-\mathrm{T}_1\right) \mathrm{K}}{\mathrm{x}}\right] \cdot \mathrm{f}$ where ' $\mathrm{f}$ ' is equal to
$1$
$\frac{2}{3}$
$\frac{1}{2}$
$\frac{1}{3}$
Solution
$\begin{array}{ll}
& \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_1+\mathrm{R}_2 \\
\because & \mathrm{R}_1=\frac{\mathrm{x}}{\mathrm{KA}}, \mathrm{R}_2=\frac{4 \mathrm{x}}{2 \mathrm{KA}} \\
\therefore & \mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{x}}{\mathrm{KA}}+\frac{2 \mathrm{x}}{\mathrm{KA}}=\frac{3 \mathrm{x}}{\mathrm{KA}}
\end{array}$
Rate of heat transfer of composite slab is given by,
$\begin{aligned}
& \quad \frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{T}_2-\mathrm{T}_1}{\mathrm{R}_{e \mathrm{q}}}=\frac{\mathrm{KA}\left(\mathrm{T}_2-\mathrm{T}_1\right)}{3 \mathrm{x}} \\
& \therefore \quad \mathrm{f}=\frac{1}{3}
\end{aligned}$