Question

A metallic spherical shell having inner and outer radii $a$ and $b$ respectively has thermal conductivity $K=\frac{K_0}{r^2}\{a\le r\le b\}$, where $r$ is the distance from the center of the sphere. The inner surface is maintained at temperature ${\theta }_1$ and outer surface is maintained at temperature ${\theta }_2$ . Calculate the rate at which heat flows radially through the material $[{\theta }_2>{\theta }_1]$.

• A. $\frac{4\pi K_0({\theta }_1-{\theta }_2)}{b-a}$
• B. $\frac{4\pi K_0({\theta }_2-{\theta }_1)}{b-a}$
• C. $\frac{4\pi K_0(b-a)}{{\theta }_2-{\theta }_1}$
• D. $\frac{8\pi K_0({\theta }_2-{\theta }_1)}{b-a}$

Answer 1

The correct option is B $\frac{4\pi K_0({\theta }_2-{\theta }_1)}{b-a}$

Let us consider an infinitesimal spherical element at radius $r$ and of thickness $dr$, which has a temperature difference of $d\theta$ across it [shown in figure below].


Rate of heat flow through this spherical element is given by
$H=\frac{dQ}{dt}=KA\frac{d\theta }{dr}=\frac{K_0}{r^2}\left(4\pi r^2\right)\frac{d\theta }{dr}$
Integrating on both sides, we get
$H{\int }_a^{b}dr=4\pi K_0{\int }_{{\theta }_1}^{{\theta }_2}d\theta$
$\Rightarrow H(b-a)=4\pi K_0({\theta }_2-{\theta }_1)$
$\Rightarrow H=\frac{4\pi K_0({\theta }_2-{\theta }_1)}{(b-a)}$
Thus, option (b) is the correct answer.