Question

Thermal resistance $\left(R\right)$ of a spherical shell having inner radius $r_1$ and outer radius $r_2$ is given by,
($K$ is the thermal conductivity of the shell.)

• A. $R=\frac{r_1{r}_2}{4\pi K(r_2-r_1)}$
• B. $R=\frac{r_2-r_1}{4\pi K\left(r_1{r}_2\right)}$
• C. $R=\frac{r_1{r}_2}{4K(r_2-r_1)}$
• D. $R=\frac{r_2-r_1}{4K\left(r_1{r}_2\right)}$

Answer 1

The correct option is B $R=\frac{r_2-r_1}{4\pi K\left(r_1{r}_2\right)}$


Given,
Thermal conductivity of the spherical shell $=K$
Inner radius$=r_1$
Outer radius$=r_2$
Consider a thin spherical shell at radius $r$ having thickness $dr$.
Let thermal resistance of the shell of thickness $dr$, is $dR)$
$\Rightarrow dR=\frac{l}{kA}=\frac{dr}{k4\pi r^2}$

Let the inner surface is at higher temperature than outer surface. So, in steady state, heat will flow from inner surface to outer surface. Therefore, we can consider the all thin spherical surface to be in series.

$\therefore$ The total thermal Resistance is,

$R={\int }_{r_1}^{r_2}dR={\int }_{r_1}^{r_2}\frac{dr}{k4\pi r^2}$

$=\frac{1}{4\pi k}{\int }_{r_1}^{r_2}\frac{dr}{r^2}$

$=\frac{1}{4\pi k}{[-\frac{1}{r}]}_{r_1}^{r_2}$

$\Rightarrow R=\frac{1}{4\pi K}(\frac{1}{r_1}-\frac{1}{r_2})$

$\therefore R=\frac{r_2-r_1}{4\pi K{r}_1{r}_2}$

Hence, $\left(B\right)$ is the correct answer.

$$\begin{array}{c}\text{Why this Question :}\\ \text{Tip : To calculate the thermal conductivity or thermal resistance of spherical shell, cylindrical shell or similar structure, consider a thin shell and calculate the resistance or conductivity of that thin shell and then use integration to calculate the overall thermal conductivity or thermal resistance}\end{array}$$