Question

A point source of heat of power $\mathrm{P}$ is placed at the center of a spherical shell of mean radius $\mathrm{R}$. The material of the shell has thermal conductivity $\mathrm{k}$. If the temperature difference between the outer and the inner surface of the shell is not to exceed $\mathrm{T}$., then the thickness of the shell should not be less than

• A. $\frac{2{\mathrm{\pi R}}^2\mathrm{kT}}{\mathrm{P}}$
• B. $\frac{4{\mathrm{\pi R}}^2\mathrm{kT}}{\mathrm{P}}$
• C. $\frac{{\mathrm{\pi R}}^2\mathrm{kT}}{\mathrm{P}}$
• D. $\frac{{\mathrm{\pi R}}^2\mathrm{kT}}{4\mathrm{P}}$

Answer 1

The correct option is B

$\frac{4{\mathrm{\pi R}}^2\mathrm{kT}}{\mathrm{P}}$

Explanation for correct option

Step 1: Given data

1. It is given that, a point source of heat of power $\mathrm{P}$ is placed at the center of a spherical shell of mean radius $\mathrm{R}$.
2. The material of the shell has thermal conductivity $\mathrm{k}$. If the temperature difference between the outer and the inner surface of the shell is not to exceed $\mathrm{T}$.

We have to find the thickness of the shell.

Step 2: Formula used.

We will find the thickness of the shell by using the term of power radiated per unit area per second in conduction.

1. The point $\mathrm{P}$ is the amount of heat obtained by the source per unit of time, this heat is dispersed across the inside surface of the spherical shell, which has an area of $4{\mathrm{\pi R}}^2$.
2. So, Rate of flow of heat $\mathrm{P}=\frac{\mathrm{kA}∆\mathrm{\theta }}{∆\mathrm{x}}$
3. Here, $\mathrm{k}$ is overall heat transfer coefficient, $\mathrm{A}$ is heat transfer area of shell, $∆\mathrm{\theta }$ is temperature difference and $∆\mathrm{x}$ is the thickness of shell.

Step 3. Calculate the thickness of shell.

By applying the given condition in the above formula, we get,

Rate of generation of heat $=$rate of heat transfer

$\mathrm{P}=\frac{\mathrm{kA}∆\mathrm{\theta }}{∆\mathrm{x}}$

Here,

$\mathrm{A}=4{\mathrm{\pi R}}^2$

$∆\mathrm{\theta }=\mathrm{T}$

So,

$\mathrm{P}=\frac{4{\mathrm{\pi R}}^2\mathrm{T}}{∆\mathrm{x}}$

$∆\mathrm{x}=\frac{4{\mathrm{\pi R}}^2\mathrm{T}}{\mathrm{P}}$

Therefore, the thickness of the shell is $\frac{4{\mathrm{\pi R}}^2\mathrm{T}}{\mathrm{P}}$

Hence option $\mathrm{B}$ is correct answer.