Question

Two thin metallic spherical shells of radii \(r_1\) and \(r_2\) \((r_1 < r_2)\) are placed with their centres coinciding. A thermal conductivity material, \(K\), fills the space between the shells. The inner shell is maintained at a temperature \(\theta_1\) and the outer shell at temperature \(\theta_2\) \((\theta_1 < \theta_2)\). The rate at which heat flows radially through the material \(\dfrac{dQ}{dt}\), is:

• A. \(\dfrac{4\pi K r_1 r_2 (\theta_2 - \theta_1)}{r_2 - r_1}\)
• B. \(\dfrac{8\pi K r_1 r_2 (\theta_2 - \theta_1)}{r_2 - r_1}\)
• C. \(\dfrac{\pi K r_1 r_2 (\theta_2 - \theta_1)}{r_2 - r_1}\)
• D. \(\dfrac{\pi K r_1 r_2 (\theta_2 - \theta_1)}{4(r_2 - r_1)}\)

Hint:

Apply Fourier's law with area \(4\pi r^2\) and integrate radially; the spherical shell resistance is \(\dfrac{r_2 - r_1}{4\pi K r_1 r_2}\).

Answer 1

The Correct Option is A

Step 1: Consider steady radial conduction. At radius \(r\), heat crosses a spherical surface of area \(A = 4\pi r^2\). Fourier's law gives the (constant) heat current
\[\frac{dQ}{dt} = -K(4\pi r^2)\frac{d\theta}{dr}.\]
Step 2: Separate variables. Let \(H = \dfrac{dQ}{dt}\) be constant in the steady state:
\[\frac{H}{4\pi K}\,\frac{dr}{r^2} = -\,d\theta.\]
Step 3: Integrate from the inner shell \((r_1,\ \theta_1)\) to the outer shell \((r_2,\ \theta_2)\):
\[\frac{H}{4\pi K}\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = \theta_2 - \theta_1.\]
Step 4: Note \(\dfrac{1}{r_1} - \dfrac{1}{r_2} = \dfrac{r_2 - r_1}{r_1 r_2}\). Solving for \(H\):
\[H = \frac{4\pi K r_1 r_2 (\theta_2 - \theta_1)}{r_2 - r_1}.\]
\[\boxed{\dfrac{dQ}{dt} = \dfrac{4\pi K r_1 r_2 (\theta_2 - \theta_1)}{r_2 - r_1}}\]