A point source of heat of power P is placed at the centre of a thin...

Consider a concentric spherical shell of radius $r$ and thickness $dr$ as shown in diagram. The radial rate of flow of heat through this shell in steady state will be,



$\text{H} = \dfrac{\text{d}\text{Q}}{\text{d}\text{t}} = - \text{KA}\dfrac{\text{d}\text{\theta}}{\text{d}\text{r}}$

[Negative sign is used because, with an increase in $r$, $\theta$ decreases.]

Now as for spherical shell $A = 4\pir^{2}$

So$\text{H} = - 4\text{\pi}r^{2}\text{K}\dfrac{\text{d}\text{\theta}}{\text{d}\text{r}}\text{or}{∫}_a^{b}\dfrac{\text{d}\text{r}}{\text{r}2} = - \dfrac{4\text{\pi}\text{K}}{H}{∫}_{\text{\theta}1}^{\text{\theta}2}\text{d}\text{\theta}$

which on integration and simplification gives

$\text{H} = \dfrac{\text{d}\text{Q}}{\text{d}\text{t}} = \text{K}4\text{\pi}ab\left({θ}_{1} - {θ}_{2}\right)\left(b - a\right)\text{...(i)}$

Now in steady state, as no heat is absorbed, the rate of loss of heat by conduction is equal to that of supply, i.e., $H = P$, and here

& ${θ}_{1} - {θ}_{2} = T$ and $\text{a} ≃ \text{b} = \text{R}$

So Equation (i) becomes,

& & $\text{P} = 4\piR^{2}\text{KT} \left(\text{b} - \text{a}\right)<br /><br />i.e thickness of shell is \left(b - a\right) = \dfrac{4\piR^{2}\text{KT}}{P}$