Question

A hole of radius r₁ is made centrally in a uniform circular disc of thickness d and radius r₂. The inner surface (a cylinder a length d and radius r₁) is maintained at a temperature θ₁ and the outer surface (a cylinder of length d and radius r₂) is maintained at a temperature θ₂ (θ₁ > θ₂). The thermal conductivity of the material of the disc is K. Calculate the heat flowing per unit time through the disc.

Answer 1

Let $\frac{d\theta }{dt}$ be the rate of flow of heat.
Consider an annular ring of radius r and thickness dr.
Rate of flow of heat is given by
$\frac{\mathrm{d\theta }}{\mathrm{d}t}=-K\left(2\mathrm{\pi }rd\right).\frac{\mathrm{d\theta }}{\mathrm{d}r}$
Rate of flow of heat is constant.
∴ $\frac{\mathrm{d\theta }}{\mathrm{d}t}=i$
$$\begin{gathered} i=-K\left(2\mathrm{\pi }r.d\right)\frac{\mathrm{d\theta }}{\mathrm{d}r} \\ {\int }_{r_1}^{r_2}\frac{\mathrm{d}r}{r}=-\frac{2Kd}{i}{\int }_{{\theta }_1}^{{\theta }_2}d\mathrm{\theta } \\ {\left[\ln \left(r\right)\right]}_{r_1}^{r_2}=-\frac{2\mathrm{\pi }Kd}{i}\left[{\theta }_2-{\theta }_1\right] \\ i=\frac{2\mathrm{\pi }Kd\left({\mathrm{\theta }}_1-{\mathrm{\theta }}_2\right)}{\ln \left({\displaystyle \frac{r_2}{r_1}}\right)} \end{gathered}$$