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

The heat flowing per unit time through the disc is $q=\frac{2 \pi k l\left(\theta_{1}-\theta_{2}\right)}{\ln \frac{r_{2}}{r_{1}}}$

Explanation:

Let $\mathrm{d} \mathrm{Q} / \mathrm{dt}$ be rate of heat flow

Consider an angular ring of radius r and thickness dr

$q=\frac{d Q}{d t}=-K(2 \pi r d)\left(\frac{d \theta}{d r}\right)$

Integrating both side

$\int_{r_{2}}^{r_{2}} \frac{d r}{r}=-\frac{2 \pi \kappa l \int_{\theta_{1}}^{\theta_{2}} d \theta}{q}$

$\ln \frac{r_{2}}{r_{1}}=-\frac{2 \pi K l\left(\theta_{2}-\theta_{1}\right)}{q}$

$q=\frac{2 \pi k l\left(\theta_{1}-\theta_{2}\right)}{\ln \frac{r_{2}}{r_{1}}}$