Question

A hole of radius $r_1$ is made centrally in a uniform circular disc of thickness d and radius $r_2$. The inner surface (a cyclinder of length d and radiud $r_1$) is maintained at a temperature ${\theta }_1$ and the outer surface (a cylinder of length d and radius $r_2$) is maintained at a temperature ${\theta }_2({\theta }_4>{\theta }_2)$. The thermal conductivity of the material of the disc is K. Calculate the heat flowing per unit time through the disc.

Answer 1

$\frac{dQ}{dt}$ = Rate of flow of heat

Let us consider a strip at a distance r from the centre of thickness dr.

$\frac{Q}{dt}=\frac{K\times 2\pi \times rd\times d\theta }{dr}$

[$d\theta$ = temperature difference across the thickness dr ]

C$=\frac{K\times 2\pi \times rd\times d\theta }{dr}$ $[c=\frac{\theta }{t}]$

$C\times \frac{dr}{r}=K2\pi dd\theta$

Integrating

$\Rightarrow C{\int }_{r_1}^{r_2}\frac{dr}{r}=K2\pi d{\int }_{{\theta }_2}^{{\theta }_1}d\theta$

$\Rightarrow C[logr{]}_{r_1}^{r_2}=K2\pi d({\theta }_1-{\theta }_2)$

$C(log{r}_2-log{r}_1)=K2\pi d({\theta }_1-{\theta }_2)$

$\Rightarrow C=\frac{K2\pi d({\theta }_1-{\theta }_2)}{log\left(r_2/r_1\right)}$