\[r_1\]
covered in insulation of radius \[r_2\]
and thermal conductivity \[k\]
.We can find the rate of heat loss per unit area by using radial symmetry. By symmetry the rate of hear loss per unit area
\[h\]
dependsa only on the area, so that \[h=h(r)\]
.If the total rate of heat loss is
\[H\]
we can write \[H= area \times h \rightarrow H=2 \pi rh \rightarrow h = \frac{H}{2 \pi r}\]
The heat flow per unit area is proportional to the temperature gradient, so that
\[h = - k \mathbf{\nabla} T=-k \frac{\partial T}{\partial r}=-k\frac{dT}{dr}\]
since
\[h\]
is a function of \[r\]
only.Hence
\[\frac{dT}{d r}=-\frac{H}{2 \pi kr} \]
Now seaparate varaibles:
\[dT= - \frac{h}{2 \pi k r} dr\]
and integrate. When \[r=r_1 , T=T_1\]
and when \[r=r_2 , T=T_2\]
so \[\int^{T_2}_{T_1} dT = \int^{r_2}_{r_1} - \frac{h}{2 \pi k r} dr\]
\[T_2 -T_1 = -\frac{H}{2 \pi k} (\ln r_2 - \ln r_1 )= -\frac{H}{2 \pi k} \ln (r_2 /r_1)\]
Then
\[H= - \frac{2 \pi k (T_1 - T_2 )}{ln(r_2 /r_1)}\]