## Rate of Heat Loss Per Unit Length From an Insulated Cylinder

Suppose we have a cylinder of radius
$r_1$
$r_2$
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
$H= - \frac{2 \pi k (T_1 - T_2 )}{ln(r_2 /r_1)}$