Rate of Heat Loss Per Unit Length From an Insulated Cylinder

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

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