Rate of Heat Loss Per Unit Length From an Insulated Cylinder

Suppose we have a cylinder of radius  
  covered in insulation of radius  
and thermal conductivity  
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  
  dependsa only on the area, so that  
If the total rate of heat loss is  
  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  
\[h = - k \mathbf{\nabla} T=-k \frac{\partial T}{\partial r}=-k\frac{dT}{dr}\]

  is a function of  
\[\frac{dT}{d r}=-\frac{H}{2 \pi kr} \]

Now seaparate varaibles:
\[dT= - \frac{h}{2 \pi k r} dr\]
  and integrate. When  
\[r=r_1 , T=T_1\]
  and when  
\[r=r_2 , T=T_2\]
\[\int^{T_2}_{T_1} dT = \int^{r_2}_{r_1} - \frac{h}{2 \pi k r} dr\]

\[T_2 -T_1 = -\frac{H}{2 \pi k} (\ln r_2 - \ln r_1 )= -\frac{H}{2 \pi k} \ln (r_2 /r_1)\]

\[H= - \frac{2 \pi k (T_1 - T_2 )}{ln(r_2 /r_1)}\]

You have no rights to post comments