Assumptions:
Explanation:
Use the appropriated heat transfer equation (cylindrical coordinates),
[tex]\frac {1}{r} \frac {d}{dr} (\ r \frac {dT}{dr}) + \frac {q}{k} \ = \ 0[/tex]
After separating variables and integrating we obtained the following expression,
[tex]r\frac{dT}{dr} \ = \ - \frac{q}{2k} \ r^{2} \ + \ C_{1}[/tex]
After repeating the procedure state above we obtained the following expression,
[tex]T(r) \ = \ - \frac{q}{2k} \ r^{2} \ + \ C_{1} ln(r) \ + \ C_{2}[/tex]
According to the image attached, the corresponding boundary values for the system are
[tex]\left \{ {{\frac{dT}{dr}=0 \ , \ r={0}} \atop {T(r_{0})=T_{s}}} \right.[/tex]
Then, the temperature distribution in the rod along r axis corresponds to,
[tex]T(r) \ = \ \frac{qr_{0}^{2}}{4k} (1 \ - \frac{r^2}{r_{0}^{2}} ) \ + \ T_{s}[/tex]
After deriving the function, we obtained the expression that follows,
[tex]\frac{dT}{dr} \ = \ - \frac{qr}{2k}[/tex]
Evaluating the previous expression at the surface of the rod,
[tex]\frac{dT}{dr} \lvert_{r = r_{0}}} \ = (- \frac{qr_{0}}{2k})[/tex]
Answer:
The heat transferred by the finite extended rod corresponds to,
[tex]q_{r} = -k(2 \pi L) \frac{dT}{dr}[/tex]
Replacing the derivate term, we obtained the expression that follows,
[tex]q_{r} = -k(2 \pi L) \ (- \frac{qr_{0}}{2k})\\q_{r} \ = \ \pi \ L \ {q \ r_{0}[/tex]
