A metal rod of constant cross-section and length L has its ends maintained at constant temperatures
t1 and
t2. Determine the temperature at the midpoint of the rod when
heat flows through it at a steady state. The side surfaces of the bar are thermally insulated.
Problem data:
- Temperature at the ends of the bar: t1 e t2;
- Bar Length: L.
Problem diagram:
The problem tells us that heat flows in a steady state, that is to say, that the heat flux through a
cross-section of the bar is constant.
The length e of the bar between the ends is e=L−0=L, and the
length between the end at temperature t1 and any cross-section is
e=x−0=x (Figure 1).
Solution
The heat flux is given by
\[ \bbox[#99CCFF,10px]
{\phi =kA\frac{\left(t_{1}-t_{2}\right)}{e}}
\]
Since the flux φ is constant, the amount of heat that passes through the ends, maintained at temperatures
t1 and
t2, is equal to the amount of heat that passes through the end at
temperature
t1 and any section at temperature
tx
\[
\begin{gather}
\phi=kA\frac{\left(t_{1}-t_{2}\right)}{L}=kA\frac{\left(t_{1}-t{_x}\right)}{x}\\
\frac{t_{1}-t_{2}}{L}=\frac{t_{1}-t_{x}}{x}\\
t_{1}-t_{x}=x\frac{\left(t_{1}-t_{2}\right)}{L}\\
t_{x}=t_{1}-x\frac{\left(t_{1}-t_{2}\right)}{L}
\end{gather}
\]
Generally, this expression gives the temperature at any point
x of the bar, in particular, we want
the temperature at the midpoint, where
\( x=\frac{1}{2}L \),
substituting this value
\[
\begin{gather}
t_{\frac{11}{2}}=t_{1}-\frac{1}{2}L\frac{\left(t_{1}-t_{2}\right)}{L}\\
t_{\frac{11}{2}}=t_{1}-\frac{t_{1}}{2}+\frac{t_{2}}{2}\\
t_{\frac{11}{2}}=\frac{2t_{1}-t_{1}}{2}+\frac{t_{2}}{2}\\
t_{\frac{11}{2}}=\frac{t_{1}}{2}+\frac{t_{2}}{2}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{t_{\frac{1}{2}}=\frac{t_{1}+t_{2}}{2}}
\]
The temperature at the midpoint will be the average of the temperatures at the ends.