root/trunk/matml/transport/problems/tijoule/tijoule-solution.tex

\documentclass{article}
\usepackage{fullpage}
\newcommand{\PSbox}[3]{\mbox{\rule{0in}{#3}\special{psfile=#1}\hspace{#2}}}
\begin{document}
\begin{enumerate}
\item Joule heating of a titanium rod

  \begin{enumerate}
  \item \label{cylsol} Start with the solution to the cylindrical heat
    conduction equation with uniform heat generation at steady state, from the
    handout ``1-D Thermal Diffusion Equation and Solutions'':
    $$T = -\frac{\dot{q}r^2}{4k} + A \ln r + B$$
    Since the temperature is finite at $r=0$, we know $A=0$.  The boundary
    condition $r=R\Rightarrow T=T_s$ (surface temperature) gives us:
    $$T = T_s + \frac{\dot{q}}{4k}(R^2-r^2)$$

    The maximum temperature is at $r=0$, and the minimum at $r=R$; the
    difference is:
    $$T_{max} - T_{min} = \frac{R^2\dot{q}}{4k}$$
    For $R=1.25\times10^{-3}$m, $\dot{q}=5\times10^6\frac{\rm W}{\rm m^3}$, and
    $k=20\frac{\rm W}{\rm m\cdot K}$, this gives
    $$T_{max} - T_{min} = 0.098{\rm K}$$

  \item The temperature difference is proportional to $\dot{q}$, so if that
    quadruples (because Joule heating goes as the current density squared), the
    temperature difference quadruples to about 0.4K.

  \item Here you had to estimate a sketch of something you've never seen
    before.  You know the initial condition at $t=0$ is uniform temperature at
    $T=T_s$, and the long-term steady-state distribution is given by part
    \ref{cylsol}.  In between, it should heat up pretty uniformly in the
    middle, until it reaches that steady-state.  So it will look something
    like:

    \begin{center}
      \PSbox{transgen.ps}{197pt}{145pt}
    \end{center}

  \item This is just the steady-state criterion:
    $$t_{SS} = \frac{R^2}{\alpha} = \frac{R^2\rho c_p}{k} = 0.26{\rm seconds}$$
  \end{enumerate}
\end{enumerate}
\end{document}