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

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\begin{enumerate}
\item Electron beam centrifugal atomization of metal

  \begin{enumerate}
  \item \label{radloss} If the chamber is black (meaning it absorbs all
    incident radiation) and cold (indicating it doesn't radiate significant
    heat back to the ingot), then the heat flux from the top surface is given
    by:
    $$q_{rad} = \epsilon\sigma T^4$$
    We are given 0.55 for $\epsilon$, and the temperature is the titanium
    melting point which is 1940K, so this heat flux is about 442 kW/m$^2$.

  \item \label{evaploss} The vapor pressure is given by the Clausius-Clapeyron
    equation, whose form and constants were given in the problem:
    $$\log_{10}p_v({\rm torr}) = -\frac{23200}{T} + 11.74 - 0.66\log_{10}T
    (+ 0\cdot T) = -2.39$$
    $$p_v = 4.1\times10^{-3}{\rm torr}$$
    To convert from torr to J/m$^3$ (a.k.a. N/m$^2$, or Pascals), which are the
    required units for the Langmuir equation, note that 760 torr = 101300
    Pascals = 1 atm, so
    $$p_v = {\rm 4.1\times10^{-3}torr\times
      \frac{101300\frac{J}{m^3}}{760 torr} = 0.545\frac{J}{m^3}}$$
    Now we put this into the Langmuir equation:
    $$J_{vap} = \frac{p_v}{\sqrt{2\pi MRT}} =
    {\rm\frac{0.545\frac{J}{m^3}}
      {\sqrt{2\pi \cdot 0.0479\frac{kg}{mol} \cdot 8.314\frac{J}{K\cdot mol}
          \cdot 1940K}} =
      \frac{0.545\frac{kg}{m\cdot s^2}}{\sqrt{4854\frac{kg^2m^2}{s^2mol^2}}} =
      7.8\times10^{-3}\frac{mol}{m^2\cdot s}}$$
    The heat flux at this evaporation rate is simply:
    $$q_{vap} = J_{vap}\Delta H_{vap} =
    {\rm 7.8\times10^{-3}\frac{mol}{m^2\cdot s} \cdot 440\frac{kJ}{mol} =
      3.44\frac{kW}{m^2}}$$

  \item \label{totalflux} The total required power density is the sum of:
    \begin{itemize}
    \item Heat loss to radiation as calculated in part \ref{radloss}:
      $q_{rad} = 442\frac{\rm kW}{\rm m^2}$.
    \item Heat loss to evaporation as calculated in part \ref{evaploss}:
      $q_{vap} = 3.44\frac{\rm kW}{\rm m^2}$.
    \item Heat required to raise the titanium temperature from 300K to its
      melting point of 1940K, which we can call $q_{c_p}$.  Assuming constant
      heat capacity, and neglecting solid-solid phase transformations (since no
      information about them was given), the required heat per unit volume is:
      $$H_{c_p} = \rho c_p \Delta T$$
      If we assign the variable $u$ to the melt interface speed, then the power
      required per unit area to heat the titanium to the melting point at that
      speed is:
      $$q_{c_p} = uH_{c_p} = u \rho c_p \Delta T$$
      Inserting our parameters gives:
      $$q_{c_p} = {\rm 0.01\frac{m}{min} \cdot \frac{1min}{60s} \cdot
        4700\frac{kg}{m^3} \cdot 700\frac{J}{kg\cdot K} \cdot (1940K - 300K) =
        900\frac{kW}{m^2}}$$
    \item Heat required to melt the titanium $q_m$.  The heat per unit volume
      is the heat of fusion multiplied by the density $\rho\Delta H_f$.  The
      power per unit area required to melt at linear velocity $u$ is simply:
      $$q_m = u \rho \Delta H_f =
      {\rm 0.01\frac{m}{min} \cdot \frac{1min}{60s} \cdot
        4700\frac{kg}{m^3} \cdot 300\frac{kJ}{kg} = 235\frac{kW}{m^2}}$$
    \end{itemize}

    No single use of heat dominates, but the evaporation loss is relatively
    insignificant.  The total power required comes to:
    $$q_{total} = q_{rad} + q_{evap} + q_{c_p} + q_m = 1580\frac{\rm kW}{\rm
      m^2}$$
    This sounds like a large amount of power, but since a typical atomization
    unit might use an ingot of 15 cm diameter, the power required for this
    small area is only about 28 kW.

  \item The $q_{c_p}$ and $q_m$ terms are both proportional to the velocity
    $u$, and thus proportional to the amount of material used.  The $q_{rad}$
    and $q_{evap}$ terms, on the other hand, are independent of $u$.  So if we
    go faster, we produce more powder using more $q_{c_p}$ and $q_m$ but the
    same amount of $q_{rad}$ and $q_{evap}$.  On a ``per unit of material
    produced'' basis, we use the same amount of energy per unit of material to
    heat and melt it, but {\em less} is wasted on radiation and evaporation.

    It is therefore more efficient to run faster.

  \item \label{tempdist} The steady-state temperature distribution as derived
    in class is:
    $$\frac{T-T_i}{T_m-T_i} = \exp\left(-\frac{u z}{\alpha}\right)$$
    where $z$ is the distance from the melt interface.  This is valid if the
    length of the ingot is much greater than the lengthscale over which the
    temperature is significantly different from the initial temperature:
    $L>>\alpha/u$.  That lengthscale is given by:
    $$\frac{\alpha}{u} = \frac{k}{u\rho c_p} =
    {\rm\frac{20\frac{W}{m\cdot K}}
      {0.01\frac{m}{min} \cdot \frac{1min}{60s} \cdot
        4700\frac{kg}{m^3} \cdot 700\frac{J}{kg\cdot K}} =
      0.036 m}$$
    So only the top few centimeters closest to the melt interface are
    significantly hotter than the initial temperature of the ingot.  Since the
    ingot is initially 1 meter long, this can be considered semi-infinite for
    most of the duration of the atomization process.

  \item The heat flux is given by:
    $$q_z = -k\frac{\partial T}{\partial z}$$
    For the temperature distribution given in part \ref{tempdist}, this is:
    $$q_z = -k\frac{\partial}{\partial z}
    \left[T_i + (T_m-T_i)\exp\left(-\frac{u z}{\alpha}\right)\right] =
    \frac{k(T_m-T_i)u}{\alpha} \exp\left(-\frac{u z}{\alpha}\right)$$
    Note that the ratio $k/\alpha$ is equal to $\rho c_p$ according to the
    definition of $\alpha$, so at $z=0$, this is equal to:
    $$\left.q_z\right|_{z=0} = u \rho c_p (T_m-T_i)$$
    This is identically equal to $q_{c_p}$ in part \ref{totalflux}, which was
    $\rm900\frac{kW}{m^2}$.  This is because it is this heat flux which
    provides the heat to the solid titanium ingot to raise it to the melting
    point.

  \item If the ingot is horizontal and rotating like a rolling pin, a fixed
    electron beam hitting its top will generate a narrow stream of atomized
    liquid droplets on a tangent line to the spinning ingot.  In this way, it
    is better than the vertical arrangement, which sends droplets flying all
    over the place.

    However, as the metal melts, atomizes and leaves, the horizontal cylinder
    is left with less material there, so the local diameter changes in that
    part of the ingot.  This looks something like a cylinder which is being
    peeled while rotating on a lathe.  Because droplet size is a function of
    the ingot diameter and rotation rate, one must change the rotation speed as
    the ingot gets chewed away by the atomization of the droplets in order to
    maintain a consistent droplet size.  In the vertical arrangement, the
    diameter never changes, so the droplet size doesn't either; this is an
    advantage of the vertical over horizontal centrifugal atomizer.

    Another possibility, suggested by a student, would be to have the ingot
    rotating horizontally, but with the beam scanning back and forth along a
    line on top of it.  This would result in a wide stream of droplets coming
    off roughly in a plane tangent to the cylinder, which could be a
    disadvantage.  On the other hand, the diameter of the cylinder decreases
    roughly uniformly as it is atomized, so it may be possible to have slightly
    more uniform droplet size than in the horizontal atomizer with fixed beam
    position.

    As you can see, this can get as complicated as you want to make it; any of
    these aspects of the various designs gets full credit for this 4-point
    subpart of the problem.
  \end{enumerate}
\end{enumerate}
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