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

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\begin{enumerate}
\item Helium diffusion through a pyrex tube

  \begin{enumerate}
  \item The helium conservation equation in cylindrical coordinates starts with
    $$\rm accumulation = in - out + generation$$
    and with no accumulation (because this is steady-state) and no generation
    (because helium is not involved in any homogeneous chemical reactions in
    the glass), this becomes:
    $$0 = -\left(J_r\cdot2\pi rL\right)_r +
    \left(J_r\cdot2\pi rL\right)_{r+\Delta r}$$
    Divide through by $-2\pi L$ and let $\Delta r$ go to zero to give
    $$\frac{\partial}{\partial r}(Jr) = 0$$
    Now substituting Fick's first law and multiplying by $-1$ gives the
    differential equation:
    \begin{equation}
      \label{eq:radial}
      \frac{\partial}{\partial r}\left(rD\frac{\partial C}{\partial r}\right)=0
    \end{equation}
    To solve this, integrate once with respect to $r$ and then divide by $r$:
    $$D\frac{dC}{dr}=\frac{A}{r}$$
    then divide by $D$ and integrate again:
    $$C=\frac{A}{D}\ln r+B=A'\ln r+B$$
    where $A'$ and $B$ are new constants ($A'=\frac{A}{D}$).

  \item Just substitute in the two boundary conditions:
    $$C_{\rm He, in}=A'\ln R_{\rm in}+B$$
    $$C_{\rm He, out}=A'\ln R_{\rm out}+B$$
    Subtracting these two gives:
    $$C_{\rm He, in}-C_{\rm He, out}=A'(\ln R_{\rm in}-\ln R_{\rm out})=
    A'\ln\frac{R_{\rm in}}{R_{\rm out}}$$
    $${\rm So,}\ A'=\frac{C_{\rm He, in}-
      C_{\rm He, out}}{\ln(R_{\rm in}/R_{\rm out})}$$
    Going back to the second of the two:
    $$B=C_{\rm He, out}-A'\ln R_{\rm out}$$
    $$B=C_{\rm He, out}-\frac{C_{\rm He, in}-
      C_{\rm He, out}}{\ln(R_{\rm in}/R_{\rm out})}\ln R_{\rm out}$$
    So insert these into the general solution above to give:
    $$C=\frac{C_{\rm He, in}-C_{\rm He, out}}{\ln(R_{\rm in}/R_{\rm out})}\ln r
    +C_{\rm He, out}-\frac{C_{\rm He, in}-
      C_{\rm He, out}}{\ln(R_{\rm in}/R_{\rm out})}\ln R_{\rm out}$$
    Subtract $C_{\rm He, out}$ from both sides and simplify:
    $$C-C_{\rm He, out}=
    \frac{(C_{\rm He, in}-C_{\rm He, out})(\ln r-\ln R_{\rm out})}
    {\ln(R_{\rm in}/R_{\rm out})}$$
    $$\frac{C-C_{\rm He, out}}{C_{\rm He, in}-C_{\rm He, out}}=
    \frac{\ln (r/R_{\rm out})}{\ln(R_{\rm in}/R_{\rm out})}$$
    
  \item First calculate the flux of helium through the inner surface, then
    multiply by the area of that surface to give the answer in grams/second.
    (You will get the same answer if you use the outer surface.)  Finally
    convert these units to m$^3$/hr at STP.
    $$J_r=-D\frac{dC}{dr}=-D\frac{d}{dr}\left[C_{\rm He, out}+
      \frac{(C_{\rm He, in}-C_{\rm He, out})(\ln r-\ln R_{\rm out})}
      {\ln(R_{\rm in}/R_{\rm out})}\right]$$
    $$J_r=-D\frac{C_{\rm He, in}-C_{\rm He, out}}{\ln(R_{\rm in}/R_{\rm out})}
    \frac{1}{r}$$
    Substitute $r=R_{\rm in}=0.5{\rm mm}=0.05$cm,
    $R_{\rm out}=1{\rm mm}=0.1$cm, $C_{\rm He, in}=
    10^{-5}\frac{\rm g}{\rm cm^3}$, $C_{\rm He, out}=0$ and
    $D=2\times 10^{-8}\frac{\rm cm^2}{\rm s}$:
    $$\left.J_r\right|_{r=R_{\rm in}}=-2\times 10^{-8}\frac{\rm cm^2}{\rm s}
    \frac{10^{-5}\frac{\rm g}{\rm cm^3}}{\ln\frac{1}{2}}\frac{1}{0.05cm}$$
    $$\left.J_r\right|_{r=R_{\rm in}}=
    5.77\times10^{-12}\frac{\rm g}{\rm cm^2\cdot s}$$
    The surface area is $2\pi rL=2\pi\cdot0.05{\rm cm}\cdot1000{\rm cm}=
    314.15{\rm cm}^2$, so helium is leaving the tubes at a rate of
    $1.81\times10^{-9}\rm\frac{g}{s}$.

    The molar density of helium is $\frac{p}{RT}$, at STP $p=101300$ Pa
    (=$\rm\frac{J}{m^3}$) and $T=298$K, so the density is
    $40.9\rm\frac{mol}{m^3}$.  Multiply by a molar mass of $\rm 2\frac{g}{mol}$
    to give a density of $81.8\rm\frac{g}{m^3}$.  If we divide the above answer
    by this density, and multiply by $\rm3600\frac{s}{hr}$, we arrive at a
    helium production rate of $7.98\times10^{-8}\rm\frac{m^3}{hr}$ at STP.
    That's pretty slow!
  \end{enumerate}
\end{enumerate}
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