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I Using a centrifuge to extract heavy water (help with the calculations please)

  1. Jul 9, 2017 #1
    Hello everyone!

    I have seen several DIY projects which successfully gathered heavy water
    from normal water. For example Cody from codys lab used electrolysis
    to "enrich" the water. This however is a messy process.

    So i became curious, if this can be done easier by centrifugation.

    Based on these data http://www1.lsbu.ac.uk/water/water_properties.html
    around 1 in 6600 water molecules are heavy water (HDO). This would
    have a density of ~1050kg/m³ compared to ~996kg/m³. This is not enough
    to seperate them under normal gravity but maybe this difference is enough
    to be able to seperate them in a centrifuge.

    To get a feel for the forces needed, i want to calculate this. But here
    is where i got stuck.

    First, we assume a cylindrical centrifuge, spinning at a constant RPM.
    We also assume that the system has reached a steady state so in the
    frame of reference of the spinning liquid, nothing is moving. In that
    frame, a fictitious force depending on the radius F(r) will act on the
    liquid. Since the system is highly symmetric, it should be equivalent
    to looking at only a 1D column along the radius.

    In this steady state, the movement of particles due to the centrifugal
    action and the movement due to diffusion will balance. Looking again at
    the data table we find the diffusion coefficients of H2O and HDO, which
    are (in SI) 2.299e-7 m²/s and 2.34e-7 m²/s respectively.

    Here is my first confusion: the diffusion coefficient requires two
    substances, is the second substance "normal water" (normal mixture of
    H2O and HDO) ?

    Second, to calculate the distribution, i am looking for a PDE i can
    integrate numerically. I found the diffusion equation:
    [tex]\frac{\partial \phi(r, t)}{\partial t} = \nabla \cdot \big[ D(r,t))\nabla \phi(r,t) \big][/tex]
    with [itex]\phi[/itex] being the density and [itex]D[/itex] the collective diffusion coefficient.

    (Can i assume the collective diffusion coefficient is the density weighted
    average of the individual diffusion coefficients?)

    Obviously, this equation does not include the radial force or the resultant
    pressure.

    I hope i can get around using the full navier-stokes equations, as they are
    quite complicated. Is there a simpler form i can use? Essentially i am looking
    for a 1D PDE.
     
  2. jcsd
  3. Jul 10, 2017 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    It is probably easier to view this from a thermodynamic perspective. The centrifuge produces an effective potential for HDO (given by the mass difference to H2O), and your HDO molecules will be in thermodynamic equilibrium in this potential.

    Rough estimate: A rim speed of 300 m/s (that is a very powerful centrifuge) leads to an effective potential of 45 kJ/kg or 0.5 meV if multiplied by the mass difference. kT at room temperature is about 25 meV. The separation will be quite weak.
     
  4. Jul 10, 2017 #3
    That is a very quick but useful estimate. Since 300m/s rim speed is way out of reach as a hobbyist, it seems like this approach will not work.

    Thanks for your answer!
     
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