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

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The discussion explores the feasibility of using centrifugation to extract heavy water (HDO) from normal water, proposing it as a potentially simpler alternative to electrolysis. The participant notes that the density difference between HDO and H2O is insufficient for separation under normal gravity but may be adequate in a centrifuge. They seek clarification on the diffusion coefficient calculations, questioning whether the second substance in the diffusion equation is normal water and if the collective diffusion coefficient can be approximated as a density-weighted average. The participant also expresses a desire to avoid complex Navier-Stokes equations, looking for a simpler 1D partial differential equation to model the system. Ultimately, they conclude that achieving the necessary rim speed of 300 m/s for effective separation is impractical for hobbyists, suggesting that this method may not be viable.
Gotbread
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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 separate them under normal gravity but maybe this difference is enough
to be able to separate 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:
\frac{\partial \phi(r, t)}{\partial t} = \nabla \cdot \big[ D(r,t))\nabla \phi(r,t) \big]
with \phi being the density and D 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.
 
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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.
 
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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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