Entropy and RO

Barry Meyers-Rice (barry@as.arizona.edu)
Tue, 22 Feb 94 09:59:19 MST

>That is why it takes pressure to "drive it backwards". The membrane
>pores are too small to allow the solvated ions through.
>But it takes added pressure to drive the water molecules back
>against the entropy. It all works out fine if you calculate
>such things with the molar Gibbs energy, ie the "chemical potential",
>that keeps track of the entropy contribution as well as the
>"mechanical" potential energy terms.

>No entropy problem, Barry. The reason why RO units are so expensive if
>they're any good is that you have to push the water REAL HARD to get
>it through the membrane against the osmotic pressure. That means you
>have to have a good pump, and the membrane has to have a VERY strong
>support.

So the ``pure''and ``waste'' waters that come out are at lower pressures than
the input water? I suppose you can interpret a pressure differential as
a difference in energy densities on either side of the membrane. And I suppose
that since the high pressure side has a greater energy density, you can use
that drop in pressure as a trade-off? i.e. Extract pure water at the cost
of losing pressure.

I think the pressure doesn't have to be too high, since most units work off
municipal water pressure.

Now that I think of it, in some ways, water purification by distillation
is not too different. You are separating out a pure component, at an energy
expenditure. It's just strange since I'm used to thinking of water as
being incompressible, and so am used to neglecting pressure.

Thanks for the info

Barry