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I have a physics problem I'd like to make sure I get correct.

The practical aspect of this problem is that the photosynthetic efficiency of algae is inhibited with dissolved O2 in the growth medium, and the economics of any algae cultivation system are very sensitive to photosynthetic efficiency.

The problem I'm trying to solve is determine the theoretic energy to separate a gas mixture of, O2, NH3 and H2O (in boiling point order) into two gas streams, one of which is 99.99% pure O2 but the other stream can have as much as 50% of the original O2 along with NH3 and H2O. The input and output temperatures and pressures are STP and the input, at least, is at saturated humidity.

Presumably, the energy is derived from the equations in this paper.

Any approximate ideas of what the formula would look like?

PS: The detailed model of the algae system is here.

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The theoretical limit is not going to be of much use to you, I don't think, since the obvious separation method is to refrigerate and boil off the O<sub>2</sub>, and the losses in this method are not going to be close to optimal. Are you interested in the theoretical optimal separation method, or in the most practical? The optimal method is probably a gas centrifuge, but this is industrial scale gas centrifuges which are probably controlled. The energy cost will be much greater than the thermodynamic limit in any method. – Ron Maimon May 11 '12 at 5:57
Those algae soup produces that mixtureincluding ammonia? I doubt that. Any solution in water giving off some ammonia would be so alkaline that all (most?) living cells would be killed instantly. – Georg Jul 10 '12 at 10:17
@ColinMcFaul: If you clarify the points in the comment, it is easy to give an answer. Are you interested in the theoretical limit, the least energy to do the separation, or real life process involving refrigerating and boiling? – Ron Maimon Aug 9 '12 at 9:10

1 Answer 1

up vote 1 down vote accepted

As I understand it, no phase change occurs. At STP, you can use ideal gas approximation. Therefore only entropy costs are in your equation. Calculate mass balance and then just calculate entropy of mixing $\sum$ R $ x ~ ln(x)$. Did I miss anything?

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I assume theoretic lower bounds on the energy would apply even to semipermiable membranes. Is that correct? – James Bowery Jun 10 '12 at 15:46

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