Q:

Hi
I would like to know what would happen in a deep (think kilometres here), still column of salty water at constant temperature.
Specifically, I'd like to know if the salt molecules would tend to sink to the bottom making the salinity at depth greater than at the surface, and if so, what the expression describing this is.

- Martin Williams (age 71)

Wirral

- Martin Williams (age 71)

Wirral

A:

That's a great question.

The easy part is: yes, in still water allowed to reach equilibrium the salt concentration does go up with depth. Saltier water has higher mass density, so the gravitational energy can be lowered that way. The concentration differences go up until the free-energy of creating that big a concentration difference balances the gravitational energy change.

We can make an approximate simple description, by pretending that the Na^{+} and Cl^{-} ions (not molecules) form an ideal solution and that the solution's mass density is a linear function of the salt concentration. Although the aqueous NaCl isn't an ideal solution, the activity coefficient (describing the non-ideality) doesn't vary too much in the concentration range of interest, around the 3.5% by weight typical of seawater. (See .)

Let's also ignore any dependence of the chemical properties on the water pressure, which goes up substantially at kilometers of depth. I'm not sure how good an approximation that is.

We then have an expression very similar to the simple law of atmospheres for how the concentration varies with depth:

n(h) =n(0) e^{-(mgh/2kT) } where m is the effective mass of each NaCl unit, g is the gravitational field, k is Boltzmann's constant, T is absolute temperature, and the factor of 2 is in there because the Na^{+} and Cl^{-} form two nearly independent particles. The depth is *-h*, if we set *h*=0 at the surface. Adding NaCl to water displaces some of the water, so that the extra weight in a given volume is only about 70% of the NaCl weight. So *m* of one NaCl unit is about 0.7*58gm/6*10^{23}, where the last number is the mass per mole divided by Avogadro's number. So that gives ~7*10^{-23} gm. The temperature of the deep ocean is mostly around 4°C so T = 277 K. *k*=1.4*10^{-16} ergs/K. Multiplying out gives that the equilibrium concentration goes up exponentially with depth, by a factor of *e* for each 10 km or so.

The actual oceans are stirred by currents, so this equilibrium concentration difference isn't present in them. In principle, that non-equilibrium salt concentration could be used to extract energy, but I haven't heard of any practical ways to do so.

Mike W.

posted without checking by Lee until he returns from Paris

*(published on 10/14/2013)*

Q:

I wanted to thank Mike W for his careful answer to my question regarding salt concentration v water depth.
He/you mention the extraction of energy from the existing constant salinity. One way of doing this would be to place a reverse osmosis filter at the bottom of a (very) long pipe full of fresh water. The differential pressure due to the depth of salt and fresh water is sufficient to cause water to emerge from the pipe at the surface at a few bar pressure if you go to the deepest part of the earth's ocean. Fresh water and energy!
However, as you say, the engineering problems of doing this effectively rule it out as a source of either!

- Martin Williams (age 71)

Wirral

- Martin Williams (age 71)

Wirral

A:

Are you sure of your calculations? I went through this roughly and get that it's marginal at best, even in principle. Say you have a *very *deep column of water, 10 km. The density difference between seawater and pure water is around 0.025 gm/cm^{3}. At the bottom of our column, the pressure difference should be about 25 atm. The osmotic pressure of seawater is around 27 atm, so it sounds as if you'd have to go even deeper to get any flow.

Aha- I just looked up and the Mariana Trench may be closer to 11 km deep. And maybe I've rounded some numbers too much, or used slightly too much approximation. So it might be right at the edge of working, in principle. We agree that in practice, this won't work.

I was a little familiar with this calculation beforehand because an irritating blog commentor had claimed it would be actually feasible in the Persian Gulf or some such place, where there's not nearly enough depth. It had been too tempting a target not to check.

Anyway, going through this more carefully was fun. Thanks for the questions!

Mike W.

posted without Lee checking, until his return from Paris

*(published on 10/16/2013)*