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According to Wikipedia, saltwater reverse osmosis occurs at 600-1200 PSI. Saltwater adds .455 PSI per foot, so 900 PSI would be at a depth of about 1980 ft or 603 meters. This is about halfway between the test depth and crush depth of a good submarine. Could some scale of effective freshwater generation happen naturally like this?

Edit: The facility has normal air pressure inside.

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  • $\begingroup$ That is, a 600-1200 PSI pressure difference across the membrane. $\endgroup$ – Solomon Slow Apr 17 '15 at 17:20
  • $\begingroup$ Yes, I inteneded for the internal pressure of the facility to be 1 atm $\endgroup$ – user78090 Apr 17 '15 at 17:58
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    $\begingroup$ The water that entered through the RO membrane would have to be pumped back out of "the facility" in order to continue the process indefinitely. Otherwise, internal pressure would increase until the process stopped. The cost of pumping the fresh water out, would be the same as the cost of pressurizing the sea-water in a surface-based facility. $\endgroup$ – Solomon Slow Apr 17 '15 at 18:08
  • $\begingroup$ Also, can you imagine a permeable membrane that withstands such pressure difference? $\endgroup$ – gigacyan Apr 17 '15 at 18:21
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    $\begingroup$ Gigacyan: Wouldn't any RO filter need to withstand that to work? $\endgroup$ – user78090 Apr 17 '15 at 18:24
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Yes, your idea would work, and it does save a bit of energy though not much.

Suppose we are making water by reverse osmosis at sea level. We have to pressurise the water to around 6.2MPa (900 psi) so the work needed to produce 1 cubic metre of fresh water is 6.2MJ.

Now let's do it your way. Let's assume already have a shaft sunk into the sea so we'll won't worry about the energy needed to create the shaft. The pressure at a depth $d$ is given by:

$$ P = \rho_s g d \tag{1} $$

where $P$ is our 6.2MPa and $\rho_s$ is the density of the salt water. We let one cubic metre of water seep into our shaft, then we have to pump it to the surface. The energy required for the pumping is:

$$ E = mgd = \rho_w g d \tag{2} $$

where $\rho_w$ is the density of pure water because it's pure water that we're pumping out, not saline. If we use equation (1) to substitute for $d$ we get:

$$ E = \frac{\rho_w}{\rho_s} = \frac{\rho_w}{\rho_s} 6.2 \text{MJ} $$

And because $\rho_g \lt \rho_s$ this energy is indeed a bit less than the energy it took to make the fresh water at the surface.

But the relative density of seawater is around 1.03, so the ratio $\rho_w/\rho_s$ is about 0.97. This means installing our shaft into the sea only reduced the energy needed to purify water by around 3%. In practice such a small saving is unlikely to pay back the cost of creating and maintaining a 600m deep shaft into the sea.

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It depends on the particular membranes that are employed. Example, if conversion rate is greater than 60 percent, I believe the savings will be small. However, if the membranes conversion rate is say only 30 percent, the savings can be large. Here is why:

A million gallons per day plant uses membranes with a 30 percent flux rate, than to produce the 1,000,000 gallons you will have to pump

Total water = 1,000,000/.30 = 3,333,333 gallons of water at RO pressure (Say 1,000 psi). Required horsepower is: HP = (3,333,333 * 1,000)/1714 = 1944.

However, the pure water is only 1,000,000 gallons, so the HP = (1,000,000 * 1,000)/1714 = 583. This is a savings of the cost of 1,361 more horsepower over the pumping schedule (normally 8 hours). In a year time, the horsepower savings is (365 days * 1,361 hp) = 496,765 yearly hp savings.

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