Reverse osmosis perpetual motion machine

In summary: Pressure = density x constant of gravity x height.The denser the liquid, the higher the pressure. The salt water column has a higher density than the fresh water column, so the pressure at the bottom of the salt water column is higher than the pressure at the bottom of the fresh water column. This pressure pushes fresh water into the fresh water column. If the freshwater feeds back into the saltwater column, after perhaps passing a waterwheel that generates power, this becomes a perpetual motion machine. Unfortunately, the van't Hoff formula only works for liquids that are completely incompressible.
  • #36
Count Iblis said:
The chemical potential of the salt will be constant as a function of the height. The gravitational potential contributes to the chemical potential, and this then leads to a formula for the concentration of the salt as a function of the height, analogous to the formula that gives you the pressure as a function of the height in the atmosphere. This means that the concentration of the salt at the bottom is much higher than at the top. So, the osmotic pressure is higher too.

Count Iblis said:
Another way to see this is to simply consider hydrostatic equilibrium. The water will only flow through the membrane if the partial pressure of the water becomes higher on the left side than on the right side if we add the salt there. But this is exactly what does not happen. The total pressure increases due to the weight of the salt, but this is accounted for by the partial presssure of the salt ions.

I agree to what you are saying: That the slight compressibility of the salt-water will serve to increase the relative concentration of the salt-water at the bottom of its column. Yet at these depths, as given before, the percentage change is much lower than the difference in density between salt-water and fresh water. This means that it is true that this will shift the equilibrium a tiny bit in the favor of water running from fresh to salt, but all you have to to is to increase to, say, 6.01 km and we are back to the same paradox again.
 
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  • #37
Mapes said:
What post are you referencing? (Are madcowswe and Einstein2nd the same person?)

Sorry, I got confused. I meant madcowswe, the OP. The other guy was someone from a different thread.
 
  • #38
The compressibility of the (salt) water is irrelevant. The salt ions can still move freely in the water.

If we start with fresh water and then add salt, the additional pressure delta P as a function of height will be exactly that of the barometric height formula. Therefore the extra pressure at the bottom is due to the salt only, the partial pressure of the water at the bottom has not been changed at all. So, no water will flow through the membrane.
 
  • #39
Count Iblis said:
If we start with fresh water and then add salt, the additional pressure delta P as a function of height will be exactly that of the barometric height formula.

Yes, that's what I meant when I said it was like the density of air at the top of Mt. Everest. At a height of 6 km the water has much less salinity than the water down at the membrane.
 
  • #40
There will be diffusion of salt ions at the top of the loop,the salt concentration will tend to equalise on both sides and the flow will slow down and stop.
 
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  • #41
conway said:
Yes, that's what I meant when I said it was like the density of air at the top of Mt. Everest. At a height of 6 km the water has much less salinity than the water down at the membrane.

I see now! I didn't read all the postings at first...
 
  • #42
Simple observation: Suppose you hold a beach ball at the bottom of a swimming pool and then release it. It rises to the top of the pool. Have you created potential energy out of nothing? No. First, the gravitational potential gained by the risen ball, mgh, is (disregarding friction, etc.) equal to the buoyant potential energy of the ball held down at the bottom of the pool -- it's just a conversion of one type of energy to another -- and second, there is vibrational/heat energy generated by the friction of the water on the rising ball and the wave-energy generated by the ball bobbing on the surface of the water. But the biggest factor is that the buoyant potential energy is converted to kinetic energy, and then comes to rest in gravitational potential energy and wave motion.

You always loose more going out than you put in. (That's the "second law".)
 
  • #43
worldrimroamr said:
Simple observation: Suppose you hold a beach ball at the bottom of a swimming pool and then release it. It rises to the top of the pool. Have you created potential energy out of nothing? No. First, the gravitational potential gained by the risen ball, mgh, is (disregarding friction, etc.) equal to the buoyant potential energy of the ball held down at the bottom of the pool -- it's just a conversion of one type of energy to another -- and second, there is vibrational/heat energy generated by the friction of the water on the rising ball and the wave-energy generated by the ball bobbing on the surface of the water. But the biggest factor is that the buoyant potential energy is converted to kinetic energy, and then comes to rest in gravitational potential energy and wave motion.
While true, you have yet to demonstrate that it is relevant to this discussion.

The beachball will rise to the surface and then stay there. We know that there is no energy in the system that would sink the beachball again, allowing the process to continue cyclically. In effect, it was merely stored potential energy that was released.

It is not clear yet that the RO system will stop when it releases some amount of potential energy and settles to equilibrium. (We're all sure it does, we're just not sure how or where.)
worldrimroamr said:
You always lose more going out than you put in. (That's the "second law".)
Yes. This is not in contention. The question at hand is: where is it lost?
 
  • #44
This is a very long thread, and I haven't read it all. But cesium frog has got it right.

Has anyone noticed that the drawing in post #1 is wrong? The salt water should be at a higher level than the fresh. I imagine that any supporting argument invloving osmotic pressure may have put the pressure on the wrong side of the equation, or droped a negative sign.
 
  • #45
Phrak said:
This is a very long thread, and I haven't read it all. But cesium frog has got it right.

Has anyone noticed that the drawing in post #1 is wrong? The salt water should be at a higher level than the fresh.
If that were true, the salt water would diffuse into the fresh. How would the fresh get into the salt uphill?
(Or is this the ideas fatal flaw?)
 
  • #46
DaveC426913 said:
If that were true, the salt water would diffuse into the fresh. How would the fresh get into the salt uphill?
(Or is this the ideas fatal flaw?)

In the drawing the fresh water is show at a higher level. This is wrong--or it is something that will not last for long. The fresh water is drawn into the salt water through the membrane. The salt water would rise, spill over, and put saltwater into the fresh water.

(to bad. and I am always so hopeful of something that violates the second law--really.)
 
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  • #47
Phrak said:
This is a very long thread, and I haven't read it all. But cesium frog has got it right.
I trust you are referring to my agreement with conway's original reply, rather than my early posts.
 
  • #48
I'm referring to your first post, #5.
 
  • #49
madcowswe said:
osmosis2.gif




some considerations noted: the saltwater will not become diluted by the fresh water pouring over at the top since the salt never leaves the column through the semi-permeable membrane

.

This is a clarification of my earlier post.Assume that once set up there is a flow from right to left at the top.If there is a flow there will be a dilution of the salt water because water and salt ions will diffuse via the water bridge at the top of the loop.The membrane at the bottom may be able to stop the flow of salt ions but there is no membrane at the top.
 
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  • #50
Phrak said:
In the drawing the fresh water is show at a higher level. This is wrong--or it is something that will not last for long. The fresh water is drawn into the salt water through the membrane. The salt water would rise, spill over, and put saltwater into the fresh water.
You lost me. If fresh water is being drawn out of the left column through the membrane, why would the left column rise?
 
  • #51
Dadface said:
This is a clarification of my earlier post.Assume that once set up there is a flow from right to left at the top.If there is a flow there will be a dilution of the salt water because water and salt ions will diffuse via the water bridge at the top of the loop.The membrane at the bottom may be able to stop the flow of salt ions but there is no membrane at the top.

Have you ever seen salt diffuse up a waterfall?
 
  • #52
Phrak said:
This is a very long thread, and I haven't read it all. But cesium frog has got it right.

Has anyone noticed that the drawing in post #1 is wrong? The salt water should be at a higher level than the fresh. I imagine that any supporting argument invloving osmotic pressure may have put the pressure on the wrong side of the equation, or droped a negative sign.

You are correct, if this would involve osmosis, but it involves reverse osmosis driven by the density of the salt water.
 
  • #53
Count Iblis said:
The compressibility of the (salt) water is irrelevant. The salt ions can still move freely in the water.

If we start with fresh water and then add salt, the additional pressure delta P as a function of height will be exactly that of the barometric height formula. Therefore the extra pressure at the bottom is due to the salt only, the partial pressure of the water at the bottom has not been changed at all. So, no water will flow through the membrane.

The change in partial perssure due to salt ions, and the change in barometric pressure due to the density change due to the salt ions are not exactly the same, which is the basis of the paradox.
 
  • #54
The reason why there exists such a thing as osmotic pressure is simply because you are letting molecules of one type pass but not of the other type. This is the relevant physics, any formulas that are derived (theoretically or from experiments) can come with their small print with assumptions that may not be alway valid.

If you have a membrane that doesn't let the salt pass then you can add as much salt as you like one one side, it won't cause the water to move through. The more salt you add, the higher the osmotic pressure will become precisely because the membrane equalizes the partial pressure of the water and doesn't care about the salt.
 
  • #55
madcowswe said:
Have you ever seen salt diffuse up a waterfall?

I can't see any reason why salt will not diffuse upwards, the thermal velocity of the salt ions will be high compared to the velocity of the falling water and each salt ion will follow a random walk path.The only difference I see is that diffusion against the flow will be slowed down.
 
  • #56
madcowswe, Dadface. Well pshaw. I guess I'm it. Oh well, there always has to be some guy slower witted than the rest. :smile:
 
  • #57
conway said:
I believe Einstein2nd has given a combination of density and osmotic pressure that is reasonably correct for some particular concentration of salt water. And if he hasn't, then his numbers are in any case consistent with some imaginary solution...all that needs to change is the atomic weight of the ions.
My concern is that he may be giving osmotic pressure for one solution and density difference for another solution. Such a combination of density difference and osmotic pressure may itself violate thermodynamics and thus be impossible.
 
  • #58
DaleSpam said:
My concern is that he may be giving osmotic pressure for one solution and density difference for another solution. Such a combination of density difference and osmotic pressure may itself violate thermodynamics and thus be impossible.

Hard to see how that could be. You can easily adjust the density while holding the osmotic pressure constant simply by varying the atomic weight of the solute particles. And even if you object that there is no actual positive ion with molecular weight 45 or whatever, I can't see a thermodynamic reason why there couldn't theoretically be such an ion.
 
  • #59
To approch the problem obliquely, what would happen at both the top interface and the membrane interface if the solvent were water and the solute a gas such as carbon dioxide? One side is a column of water, and the other is a column of CO2.
 
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  • #60
madcowswe said:
Water is not nearly as compressible as air, it is almost completely incompressible. Yet it's true that neither water nor salt-water is truly incompressible, but it doesn't really play a role in this problem since water and salt-water both increase their densities with pressure at the same relative rate.

"WORNG!" Salt water is around 10ppm/atm more compressible than fresh water.
Furthermore, the 2.5% difference in density is never countered, even if the compressiveness of the liquids were to prove to be very slightly different:
"The low compressibility of water means that even in the deep oceans at 4000 m depth, where pressures are 4×10^7 Pa, there is only a 1.8% decrease in volume.", from Wikipedia.
(snip)

Apply the Poynting correction to water activities in salt water and in fresh water at the pressure for whatever depth you're examining (use the different compressibilities for the two to calculate the partial molal volumes for water, fresh and saline, and the differences), and you will see a difference.
 
  • #61
Bystander said:
"WORNG!" Salt water is around 10ppm/atm more compressible than fresh water.

Bystander, could you give a reference for this please?
 
  • #62
DaveC426913 said:
It is not clear yet that the RO system will stop when it releases some amount of potential energy and settles to equilibrium. (We're all sure it does, we're just not sure how or where.)

For a dilute solution, letting [itex]r[/itex] be the ratio between the column height and the fundamental length scale [itex]{i_H k_B T}({\kappa g})^{-1}[/itex] (where [itex]\kappa[/itex] is the difference in mass between a solute molecule and the solvent it displaces), reverse osmosis occurs iff the concentration at the semipermeable membrane (as a fraction of the mean concentration) is less than [itex]r[/itex]. Thus the solvent will start cycling (provided [itex]r[/itex]>1 and that the solution is initially mixed and not yet settled), however, at equilibrium (when the Archimedean force balances the osmotic potential gradient) the concentration increases exponentially with depth and the reverse osmosis cycle halts because the concentration at the membrane is [tex]\frac{r}{1-e^{-r}}[/tex].

Not sure yet how to generalise this to non-dilute solutions.
 
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  • #63
Here is a detailed explanation: http://www.lhup.edu/~dsimanek/museum/osmosis.htm

The tone of the linked webpage is condescending at times (and starts off that way), but the information is useful.
 
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