# Reverse osmosis perpetual motion machine

The original problem was as follows:

Closed-loop system, with complete isolation.
Salt water is green, fresh water blue,
porous plug (AB) is red.

Fresh water spills into the salt water column
at C if the fresh water is lifted
higher than the salt water,
maintaining continuous flow.

Can this really happen?

The theory that i worked out is that the pressure in the bottom of any of the columns is Pressure = density x constant of gravity x height

The density of pure water is 1000 kg/m^3, while the density of salt water is 1025 kg/m^3. This means that the pressure at the bottom of a salt water column is 2.5% higher than the fresh water.

The osmotic pressure needed to drive water from the salty side to the fresh side is 1450kPa, or around 14.5 bar.

If both columns are around 6km high, the osmotic pressure at the bottom of:
the salt water tank is 1025 * 9.8 * 6000 = 60.27 MPa, or 602.7 bar
the fresh water tank is 1000 * 9.8 * 6000 = 58.8 MPa, or 588 bar

The difference in pressure is 14.7 bar, enough to counter the 14.5bar requirement for reverse osmosis. The salt column's pressure now pumps 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.

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

I know it doesn't really work due to thermodynamics, but i can't figure out why in terms of fluid dynamics and statics. Please tell me why it doesn't work, i cant figure it out.

conway
This is surely the best perpetual motion machine I have ever seen. I have puzzled over it for a long time and sadly I must report that I have uncovered a flaw.

The column is very high and therefore the density of salt cannot be a constant. Just like the density of air is only 35% at the top of Mount Everest, the density of salt is accordingly lower at the top of your column. At the same time, the density is higher at the bottom of the column...high enough to overcome the gravitational differential and suck fresh water in through the porous plug, rather than the other way around.

This is surely the best perpetual motion machine I have ever seen. I have puzzled over it for a long time and sadly I must report that I have uncovered a flaw.

The column is very high and therefore the density of salt cannot be a constant. Just like the density of air is only 35% at the top of Mount Everest, the density of salt is accordingly lower at the top of your column. At the same time, the density is higher at the bottom of the column...high enough to overcome the gravitational differential and suck fresh water in through the porous plug, rather than the other way around.

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.

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.

I'm sorry to say, but you haven't managed to convince me.

Gold Member
Deos this machine do any useful work? Is that not a requirement of at least one class of PMMs?

cesiumfrog
Uh, the water is going to diffuse in the opposite direction.

Deos this machine do any useful work? Is that not a requirement of at least one class of PMMs?

When the water from the pure side spills over at C, it can perhaps turn a waterwheel.

Uh, the water is going to diffuse in the opposite direction.

Well, the idea is that the increased density of the salt-water will perform reverse osmosis. Have a look at the numbers.

Mentor
This problem appears to have stumped the author of the site you got that pic from. I suspect the answer has something to do with the energy associated with the chemical process of osmosis, which unfortunately, I don't fully understand. Perhaps a chemist can verify this, but a temperature change across the membrane either due to the chemical process or the thermodynamics of flow through a nozzle could provide the answer.

This hints at it, but unfortunately, I can only get the abstract:
The van't Hoff formula for osmotic pressure, which is identical in form to the formula for ideal gas pressure, is a direct outcome of the second law of thermodynamics. The formula is derived by applying a closed cycle reversible and isothermal process, following an argument in Fermi's book on thermodynamics.
http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000075000011000997000001&idtype=cvips&gifs=yes [Broken]

This implies to me that if you don't return that potential energy to the salt water (ie, you extract it with a turbine), the entire system will cool continuously until the water freezes.

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cesiumfrog
Where are you getting your numbers (for the osmotic pressure between water at high compression and brine at high compression)?

conway
His numbers are fine.

This problem appears to have stumped the author of the site you got that pic from. I suspect the answer has something to do with the energy associated with the chemical process of osmosis, which unfortunately, I don't fully understand. Perhaps a chemist can verify this, but a temperature change across the membrane either due to the chemical process or the thermodynamics of flow through a nozzle could provide the answer.

This hints at it, but unfortunately, I can only get the abstract: http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000075000011000997000001&idtype=cvips&gifs=yes [Broken]

This implies to me that if you don't return that potential energy to the salt water (ie, you extract it with a turbine), the entire system will cool continuously until the water freezes.

This I understand. I know the laws of thermodynamics will not allow a perpetual motion machine. What i want to know is the reasons why this will not work in terms of what is really going on, that is, in terms of fluid statics and osmosis.

Where are you getting your numbers (for the osmotic pressure between water at high compression and brine at high compression)?

Well, according to http://en.wikipedia.org/wiki/Osmotic_pressure, the osmotic pressure should simply be linear to the molarity (and temperature, but that is constant anyway). I looked up the water potential pressures in a table that i cannot find for the moment, but here is a diffrent example: http://books.google.se/books?id=oAF...5uWhCw&sa=X&oi=book_result&ct=result&resnum=2. This is for sucrose, but all that it might be different from salt is in its van't Hoff factor, but since we are considering salt-water with pure water, neither this plays a role.

Simply put, the molarity is the only thing that plays a role, and it serves to put a relatively lower osmotic pressure on the side with higher molarity.

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Gold Member
This implies to me that if you don't return that potential energy to the salt water (ie, you extract it with a turbine), the entire system will cool continuously until the water freezes.
Free refrigeration!

cesiumfrog
If osmotic pressure is independent of absolute pressure, then I would guess conway is right: that the energy extracted will be insufficient to mix the saline gradient.

Free refrigeration!

Well, yeah, even if this machine ends up freezing over, it still did free useful work, which is also thermodynamically impossible. Hence, we have a problem.

Yet I am still more concerned with the explanation that foils this plan without simply stating "There is no free lunch".

Mentor
This I understand. I know the laws of thermodynamics will not allow a perpetual motion machine. What i want to know is the reasons why this will not work in terms of what is really going on, that is, in terms of fluid statics and osmosis.
You're asking for an explanation without an explanation. Turbines and nozzles are thermodynamic devices, so the explanation of how they affect the system has to include thermodynamics. And I gave the explanation (I'm just not certain it is correct! ):

-When a fluid passes through a nozzle or a turbine, it loses energy, so unless that energy is returned to the system, it will continuously cool.

Now that I think about it, though, I'm not sure you even really need the chemistry of it - the thermodynamics would seem to be enough (though I haven't done the math to see if the energies are really equal...).

If osmotic pressure is independent of absolute pressure, then I would guess conway is right: that the energy extracted will be insufficient to mix the saline gradient.

Yes it seems to be that osmotic pressure is independent of absolute pressure, it was one of my first ideas to try to uncover the answer, but to no avail. You mean that the pressure exerted will not be enough to work backwards against the saline gradient, or did i misunderstand you?

Anyway it seems that, given that the osmotic pressure is independent, the pressure difference at those depths generated by difference in densities should be enough to make the process run in the correct (impossible) direction.

You're asking for an explanation without an explanation. Turbines and nozzles are thermodynamic devices, so the explanation of how they affect the system has to include thermodynamics. And I gave the explanation (I'm just not certain it is correct! ):

-When a fluid passes through a nozzle or a turbine, it loses energy, so unless that energy is returned to the system, it will continuously cool.

Now that I think about it, though, I'm not sure you even really need the chemistry of it - the thermodynamics would seem to be enough (though I haven't done the math to see if the energies are really equal...).

Hehe, well, i might be into a hopeless pursuit, but simply saying "No device can make free energy" feels kind of an unsatisfactory explanation, when the hydro-statics and osmotic mechanisms all declare that the pure water should rise above the salt-water, period. And thus generate potential energy out of nowhere.

There must be something in terms of fluid statics or the mechanisms of osmosis that I have neglected for the theory to converge with thermodynamics. And this neglected theory is what I'm trying to recover here.

Mentor
Well, yeah, even if this machine ends up freezing over, it still did free useful work, which is also thermodynamically impossible. Hence, we have a problem.
No, it isn't. The system is then essentially just a charged battery.

No, it isn't. The system is then essentially just a charged battery.

The problem becomes that when the risen water at point C goes to the level where it pours back into the salt-water, it is still in the same "charged" state, while water is pouring.

Mentor
Hehe, well, i might be into a hopeless pursuit, but simply saying "No device can make free energy" feels kind of an unsatisfactory explanation....
I didn't say that! I didn't specifically invoke any of the laws of thermodynamics in my explanation!
...when the hydro-statics and osmotic mechanisms all declare that the pure water should rise above the salt-water, period. And thus generate potential energy out of nowhere.
Well that's kinda the point: if the system is cooling, then they are converting heat energy to potential energy. There is no violation implied in that unless you try to extract energy from it indefinitely. It looks like this:

Without a turbine:

heat energy input - mechanical potential energy output = 0
That's a conservation of energy statement

With a turbine:
heat energy input - mechanical potential energy output - mechanical kinetic energy output < 0
That says the energy will decrease continually, a violation of conservation of energy. The water freezing is a manifestation of it.

cesiumfrog
the entire system will cool continuously until the water freezes.
No, it isn't [a problem]. The system is then essentially just a charged battery.
Again with the Clausius statement Russ? A battery that can recharge itself from ambient thermal energy is unacceptable.
You're asking for an explanation without an explanation. Turbines and nozzles are thermodynamic devices, so the explanation of how they affect the system has to include thermodynamics.
That's just being obtuse; The fact that a problem can sometimes be solved by parroting global arguments does not alter the fact it always can also be solved purely by integration of local forces.

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I didn't say that! I didn't specifically invoke any of the laws of thermodynamics in my explanation!
Well that's kinda the point: if the system is cooling, then they are converting heat energy to potential energy. There is no violation implied in that unless you try to extract energy from it indefinitely. It looks like this:

Without a turbine:

heat energy input - mechanical potential energy output = 0
That's a conservation of energy statement

With a turbine:
heat energy input - mechanical potential energy output - mechanical kinetic energy output < 0
That says the energy will decrease continually, a violation of conservation of energy. The water freezing is a manifestation of it.

The 1st law of thermodynamics, the law of conservation of energy is what you describe. There is the second law, the law I am referring to, that makes this freezing over, charged battery, argument impossible. As you might know, this law deals with work, not just energy. You are correct in assuming that the 1st cycle of the freezing-over of the system might represent energy stored in the initialization, but if we assume that the waste heat of whatever appliance we used the useful work from the turbine for, returns back to heat both water columns again, then we still get useful work without freezing the columns.
Hence, the system still represents free energy (or rather work), which is of course impossible.

Still, my main aim here is not to state that this will work or that it will not work, what I am looking for is the missing osmosis / fluid-static theory that will make the math converge with thermodynamics.

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The fact that a problem can sometimes be solved by global energetic arguments does not alter the fact it always can also be solved purely by integration of forces.

Thank you cesiumfrog, we are getting back to the right track. It is trying to get the differential density powered reverse osmosis to "not work" in terms of density and osmosis that I'm really looking for.

Where I live, in Stockholm, Sweden, it is approaching 5 AM and I have school tomorrow, but keep it up, Ill join you tomorrow.

Homework Helper
Gold Member
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.

madcowswe, can you point me to a reference for this claim? I noticed that you assumed incompressibility in your calculations, and equal compressibility would also support device operation, but I'm not convinced that a difference in compressibility doesn't play a factor in the actual non-operation of the device.

conway
The differential compressibility of salt water vs fresh water is not relevant.

Homework Helper
Gold Member
The differential compressibility of salt water vs fresh water is not relevant.

Please explain; to me it looks like this factor affects the differential pressure at the semipermeable membrane.

EDIT: Specifically, the change in pressure with height is

$$\frac{dP}{dh} = \rho g+\frac{d\rho}{dh}gh$$

where $\frac{d\rho}{dh}$ is assumed to be zero in madcowswe's first post; in post #3, $\frac{d\rho}{dh}$ is assumed to be equal for salt water and pure water. I'm asking what happens if we drop this assumption.

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conway
Please explain; to me it looks like this factor affects the differential pressure at the semipermeable membrane.

EDIT: Specifically, the change in pressure with height is

$$\frac{dP}{dh} = \rho g+\frac{d\rho}{dh}gh$$

where $\frac{d\rho}{dh}$ is assumed to be zero in madcowswe's first post; in post #3, $\frac{d\rho}{dh}$ is assumed to be equal for salt water and pure water. I'm asking what happens if we drop this assumption.

If we drop this assumption it only makes the problem more complicated. What I am saying is that there is no need to worry about the liquid compressibilities...we can take both fluids to be totally incompressible.

If we drop this assumption it only makes the problem more complicated. What I am saying is that there is no need to worry about the liquid compressibilities...we can take both fluids to be totally incompressible.

Exactly. While they are not totally incompressible, the density increases with such a tiny amount for both salt-water and regular water, that it is negligible. Furthermore, as it increases pretty much exactly with the same factor for both of them, we can neglect this change.

Mentor
2021 Award
Be careful here, "salt water" is not well defined. What you need is something more like the density and osmotic pressure of 1 M NaCl @ 20°C and the density of pure water @ the same temperature. It is possible that your listed osmotic pressure and density are for different solutions.

I am also not sure that the compressibility is the same for both, although I agree that the compressibility is low for both.

Homework Helper
Gold Member
If we drop this assumption it only makes the problem more complicated. What I am saying is that there is no need to worry about the liquid compressibilities...we can take both fluids to be totally incompressible.

Well, the point of the thread, as created by madcowswe, would seem to be to look deeper into the physics of the problem to see specifically what prevents the perpetual motion machine from working. The movement of water from the salty side to the fresh side at the membrane is controlled by the gradient in chemical potential, itself a function of the pressures. The visualized machine purportedly operates by a relatively small pressure differential between two 6km towers of liquid, so it doesn't seem unreasonable to ask for the pressure to be calculated more rigorously.

But since I'm the one bringing it up, I realize that the onus is on me (and any other interested parties) to refine the calculation and see if the difference is indeed negligible.

conway
Be careful here, "salt water" is not well defined. What you need is something more like the density and osmotic pressure of 1 M NaCl @ 20°C and the density of pure water @ the same temperature. It is possible that your listed osmotic pressure and density are for different solutions.

It doesn't matter. First, 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.

The laws of thermodynamics must work for theoretical ideal gasses the same as for real gasses. Similarly, they must apply equal well to Einstein2nd's imaginary water solution as they do to real saline water.

Gold Member
Would the flow through the membrane really increase the volume of a 6 km tall column of liquid? My intuition is that the mass flow at the bottom would not increase the volume at the top, but only add a density gradient to the fresh water side.

Homework Helper
Gold Member
It doesn't matter. First, 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.

The laws of thermodynamics must work for theoretical ideal gasses the same as for real gasses. Similarly, they must apply equal well to Einstein2nd's imaginary water solution as they do to real saline water.

What post are you referencing? (Are madcowswe and Einstein2nd the same person?)

Count Iblis
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
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.