# Would this pump water up? and if so, how far?

I had this idea of an osmotic pump way back in high school and I never got a satisfactory answer if it would work. If I had this configuration:

Would it continually pump water up given ambient heat so long as the bottom reservoir is full?

EDIT To explain what is happening, there is pure water in each of the dark blue reservoirs, saline in the cyan containers (same concentration of saline in each container) and a semipermeable membrane at the bottom and near the top of each container.

The pure water in each reservoir would be sucked up into each container directly above it due to osmotic pressure (high water concentration flows to lower water concentration), and then dumped out at the top also due to osmotic pressure (saline to air which is almost 0% water concentration). Since the membrane is not permeable to salt, only the water is released from the container into the next higher reservoir.

NOTE that the membrane at the top of each saline container doesn't touch the pure water in the reservoir it empties into. I'm also thinking that the saline containers may have to be completely filled with extra osmotic pressure to spare to counteract the pure water that sticks to the outside of the upper membrane and cause a reverse osmosis effect. Other possible tricks relying on surface tension and gravity might also aid in pulling the water away from the membrane. END EDIT

If enough were stacked, would this allow for transporting water higher than the maximum that trees can transport (about 138m according to this article)?

• It is tough to understand what is happening. If you could provide a more detailed explanation of your setup, it would help. Aug 24, 2013 at 16:05
• Added further explanation. Aug 24, 2013 at 16:13
• How is it dumped out at the top due to osmotic pressure? The semi-permeable membrane separetes what at the top where water is dumped? Aug 24, 2013 at 16:49
• It is dumped out into the air (~0% water concentration) and then falls into the reservoir. The main problem I've thought of is trying to get the pure water from sticking next to the semipermeable membrane. Since it is 100% water concentration, it would impede the water transport across the membrane. Aug 24, 2013 at 16:54
• @Adrian that dictionary definition does not go into enough depth to get the concept across. Please read up on "perpetual motion machines of the second kind." These are machines that take energy from the environment in the form of heat and turn it into work. They are just as impossible as perpetual motion machines of the first kind, which is what your link describes. In general it is a bad idea to rely on dictionary definitions for physics concepts - you should check Wikipedia at the very least, and preferably also a good text book. Aug 25, 2013 at 6:40

A quick sanity test for this sort of idea is this: "if it worked, could I use it to construct a perpetual motion machine?" In this case, yes, you could - all you need to do is let the water flow back down from the top reservoir to the bottom again, via a waterwheel, and you'd have an endless source of work without putting any energy into the system. This means that your idea breaks the first or second law of thermodynamics somewhere along the way, and the only remaining task is to figure out where exactly this happens.

The problem in this case is that you're relying on the air being "almost 0% water concentration" (i.e. this machine is only supposed to work on dry days, when it's not very humid.) You're expecting that when the water moves out of the upper membranes it will stay in a liquid phase and drip down into the reservoirs below.

However, when the water molecules move from region of liquid water into a region of dry air, they don't stay liquid but become vapour. We call this evaporation. Whether water can evaporate through the membrane depends on the balance between the vapour pressure of water in the air and the osmotic pressure of water in the solution. If the air is dry then water molecules will indeed move out (very slowly) from the upper membranes, but they will turn into water vapour. They won't condense back into the upper reservoirs, because the air is dry and doesn't want to give up that moisture. In fact, the upper reservoir will be losing water to evaporation itself, at a much faster rate.

So when the air is dry enough for water to pass through the upper membranes, all that will happen is that the water in all of the reservoirs will evaporate. You will not see an accumulation of water in the upper reservoir.

• Actually, the energy was to come from the thermal ambient energy. When I thought of this idea, I knew that it wouldn't work at absolute zero, or even at zero Celsius. Also, this was supposed to be a closed system (not thermally, but physically). Water wouldn't be able to enter or leave the system, and once at the top, it would fall back to the bottom reservoir. Though, I hadn't considered the air/water vs the air/salt concentrations, which might cause failure. Aug 25, 2015 at 23:28
• If the energy is to come from thermal ambient energy then you are breaking the second law. This is called a "perpetual motion machine of the second kind," and it's just as fundamentally impossible as creating energy from nothing (which breaks the first law and is called a perpetual motion machine of the first kind). Aug 26, 2015 at 10:52
• It doesn't have to be 100% efficient. The energy comes from the ambient thermal energy like for a Stirling Engine. This is not getting it's energy from itself, but from the environment around it. Thus, it would not be a perpetual motion machine of the second kind. Aug 27, 2015 at 13:01
• @adrian a Stirling engine does not turn ambient heat into work, it extracts work from the temperature difference between two thermal reservoirs. It is impossible to do it with only one. That is exactly what "perpetual motion machine of the second kind" means. Aug 31, 2015 at 0:07
• Looking at this again, a "quick sanity check" would show that this would not create a perpetual motion machine as, when the ambient thermal energy disappears, the machine would stop working, like a solar panel stops generating a potential difference when there is no sun. A Stirling engine extracts work from the temperature difference between two thermal reservoirs, but this would extract work from the difference between two gravitational potentials. I guess reservoirs and potentials are the keywords that would keep this from working? Jul 30, 2018 at 15:57

It won't work. It is true that a semi-permeable membrane can raise a column of salt water until the pressure (due to the column) matches the osmotic pressure across the semi-permeable membrane.

The problem comes at the top of each level: how do you envision that water gets out of the salt column into the next higher pool?

If the fresh water in the upper pool actually touches the membrane at the top of the column, it will flow into the column, increasing the water level, increasing the pressure at the bottom, forcing water down into the bottom pool. Thus you end up draining the upper pool into the lower one.

If it doesn't actually touch the membrane, you seem to be assuming that for some reason water will drip out. But it won't. If you have a film of water on the other side of the membrane, the osmotic pressure will suck it into the saline. (In fact, water will be sucked into the saline from the air, as the vapor pressure above the membrane will be lower than the vapor pressure above water, creating net flow of water into the saline.)

• Yes, I was thinking that the problem would most likely occur at the top membrane. I tried to show in the diagram that the upper water reservoir doesn't allow the water to touch the membrane. Your last paragraph is a concern though. An idea would be get the saline to entirely fill the container with osmotic pressure to spare. If the lower membrane is deep enough in it's pool, the water would tend to leave through the upper membrane. This might be augmented by somehow pulling the water away from the membrane using some surface tension and gravity tricks. Possible? Aug 25, 2013 at 5:55
• @Adrian - No, not possible, due to conservation of energy. You can verify by working out the effective pressure for this new scheme (that is, water will go in at the upper membrane, as the inward pressure there will be stronger than at the bottom). Aug 25, 2013 at 9:46