# Using capillary action to generate energy

Hello. I've had this idea floating around in my head for a long time and lately it's really been bugging me.

Say you had a tray of water, and suspended another tray above that, and then you connected the two trays vertically with some very thin tubes, say, 0.2mm in diameter.

Now, due to capillary action, water should travel up the tubes in to the top tray. If we have an outlet that allows the water in the top tray to freely pour back in to the bottom tray, the water would travel back down.

My first question is: with this set up, would this create a perpetual cycle of water between the trays?

Next, suppose that it does create such a continuous waterfall, and we put a waterwheel in the path of the falling water, making it turn. Now we attach an electric generator to the wheel, we would produce energy.

My second question is: given the number of tubes, their diameter, the diameter of the waterwheel and the electricity generation capacity of the generator, how do you calculate how much electricity the setup would produce?

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Mapes
Homework Helper
Gold Member
This setup won't produce any energy. The water in the top tray will run down the capillary tube, which is the opposite direction to what you're assuming will happen.

And unfortunately, no augmentation will make this device work. What you're proposing would effectively create energy, which just isn't possible.

Dale
Mentor
Capillary action relies on surface tension. The water will not go out of the tube and into the top tray because that would require breaking the surface tension which would prevent the capillary action in the first place.

IBTL!

This would be a nifty idea if it were possible! :)

Andy Resnick
Hello. I've had this idea floating around in my head for a long time and lately it's really been bugging me.

Say you had a tray of water, and suspended another tray above that, and then you connected the two trays vertically with some very thin tubes, say, 0.2mm in diameter.

Now, due to capillary action, water should travel up the tubes in to the top tray. If we have an outlet that allows the water in the top tray to freely pour back in to the bottom tray, the water would travel back down.

My first question is: with this set up, would this create a perpetual cycle of water between the trays?

Next, suppose that it does create such a continuous waterfall, and we put a waterwheel in the path of the falling water, making it turn. Now we attach an electric generator to the wheel, we would produce energy.

My second question is: given the number of tubes, their diameter, the diameter of the waterwheel and the electricity generation capacity of the generator, how do you calculate how much electricity the setup would produce?
I think this is actually a very clever idea, because it's not obvious where the problem is: clearly, water (for example) will wick up a thin capillary, and if the upper tray is hydrophilic, the water will then be pulled away from the capillary mouth. Or, if the capillary mouth is slightly elevated from the tray surface, the water could simply flow down into the upper tray. Another solution could be smoothly joining end of the tube to the tray so the contact line is not pinned, or by putting the capillary into a porous substance which pulls the water out of the capillary.

It seems logical: the reason the water wicks up the tube is wetting- the interfacial energy of the glass-air is lowered by being glass-water. The water will wick upward until the increase in potential energy exactly counterbalances the decrease in surface energy. So it seems that raising the water *lowers* the total energy, creating a return path for this cycle.

What is missing is the curvature of the interface- there is a pressure jump across the interface due to Laplace's equation dP = -2sk, where 's' is the air-water interfacial energy, and k the surface curvature. While the water is rising, the surface is concave (since the water wets the glass) and so the pressure on the water side is *lower* than the air side. In order for the water to flow out of the tube, the curvature of the interface must decrease (the surface must 'spread out') and so the pressure difference equalizes, stopping the flow.

That's easy to test- get a capillary tube, stick it into a pan of water, and see if water squirts out of the tube or not. I think the water will stop rising when the interface becomes flat with the glass.

I had the self same idea. I think that with a little experimentation it will work.

This is how

1) you need multiple trays with multiple stages (e.g. 100) that elevate the water up to a good height, 5 metres is acceptable in a residential area.

2) you need multiple capillary tubes in a bank for each stage, the outlet of which must be kept clear of the upper trays water surface to allow the water to break out into the tray.

3) the device should be designed to generate power for a tyipcal household, say 3KWatts.
start with the quantity of water required and then derive the tray sizes, height and capillary tube from this.

The key is the flow back up needs to be sufficient to provide the flow back down.

I would then feed the turbine generator's electricity into a UPS so that a good clean source of electrical power is available to the house user.

Once you have started to get a handle on the power you can generate you can see if it is practical to build commercially.

Of course water flows through our water reticulation systems all the time but nobody is harnessing this water flow power, yet.

Multivits

Dale
Mentor
Even a single stage won't work

GT1
I think this is actually a very clever idea, because it's not obvious where the problem is: clearly, water (for example) will wick up a thin capillary, and if the upper tray is hydrophilic, the water will then be pulled away from the capillary mouth. Or, if the capillary mouth is slightly elevated from the tray surface, the water could simply flow down into the upper tray. Another solution could be smoothly joining end of the tube to the tray so the contact line is not pinned, or by putting the capillary into a porous substance which pulls the water out of the capillary.

It seems logical: the reason the water wicks up the tube is wetting- the interfacial energy of the glass-air is lowered by being glass-water. The water will wick upward until the increase in potential energy exactly counterbalances the decrease in surface energy. So it seems that raising the water *lowers* the total energy, creating a return path for this cycle.

What is missing is the curvature of the interface- there is a pressure jump across the interface due to Laplace's equation dP = -2sk, where 's' is the air-water interfacial energy, and k the surface curvature. While the water is rising, the surface is concave (since the water wets the glass) and so the pressure on the water side is *lower* than the air side. In order for the water to flow out of the tube, the curvature of the interface must decrease (the surface must 'spread out') and so the pressure difference equalizes, stopping the flow.

That's easy to test- get a capillary tube, stick it into a pan of water, and see if water squirts out of the tube or not. I think the water will stop rising when the interface becomes flat with the glass.
Why the attached configuration will not work? why can't you use gravity to overcome the pressure jump at the interface?

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Dale
Mentor
Why the attached configuration will not work? why can't you use gravity to overcome the pressure jump at the interface?
Because if gravity could overcome the pressure jump at the interface then gravity would be too strong for it to wick up the capillary in the first place.

GT1
Because if gravity could overcome the pressure jump at the interface then gravity would be too strong for it to wick up the capillary in the first place.
Ok-I understand, but what if at the outlet we will use pipe with variable cross section so capillary won't be the dominant force and the liquid will fall due to gravity, while at the inlet the liquid will raise due to capillary forces.
Why now it won't work?

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The answer has to do with the angle of conact.Water in glass has an acute angle of contact(which is zero with very clean glass and pure water) and the pressure just above the meniscus is greater than that just below by an amount equal to 2 gamma/r (where gamma is the surface tension and r is the radius of the meniscus).This pressure difference causes the liquid to rise in the tube to a height(h) where the pressures due to surface tension and the hydrostatic pressure balance;

h*pho* g=2 gamma/r

Here comes the clever bit.If the tube is pushed down in an attempt to have water continually flowing out then when the top of the tube reaches the top of the meniscus the angle of contact starts to increase and the meniscus starts to flatten out with its radius increasing.This results in a reducing pressure difference across the meniscus and a corresponding reducing capillary rise.When the top of the tube reaches the top of the water in the container the meniscus is perfectly flat and the pressure difference across it is zero

Dale
Mentor
Ok-I understand, but what if at the outlet we will use pipe with variable cross section so capillary won't be the dominant force and the liquid will fall due to gravity, while at the inlet the liquid will raise due to capillary forces.
Why now it won't work?
In addition to Dadface's comments, if gravity is the dominant force at the outlet then all you have is a siphon. You could probably design a self-starting siphon in this way, which could be nice in the case of toxic liquids with a high surface tension, but would not violate conservation of energy in any way.

GT1
The answer has to do with the angle of conact.Water in glass has an acute angle of contact(which is zero with very clean glass and pure water) and the pressure just above the meniscus is greater than that just below by an amount equal to 2 gamma/r (where gamma is the surface tension and r is the radius of the meniscus).This pressure difference causes the liquid to rise in the tube to a height(h) where the pressures due to surface tension and the hydrostatic pressure balance;

h*pho* g=2 gamma/r

Here comes the clever bit.If the tube is pushed down in an attempt to have water continually flowing out then when the top of the tube reaches the top of the meniscus the angle of contact starts to increase and the meniscus starts to flatten out with its radius increasing.This results in a reducing pressure difference across the meniscus and a corresponding reducing capillary rise.When the top of the tube reaches the top of the water in the container the meniscus is perfectly flat and the pressure difference across it is zero
Why at the outlet the hydrostatic pressure needs to balance the surface tension? - gravity is at the direction of capillary force- they are both facing down.

In addition to Dadface's comments, if gravity is the dominant force at the outlet then all you have is a siphon. You could probably design a self-starting siphon in this way, which could be nice in the case of toxic liquids with a high surface tension, but would not violate conservation of energy in any way.
If self-starting siphon is possible and both tanks are at the same height I think it violates the conversation of energy because once it started what will stop it following?

Dale
Mentor
If self-starting siphon is possible and both tanks are at the same height I think it violates the conversation of energy because once it started what will stop it following?
Do you understand how a siphon works? If both tanks are at the same height then the siphon doesn't move any fluid. Siphons only move fluid downhill from a higher level to a lower level, as you have drawn above.

You cannot have it both ways. Either the surface tension is the dominant force in which case the fluid will not leave the capillary or gravity is the dominant force in which case the fluid will only flow downhill. Either way energy is conserved.

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GT1
Do you understand how a siphon works? If both tanks are at the same height then the siphon doesn't move any fluid. Siphons only move fluid downhill from a higher level to a lower level, as you have drawn above.
I misunderstood you. the drawing is wrong-both tanks are at the same height and connected- there can't be any level difference.

Dale
Mentor
I misunderstood you. the drawing is wrong-both tanks are at the same height and connected- there can't be any level difference.
Then the force due to gravity always acts to pull the liquid back to the left tank, and we are back with the original situation and explanation. The water will not flow out of the capillary due to the same surface tension that drew it up in the first place. See Andy Resnick's and Dadface's detailed explanations of the original situation.

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Redbelly98
Staff Emeritus