# Harvesting the energy of sea waves using capacitance

Summary:
capacity changes as waves move up and down
can the changing current be rectified?
Hello. I have a question related to capacitance.

here is another mechanical solution to harness wave energy.

Can changing capacity of the capacitor be used to harness wave energy?

Let's say we envelope conductive plates in plastic and submerge them
in the ocean and the water in between them will act as the other set of plates.

As the water level moves up and down, can this generate movement
of current that would then be sent to bridge rectifiers and capacitors?

This Ukrainian experimenter shows it with capacitor plates he made himself.
He can power a neon bulb with the charge.

In my case, the water acts as the second capacitor plate.
I made two arrays to eliminate the need for exposed sea water ground that will corrode.
This is a very simplified illustration.

Is this however possible to use to produce electricity cost-effectively?
Metal sheets that are enveloped in plastic can be made very thin.

Would this be cost-effective (with no moving parts, other than the sea water itself)
What would define the thermodynamic limit?

Baluncore
Is this however possible to use to produce electricity cost-effectively?
Your idea is similar to a parametric amplifier or a capacitance multiplier. Both systems would need to be more complex, and require external excitation. Although the systems can produce high voltages from low voltages, neither system will generate significant energy.

The work done, or energy transferred, will be proportional to the force between the water and the capacitor multiplied by a distance moved while subjected to that force. To extract mechanical energy from the wave, the water must be displaced from it's natural motion by the electrostatic force of the capacitor plates. The water will be little influenced by the presence of the charged capacitor unless you have tens of thousands of volts on the capacitor.

Your idea is similar to a parametric amplifier or a capacitance multiplier. Both systems would need to be more complex, and require external excitation. Although the systems can produce high voltages from low voltages, neither system will generate significant energy.

The work done, or energy transferred, will be proportional to the force between the water and the capacitor multiplied by a distance moved while subjected to that force. To extract mechanical energy from the wave, the water must be displaced from it's natural motion by the electrostatic force of the capacitor plates. The water will be little influenced by the presence of the charged capacitor unless you have tens of thousands of volts on the capacitor.

I understand that the capacitor will push back on the water trying to move through the plates, as the charge is building up.

I want to know if the movement of the water will cause any alternating/back and forth currents to flow between the two halves of the system. Every time the water level changes, the capacity of the capacitor also changes. The sealed plate attracts more of the opposite charge, which creates current.
I assume that the net change in charge can be rectified into DC.

Why would it require a seed charge of any kind? The capacitor in the video does not get pre charged with anything. Charge happens from change in capacity, as I understand.

Baluncore
You do not show the whole capacitor. Are the plates interleaved and subjected to a changing dielectric level?
Is there one capacitor connected through the seawater, or are there two with capacitance to the water only?
Draw an electrical circuit that can then be analysed.

Baluncore
Here are a couple of possible circuits.

In the first circuit the C1 capacitance is loaded by R and will discharge. Without an excitation voltage there can be no charge to pump, V, Q and I, will settle to and remain at zero, even with changing capacitance.

In the second circuit C2 and C3 alternate in capacitance. Excitation is provided by V1. As capacitance changes current flows back and forth through the R loads.

In the OP you show two capacitors, because they are separated by the wavelength, they are always immersed to the same depth. Change the separation to half a wavelength and they will alternate in capacitance and charge. Then circuit 2 may be applicable.

Baluncore
Why would it require a seed charge of any kind? The capacitor in the video does not get pre charged with anything.
He is contacting the moving plate to a different electrode at each end of the movement. There is a spark in each case. The plates are insulated by clear film. I believe those are storing and inducing the seed charge that is exciting the capacitive voltage multiplication. What are the two terminals that spark as he moves the plate?

Charge happens from change in capacity, as I understand.
The charge on the plates of an insulated capacitor stays the same while the capacitor is isolated. The voltage will change if the capacitance changes.

The dielectric constant of liquid water is about 80, allowing for insulation you might get a change in capacitance by a factor of maybe 20. That could make the capacitance change from 50 mF when the water is low, to 1F when the water is high. Unfortunately the change is less than the possible factor of 80 hoped for, because the thin paint or layer of insulation on the plates is in series with the water dielectric of the capacitor.
The definition of capacitance is the ratio; C = Q / V;
∴ Q = C * V; and V = Q / C.

Start with a 10 VDC supply. Charge the capacitor while the water is high.
The charge is then; Q = 1 F * 10 V = 10 coulombs.
The energy stored in a capacitor is; E = ½·C·V². So to get it working you have invested; E = 0.5 * 1.00 * 10² = 50 joules.

The water level then falls, so the capacitance becomes 50 mF.
Q remains at 10 coulombs since the terminals are isolated.
V = 10 C / 0.05 F = 200 volts. The voltage has increased.

The energy stored in a capacitor is then ; E = 0.5 * 0.05 * 200² = 1000 joules.
The difference in energy is the work it takes to pull the plates apart, or in this case to lower the water away from the plates.
If you remove some of that charge near 200 volts. Then on the next cycle the capacitor voltage will fall below 10 V and you will need to top up the capacitor to 10 V from the DC supply.

What you have is a charge pump. The same average current flows in as flows out, only the voltage is different as physical work has been done against the electrostatic forces.

If you did not excite the capacitor with an initial voltage, then the charge would be zero and the voltage would stay at zero.

Thank you for answering some of my questions.

I used two half-capacitors because I didn't want any metal contacting water.
How does this factor of 80 relate to the amount of plastic and aluminum that would be required to produce one kilowatt of power from oscillations of the ocean?

This circuit can use more then two "half-capacitor brushes" that work as capacitors together with the ocean.
That way you can have an average and use it as a reference for individual units so there would be no need for water "ground" to convert changes in water level to electricity. I don't want to go into detail since we can ignore the sea water "ground" corrosion in our calculations.

How much should the total surface area of the plates of my half-capacitor be to get me 1 kilowatt of power?
The total surface area that is exposed to change in sea water levels.

If I use a "joule thief" kind of inverter, would I give myself the ability to harvest energy before and after the peak of the cycle of the wave, to stretch the harvesting in time?

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Baluncore
This circuit can use more then two "half-capacitor brushes" that work as capacitors together with the ocean.
You need the second half of the capacitor so you can close the circuit without the resistance of sea water or the huge distance to the background.
How much should the total surface area of the plates of my half-capacitor be to get me 1 kilowatt of power?
Capacitance does not work that way. The ratio of the dielectric constant of water = 80, to air = 1, gives you the maximum capacitance change possible. You must have plates or chicken mesh screens for all interfingered plates of the capacitor. When it is out of the water it has air as the dielectric. When it is underwater it has water as the dielectric. That is why the capacitance changes.

You need the second half of the capacitor so you can close the circuit without the resistance of sea water or the huge distance to the background.

Capacitance does not work that way. The ratio of the dielectric constant of water = 80, to air = 1, gives you the maximum capacitance change possible. You must have plates or chicken mesh screens for all interfingered plates of the capacitor. When it is out of the water it has air as the dielectric. When it is underwater it has water as the dielectric. That is why the capacitance changes.

Is sea water a dielectric?

Baluncore
Is sea water a dielectric?
Yes. But it is not a good insulator because of the dissolved salt.

256bits
Gold Member
He is contacting the moving plate to a different electrode at each end of the movement.
To me it looks as if he is sparking the same electrode, which should mean the plates are alternating +/- .
The plates should completely discharge, as air doesn't have much of, if any worth mentioning, a hysteresis.
Perhaps the plastic covering then.

Yes. But it is not a good insulator because of the dissolved salt.
I want sea water to act as a "liquid capacitor plate" that is also connected to the "ground".
The plastic will act as a dielectric.

How much power would this produce?

There is already an insulator - all metal is covered with a layer of plastic.
Sea water and metal are separated.

How much power can I produce from 1 m^2 of this capacitor plate, insulated in plastic, surrounded by sea water that moves up and down? This means all submerged or all exposed. Salt water acts as a capacitor plate and plastic acts as an insulator between the plate and the sea water.

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Baluncore
How much power can I produce from 1 m^2 of this capacitor plate, insulated in plastic, surrounded by sea water?
None. It would cost more than it would generate.

I want sea water to act as a "liquid capacitor plate" that is also connected to the "ground".
The plastic will act as a dielectric.
Then the plastic will reduce the power you can extract because it is a fixed capacitor in series with the variable capacitance component.

The “comb” or “fingers” dipped in water will not have high capacitance. The middle fingers have no electrostatic exposure to the other electrode so are a total waste of time and effort.

Power will be proportional to capacitance and ratio of change. You must have close separation of the mesh plates, with the highest possible area of opposed plates. Waves flooding that volume with alternating water and air will cause the greatest possible change in capacitance, so the greatest power. Anything less than that should be replaced with a buoyant generator.

berkeman
None. It would cost more than it would generate.

Then the plastic will reduce the power you can extract because it is a fixed capacitor in series with the variable capacitance component.

The “comb” or “fingers” dipped in water will not have high capacitance. The middle fingers have no electrostatic exposure to the other electrode so are a total waste of time and effort.

Power will be proportional to capacitance and ratio of change. You must have close separation of the mesh plates, with the highest possible area of opposed plates. Waves flooding that volume with alternating water and air will cause the greatest possible change in capacitance, so the greatest power. Anything less than that should be replaced with a buoyant generator.

Why wouldn't middle plates add to capacitance? Look at ordinary capacitor. Imagine that plates on one side are replaced by water.

Yes, I am talking about water and air alternating.

How do I calculate the amount energy a changing capacitor like this can produce?

Baluncore
Why wouldn't middle plates add to capacitance?
Because the electrostatic field will not penetrate between the two connected plates. Penetration of the electric field is necessary to get high plate area, so it needs a metal conductive plate, probably made from chicken wire so the water can flow in and out.

How do I calculate the amount energy a changing capacitor like this can produce?
You calculate the area and separation of the plates to optimise the parameters.
Then you go through the calculations like I did in post #6.
You will be lucky to get 250 watts per wave period of 10 seconds = 25 watts.

Imagine that there is a dielectric and one metal "half" of the capacitor, yet water flows instead of the other half of the capacitor plates. How does it not penetrate electrostatically?

Water meshed in with metal, separated by a dielectric vs metal meshed in with metal, separated by a dielectric. What is the difference?

What difference would it make if capacitor plates on one side were made of metal or salty water?

Why would it serve me to use chicken wire?
It would be a lot harder to surround it with plastic then plates or sheets of aluminum foil.

All capacitors are made with some kind of stacked electrodes, connected to a bus bar. All I suggest is that we replace electrodes on one side of the capacitor with sea water that is subjected to waves and sometimes enters the capacitor and sometimes leaves it.

The other side is still the same set of plates, connected to a bus bar, all coated in a dielectric.

25 watts per 1m^3 of capacitor plate?

Permittivity of dielectric is some plastic that can withstand sea water corrosion.
Distance between plates is the thickness of this material that is required.

I still don't see how the change in capacitance from 1% to 100% can allow me to predict the amount of power this device can put out. I am not a physicist.

This change transfers into some flow of electricity, as opposite charges attract across the dielectric(plastic).

Capacitance can be measured in square meters.
What would be the ratio of capacitor dimensions to the maximum power that it can generate at described conditions?

How much charge can water displace?

Is there a good video that is in English that is similar to that one with the capacitor plate experiment?
So we can discuss it.

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Baluncore
What difference would it make if capacitor plates on one side were made of metal or salty water?
Metal guides the electric field, salt water does not, so the field takes a short cut.
Why would it serve me to use chicken wire?
It would be a lot harder to surround it with plastic then plates or sheets of aluminum foil.
The force of a wave impact would destroy a flat sheet if it did not allow the water to flow through.
I still don't see how the change in capacitance from 1% to 100% can allow me to predict the amount of power this device can put out. I am not a physicist.
You are not a physicist so you don't understand the principles necessary to design it. You also have unshakable false beliefs that cripple your acceptance of the science and technology. You are dreaming of irreality.
Is there a good video that is in English that is similar to that one with the capacitor plate experiment?
So we can discuss it.
The video of the capacitor plate experiment is a waste of time because it is a trick, made to appear like magic. You will not learn the needed fundamentals by watching videos.
If you work hard and study for 3 to 5 years, you will then wonder why you ever thought variable capacitance would work efficiently in the first place.

russ_watters and anorlunda
Metal is guides the electric field. Salt water does not, the field takes a short cut.

The force of a wave impact would destroy a flat sheet if it did not allow the water to flow through.

You are not a physicist so you don't understand the principles necessary to design it. You also have unshakable false beliefs that cripple your acceptance of the science and technology. You are dreaming of irreality.

The video of the capacitor plate experiment is a waste of time because it is a trick, made to appear like magic. You will not learn the needed fundamentals by watching videos.

What irreality? I am not suggesting to build a commercial powerplant. I want to know how it would compare to other technology available.

I asked you how much power can such capacitor device produce from each change of "wet" to "dry" or full to empty.

What don't I understand? It took me several attempts to remind you that salt water is a conductor and that the role of salt water in this device is to serve as a conductor, a capacitor plate.

I understand that there are structural limitations that would define how this device can be built, if possible.
That would be a concern if I were to actually build one.

How does the field not take a shortcut in regular capacitors?

I hope that somebody else answers my question.

If I was a physicist, I would not have asked how to calculate this device.

Baluncore
What don't I understand? It took me several attempts to remind you that salt water is a conductor and that the role of salt water in this device is to serve as a conductor, a capacitor plate.
Don't kid yourself. I first started work in the Electronics Instrument Lab of a University Chemistry Department in 1978.
You don't appreciate the resource you are wasting.
https://en.wikipedia.org/wiki/Dunning–Kruger_effect

russ_watters
Don't kid yourself. I first started work in the Electronics Instrument Lab of a University Chemistry Department in 1978.
You don't appreciate the resource you are wasting.
https://en.wikipedia.org/wiki/Dunning–Kruger_effect

Why did it take me so much energy to explain to you that salt water is not a dielectric in this design?
That water is another "half" of the capacitor.

I have a theoretical, abstract 1m^2 sheet of aluminum of negligible thickness.
It is sandwiched between two layers of 0.2mm of polyethylene.

The plate is submerged in salt water that is also connected to the circuit using some water "ground".
Then it is lifted out of the water.

Plate is submerged in sea water cyclically every ten seconds.
How much power would this produce?

I did not say that it has to be a commercial amount of power and that it would be commercially viable to do this.
I would simply like to know how much so I can compare it to other energy sources.

Can you answer my question? Thank you.

Thank you.

sophiecentaur
Gold Member
2020 Award
Anyone who has owned a seagoing boat or used a jetty or pontoon etc. will be aware that marine life will grow almost anywhere. Any structure with the sort of spacing that’s proposed would be an attractive home for many organisms. I’d give any system about six months before the fields and even the effective wave height would be. Severly compromised. Maintenance costs could be prohibitive.
The principle could be useful for measuring waves over a short time, though.

berkeman
I am not talking about the real thing, only about a thought experiment with this 1m^3 plate.

sophiecentaur
Gold Member
2020 Award
I am not talking about the real thing, only about a thought experiment with this 1m^3 plate.
The subject was ‘harvesting’. That implies more than just ‘detecting’ in an experiment.!

Maybe I only need to harvest enough to top off a very small battery on an all-plastic research buoy that floats in the middle of the ocean and this system suits my requirements in many ways?

Also, I understand that if I use two sets of sealed electrodes, that they will need to be submerged at different times for electricity to flow. I made a very crude illustration.

I don't want to build a power plant. And I don't want to give too many unnecessary details in this post.

I don't know how to calculate 1 meter surface area of a metal plate, sealed in dielectric in air vs in salt water.
How much current flow in either direction can I get out of this?

What would be the most efficient way to utilize this to power anything?