Make energy out of a capacitor in the sea
$begingroup$
First of all sorry about my english it's not my mother tongue.
Problem statement:
A cylindrical capacitor is placed in the sea so that when a wave comes (the water goes up), the water becomes the capacitors dielectric, when the wave has passed (the water goes down) the air becomes the dielectric.
So now when the water is at its highest, i attach a battery to the capacitor.
When the water goes down i detach the capacitor from the battery, this creates an isolated system in which the charge remains constant and can change the potential difference between the capacitors plates.
The force of gravity makes the water that was between the capacitors plates go down, and so it reduces the capacitors capacitance. The dielectric constant of water is quite high (about 81) and the change of capacitance is also quite high.
The charge remained the same but the potential difference between the capacitors plates has increased.
This means that the energy in the system has increased.
So i found a way to harvest energy from the waves movement without moving bodies.... but....
- Explain why this system in practice can not really work.
- Think about how to make it work (even if with reduced performance, how to make the capacitor stay at the same height is not to be considered).
- Estimate how much energy could be made out of such a device.
The attempt at a solution
I think this does not work because sea water is not distilled which means it conducts current. So now there's a current flowing through the capacitors plates which i think would damage the capacitor itself.
Or maybe the battery would get damaged because the capacitor with sea water inside makes a short circuit?
To make it work i think one would need to put distilled water in the capacitor or another insulating material and make it go up and down the capacitor thanks to the waves movement.
I have absolutely no idea how to estimate how much energy could be made out of such a device.
If i'm not mistaken energy in a cylindrical capacitor is calculated as:
$$U = frac{1}{2}CDelta V^2$$
So i thought about calculating
Uw = energy when the capacitor is filled with water
and then
Ua = energy when the capacitor is filled with air
and then doing
Uw-Ua = energy generated by the device
But i'm not sure this is right.
What do you think?
capacitor
asked Dec 8 '18 at 12:44
MaialMaial
1584
$endgroup$
|
show 1 more comment
$begingroup$
First of all sorry about my english it's not my mother tongue.
Problem statement:
A cylindrical capacitor is placed in the sea so that when a wave comes (the water goes up), the water becomes the capacitors dielectric, when the wave has passed (the water goes down) the air becomes the dielectric.
So now when the water is at its highest, i attach a battery to the capacitor.
When the water goes down i detach the capacitor from the battery, this creates an isolated system in which the charge remains constant and can change the potential difference between the capacitors plates.
The force of gravity makes the water that was between the capacitors plates go down, and so it reduces the capacitors capacitance. The dielectric constant of water is quite high (about 81) and the change of capacitance is also quite high.
The charge remained the same but the potential difference between the capacitors plates has increased.
This means that the energy in the system has increased.
So i found a way to harvest energy from the waves movement without moving bodies.... but....
- Explain why this system in practice can not really work.
- Think about how to make it work (even if with reduced performance, how to make the capacitor stay at the same height is not to be considered).
- Estimate how much energy could be made out of such a device.
The attempt at a solution
I think this does not work because sea water is not distilled which means it conducts current. So now there's a current flowing through the capacitors plates which i think would damage the capacitor itself.
Or maybe the battery would get damaged because the capacitor with sea water inside makes a short circuit?
To make it work i think one would need to put distilled water in the capacitor or another insulating material and make it go up and down the capacitor thanks to the waves movement.
I have absolutely no idea how to estimate how much energy could be made out of such a device.
If i'm not mistaken energy in a cylindrical capacitor is calculated as:
$$U = frac{1}{2}CDelta V^2$$
So i thought about calculating
Uw = energy when the capacitor is filled with water
and then
Ua = energy when the capacitor is filled with air
and then doing
Uw-Ua = energy generated by the device
But i'm not sure this is right.
What do you think?
capacitor
asked Dec 8 '18 at 12:44
MaialMaial
1584
$endgroup$
3
$begingroup$
Sounds like a perfect model for a Kickstarter scam.
$endgroup$
– winny
Dec 8 '18 at 13:09
$begingroup$
Distilled water or insulated electrodes.
$endgroup$
– JRE
Dec 8 '18 at 13:15
$begingroup$
But, as I recall, the energy is stored in the plates. Changing the dielectric will change the voltage but not the energy (I think.) Also, you will have to start with charged capacitors. You'd have to be careful not to discharge the capacitor too much when collecting energy.
$endgroup$
– JRE
Dec 8 '18 at 13:18
1
$begingroup$
The principle works, plenty of physics notes on the web regarding work done removing/inserting a dielectric from a charged capacitor. The problem regarding seawater could be solved using a mechanical buffer. Instead of seawater the wave would push/pull distilled water between the plates (general idea). Then you get to deal with the vagaries of the ocean, same as wind farms deal with wind. As to the amount of energy extracted, that is relative to your application.
$endgroup$
– isdi
Dec 8 '18 at 13:50
1
$begingroup$
The plates will stay wet while the wave drops, so it would be a relatively long time before the capacitance dropped off. Gut feeling of minutes at least before a significant change in capacitance. Whereas the period of a wave is about 5-10 seconds, so there's not going to be a great difference between high and low. But +1 for a great writeup and description.
$endgroup$
– Criggie
Dec 8 '18 at 21:35
|
show 1 more comment
$begingroup$
First of all sorry about my english it's not my mother tongue.
Problem statement:
A cylindrical capacitor is placed in the sea so that when a wave comes (the water goes up), the water becomes the capacitors dielectric, when the wave has passed (the water goes down) the air becomes the dielectric.
So now when the water is at its highest, i attach a battery to the capacitor.
When the water goes down i detach the capacitor from the battery, this creates an isolated system in which the charge remains constant and can change the potential difference between the capacitors plates.
The force of gravity makes the water that was between the capacitors plates go down, and so it reduces the capacitors capacitance. The dielectric constant of water is quite high (about 81) and the change of capacitance is also quite high.
The charge remained the same but the potential difference between the capacitors plates has increased.
This means that the energy in the system has increased.
So i found a way to harvest energy from the waves movement without moving bodies.... but....
- Explain why this system in practice can not really work.
- Think about how to make it work (even if with reduced performance, how to make the capacitor stay at the same height is not to be considered).
- Estimate how much energy could be made out of such a device.
The attempt at a solution
I think this does not work because sea water is not distilled which means it conducts current. So now there's a current flowing through the capacitors plates which i think would damage the capacitor itself.
Or maybe the battery would get damaged because the capacitor with sea water inside makes a short circuit?
To make it work i think one would need to put distilled water in the capacitor or another insulating material and make it go up and down the capacitor thanks to the waves movement.
I have absolutely no idea how to estimate how much energy could be made out of such a device.
If i'm not mistaken energy in a cylindrical capacitor is calculated as:
$$U = frac{1}{2}CDelta V^2$$
So i thought about calculating
Uw = energy when the capacitor is filled with water
and then
Ua = energy when the capacitor is filled with air
and then doing
Uw-Ua = energy generated by the device
But i'm not sure this is right.
What do you think?
capacitor
asked Dec 8 '18 at 12:44
MaialMaial
1584
$endgroup$
First of all sorry about my english it's not my mother tongue.
Problem statement:
A cylindrical capacitor is placed in the sea so that when a wave comes (the water goes up), the water becomes the capacitors dielectric, when the wave has passed (the water goes down) the air becomes the dielectric.
So now when the water is at its highest, i attach a battery to the capacitor.
When the water goes down i detach the capacitor from the battery, this creates an isolated system in which the charge remains constant and can change the potential difference between the capacitors plates.
The force of gravity makes the water that was between the capacitors plates go down, and so it reduces the capacitors capacitance. The dielectric constant of water is quite high (about 81) and the change of capacitance is also quite high.
The charge remained the same but the potential difference between the capacitors plates has increased.
This means that the energy in the system has increased.
So i found a way to harvest energy from the waves movement without moving bodies.... but....
- Explain why this system in practice can not really work.
- Think about how to make it work (even if with reduced performance, how to make the capacitor stay at the same height is not to be considered).
- Estimate how much energy could be made out of such a device.
The attempt at a solution
I think this does not work because sea water is not distilled which means it conducts current. So now there's a current flowing through the capacitors plates which i think would damage the capacitor itself.
Or maybe the battery would get damaged because the capacitor with sea water inside makes a short circuit?
To make it work i think one would need to put distilled water in the capacitor or another insulating material and make it go up and down the capacitor thanks to the waves movement.
I have absolutely no idea how to estimate how much energy could be made out of such a device.
If i'm not mistaken energy in a cylindrical capacitor is calculated as:
$$U = frac{1}{2}CDelta V^2$$
So i thought about calculating
Uw = energy when the capacitor is filled with water
and then
Ua = energy when the capacitor is filled with air
and then doing
Uw-Ua = energy generated by the device
But i'm not sure this is right.
What do you think?
capacitor
capacitor
asked Dec 8 '18 at 12:44
MaialMaial
1584
asked Dec 8 '18 at 12:44
MaialMaial
1584
asked Dec 8 '18 at 12:44
MaialMaial
1584
asked Dec 8 '18 at 12:44
MaialMaial
1584
asked Dec 8 '18 at 12:44
MaialMaial
1584
1584
3
$begingroup$
Sounds like a perfect model for a Kickstarter scam.
$endgroup$
– winny
Dec 8 '18 at 13:09
$begingroup$
Distilled water or insulated electrodes.
$endgroup$
– JRE
Dec 8 '18 at 13:15
$begingroup$
But, as I recall, the energy is stored in the plates. Changing the dielectric will change the voltage but not the energy (I think.) Also, you will have to start with charged capacitors. You'd have to be careful not to discharge the capacitor too much when collecting energy.
$endgroup$
– JRE
Dec 8 '18 at 13:18
1
$begingroup$
The principle works, plenty of physics notes on the web regarding work done removing/inserting a dielectric from a charged capacitor. The problem regarding seawater could be solved using a mechanical buffer. Instead of seawater the wave would push/pull distilled water between the plates (general idea). Then you get to deal with the vagaries of the ocean, same as wind farms deal with wind. As to the amount of energy extracted, that is relative to your application.
$endgroup$
– isdi
Dec 8 '18 at 13:50
1
$begingroup$
The plates will stay wet while the wave drops, so it would be a relatively long time before the capacitance dropped off. Gut feeling of minutes at least before a significant change in capacitance. Whereas the period of a wave is about 5-10 seconds, so there's not going to be a great difference between high and low. But +1 for a great writeup and description.
$endgroup$
– Criggie
Dec 8 '18 at 21:35
|
show 1 more comment
3
$begingroup$
Sounds like a perfect model for a Kickstarter scam.
$endgroup$
– winny
Dec 8 '18 at 13:09
$begingroup$
Distilled water or insulated electrodes.
$endgroup$
– JRE
Dec 8 '18 at 13:15
$begingroup$
But, as I recall, the energy is stored in the plates. Changing the dielectric will change the voltage but not the energy (I think.) Also, you will have to start with charged capacitors. You'd have to be careful not to discharge the capacitor too much when collecting energy.
$endgroup$
– JRE
Dec 8 '18 at 13:18
1
$begingroup$
The principle works, plenty of physics notes on the web regarding work done removing/inserting a dielectric from a charged capacitor. The problem regarding seawater could be solved using a mechanical buffer. Instead of seawater the wave would push/pull distilled water between the plates (general idea). Then you get to deal with the vagaries of the ocean, same as wind farms deal with wind. As to the amount of energy extracted, that is relative to your application.
$endgroup$
– isdi
Dec 8 '18 at 13:50
1
$begingroup$
The plates will stay wet while the wave drops, so it would be a relatively long time before the capacitance dropped off. Gut feeling of minutes at least before a significant change in capacitance. Whereas the period of a wave is about 5-10 seconds, so there's not going to be a great difference between high and low. But +1 for a great writeup and description.
$endgroup$
– Criggie
Dec 8 '18 at 21:35
3
3
$begingroup$
Sounds like a perfect model for a Kickstarter scam.
$endgroup$
– winny
Dec 8 '18 at 13:09
$begingroup$
Sounds like a perfect model for a Kickstarter scam.
$endgroup$
– winny
Dec 8 '18 at 13:09
$begingroup$
Distilled water or insulated electrodes.
$endgroup$
– JRE
Dec 8 '18 at 13:15
$begingroup$
Distilled water or insulated electrodes.
$endgroup$
– JRE
Dec 8 '18 at 13:15
$begingroup$
But, as I recall, the energy is stored in the plates. Changing the dielectric will change the voltage but not the energy (I think.) Also, you will have to start with charged capacitors. You'd have to be careful not to discharge the capacitor too much when collecting energy.
$endgroup$
– JRE
Dec 8 '18 at 13:18
$begingroup$
But, as I recall, the energy is stored in the plates. Changing the dielectric will change the voltage but not the energy (I think.) Also, you will have to start with charged capacitors. You'd have to be careful not to discharge the capacitor too much when collecting energy.
$endgroup$
– JRE
Dec 8 '18 at 13:18
1
1
$begingroup$
The principle works, plenty of physics notes on the web regarding work done removing/inserting a dielectric from a charged capacitor. The problem regarding seawater could be solved using a mechanical buffer. Instead of seawater the wave would push/pull distilled water between the plates (general idea). Then you get to deal with the vagaries of the ocean, same as wind farms deal with wind. As to the amount of energy extracted, that is relative to your application.
$endgroup$
– isdi
Dec 8 '18 at 13:50
$begingroup$
The principle works, plenty of physics notes on the web regarding work done removing/inserting a dielectric from a charged capacitor. The problem regarding seawater could be solved using a mechanical buffer. Instead of seawater the wave would push/pull distilled water between the plates (general idea). Then you get to deal with the vagaries of the ocean, same as wind farms deal with wind. As to the amount of energy extracted, that is relative to your application.
$endgroup$
– isdi
Dec 8 '18 at 13:50
1
1
$begingroup$
The plates will stay wet while the wave drops, so it would be a relatively long time before the capacitance dropped off. Gut feeling of minutes at least before a significant change in capacitance. Whereas the period of a wave is about 5-10 seconds, so there's not going to be a great difference between high and low. But +1 for a great writeup and description.
$endgroup$
– Criggie
Dec 8 '18 at 21:35
$begingroup$
The plates will stay wet while the wave drops, so it would be a relatively long time before the capacitance dropped off. Gut feeling of minutes at least before a significant change in capacitance. Whereas the period of a wave is about 5-10 seconds, so there's not going to be a great difference between high and low. But +1 for a great writeup and description.
$endgroup$
– Criggie
Dec 8 '18 at 21:35
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Yes, you can convert wave energy to electrical energy in this manner (but it's horribly inefficient — there are much better ways to extract energy from waves).
Two equations cover what you're proposing. The first is the relationship between charge, capacitance and voltage:
$$Q = C cdot V$$
The second is the relationship between energy, capacitance and voltage:
$$E = frac{1}{2}C cdot V^2$$
Start by putting a charge on a capacitor that's filled with water, and then removing the water. The first equation tells us that if the capacitance goes down by a factor of 81, then the voltage must go up by the same factor, because at this point, the charge can't go anywhere.
But the second equation says that if the capacitor had a certain amount of energy with the water in it, afterward, it has 81 times as much! Capacitance went down by a factor of 81, but $V^2$ went up by a factor of 812. E must rise by a factor of 81 in order to keep the equation balanced.
Where did this energy come from? It takes physical work to remove a dielectric from a charged capacitor. In this case, the potential energy of the mass of water within the capacitor was reduced when gravity pulled it out during the low part of the wave. And when the next peak comes along, the energy in the wave pushes the water back up against gravity.
When you put real, physical numbers on the values for C and V that you can realistically work with, you'll soon realize that the power (energy per time) that you can extract is miniscule.
answered Dec 8 '18 at 14:04
Dave Tweed♦Dave Tweed
118k9145256
$endgroup$
add a comment |
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$begingroup$
Yes, you can convert wave energy to electrical energy in this manner (but it's horribly inefficient — there are much better ways to extract energy from waves).
Two equations cover what you're proposing. The first is the relationship between charge, capacitance and voltage:
$$Q = C cdot V$$
The second is the relationship between energy, capacitance and voltage:
$$E = frac{1}{2}C cdot V^2$$
Start by putting a charge on a capacitor that's filled with water, and then removing the water. The first equation tells us that if the capacitance goes down by a factor of 81, then the voltage must go up by the same factor, because at this point, the charge can't go anywhere.
But the second equation says that if the capacitor had a certain amount of energy with the water in it, afterward, it has 81 times as much! Capacitance went down by a factor of 81, but $V^2$ went up by a factor of 812. E must rise by a factor of 81 in order to keep the equation balanced.
Where did this energy come from? It takes physical work to remove a dielectric from a charged capacitor. In this case, the potential energy of the mass of water within the capacitor was reduced when gravity pulled it out during the low part of the wave. And when the next peak comes along, the energy in the wave pushes the water back up against gravity.
When you put real, physical numbers on the values for C and V that you can realistically work with, you'll soon realize that the power (energy per time) that you can extract is miniscule.
answered Dec 8 '18 at 14:04
Dave Tweed♦Dave Tweed
118k9145256
$endgroup$
add a comment |
$begingroup$
Yes, you can convert wave energy to electrical energy in this manner (but it's horribly inefficient — there are much better ways to extract energy from waves).
Two equations cover what you're proposing. The first is the relationship between charge, capacitance and voltage:
$$Q = C cdot V$$
The second is the relationship between energy, capacitance and voltage:
$$E = frac{1}{2}C cdot V^2$$
Start by putting a charge on a capacitor that's filled with water, and then removing the water. The first equation tells us that if the capacitance goes down by a factor of 81, then the voltage must go up by the same factor, because at this point, the charge can't go anywhere.
But the second equation says that if the capacitor had a certain amount of energy with the water in it, afterward, it has 81 times as much! Capacitance went down by a factor of 81, but $V^2$ went up by a factor of 812. E must rise by a factor of 81 in order to keep the equation balanced.
Where did this energy come from? It takes physical work to remove a dielectric from a charged capacitor. In this case, the potential energy of the mass of water within the capacitor was reduced when gravity pulled it out during the low part of the wave. And when the next peak comes along, the energy in the wave pushes the water back up against gravity.
When you put real, physical numbers on the values for C and V that you can realistically work with, you'll soon realize that the power (energy per time) that you can extract is miniscule.
answered Dec 8 '18 at 14:04
Dave Tweed♦Dave Tweed
118k9145256
$endgroup$
add a comment |
$begingroup$
Yes, you can convert wave energy to electrical energy in this manner (but it's horribly inefficient — there are much better ways to extract energy from waves).
Two equations cover what you're proposing. The first is the relationship between charge, capacitance and voltage:
$$Q = C cdot V$$
The second is the relationship between energy, capacitance and voltage:
$$E = frac{1}{2}C cdot V^2$$
Start by putting a charge on a capacitor that's filled with water, and then removing the water. The first equation tells us that if the capacitance goes down by a factor of 81, then the voltage must go up by the same factor, because at this point, the charge can't go anywhere.
But the second equation says that if the capacitor had a certain amount of energy with the water in it, afterward, it has 81 times as much! Capacitance went down by a factor of 81, but $V^2$ went up by a factor of 812. E must rise by a factor of 81 in order to keep the equation balanced.
Where did this energy come from? It takes physical work to remove a dielectric from a charged capacitor. In this case, the potential energy of the mass of water within the capacitor was reduced when gravity pulled it out during the low part of the wave. And when the next peak comes along, the energy in the wave pushes the water back up against gravity.
When you put real, physical numbers on the values for C and V that you can realistically work with, you'll soon realize that the power (energy per time) that you can extract is miniscule.
answered Dec 8 '18 at 14:04
Dave Tweed♦Dave Tweed
118k9145256
$endgroup$
Yes, you can convert wave energy to electrical energy in this manner (but it's horribly inefficient — there are much better ways to extract energy from waves).
Two equations cover what you're proposing. The first is the relationship between charge, capacitance and voltage:
$$Q = C cdot V$$
The second is the relationship between energy, capacitance and voltage:
$$E = frac{1}{2}C cdot V^2$$
Start by putting a charge on a capacitor that's filled with water, and then removing the water. The first equation tells us that if the capacitance goes down by a factor of 81, then the voltage must go up by the same factor, because at this point, the charge can't go anywhere.
But the second equation says that if the capacitor had a certain amount of energy with the water in it, afterward, it has 81 times as much! Capacitance went down by a factor of 81, but $V^2$ went up by a factor of 812. E must rise by a factor of 81 in order to keep the equation balanced.
Where did this energy come from? It takes physical work to remove a dielectric from a charged capacitor. In this case, the potential energy of the mass of water within the capacitor was reduced when gravity pulled it out during the low part of the wave. And when the next peak comes along, the energy in the wave pushes the water back up against gravity.
When you put real, physical numbers on the values for C and V that you can realistically work with, you'll soon realize that the power (energy per time) that you can extract is miniscule.
answered Dec 8 '18 at 14:04
Dave Tweed♦Dave Tweed
118k9145256
answered Dec 8 '18 at 14:04
Dave Tweed♦Dave Tweed
118k9145256
answered Dec 8 '18 at 14:04
Dave Tweed♦Dave Tweed
118k9145256
answered Dec 8 '18 at 14:04
Dave Tweed♦Dave Tweed
118k9145256
118k9145256
add a comment |
add a comment |
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3
$begingroup$
Sounds like a perfect model for a Kickstarter scam.
$endgroup$
– winny
Dec 8 '18 at 13:09
$begingroup$
Distilled water or insulated electrodes.
$endgroup$
– JRE
Dec 8 '18 at 13:15
$begingroup$
But, as I recall, the energy is stored in the plates. Changing the dielectric will change the voltage but not the energy (I think.) Also, you will have to start with charged capacitors. You'd have to be careful not to discharge the capacitor too much when collecting energy.
$endgroup$
– JRE
Dec 8 '18 at 13:18
1
$begingroup$
The principle works, plenty of physics notes on the web regarding work done removing/inserting a dielectric from a charged capacitor. The problem regarding seawater could be solved using a mechanical buffer. Instead of seawater the wave would push/pull distilled water between the plates (general idea). Then you get to deal with the vagaries of the ocean, same as wind farms deal with wind. As to the amount of energy extracted, that is relative to your application.
$endgroup$
– isdi
Dec 8 '18 at 13:50
1
$begingroup$
The plates will stay wet while the wave drops, so it would be a relatively long time before the capacitance dropped off. Gut feeling of minutes at least before a significant change in capacitance. Whereas the period of a wave is about 5-10 seconds, so there's not going to be a great difference between high and low. But +1 for a great writeup and description.
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– Criggie
Dec 8 '18 at 21:35