Potential energy of water from electrolysis

[Grombely]

Could I please have help with something: If I generated hydrogen via electrolysis from water, sent it high up a long tube then converted it back to water (via a cell or motor to reclaim energy) then, taking into account inefficiencies (or assume no energy loss), where does the energy come from to give the potential energy from? I can't work this out from applying gas laws (eg PV=nRt)and force/work/mass and buoyancy relationships. Please help. Thank you.

 

[Simon Bridge]

Welcome to PF;

You have to add energy to the water to electrolyze it.
You get hydrogen and oxygen.

The hydrogen is less dense than the air so it rises - gaining gravitational potential energy.
Do you understand why things float - in terms of energy?

Convert it back to water by reacting it with atmospheric oxygen - or do you imagine using the oxygen off the electrolysis?

It looks like you could have the water flow back to the original reservoir, driving a water-wheel on it's way, and so you'd have a potential closed mill.

So your question boils down to: where does the energy come from to get the separate gasses to float up the tube in the first place. Instead of a tube - imagine the gasses are caught in balloons and the balloons float to where they can recombine.

So I think I need to see how you are handling the buoyancy part.
Also see the bouyancy motors in http://www.lhup.edu/~dsimanek/museum/unwork.htm - they all use water, but the principles are the same.

 

[Grombely]

Thank you Simon -I found the Dr Simanek's Museum site fascinating and would probably need to look into this more.
I suppose my question could be simplified to say, If a mass of H2 (say 10M/224L) were converted to water at sea level, then the same mass of H2 at say 100m elevation (V would be >224L) were converted, how would the energy produced + potential energy of water(180mL @ 100m)+ buoyancy input(of H2) be the same (as energy from water produced at sea level)?
thanks again-
Grombely

 

[russ_watters]

As the gases rise, the pressure drops, losing pressure energy.

 

[Simon Bridge]

Seen this bit?
http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm
Simply put - as something rising to a height by buoyancy effectively means that the same volume of the fluid it is immersed in must fall the same height. That's where you extra energy comes from.

Does not look, at first glance, like it disobeys conservation of energy - but there are probably major practical difficulties.

May have similar limits to the familiar: boiling water to make steam - steam rises, condenses, falls to earth, flows downhill, turns a turbine, generates energy to boil the water.

 

[Buckleymanor]

Seen this bit?
http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm
Simply put - as something rising to a height by buoyancy effectively means that the same volume of the fluid it is immersed in must fall the same height. That's where you extra energy comes from.

Does not look, at first glance, like it disobeys conservation of energy - but there are probably major practical difficulties.

May have similar limits to the familiar: boiling water to make steam - steam rises, condenses, falls to earth, flows downhill, turns a turbine, generates energy to boil the water.

The something rising must at first raise the fluid it is immersed in.
So does the amount of energy used to produce the electricity which makes the hydrogen equal the amount of energy required to lift a column of air to the top of the upper atmosphere.
When you consider that each hydrogen bubble produced must displace a column of air above it, that is a lot of weight.

 

[Simon Bridge]

The something rising must at first raise the fluid it is immersed in.
So does the amount of energy used to produce the electricity which makes the hydrogen equal the amount of energy required to lift a column of air to the top of the upper atmosphere.
When you consider that each hydrogen bubble produced must displace a column of air above it, that is a lot of weight.

The hydrogen and the air, being gasses, get to pass through each other. It's not a matter of a bubble of hydrogen forming in the tube and moving intact all the way.
So the hydrogen in the delivery tube does not need to lift the entire column of air, all that has to happen is for a volume of air displaced by the hydrogen to drop.

I can see an issue for a continuous operation where the tube fills with hydrogen ... you run the risk that, when the entire tube is filled, there may not be much to be buoyant with. So long as the density of H2 close to the top of the tube is less that the air close to the top, it may still flow from the pressure difference.

If you collected the hydrogen in balloons, energy must be expended inflating the balloon against the surrounding air - but the operation is easier to imagine.

People seem to want me to produce actual numbers ... It's not a trivial calculation: I charge a consultancy fee for that sort of thing :)

I'm not saying it is a practical way to get energy, there are all kinds of problems - one of which being whether there is a height available high enough for this to break even for reasonable quantities of water - only that the concept does not appear to violate any laws.

 

[Buckleymanor]

The hydrogen and the air, being gasses, get to pass through each other. It's not a matter of a bubble of hydrogen forming in the tube and moving intact all the way.
So the hydrogen in the delivery tube does not need to lift the entire column of air, all that has to happen is for a volume of air displaced by the hydrogen to drop.



People seem to want me to produce actual numbers ... It's not a trivial calculation: I charge a consultancy fee for that sort of thing :)

I can't see how the volume (column) of air displaced by the hydrogen is not entirely lifted.
If you introduced a pebble into the bottom of a tank of water surely the water level rises slightly.
Unless it was sealed from it's surroundings the column of air not the tank.
Would the calculation start at a about 14lbs per square inch.

 

[CWatters]

I believe you can look at it two ways. Either..

1) A volume of water equal to the volume of the bubble has been raised the full depth of the water or
2) The whole water column has been raised by the height of the bubble.

I think both points of view should give the same answer?

Edit: Here you go..

Consider a 0.1m deep layer of air introduced at the bottom of a 1m cube of water.

The volume of air is 0.1 cubic meters.
The mass of water is 1000kg
The mass of displaced water is 100kg.

According to 1) the energy gain is 100kg * 1 * 9.8 = 980J
According to 2) the energy gain is 1000kg * 0.1 * 9.8 = 980J

 

[CWatters]

Could I please have help with something: If I generated hydrogen via electrolysis from water, sent it high up a long tube then converted it back to water (via a cell or motor to reclaim energy) then, taking into account inefficiencies (or assume no energy loss), where does the energy come from to give the potential energy from? I can't work this out from applying gas laws (eg PV=nRt)and force/work/mass and buoyancy relationships. Please help. Thank you.

The energy required to electrolize water is proportional to pressure. So it's harder and takes more energy to do it at sea level than at altitude. That extra energy takes into account the energy needed to "raise the air" above the bubble giving the air PE.

Then as the bubble rises the air falls and looses PE as others have said.

I think I read it's more efficient to electrolize water under pressure but the overall efficiency of the system you propose can't be more than 100%.

 

[Buckleymanor]

The energy required to electrolize water is proportional to pressure. So it's harder and takes more energy to do it at sea level than at altitude. That extra energy takes into account the energy needed to "raise the air" above the bubble giving the air PE.

Then as the bubble rises the air falls and looses PE as others have said.

I think I read it's more efficient to electrolize water under pressure but the overall efficiency of the system you propose can't be more than 100%.

There is a contradiction in the above it can not be more efficient under pressure if it takes more energy to do it at sea level than altitude.It would be under more pressure at sea levell.
So which one is more efficient?

 

[russ_watters]

There is a contradiction in the above it can not be more efficient under pressure if it takes more energy to do it at sea level than altitude.It would be under more pressure at sea levell.
So which one is more efficient?

If the goal is COMPRESSED hydrogen, it is more efficient to compress the water first even if the electrolysis is less efficient.

There's really nothing to see here. The there cannot be a net energy output for the device described by the OP. You may note that old single-pipe steam systems utilize this principle for transporting steam without a pump. The boiler provides the energy to move the steam up against gravity and gravity gives the energy back when returning the condensate to the boiler.

 

[Buckleymanor]

If the goal is COMPRESSED hydrogen, it is more efficient to compress the water first even if the electrolysis is less efficient.

There's really nothing to see here. The there cannot be a net energy output for the device described by the OP. You may note that old single-pipe steam systems utilize this principle for transporting steam without a pump. The boiler provides the energy to move the steam up against gravity and gravity gives the energy back when returning the condensate to the boiler.

Nothing! well if what you say is true then it might be more efficient to electrolize water in a vacuum.

 

[russ_watters]

Nothing! well if what you say is true then it might be more efficient to electrolize water in a vacuum.

Sure, but why would you want to? You can't do anything with the hydrogen you recover unless you compress it.

 

[Simon Bridge]

I've had a chance to revisit the proposal while I'm less tired ... see next post.

I believe you can look at it two ways. Either..

1) A volume of water equal to the volume of the bubble has been raised the full depth of the water or
2) The whole water column has been raised by the height of the bubble.

That would work the same if you lifted the volume in option 1 above the surface of the water.
However, we are talking about the air bubble going up requiring a corresponding bubble of water going down.

I can't see how the volume (column) of air displaced by the hydrogen is not entirely lifted.
If you introduced a pebble into the bottom of a tank of water surely the water level rises slightly.

If you introduced the pebble from outside the tank sure. If you started with the pebble inside the tank, and lowered it the water level remains the same. If the pebble started on a platform floating on the water in the tank, then the water level goes down.

It is easier to think about bubbles rising through water.
The situation I was talking about, the bubble is already at the bottom of the tank and rises to another level, still under water. The height of the water in the tank dos not change.

You do have to do work to introduce the bubble into the bottom of the tank in the first place though. There's an example of this on Simanek's buoyancy page linked earlier, and it is why I have pointed out the energy needed to inflate a balloon ... even without the balloon the released gas must displace (and mix with) the surrounding air somewhat.

As the others point out, it is trickier with hydrogen in air.

=== [edit] ================

After writing it occurs to me there is a risk of the thread going astray by discussing the "ways of getting water to the top of a hill" ... that is not the intention: I know they are seductive. I wanted them as examples of the seductive nature of these things.
I've deleted the exampes ... apologies to those who received them.

Please, anyone who wants to know more, refer to the Museum of Unworkable Devices.

 

[Simon Bridge]

The main question was about where the energy from buoyancy comes from - and that has been answered.
I suspect that this may be a red-herring when considering whether this proposal could produce a surplus of energy. What about the energy needed to make the "bubble" in the first place?

I think the closest example in the Museum of Unworkable Devices is the one at the bottom of http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm.

The idea is to inject air into the bottom of the water tank, which rises into the wheel's chambers, and causes the wheel to rotate until the air is released near the top. Of course energy is to be extracted from the turning of the wheel.

This was not proposed as an over-unity device, only as a method for extracting energy (that would otherwise be wasted) of pressurized exhaust gases from manufacturing processes.​

The energy needed to introduce the gas at the bottom of the tank is, at best, equal to the energy it gives you from floating to the surface.

For the proposal, this energy comes during the electrolysis process when the gas bubbles are first made.

I suspect the energy needed to make a bubble (whatever) capable of floating up to a particular height will be more than what could be returned from water flowing the opposite way ... so any working version of the proposal requires net input energy someplace.

Possibly just transporting the hydrogen ... use ambient oxygen - which gets there powered by the Sun.

I submit that this is clever.
It hides the usual problems with returning masses in closed mills by changing state.

 

[CWatters]

If the goal is COMPRESSED hydrogen, it is more efficient to compress the water first even if the electrolysis is less efficient.

and the goal was indeed to create slightly compressed hydrogen.. his idea was to create it at low altitude instead of at high altitude.

There's really nothing to see here.

I agree. Surprised the thread has been allowed to run.

 

[Grombely]

I think Simon seems to understand more so what I'm trying to ask.

To CWatters, I also posted this earlier on:

I suppose my question could be simplified to say, If a mass of H2 (say 10M/224L) were converted to water at sea level, then the same mass of H2 at say 100m elevation (V would be >224L) were converted, how would the energy produced + potential energy of water(180mL @ 100m)+ buoyancy input(of H2) be the same (as energy from water produced at sea level)?
thanks again-
Grombely

So the Hydrogen pressure would not be great due to buoyancy and weight of H2.(which for 10M would be 20.2g) I am not proposing compressed hydrogen, and this would be operational in air (not bubbles under water). I'm also not necessarily proposing an over-unity device, just trying to reconcile the energies of the system.

For the proposal, this energy comes during the electrolysis process when the gas bubbles are first made.

I suspect the energy needed to make a bubble (whatever) capable of floating up to a particular height will be more than what could be returned from water flowing the opposite way ... so any working version of the proposal requires net input energy someplace.

Possibly just transporting the hydrogen ... use ambient oxygen - which gets there powered by the Sun.

I submit that this is clever.
It hides the usual problems with returning masses in closed mills by changing state.
.......................
You do have to do work to introduce the bubble into the bottom of the tank in the first place though. There's an example of this on Simanek's buoyancy page linked earlier, and it is why I have pointed out the energy needed to inflate a balloon ... even without the balloon the released gas must displace (and mix with) the surrounding air somewhat.

As the others point out, it is trickier with hydrogen in air.

I've put these quotes from Simon to help explain-

Pot. energy of H2O @ 100m =.180kg x 9.8 x 100m = 176.4J
Pot. energy of H2 =20.16g x 9.8 x 100m = 19.76J
∴ energy gain from this would be 156.6J by using ambient oxygen.
so Simon is correct (thank you)

I would still like to understand the work from the H2 buoyancy & state change @ 100m elv.
Also could someone please determine the amount( or lack thereof) of extra work for a cell to produce 224L H2 at 100m difference in height?

Thank you for your input on this

 

[CWatters]

All I know is that electrolysis is more complicated than people expect. A quick look at articles such as this one below suggest working out the energy required to electrolize the water is tricky unless you happen to be an expert in the field..

http://www.lsbu.ac.uk/water/electrolysis.html

Although theoretically as above, the current passing should determine the amounts of hydrogen and oxygen formed, several factors ensure that somewhat lower amounts of gas are actually found. Some electrons (and product) are used up in side reactions, some of the products are catalytically reconverted to water at the electrodes particularly if there is no membrane dividing the electrolysis compartments, some hydrogen may absorb into the cathode (particularly if palladium is used) and some oxygen oxidizes the anode. Finally some gas remains held up in the nanobubbles for a considerable time and some gas may escape measurement.

The above description hides much important science and grossly over-simplifies the system.

 

[Simon Bridge]

I would still like to understand the work from the H2 buoyancy & state change @ 100m elv.
Also could someone please determine the amount( or lack thereof) of extra work for a cell to produce 224L H2 at 100m difference in height?

It's a hard calculation - if it is important to you, you should hire someone familiar with different electrolysis methods. Perhaps a simpler, but inaccurate, model will be good enough for you? Your numbers seem really specific.

 

[russ_watters]

The problem can be made easier by ignoring the electrolysis and combustion. It doesn't really matter here anyway. When talking about the energy for them, people actually lump together a bunch of different processes. But the chemical bond strength is constant, so the two reactions are mirror images of each other. Strip that away and you have a single-pipe boiler problem to solve with a steam table.

 

[russ_watters]

Actually, that's probably the problem with such problems. Ultimately you are just proving that 0=0. And while you can build a big, long equation to prove that, without even plugging in any numbers you can simplify most of the equation away, including the part you are most curious about, leaving nothing to do!

 

[Buckleymanor]

The problem can be made easier by ignoring the electrolysis and combustion. It doesn't really matter here anyway. When talking about the energy for them, people actually lump together a bunch of different processes. But the chemical bond strength is constant, so the two reactions are mirror images of each other. Strip that away and you have a single-pipe boiler problem to solve with a steam table.

That seems an over simplification a single-pipe boiler problem is limited in comparison.
In effect your hydrogen has the whole atmosphere to play with.
It does not have to be limited to returning as just water it could carry a device to the upper atmosphere say a magnet and fall back through a coil. A boiler is limited to the length of pipe used and when the water condenses.
Hydrogen use could be extended for many miles of atmosphere.Once it is released or a device inflated more usefull work will be recovered the higher it goes the longer the fall back to earth.
It takes the same amount of energy to raise it a foot as 10 miles.
Surely the similarities stop at the length of pipe used in a boiler.

 

[Simon Bridge]

It takes the same amount of energy to raise it a foot as 10 miles.

No it doesn't.
Clearly each kg of hydrogen gas needs ore than 17J of energy for each mile risen.

But the boiler analogy does oversimplify - the created gas has to displace (or otherwise interact with) the immediately surrounding air, which requires additional energy - do we get that energy back at the other end?

A hydrogen balloon could lift a magnet and then drop it through a coil ... for that matter it could lift the magnet through a coil too. Then you get to asking how long the coil has to be to get enough hydrogen to do it again ... you cannot reuse the balloon because the energy needed to pull the balloon back down will be less than that generated by the falling magnet.

Actually, that's probably the problem with such problems. Ultimately you are just proving that 0=0. And while you can build a big, long equation to prove that, without even plugging in any numbers you can simplify most of the equation away, including the part you are most curious about, leaving nothing to do!

OTOH: the long equations are important details - without them we cannot figure how to build actual working generators ... their equations all balance too.

I agree that the exact process of getting the H2 and O2 out of water is a red herring here - we can set it up according to the Simanek guidelines by allowing a black-box process that inputs water and energy and outputs hydrogen and oxygen gas in a reversible process that outputs the same energy. Put one at each end of the process .. but still rely on buoyancy to carry the gas between boxes.

If that works out - then one examines about what goes on inside the box.

But I suspect the discussion is headed in a direction not permitted under the rules.

 

[russ_watters]

In effect your hydrogen has the whole atmosphere to play with.
It does not have to be limited to returning as just water it could carry a device to the upper atmosphere say a magnet and fall back through a coil. A boiler is limited to the length of pipe used and when the water condenses.
Hydrogen use could be extended for many miles of atmosphere.Once it is released or a device inflated more usefull work will be recovered the higher it goes the longer the fall back to earth.
It takes the same amount of energy to raise it a foot as 10 miles.
Surely the similarities stop at the length of pipe used in a boiler.

That's not the situation the OP described: the OP described a closed cycle.

 

[Buckleymanor]

Quote by Simon Bridge,

No it doesn't.
Clearly each kg of hydrogen gas needs ore than 17J of energy for each mile risen.

Why once an object is buoyant why do you have to keep adding energy.
Once it's density is less than the fluid or gas surrounding it just floats or rises.
The atmosphere gets thiner but surely you would account for this at the start so no need to add any more energy on the way up.

 

[russ_watters]

Why once an object is buoyant why do you have to keep adding energy. Once it's density is less than the fluid or gas surrounding it just floats or rises.

Think of it the other way around: in order to sink a buoyant object, you have to apply a force on it and push it all the way down. It is like a potential energy.

You aren't adding energy to a rising object, you are taking it away.

 

[Dale]

To the OP. The higher the column of fluid that you want to send your gas up the greater the gauge pressure at the bottom. It is well known that electrolysis "must provide the energy for the dissociation plus the energy to expand the produced gases" ( http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html ). Increasing the gauge pressure increases the energy required to expand the gasses. So you cannot get any extra energy this way.

 

[Buckleymanor]

Think of it the other way around: in order to sink a buoyant object, you have to apply a force on it and push it all the way down. It is like a potential energy.

You aren't adding energy to a rising object, you are taking it away.

Which is what I said,"Why once an object is buoyant why do you have to keep adding energy".
There seems to be some confusion with regards floating bodies and there relative densities.
In order to sink a bouyant object all you have to do is make it slightly denser than the fluid or gas it is immersed in.You don't have to keep on applying a force, throw a rock into water and it sinks.
Make your object less dense than water and it floats.
So I still am not sure where Simon gets his 17J of energy for each mile risen.
Unless it is to do with a pressure gradient which I have already mentioned.

 

[Buckleymanor]

Quote by Simon Bridge.
A hydrogen balloon could lift a magnet and then drop it through a coil ... for that matter it could lift the magnet through a coil too. Then you get to asking how long the coil has to be to get enough hydrogen to do it again ... you cannot reuse the balloon because the energy needed to pull the balloon back down will be less than that generated by the falling magnet.

You could then do what the OP said and use the hydrogen at the top of the cycle in a cell.
Then the balloon would fall back down.

 

[russ_watters]

You need to be much more precise and consistent in defining the examples/cases. Your case was the rising of the buoyant object. You can continuously extract energy as it rises. My case was pushing the same object back down: you're not allowed to change its buoyancy because its my example and I get to decide its definition. You only need to recognize it is an exact mirror image of yours, so you need to continuously push it down to make it sink.

And regarding the 17 J: do you know the equation for gravitational potential energy?

 

[Buckleymanor]

Quote by russ-watters
And regarding the 17 J: do you know the equation for gravitational potential energy?
Thanks Report

Oh! so is Simon talking about the PE gained by the gas or object as it rises if it is it is not what I described.
If he is it's a gain as it rises I was talking about the start point.

 

[Simon Bridge]

Which is what I said,"Why once an object is buoyant why do you have to keep adding energy".

A rising object does work against the gravitational field. Energy has to be added to keep it rising. Does not matter what it is.

(BTW: there's a typo - it should have been 17kJ per mile, not 17J.)

Oh! so is Simon talking about the PE gained by the gas or object as it rises if it is it is not what I described.

That may not be what you intended to say but it is what you wrote down. (Is English your second language perhaps?) You wrote:

It takes the same amount of energy to raise it a foot as 10 miles.

... you didn;t say where the energy came from. It takes more energy to rise the 10 miles and it has to be added somewhere and it is a factor when examining these things - as you know from reading the links I have included.

It is a common misconception that buoyancy is a source of energy somehow.

If he is it's a gain as it rises I was talking about the start point.

You make an inflated balloon buoyant by reducing the amount of air inside for a given volume ... most commonly by heating it. Once buoyant, it will rise at no further input from the burners - the energy comes from the weight of cool air descending.

I've found you have to be very careful in discussions where a possible closed mill is being discussed... people can get the wrong end of the stick real easy. Look through D Simanek's whole site and you will quickly see why it's important to be precise and consistent in your examples - ordinary, conversational, English does not work well for these sorts of discussions.

 

[Simon Bridge]

A hydrogen balloon could lift a magnet and then drop it through a coil ... for that matter it could lift the magnet through a coil too. Then you get to asking how long the coil has to be to get enough hydrogen to do it again ... you cannot reuse the balloon because the energy needed to pull the balloon back down will be less than that generated by the falling magnet.

You could then do what the OP said and use the hydrogen at the top of the cycle in a cell.
Then the balloon would fall back down.

... OK - you can reuse the skin of the balloon that way - but you cannot reuse the whole balloon+hydrogen that way. Perhaps I should have been clearer - in common language the "balloon" is the rubber bag.

Anyway this was also what I suggested ... and it was me that pointed out that of the upwards side of the cycle could be recovered on the downwards side, you'd get a closed mill ... and this means looking at the situation more closely.

Did you also follow analysis of the process that came later?

OP wanted to know about the energy for gaining gravitational potential energy through buoyant floating - which is why I didn't consider DaleSpam's point about piping the gasses under pressure. The idea is to exploit their natural buoyancy... and it's a red herring: it's the energy needed to get/produce the gas at the bottom that is key to this ... you have to expand some gas against the surrounding air as well as break apart the molecular bonds of the water molecule. (In fairness DaleSpam got that part too.)

I don't think it adds anything to think about the magnet approach - it's another red herring.
You still need a net input of energy - which has to come from someplace else.

 

[Simon Bridge]

I suspect it is easier to think about this when the entire thing is done under water.

Lets say you inflate a balloon with a buoyant gas at the bottom of some idealized sea[1], and let it rise, but before it reaches the surface[2], use the gas[3], so that the skin of the balloon sinks to be reused[4].

Could we, in principle, get a closed mill from this process?
To do this, the energy from the sinking balloon skin needs to be equal to or greater than the energy needed to inflate the balloon at the bottom (plus other unavoidable losses.)

The key is that the amount of work needed to inflate the balloon at the bottom is the same as the amount of work needed to push the balloon to the bottom from the surface[5]. This is in addition to the amount of energy needed to make the gas in the first place.
(Remember that the gas changed volume too!)

Now you lot can do some math.
Be careful though - when we give high school students a conservation of energy exercise we usually neglect the effect of the air - this is so that the loss in GPE is gained in KE after a fall. As soon as buoyancy becomes an issue, the effect of the air (or other fluid) cannot be neglected.

------------------------

[1] see the Museum of Unworkable devices for the kinds of idealizations to use for analyzing proposed closed mills ... for fluids, we usually allow them to be in-compressible with zero viscosity. It's probably safe to allow the gas-generating and recovery mechanism to be lossless - and, what the heck, the gas can be ideal too. These are very generous idealizations indeed!

[2] this is to make it more analogous to the hydrogen balloon strategy of Buckleymanor... let the balloon go to the surface if you like. In fact, the analysis is easier of you do, and specify that the gas-production/recovery method converts water.

[3] perhaps in a way that recovers the energy used to make the gas in the first place?

[4] this means we cannot use a common idealization that the balloon skin is massless - if we did, the skin would not sink. We can let it have an arbitrarily small volume though ... so, at any mass, it's density can be very high, so it sinks quickly. We can also let it have very high strength so it won't burst and the ability to expand to any volume without exerting extra pressure on the gas inside or using additional energy stretching it ... truly a wondrous material!

[5] Or is it? ;)

 

[Buckleymanor]

Quote by Simon Bridge.
I don't think it adds anything to think about the magnet approach - it's another red herring.
You still need a net input of energy - which has to come from someplace else.

With regards to the magnet approach it is a red herring if you have to input the extra energy to raise it in the first place.
The above link might be of interest if it works, the link that is, as it might be possible to use the
water directly on it's return to produce electricity instead of a water wheel device.
The whole cycle should be more efficient.

 

[Simon Bridge]

That's also a red herring - the energy generation part of the cycle doesn't matter.
Do you really think that this can be made into a closed mill?

 

[CWatters]

The idea is to exploit their natural buoyancy... and it's a red herring: it's the energy needed to get/produce the gas at the bottom that is key to this ... you have to expand some gas against the surrounding air as well as break apart the molecular bonds of the water molecule.

I agree, but I have a question...

Imagine two different chemical reactions, one creating and expanding a bubble of Nitrogen and the other an equal size bubble of hydrogen. The pressure inside each bubble should be the same because the same volume of air is displaced (I think).

However the resulting hydrogen bubble has greater bouyancy than that of the Nitrogen bubble, so does that mean it would take more energy to expand the bubble of hydrogen than the bubble of nitrogen? It seems to me as if it would but I can't quite see why if the pressure and change of volume is the same.

 

[Buckleymanor]

You still need a net input of energy - which has to come from someplace else.
Jun6-13 11:39 PM


That's also a red herring - the energy generation part of the cycle doesn't matter.


Do you really think that this can be made into a closed mill?

I gave you a net input of energy from someplace else and it turns out now, to be a red herring as the energy part of the cycle according to you doesn't matter.
So there is no chance it being made into a closed mill.

 

[Simon Bridge]

I gave you a net input of energy from someplace else and it turns out now, to be a red herring as the energy part of the cycle according to you doesn't matter.

The energy generation phase is not a net input of energy, it's a net extraction of energy. The net input has to come from outside the cycle ... the details of how any energy is generated from inside the cycle is a red herring and it is not just my opinion, it's the law of conservation of energy.

Do you understand how the "water battery" in the video works? It's called a "Kelvin's Thunderstorm".
http://amasci.com/emotor/kelvin.html
... you have to separate the charges in the water somehow - which is where I'm guessing you think the external energy comes in? Aside from that, the energy for the electricity comes from the gravitational potential pulling the charged droplets from the end of the dripper - that does come from inside the cycle.... so: is it your contention that energy generated from within the cycle could, in principle, be used to keep the cycle going as well as provide useful work?

 

[Simon Bridge]

I agree, but I have a question...

Imagine two different chemical reactions, one creating and expanding a bubble of Nitrogen and the other an equal size bubble of hydrogen. The pressure inside each bubble should be the same because the same volume of air is displaced (I think).

It's easier to picture s you do this in water rather than in air - I made a long post about this.

The pressures would be the same - at all volumes.
The quantity of hydrogen and nitrogen in the bubble for the same volume and pressure may be different - though they are both diatomic. But if we use the ideal gas approximation, a bubble of hydrogen will be more buoyant than a bubble of nutrogen because the mass of hydrogen is much less.

However the resulting hydrogen bubble has greater bouyancy than that of the Nitrogen bubble, so does that mean it would take more energy to expand the bubble of hydrogen than the bubble of nitrogen?

No. The buoyancy force comes from the difference in pressure around the bubble vs the mass inside - not the energy used to make the bubble. In each case, the energy to expand the bubble is at least the same as the gravitational potential difference between the place the bubble is made and the top of the fluid.

It seems to me as if it would but I can't quite see why if the pressure and change of volume is the same.

It seems intuitive that something accelerating more must have had more energy put into it - but your analysis is correct: same volume, same pressure, same energy to expand.

The energy for the faster rise of the hydrogen comes from the descent of the same volume of regular air (or whatever the fluid is) - since the mass difference between hydrogen and air is greater than that for nitrogen and air, the hydrogen's ascent is faster.

You can model this for two ideal gasses with different mass per particle. PV=nRT ... creating the gas bubble involves changing the value of n in such a way that the pressure is constant ... and dW=P(V)dV

There are also wrinkles to do with gasses being compressable and pressure changing with altitude etc.

I'm half-expecting this thread to get locked real soon.
I cannot over-stress the usefulness of Simanek's buoyancy pages ... read through them several times.

 

[CWatters]

Thanks.

 

[Buckleymanor]

The energy generation phase is not a net input of energy, it's a net extraction of energy. The net input has to come from outside the cycle ... the details of how any energy is generated from inside the cycle is a red herring and it is not just my opinion, it's the law of conservation of energy.

Do you understand how the "water battery" in the video works? It's called a "Kelvin's Thunderstorm".
http://amasci.com/emotor/kelvin.html
... you have to separate the charges in the water somehow - which is where I'm guessing you think the external energy comes in? Aside from that, the energy for the electricity comes from the gravitational potential pulling the charged droplets from the end of the dripper - that does come from inside the cycle.


... so: is it your contention that energy generated from within the cycle could, in principle, be used to keep the cycle going as well as provide useful work?

Energy generated from within the cycle would not be able to keep the cycle going as well as provide usefull work, without breaking conservation laws.
If you were to combine the two main features of the proposed mill the use of hydrogen and it's use in a cell and the recovery of electricity using Kelvin's Thunderstorm.

A hydrogen fuel cell operating at 25°C has a maximum theoretical efficiency of 83%, even though the fuel cell is extracting all the electrical energy possible

So presumably you would have to find around 20% from Kelvin's Thunderstorm to keep the cycle going and provide usefull work, which does not seem like much but I am not sure how efficeint the Thunderstorm can be or how reliable the theoretical efficeincy of the fuel cell is.

 

[russ_watters]

So presumably you would have to find around 20% from Kelvin's Thunderstorm to keep the cycle going and provide usefull work, which does not seem like much but I am not sure how efficeint the Thunderstorm can be or how reliable the theoretical efficeincy of the fuel cell is.

Ok, that's enough of this thread. You are looking for perpetual motion here because you are assuming Kelvin's Thunderstorm itself is a free energy device by producing electricity without an input of mechanical energy. You should stop looking for something that doesn't exist. Thread locked.