Potential energy of water from electrolysis

Main Question or Discussion Point

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.

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Simon Bridge
Homework Helper
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 D Simanek's Museum - they all use water, but the principles are the same.

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

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

Simon Bridge
Homework Helper
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.

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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
Homework Helper
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.

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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.

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CWatters
Homework Helper
Gold Member
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

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CWatters
Homework Helper
Gold Member
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%.

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
Mentor
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.

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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
Mentor
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
Homework Helper
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.

===  ================

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.

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Simon Bridge
Homework Helper
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 the Buoyancy Misconceptions Exhibit.
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
Homework Helper
Gold Member
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.

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
Homework Helper
Gold Member
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
Homework Helper
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
Mentor
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
Mentor
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!

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 amout 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
Homework Helper
Buckleymanor said:
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
Mentor
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 amout 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.