Basic Hydrogen/gravity/thermodynamics question from a noob.

[willdo]

What altitude should a weight drop from to produce enough energy to electrolyse the hydrogen needed to lift it? Please keep it simple, i just want to know if it would work in theory.

 

[russ_watters]

Conservation of energy tells us this can't be done at all. Differing elevations doesn't change anything.

 

[jfizzix]

the simplest example would be to consider the energy it takes to break apart a single hydrogen molecule, its bond energy, and compare it to the energy it takes to lift the molecule a certain height off of the earth.

It turns out that the bond energy (about 724 zeptojoules) is actually larger than its total gravitational potential energy (about 209 zeptojoules), so even if you lifted it arbitrarily high above the earth, it would not gain more kinetic energy than its bond energy.

The bond energy of H_{2}, I looked up.

The gravitational potential energy I calculated from Newton's law of gravity

|U|= G\frac{M_{H_{2}}M_{E}}{R_{E}}

It would be a different story if you were lifting it off of a more massive or more dense body, but Earth just isn't massive enough in this case.

Hope this helps:)

 

[willdo]

Ok, i just read the basics of "conservation of energy" yet, wether i drop a weight from 1 metre or 10 metres the energy produced is not the same.I am not knowledgeable enough to explain it but in Hydroelectricity the height of the waterfall matters very much...So, does it mean that the weight speeds up over a certain distance and then stops accelerating at some point? In which case, How can i figure that distance?

Edit:sorry jfizzix, we were writing at the same time...somehow this makes sense...

Edit2: Ok, i thought about it again and...what's stopping me from harnessing the energy produced more than once? (in theory of course i know it would be impractical)
example: if a weight attains it's maximum level of kinetic energy after 1000 metres (a wild guess, i have no clue what the real number would be like) and you carry this weight to 30000 metres, then you could harness the kinetic energy 30 times...no?

 

[pbuk]

Conservation of energy tells us this can't be done at all.

This is not correct, work is done by the atmosphere raising the weight attached to the hydrogen balloon.

 

[pbuk]

Unfortunately electrolysis of water requires huge amounts of energy - to get a cubic metre of hydrogen needs 11.67MJ even at 100% efficiency.

Now 1m³ of hydrogen should lift about 1kg. A 1kg mass at 50km up (about as high as a balloon will go) has about 0.5MJ of potential energy.

So even if we could convert all of the potential energy to electricity, and all of the electrical energy to extracting hydrogen we would still not have enough by a factor of more than 20x.

You need a more massive planet.

 

[Nugatory]

example: if a weight attains it's maximum level of kinetic energy after 1000 metres (a wild guess, i have no clue what the real number would be like) and you carry this weight to 30000 metres, then you could harness the kinetic energy 30 times...no?

The maximum kinetic energy is achieved by dropping the weight from infinity... so you only get to do it once.

 

[pbuk]

The maximum kinetic energy is achieved by dropping the weight from infinity... so you only get to do it once.

Yes but if you could this would be plenty enough energy - escape velocity is 11.2km/s which would give 1kg 188MJ. The problem is that you can only use the Earth's atmosphere to power the ascent to the limits of the Earth's atmosphere, not infinity.

 

[Khashishi]

This is not possible even on a hypothetical massive planet. If you increase the gravity, you increase the pressure which will make it more difficult to electrolyze water.

 

[mfb]

Ok, i just read the basics of "conservation of energy" yet, wether i drop a weight from 1 metre or 10 metres the energy produced is not the same.I am not knowledgeable enough to explain it but in Hydroelectricity the height of the waterfall matters very much...

Right, and the difference is a factor of 10 in energy.
The basic question is: how do you get your mass to that height? Lifting it by 10m needs also ten times the energy needed to lift it by 1 meter.

Edit2: Ok, i thought about it again and...what's stopping me from harnessing the energy produced more than once?

It is pointless, as you will need the same energy to lift it before you can drop it again.

 

[willdo]

I must say, this is great, this question has bothered me for a while, thank you all for making it easy to understand!
Just to make sure, how would that change if we started in water, say 10km deep? (assuming no engineering issue)

 

[willdo]

Right, and the difference is a factor of 10 in energy.
The basic question is: how do you get your mass to that height? Lifting it by 10m needs also ten times the energy needed to lift it by 1 meter.

It is pointless, as you will need the same energy to lift it before you can drop it again.

Edit2: Ok, i thought about it again and...what's stopping me from harnessing the energy produced more than once? (in theory of course i know it would be impractical)
exemple: if a weight attains it's maximum level of kinetic energy after 1000 metres (a wild guess, i have no clue what the real number would be like) and you carry this weight to 30000 metres, then you could harness the kinetic energy 30 times...no?
T 02:52 PM

My idea was in the exemple, since the hydrogen keeps going up, there wouldn't be any need to get "back up". i since understood that it would be pointless regardless since the energy produced would be vastly insufficient unless i could reach some 1200km and there just isn't that much atmoshere to go

 

[pbuk]

Right, and the difference is a factor of 10 in energy.
The basic question is: how do you get your mass to that height? Lifting it by 10m needs also ten times the energy needed to lift it by 1 meter.

It is pointless, as you will need the same energy to lift it before you can drop it again.

You are missing the point mfb, the energy to raise the weight is provided by the buoyant force of the atmosphere.

 

[willdo]

The following is a quote from wikipédia:
Based on wind resistance, for example, the terminal velocity of a skydiver in a belly-to-earth (i.e., face down) free-fall position is about 195 km/h (122 mph or 54 m/s).[2] This velocity is the asymptotic limiting value of the acceleration process, because the effective forces on the body balance each other more and more closely as the terminal velocity is approached.In this example, a speed of 50% of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on.[end of quote]

If this is true then there should be reason to break down the harnessing of power into multiple stages (as always, no engineering issue) since every 3s one could harness 50% of terminal velocity 5x50% in 15s rather than 1x99% if we let it go...Basically, keep the acceleration at it's maximum the whole time so as to accumulate as much kinetic energy as possible.
Even for a properly shaped weight, the fall from 50km will last several minutes giving the oportunity to harness the energy many times, wouldn't the total energy harnessed be more important than the sole value of the object dropping without interference?

 

[mfb]

exemple: if a weight attains it's maximum level of kinetic energy after 1000 metres

It does not. There is no such number at all, if you can neglect effects of friction. And friction just makes it worse. The calculations here all assume no friction, the ideal case to extract energy.


Concerning buoyancy: Electrolysis at ground level needs a bit more energy than combustion of hydrogen releases if you do it a high altitude - the difference is exactly the reason why buoyancy exists at all.

 

[pbuk]

This is not possible even on a hypothetical massive planet. If you increase the gravity, you increase the pressure which will make it more difficult to electrolyze water.

I don't think that's right - by far the largest energy requirement for electrolysis is breaking the chemical bond (which is invariant with pressure), not releasing the gas from solution. At what pressure would the latter become significant?

It's true that you would need a much larger planet and/or much thicker atmosphere which would create surface gravity (and pressure) which would not be survivable.

 

[pbuk]

Note of course that I have not taken into account the recovery of chemical energy from the hydrogen by burning it in the upper atmosphere, but I think it is reasonable to assume that at best this would compensate for the inefficiencies within the system.

 

[willdo]

It does not. There is no such number at all, if you can neglect effects of friction. And friction just makes it worse. The calculations here all assume no friction, the ideal case to extract energy.

Isn't that what terminal velocity is? (sorry i just learned the name)

 

[willdo]

Note of course that I have not taken into account the recovery of chemical energy from the hydrogen by burning it in the upper atmosphere, but I think it is reasonable to assume that at best this would compensate for the inefficiencies within the system.

That's funny i wasn't counting it for the same reason (plus i woudn't know how)

 

[mfb]

Isn't that what terminal velocity is? (sorry i just learned the name)

Friction will lower the velocity, so you heat up the air a bit and can extract less energy in a useful way.
If you have too much friction, the velocity won't increase any more after a while, and you just waste energy.

 

[pbuk]

Concerning buoyancy: Electrolysis at ground level needs a bit more energy than combustion of hydrogen releases if you do it a high altitude - the difference is exactly the reason why buoyancy exists at all.

Only a bit? We might be on to a winner here then, as long as we can get that combustion energy back down to ground level efficiently (how about spinning the weight that is dropped)?

 

[pbuk]

Isn't that what terminal velocity is? (sorry i just learned the name)

Yes, but dropping a weight and catching it again at near terminal velocity is a very inefficient way of extracting potential energy. I have all along assumed that a more efficient method would be used, perhaps using an auto-gyro/turbine/flywheel kind of thing.

 

[willdo]

Yes, but dropping a weight and catching it again at near terminal velocity is a very inefficient way of extracting potential energy. I have all along assumed that a more efficient method would be used, perhaps using an auto-gyro/turbine/flywheel kind of thing.

Ah! Could this be the answer to the following?

The following is a quote from wikipédia:
Based on wind resistance, for example, the terminal velocity of a skydiver in a belly-to-earth (i.e., face down) free-fall position is about 195 km/h (122 mph or 54 m/s).[2] This velocity is the asymptotic limiting value of the acceleration process, because the effective forces on the body balance each other more and more closely as the terminal velocity is approached.In this example, a speed of 50% of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on.[end of quote]

If this is true then there should be reason to break down the harnessing of power into multiple stages (as always, no engineering issue) since every 3s one could harness 50% of terminal velocity 5x50% in 15s rather than 1x99% if we let it go...Basically, keep the acceleration at it's maximum the whole time so as to accumulate as much kinetic energy as possible.
Even for a properly shaped weight, the fall from 50km will last several minutes giving the oportunity to harness the energy many times, wouldn't the total energy harnessed be more important than the sole value of the object dropping without interference?

So the numbers you proposed earlier (0.5MJ produced from a 50km drop) were taking this into account and keeping the object in a constant accelerating mode?! If so, it's a little disapointing that the production result is so far below "the cost" but at least, it's a clear answer.

 

[mfb]

Only a bit? We might be on to a winner here then, as long as we can get that combustion energy back down to ground level efficiently (how about spinning the weight that is dropped)?

Any type of compression would need at least the energy difference you gain in the combustion process.

You cannot win against energy conservation. You cannot even get a draw.

 

[pbuk]

Ah! Could this be the answer to the following?


So the numbers you proposed earlier (0.5MJ produced from a 50km drop) were taking this into account and keeping the object in a constant accelerating mode?! If so, it's a little disapointing that the production result is so far below "the cost" but at least, it's a clear answer.

Yes I'm afraid so: 1kg x 9.8ms^-2 x 50,000m ≈ 500,000J is the gravitational potential energy of 1kg at 50km altitude so is the maximum you could ever extract from its descent to the Earth at 100% efficiency.

I could see this working on Jupiter perhaps - much larger mass, deeper atmosphere, although the high winds would be a problem (hmmm its decades since I read Larry Niven's The Integral Trees - I can't remember how the alien inhabitants of that planet got their power).

 

[pbuk]

Any type of compression would need at least the energy difference you gain in the combustion process.

I don't think there is anything being compressed?

You cannot win against energy conservation. You cannot even get a draw.

Of course not. This system gains energy from the atmosphere; each cycle the buoyant force does ALL the work raising the mass m to height h, inputting $\frac{GM}{r^2}mh$ of potential energy.

 

[Khashishi]

I already said it, but I'll make it more clear. The atmospheric pressure increases as you go down. You can't have buoyancy without atmospheric pressure. Electrolysis of water takes two molecules and splits it up into three, so it increases the pressure. The Gibbs free energy for the reaction increases with pressure, so you need to supply more energy at the bottom. You will spend more energy in electrolysis than you can recover by running a turbine on the hydrogen going up, burning it at the top, and running a water turbine on the way down.
https://en.wikipedia.org/wiki/Gibbs_free_energy

 

[mfb]

I don't think there is anything being compressed?

What do you want to spin then, and why?

Of course not. This system gains energy from the atmosphere; each cycle the buoyant force does ALL the work raising the mass m to height h, inputting $\frac{GM}{r^2}mh$ of potential energy.

You cannot gain energy from the atmosphere (apart from wind energy and similar). Neglecting wind, solar heating and so on, the atmosphere is in its lowest energy state. There is no energy you can extract, unless you find a magical vacuum at sea level or a magical source of air in space.

 

[pbuk]

What do you want to spin then, and why?

A flywheel to store energy generated by a falling auto-gyro.

You cannot gain energy from the atmosphere (apart from wind energy and similar). Neglecting wind, solar heating and so on, the atmosphere is in its lowest energy state. There is no energy you can extract, unless you find a magical vacuum at sea level or a magical source of air in space.

But the atmosphere with a hydrogen balloon at ground level is not in its lowest energy state.

 

[mfb]

A flywheel to store energy generated by a falling auto-gyro.

Don't worry about energy storage, that is not the issue here.

But the atmosphere with a hydrogen balloon at ground level is not in its lowest energy state.

Right, but you can extract this energy exactly once.

It is not necessary to consider the details. Just consider the initial state (with the hydrogen balloon, if you like) and the final state. The largest difference in potential energies you can get is to let the balloon rise up. Afterwards, there is no way to extract more energy from the atmosphere.

 

[pbuk]

Don't worry about energy storage, that is not the issue here.

Agreed.

Right, but you can extract this energy exactly once.

That's a good point. I need to sleep on it (0130 here).

 

[pbuk]

You can't have buoyancy without atmospheric pressure. Electrolysis of water takes two molecules and splits it up into three, so it increases the pressure. The Gibbs free energy for the reaction increases with pressure, so you need to supply more energy at the bottom. You will spend more energy in electrolysis than you can recover...

... you can extract this energy exactly once.

It is not necessary to consider the details. Just consider the initial state (with the hydrogen balloon, if you like) and the final state. The largest difference in potential energies you can get is to let the balloon rise up. Afterwards, there is no way to extract more energy from the atmosphere.

OK, these two are basically the same point - in order for the buoyant force (which is of course just gravity) to do work raising an object it is necessary to do work against atmospheric pressure (gravity again) to create the initial state of the buoyant object at ground level - this is the reason your average buoyancy-driven perpetual motion machine isn't.

In this case, partially inflating a weather balloon to 1m³ volume at ground level (pressure ≈ 100,000Nm^-2) costs 100kJ.

Now a cubic metre of air at ground level has mass about 1.2kg, whereas a cubic metre of hydrogen has mass about 90g so this should just lift a gross payload of 1kg. Let's assume the lifting force is 10N; this force will do 100kJ of work in only 10km but weather balloons regularly fly much higher than that, and the record is I believe about 50km. What have I missed? Is it:

1. does the lifting force decrease with altitude - surely not because 90g of hydrogen will always displace 1.2kg of air?
2. is there energy input from somewhere else - the only thing I can think of is atmospheric heat, but how?
3. going back to basics, on the way up the buoyant force (i.e. gravity) has moved 1.2kg of air down 50km and 90g of hydrogen up 50km, doing about 550kJ of net work. Surely when inflating the balloon on the ground you need to move the same column of air back up again which should cost as least as much, not just 100kJ?

 

[willdo]

I thought air was getting thinner in altitude, shouldn't it also be lighter?

 

[pbuk]

I thought air was getting thinner in altitude, shouldn't it also be lighter?

It does get less dense as the pressure decreases, but the hydrogen also expands by the same amount so it still displaces 1.2kg of air.

 

[mfb]

In this case, partially inflating a weather balloon to 1m³ volume at ground level (pressure ≈ 100,000Nm^-2) costs 100kJ.

Creating a vacuum costs 100kJ, but that's not what you want to do. The hydrogen does not appear out of nowhere. And bringing it back to the ground (to re-use it) will cost those 500kJ again.

There is a nice related thought experiment:
Imagine a closed box with gas at a pressure p in a perfect vacuum, thermally coupled to some very large object (so the gas is always at the same temperature).
As the temperature and the amount of gas is constant, p*V is constant. If you slowly expand it from volume V₁ to V₂, you get an energy of $E=pV_1 \ln\left( \frac{V_2}{V_1}\right)$. This has no upper limit - you can get as much energy as you want!

What's wrong? Well, this experiment is not practical - the divergence is only logarithmic, to get 20 pV₁ you would have to expand 1m^3 to 1 km^3. Somewhere around that point you also hit the pressure of interplanetary space.
Apart from practical considerations, it shows that assigning pV to the energy of a balloon is not useful - if you do that, you don't start at zero.

 

[pbuk]

Creating a vacuum costs 100kJ, but that's not what you want to do. The hydrogen does not appear out of nowhere.

No, it is created by electrolysis. I am assuming that this work will have to be done, in addition to breaking the chemical bonds (presumably half as much energy again will need to be used in forming the oxygen too, but I'm thinking maybe you could use a different electrolyte to avoid this: it is a detail less relevant at the moment).

And bringing it back to the ground (to re-use it) will cost those 500kJ again.

Obviously, but I'm not bringing it back to the ground, I'm burning it at altitude and reclaiming the chemical energy to help power the next cycle of electrolysis.

There is a nice related thought experiment:
Imagine a closed box with gas at a pressure p in a perfect vacuum, thermally coupled to some very large object (so the gas is always at the same temperature).
As the temperature and the amount of gas is constant, p*V is constant. If you slowly expand it from volume V₁ to V₂, you get an energy of $E=pV_1 \ln\left( \frac{V_2}{V_1}\right)$. This has no upper limit - you can get as much energy as you want!

Isn't that pretty much how an air source heat pump works to extract thermal energy from the atmosphere? Commercial pumps operate at about 250% "efficiency" I believe.

 

[mfb]

No, it is created by electrolysis. I am assuming that this work will have to be done, in addition to breaking the chemical bonds (presumably half as much energy again will need to be used in forming the oxygen too, but I'm thinking maybe you could use a different electrolyte to avoid this: it is a detail less relevant at the moment).

Okay.

Obviously, but I'm not bringing it back to the ground, I'm burning it at altitude and reclaiming the chemical energy to help power the next cycle of electrolysis.

There you lose your 500kJ compared to electrolysis on the ground.

Isn't that pretty much how an air source heat pump works to extract thermal energy from the atmosphere? Commercial pumps operate at about 250% "efficiency" I believe.

Heat pumps create temperature differences - they cool one side and heat the other side. If you add the effects on both sides, you get exactly 100% "efficiency".
I don't see what you plan to cool down here. And heat output is not the result you want, right?

 

[pbuk]

There you lose your 500kJ compared to electrolysis on the ground.

? The bond enthalpies are the same at any altitude, I have taken into account the work done in expanding the gas products (well the hydrogen anyway): what else is there?

Heat pumps create temperature differences - they cool one side and heat the other side. If you add the effects on both sides, you get exactly 100% "efficiency".

Yes, and if the side that is cooled is reheated by the atmosphere you typically get out 250% of the energy you put in. That's why I put "efficiency" in quotes as obviously true efficiency is always less than 100%.

I don't see what you plan to cool down here. And heat output is not the result you want, right?

No it isn't, and I don't want to cool anything - this was your analogy, I simply showed that it actually supports the idea of extracting energy from the atmosphere to provide a power gain.

 

[mfb]

? The bond enthalpies are the same at any altitude, I have taken into account the work done in expanding the gas products (well the hydrogen anyway): what else is there?

To reverse the process at high altitude, you first have to re-compress your hydrogen again (or get less energy out of the reaction). Which needs 500kJ.

Yes, and if the side that is cooled is reheated by the atmosphere you typically get out 250% of the energy you put in. That's why I put "efficiency" in quotes as obviously true efficiency is always less than 100%.

I don't see how this is related to this thread. Heat energy has a lower "worth" in terms of entropy. This is not relevant if we do not have temperature differences.

No it isn't, and I don't want to cool anything - this was your analogy, I simply showed that it actually supports the idea of extracting energy from the atmosphere to provide a power gain.

You want more - you want the atmosphere to be the same after the process, this would violate energy conservation (and would allow to reduce entropy, again an impossible thing).

 

[pbuk]

To reverse the process at high altitude, you first have to re-compress your hydrogen again

No, to reverse the process of decomposing water into hydrogen and oxygen I simply set light to the hydrogen. I acknowledge that this will only get me back the bonding energy (140MJ/kg) and the work done against the atmosphere in electrolysis (100kJ/m³) is lost.

I don't see how this is related to this thread. Heat energy has a lower "worth" in terms of entropy. This is not relevant if we do not have temperature differences.

The heat pump cycle is not related to this thread, let's drop it.

You want more - you want the atmosphere to be the same after the process, this would violate energy conservation (and would allow to reduce entropy, again an impossible thing).

No I don't want the atmosphere to be the same, I want it to provide energy into the process in order not to violate energy conservation (I'm not a fool), but I can't see how it can. Alternatively I want to know where the reasoning is wrong - where do you have to put that 500kJ of work done by the buoyant force back in? Khashishi seems to think it is in the electrolysis, but until I see a calculation of where that energy is going (on top of the bonding energy reclaimed through combustion - 140MJ/kg, or about 12.6MJ for the 90g of hyrdogen to fill 1m³ - and the 100kJ of work against the atmosphere I have already allowed for, and the 50kJ expanding the oxygen I have ignored) I am very much not convinced. Hand waving "it must be so for energy conservation" arguments aren't going to help when there is a whole atmosphere of heat energy out there to make up the balance.

 

[willdo]

In this case, partially inflating a weather balloon to 1m³ volume at ground level (pressure ≈ 100,000Nm^-2) costs 100kJ.

Now a cubic metre of air at ground level has mass about 1.2kg, whereas a cubic metre of hydrogen has mass about 90g so this should just lift a gross payload of 1kg. Let's assume the lifting force is 10N; this force will do 100kJ of work in only 10km but weather balloons regularly fly much higher than that, and the record is I believe about 50km. What have I missed? Is it:

1. does the lifting force decrease with altitude - surely not because 90g of hydrogen will always displace 1.2kg of air?
2. is there energy input from somewhere else - the only thing I can think of is atmospheric heat, but how?
3. going back to basics, on the way up the buoyant force (i.e. gravity) has moved 1.2kg of air down 50km and 90g of hydrogen up 50km, doing about 550kJ of net work. Surely when inflating the balloon on the ground you need to move the same column of air back up again which should cost as least as much, not just 100kJ?

No I don't want the atmosphere to be the same, I want it to provide energy into the process in order not to violate energy conservation (I'm not a fool), but I can't see how it can. Alternatively I want to know where the reasoning is wrong - where do you have to put that 500kJ of work done by the buoyant force back in? Khashishi seems to think it is in the electrolysis, but until I see a calculation of where that energy is going (on top of the bonding energy reclaimed through combustion - 140MJ/kg, or about 12.6MJ for the 90g of hyrdogen to fill 1m³ - and the 100kJ of work against the atmosphere I have already allowed for, and the 50kJ expanding the oxygen I have ignored) I am very much not convinced. Hand waving "it must be so for energy conservation" arguments aren't going to help when there is a whole atmosphere of heat energy out there to make up the balance.

I thought about it and i fail to see what bothers you.

For one thing, Hydrogen is not readily available in this form at ground level, it would go up...and that is the energy source you're looking for:
As you yourself explained to me earlier the floatability (buyancy?) potential of 1m3 of hydrogen on Earth is 500KJ which equals the 500KJ one might harness at 100% efficiency from drop the maximum payload from a 50km altitude.
the 100KJ is just a waste of the mechanical process (one among others)while the energy released from combustion is to be compared to the cost of producing the hydrogen in the first place.

 

[mfb]

No, to reverse the process of decomposing water into hydrogen and oxygen I simply set light to the hydrogen. I acknowledge that this will only get me back the bonding energy (140MJ/kg) and the work done against the atmosphere in electrolysis (100kJ/m³) is lost.

It will give you less. The energy released in the burning depends on the pressure of the gases.

No I don't want the atmosphere to be the same

So what changed after your process?
Cooling the atmosphere would violate the second law of thermodynamics, you do not have a colder reservoir.

 

[russ_watters]

MrAnchovy, the combustion energy really is different at different starting temperatures and pressures (and final water temperatures and pressures). It doesn't matter if you believe it or not: it is the truth.

 

[jim hardy]

Of course not. This system gains energy from the atmosphere; each cycle the buoyant force does ALL the work raising the mass m to height h, inputting GM r 2 mh of potential energy.

EDIT ------- oops never mind I didn't read whole thread close enough... first post bilged.

 

[jim hardy]

Coffee percolator uses similar principle - lower density inside the center tube lifts a coffee-steam mix up into the basket by force of buoyancy.
Of course the energy comes from a burner ...

Better analogy might be a hydro power plant
Seawater is vaporized by sunshine, lifted and transferred to a reservoir as rain.
Heat, and lots of it, was rejected to upper atmosphere when the raindrops condensed and we capture a teeny bit of gravitational energy as the working fluid returns to ocean. But it's minute compared to the heat of vaporization required to make the cycle go...
That energy comes from sun.

Your weather-balloon gizmo , if I understand, let's water vapor do essentially same thing as in analogy just above ? Except instead of a simple phase change you remove 16/18ths of the mass by stripping off the Oxygen atoms with electricity ??
And the question is can this break even?
I think Mr Anchovy answered that in post 6 comparing H2's bonding energy to gravitational (elevation) energy
I'm with Carnot on this one.


But being a bit Asperger's I'm usually a couple paragraphs behind the conversation. Did I miss this one too?

 

[Dale]

We do not discuss perpetual motion machines on this forum, not even just to debunk them.