Basic Hydrogen/gravity/thermodynamics question from a noob.

[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.