Potential energy of water from electrolysis

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Generating hydrogen through electrolysis requires energy input, and when the hydrogen rises, it gains gravitational potential energy. The energy to lift the hydrogen comes from the energy used to electrolyze the water, which is proportional to the pressure at which the process occurs. As the hydrogen rises, it displaces air, causing that air to lose potential energy, which can be harnessed. The discussion highlights that while the concept does not violate conservation of energy, practical challenges exist, and the overall efficiency of such a system cannot exceed 100%. Ultimately, the energy dynamics involve complex interactions between buoyancy, pressure, and energy conversion.
  • #31
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?
 
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  • #32
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.
 
  • #33
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.
 
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  • #34
Buckleymanor said:
Simon Bridge said:
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.
 
  • #35
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? ;)
 
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  • #36

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.
 
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  • #37
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?
 
  • #38
Simon Bridge said:
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.
 
  • #39
Simon Bridge said:
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.
 
  • #40
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?
 
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  • #41
CWatters said:
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.
 
  • #42
Thanks.
 
  • #43
Simon Bridge said:
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.
 
  • #44
Buckleymanor said:
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.
 

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