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willdo
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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 said:Conservation of energy tells us this can't be done at all.
willdo said: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?
Nugatory said:The maximum kinetic energy is achieved by dropping the weight from infinity... so you only get to do it once.
Right, and the difference is a factor of 10 in energy.willdo said: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...
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?
mfb said: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.
mfb said: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.
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.willdo said:exemple: if a weight attains it's maximum level of kinetic energy after 1000 metres
Khashishi said: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.
Isn't that what terminal velocity is? (sorry i just learned the name)mfb said: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.
That's funny i wasn't counting it for the same reason (plus i woudn't know how)MrAnchovy said: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.
Friction will lower the velocity, so you heat up the air a bit and can extract less energy in a useful way.willdo said:Isn't that what terminal velocity is? (sorry i just learned the name)
mfb said: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.
willdo said:Isn't that what terminal velocity is? (sorry i just learned the name)
Ah! Could this be the answer to the following?MrAnchovy said: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 said: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?
Any type of compression would need at least the energy difference you gain in the combustion process.MrAnchovy said: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)?
willdo said: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.
mfb said:Any type of compression would need at least the energy difference you gain in the combustion process.
mfb said:You cannot win against energy conservation. You cannot even get a draw.
What do you want to spin then, and why?MrAnchovy said:I don't think there is anything being compressed?
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.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.
A flywheel to store energy generated by a falling auto-gyro.mfb said:What do you want to spin then, and why?
But the atmosphere with a hydrogen balloon at ground level is not in its lowest energy state.mfb said: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.
Don't worry about energy storage, that is not the issue here.MrAnchovy said:A flywheel to store energy generated by a falling auto-gyro.
Right, but you can extract this energy exactly once.But the atmosphere with a hydrogen balloon at ground level is not in its lowest energy state.
Agreed.mfb said:Don't worry about energy storage, that is not the issue here.
That's a good point. I need to sleep on it (0130 here).mfb said:Right, but you can extract this energy exactly once.
Khashishi said: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...
mfb said:... 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.
willdo said:I thought air was getting thinner in altitude, shouldn't it also be lighter?
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.MrAnchovy said:In this case, partially inflating a weather balloon to 1m3 volume at ground level (pressure ≈ 100,000Nm-2) costs 100kJ.