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- Thread starter willdo
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russ_watters

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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 [itex]H_{2}[/itex], I looked up.

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

[itex]|U|= G\frac{M_{H_{2}}M_{E}}{R_{E}}[/itex]

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:)

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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 accelarating at some point? In wich 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?

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?

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pbuk

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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.

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pbuk

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

Now 1m^{3} 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.

Now 1m

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.

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Nugatory

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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.

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pbuk

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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.

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Khashishi

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mfb

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Right, and the difference is a factor of 10 in energy.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...

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?

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Just to make sure, how would that change if we started in water, say 10km deep? (assuming no engineering issue)

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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

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pbuk

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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.

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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?

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?

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mfb

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

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.

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pbuk

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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.

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pbuk

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

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mfb

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Friction will lower the velocity, so you heat up the air a bit and can extract less energy in a useful way.Isn't that what terminal velocity is? (sorry i just learned the name)

If you have too much friction, the velocity won't increase any more after a while, and you just waste energy.

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pbuk

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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)?

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pbuk

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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.

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Ah! Could this be the answer to the following?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.

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 accelarating 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.

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mfb

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

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

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pbuk

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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 accelarating 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

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).

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