# Dumb Question

but the vast majority can be provided. the falling water genertaing electricity would provide the rest.

Electrolysis doesnt use up lots of energy... it can be done with 6v

Danger
Gold Member
Electrolysis doesnt use up lots of energy... it can be done with 6v

Voltage isn't the only factor. You could discharge a Van de Graf generator at a couple of million volts into the tank and get nothing. A 12V car battery at 1,000 amps, however, could produce enough to create a serious explosion hazard.
And I dispute your assertion that 'the vast majority' of the electrolysis energy could be created by a fuel cell powered from that same water. This is once more, though, in the realm of math.

in a perfect world it would create all of the energy. bonds broken and bonds formed. When elctrolysing it, you have effectively broken 2 hydrogen water bonds and formed 1 hydrogen hydrogen bond. in combustion the opposite happens. obvioudly its not a perfect world however im sure the energy produced would go most of the way to electrolysing it.

Danger
Gold Member
In a perfect world, I'd be married to Sandra Bullock, have Angie Harmon for a mistress, and never have to work for a living...
Again, it comes down to the numbers. You have to work it out and post actual figures in order to make your point.

from howstuffworks: "If the fuel cell is powered with pure hydrogen, it has the potential to be up to 80-percent efficient. That is, it converts 80 percent of the energy content of the hydrogen into electrical energy. "

however im not sure if this is pressurised hydrogen...

edit: here is another source from wikipedia: "A cell running at 0.6V has an efficiency of about 50%, meaning that 50% of the available energy content of the hydrogen is converted into electrical energy; the remaining 50% will be converted into heat."

However im pretty sure many methods exist to make use of the heat energy (steam turbines etc). Just as in many power plants, the steam produced can passed close to water pipes, heating it up and producing high pressure steam which drives turbines.

edit2: this is pretty interesting: http://en.wikipedia.org/wiki/Reversible_fuel_cell

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Astronuc
Staff Emeritus
dont quite understand the importance of this? do u mean nitrogen dioxide will be produced (as in car engines)? Any car can be run on hydrogen so the actual combustion would be no problem, the combustion could turn a dynamo etc
NOx are consequences of combustion in air, but that can be addressed. The issue is that any energy released from combustion will be somewhat dissipated, in this case through N2 which is relatively inert. At the other end, one has to put more energy into the electrolysis process than is required to break each bond. So in either step, there is a loss of useful energy.

Im starting to want this to work lol... But now that the source of energy has been identified as convection, it seems a bit more pheasable...
One has to look at the energy transfer in various processes to see how feasible.

Basically, hydropower utilizes some of the steps you've mentioned, but without electrolysis and separation of H from O in H2O. The sun evaporates moisture, which is transported from a lower level (e.g. sea level) to higher level (mountains) - there is a change in the gpe of that water. The atmosphere (wind, water vapor and clouds) does the mass transport. Condensation causes vapor to liquid, which then runs down into streams, rivers, into lakes/reservoirs. The water is allowed to fall through hydroturbines which drives generators, which generate electricity. The mass transport system is huge - and the solar energy is one the order of 1 kW/m2.

Then there is wind power, which extacts energy from wind, which is driven by large mass flows (convection) in the atmosphere.

Danger
Gold Member
Pressurization is irrelevant; it has to be de-pressurized for use. 80% return sounds like a reasonable figure, but that's a far cry from 'vast majority'.

edit: Oops! A bunch of bloody customers came in while I was composing, so I've been gone for a while. Missed Astro's post until now. In fact, I've still missed it. Have to go back and read it now.

edit#2: And now that I've read it, I'll add only one comment. Wind power is still solar-driven, since the heat energy that causes convection comes from the sun.
Thanks for re-joining, Astro. I'm at the end of my knowledge here.

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Astronuc
Staff Emeritus
Don't confuse energy conversion (and it's attendant efficiencies) with energy creation.

Bullock and Harmon, eh? Hmmmm. lol

Danger
Gold Member
Bullock and Harmon, eh? Hmmmm. lol

Yeah, well... you know...

Astronuc
Staff Emeritus
Yeah, well... you know...
:rofl: I had the same thought. Could be detrimental to one's longevity. :rofl:

ok, so is there anyway of getting some estimates on this? not too sure where to go for estimates of energy loss etc

good to know that it doesnt need to be pressurised, that could have used quite a lot of energy.

so, we have ~20% energy loss to make up for the combustion process (dont know if this inlcludes getting energy back from the exhaust gas heat) and how much in the electrolysis process would you say is it in thr order of about 20% again? As well as other things, such as waste heat (if that wasnt included in the 20%above). When making this up, any turbines turned by the water also have frictional loss... do you have anyywhere i could get numbers for this? i imagine it depends of turbine size.

Yes i can see the similarity between this and hydropower... however its disadvantages are that you cant just turn rain on and off. And the same for wind power... This seems to be much more reliable (although more complicated than just a large turbine/impellor (i think thats the name for it)...

anyway, its getting late here in merry england so i'll call it a night!

thanks very much both of you...

sam

Hey, i've started a few calculations. I got the overall energy input to to break 2 mol of water as 469.6kJ. There are 4 moles of O-H bonds to break @ 458.9kJ/mol each (source: http://en.wikipedia.org/wiki/Bond_energy)

You form 2 H-H bonds @ 436kJ/mol each (same source). And you form one O=O bond @ 494kJ/mol.

458.9 * 4 - 436 * 2 - 494 = 1835.6 - 872 - 494 = 469.6

at the combustion side im assuming you would get this energy out (in a perfect world)

At the combustion side, assuming 80% efficiency, the energy output is
0.8*469.6 = 375.68kJ/2mol = 187.84kJ/mol

at the electrolysis side, if i have 80% efficiency (this not a real number), is it 1.2 *469.6 = 563.52kJ/2mol = 281.76kJ/mol

therefore, total energy to make up/mol = 281.76-187.84=93.92kJ/mol

OK, i found this about water on (http://en.wikipedia.org/wiki/Pumped-storage_hydroelectricity): [Broken] "For example, 1000 kilograms of water (1 cubic meter) at the top of a 100 meter tower has a potential energy of about 0.272 kW·h."

That's 979.2kJ so thats 979.2J per 1000grams, 1mol water = 18g. 0.9792 * 18 = 17.6256J

This is getting rediculous! per meter that's 0.176256J. we want 93.92kJ. so total meters to fall is 93920/0.176256 = 532861.29m = 533km

LMAO.

I think this is over...

sam

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no wait i think we're ok:
"However, the exosphere can extend from 500 up to 10,000 km above the surface,"

lmao :P

Danger
Gold Member
Okay, so we're agreed that it's over. But isn't it far more satisfying and eductional to have come to that conclusion yourself through logical analysis rather than if you'd just been told to pack it in?

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yea its definitely much better.

further musings:

if the drop was 500m, the total generated would be 1.36kwh = 4896kJ per 1000 kg of water = 4.896j/g. Times 18 is 88.128J/mol lost. Tot energy input = 88.128+469600 = 469688.128J Tot energy out = 469600j

469600/469688.128 * 100 = 99.98123% efficiency

hmmm... interesting

Danger
Gold Member
I'll have to wait for Astro to verify those numbers; it's way over my head. Good on ya for going through all of that.

hehe.. i was surprised i could actually work it out with my knowledge of chemistry and physics...

Thanks both of you for your help... it was nice to get to the bottom of it