Exploring Energy Conversion with Hydrogen and Water

[samski]

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 doesn't 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 can't 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 that's 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

 

[samski]

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 I am 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): "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 that's 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

 

[samski]

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]

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?

 

[samski]

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]

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

 

[samski]

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