Originally Posted By: paul

the molar mass of 1 molecule of water = 1 H + 1 H + 16 O = 18 g * mol^-1

Originally Posted By: paul

Michael Farraday has shown that 1 liter of HHO can be produced in 1 hour using 2.34 watts or 2.34 watt hours.
[...]
but 2.34 watts delivered over a 1 hour time period is only 8424 J


Let's see:

2.34 watt hours ~ 8424 Joule. So far ok.

Originally Posted By: paul

it would take 1800 hours to convert 1 liter of water into HHO using 2.34 watts


Didn't understand that one. Why 1800 hours?
I tried a different approach:

1 liter of an ideal gas, at standard temperature and pressure, will contain 0,0446 moles of molecules.
= 6.022 * 10^23 / 22.414 molecules.

The gas will be in molecular state, meaning H2 + O2 molecules in a 2:1 ratio.
= 0.272 * 10^23 molecules (H2 and O2) per liter
means: 0.091 * 10^23 of O2 and 0.181 * 10^23 of H2
or: 0,015 mole of O2 and 0,03 mole of H2

= they could recombine to 181 * 10^20 molecules of water.
= 181 * 10^20 * 3 * 10^-23g
= 0.543g

this is pretty much the same as your figures:

Originally Posted By: paul

0.555 g of water / 0.00000000000000000000000002991507 kg 1 molecule of water =
= 185... * 10^20 water molecules


How much energy can we get back of our one liter of HHO?

Originally Posted By: paul

Originally Posted By: Momos

The enthalpy of combustion for hydrogen is 286 kJ/mol




This figure is from http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/electrol.html,
while it is the energy input it is NOT the energy output (due to excess heat) and far less the usable energy in a combustion engine.

But nevertheless, lets see what we get:

286 kJ/mol of hydrogen
0.272 * 10^23 molecules (H2 and O2 in 2:1 ratio).
means: 0,091 * 10^23 of O2 and 0.181 * 10^23 of H2
= 0,03 mol of H2

0,03 * 286 kJ/mol = 8,58 kJ

Nearly the same as the 8424 J at the beginning (probably due to rounding and because my initial assumption about HHO as an ideal gas is wrong).

But why are you so far off?

Originally Posted By: paul

= 18552565602540531505933.4781882 water molecules
by multiplying the number of water molecules by the 286 kJ/mol
we get 5306033762326592010696974761.8252 J


Well: the energy is in Joule PER Mol.
You multiplied by the number of water molecules, instead you have to divide again by Avogadro's constant.
Doing this we get:
5306033762326592010696974761.8252 J / 6.0221415 1023
= 5306 * 10^24 J / 6.022 * 10^23
= 8810 J

There you are.
Still you could argue in this calculation you still have some energy gained (8424 J input and 8580 J respective 8810 J output). But we haven't consider the surrounding heat which is reducing the necessary input energy for the electrolysis (but also reduces the energy we can gain).
Furthermore this 8424 J of input energy are disputable, since I couldn't find a credible source for this figure.

And all this without even thinking about losses in alternator, excess heat, wheel, sub-optimal chamber design, etc. ...

[quote=paul]
thats probably why they use hydrogen and oxygen to launch the space shuttle instead of gasoline.
[quote]

Nobody has ever doubted hydrogen and oxygen are a great way to store energy. The potential energy in HHO might be 3 or 4 times larger as for gasoline. The same is true for tnt or plutonium.
STORING Energy is nice, but you can't get energy out of nothing.

Last edited by Momos; 05/27/10 12:05 AM.