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Limit to gas production by electrolysis of water?

Unread postPosted: Sat Mar 26, 2011 3:34 pm
Author: saefroch
So being that I just finished a short review of electrochem, I decided it'd be interesting to calculate how much gas, as a mixture of H<sub>2</sub> and O<sub>2</sub> can be generated, in Litres per minute per amp. I came out with a results of about 0.0114, which seems terribly low... can anyone confirm or prove me wrong?

Unread postPosted: Sat Mar 26, 2011 5:57 pm
Author: Technician1002
Most likely most spudders have not gone far into efficiency. There are forums in the HHO groups, brown gas, and electrolysis.

Have you searched those threads?

The amount of power required for the conversion catches many of the over unity believers off guard.

Free energy from water isn't free. Water is burnt Hydrogen. The energy you get from burning Hydrogen needs be supplied to un burn water to get the Hydrogen and Oxygen back.

Unread postPosted: Sat Mar 26, 2011 7:12 pm
Author: saefroch
I have looked at some of those forums, and I thought I might be able to get a more educated answer here than on them, as most of the people on many of those sites, especially HHOforums, aren't keen on calculations.

On one of the links you posted I saw an answer of 11.4milliliters/minute/watt, which seems somewhat like the answer I got but it's in mL, minutes, and watts, and I can't figure out how to convert it. I believe I also saw other results of different answers.

It seems every time I ask a question about electrolysis of water, people assume I'm trying to use the mixture of H<sub>2</sub> and O<sub>2</sub> to power something, which I'm most certainly not. I am nearly at the end of AP Chemistry and Physics B, which means I am familiar with the conservation of energy and what Gibbs free energy is.

I am also familiar with entropy, which is why I don't see why there is so much research into "HHO." My intention is to produce H<sub>2</sub> for use as a propellant gas.

Unread postPosted: Sun Mar 27, 2011 11:09 am
Author: DYI
The reason you can't figure out how to convert it is that a conversion is not possible. You're looking at an I<sup>2</sup> dependency here, not a linear one. As such, L*s<sup>-1</sup>*W<sup>-1</sup> is the correct dimensionality. An excessively simplified analysis of the situation shows that the absolute maximum efficiency (with no regard for electrochemistry concerns, which I'm not very familiar with) is 0.39L/(W*s). This is measured in power driving the reaction, not power draw of the cell.

Expect roughly twice that power draw from the cell, with a very well designed cell, per liter gas output. Efficiency varies based on cell geometry, voltage, and use of electrolyte. figuring out how to optimize those parameters is your problem.

Unread postPosted: Sun Mar 27, 2011 1:26 pm
Author: saefroch
How did you get that answer? I used electrochemistry and dimensional analysis to get mine, with a tiny bit of physics. I'll explain below, and maybe you can show me how you got your answer?

1 amp for 1 min is 60C. Convert that to moles of electrons with the Faraday(1/96,500), convert moles of electrons to moles of reaction(1/4), convert moles of reaction to moles of gas(3/1), convert moles of gas to Litres of gas (22.4/1), so that gives an answer in L*min<sup>-1</sup>*amp<sup>-1</sup>, then convert that using ideal gas relationships , final answer is 0.0114L*min<sup>-1</sup>*amp<sup>-1</sup>.

I understand that a real cell will be far less efficient, one of the efficiency calculations I saw came out to only about 44% (not all that bad in the grand scheme of things).

Unread postPosted: Sun Mar 27, 2011 2:17 pm
Author: DYI
My result is simply based on the enthalpy of formation of water. It yields a result in joules per mol of gas production which is a lower bound. This is equivalent to W/(s * mol), which is obviously not linear with current.

I'd like to stress that I've got no experience in the field of electrochemistry, but there's no way it could possibly be more efficient than that.