Meyer type WFC - from design and fabrication to test and development.

[Farrah Day]

Left my test cell in the de-ionised water overnight and it still read 0.7 volts this morning.

Took my test cell apart today.

I roughed it up, inside and out with a fine sandpaper, cleaned all the residues off and put it back together.

Now, under the same conditions as previously, in de-ionised water with 31 volts, I draw a current of 120mA from the PSU. Twice that I did before.

Discharge times/voltages remain unaltered.

I assume that I have increased the surface area of my tubes by around double by my etching action, and hence the cells capacitance too.  So I would now expect my cell to be capable of holding twice the charge it did previously.

I also made up a smaller test cell, 5" long with a 3/4" dia tube inside a 1" dia tube.  This cell has only 1/16th inch space between the tubes.

When placed in my de-ionised water, and fed 31 volts, it draws 0.5 Amp, and produces a considerable amount of gas from the deionised water.

With PSU removed, my discharge voltages are much lower than the larger test cell, but it still holds its charge.

After 1 min = 1.4 volts, 2 mins = 1.1 volts and 5 mins = 0.7 volts

What I am thinking now is, if the potential required to initiate electrolysis of my small test cell is, let's say, 2 volts, then I will only need to apply a dc pulse periodically to keep the voltage topped up. If after 1 minute my cell still exhibits a voltage of 1.4 volts, then I only need to effectively add 0.6 of a volt to initiate electrolysis, rather than the full 2 volts.

Of course, ideally we would not want it to drop below the 2 volts at all, in order that electrolysis was continually taking place. So the most efficient scenario would be to find the correct pulse frequency to maintain this 2 volts continually, but no more than necessary.

Now, although there is no relationship between the size of my larger test cell or the smaller one, one thing immediately strikes me. After I minute of discharging, my small 2 tube test cell still shows a voltage of 1.4 volts.  My much larger test cell, with its 2 floating plates shows 4.2 volts after 1 minute of discharge. If I divide 4.2 volts by the number of cells (3) I have exactly 1.4 volts... the same as my small test cell!  I can not see this as being a coincidence.

I need to make a large 2 tube cell to see if I can corroborate this 1.4 volts per cell pd after 1 minute.  I also want to make a copper test cell up to see if it too exhibits this charge retention capability.

Incidentally, I find that if, for example I'm discharging my large test cell and leave it for an hour, I only have to momentarily touch the PSU power leads to the test cell to get the discharge voltage way back up.

Basic stuff, I know, but nevertheless very intriguing.

[Farrah Day]

Not yet had chance to do any tests with a copper tube cell, hopefully do this tomorrow.

Today I upped the voltage across my large test cell to 70 volts, which drew approximately 200 mA from the supply.

Even in the de-ionised water, at this voltage, it was giving off gas considerably.

However, it is clear that upping the voltage does not effect the discharge time/voltage, as the results were the same as for 31 volt tests.

It would appear then that my test cells will hold the same amount of charge irrelevant of how much voltage I apply or for how long, once this charge threshold for my cell is reached.  The extra charge of the higher voltages and respective current flow is actively used up in electrolysis, so that once I remove the PSU, the same remaining amount of charge is left every time, to discharge at its own pace.

Why? 

Of course... I'm below the electrolysis threshold voltage!  How did I not see this??

So, once I remove my PSU, electrolysis takes place for a finite amount of time, but then the voltage rapidly drops below the voltage level that is needed to maintain electrolysis. However, there are still charges within the electrodes (-ve electrons and +ve holes), and -ve and +ve ions waiting on the surface. From this point on, only very slow leakage of charges occurs, and as the cell slowly discharges, the lower the voltage drops, so the slower and slower discharges, etc, etc.  Makes sense now. 

That is of course why it doesn't matter what size or shape my electrodes/tubes are.  Now I always give the figure for the cell discharging against time, but if I look at the voltage across my large test cell at the instant I disconnect the PSU from the cell, then it momentarily reads 5.8/5.9 volts.  This will be just under the electrolysis threshold voltage, which for my large test cell will then be around 6 volts, or 2 volts per cell. All sounds right now, doesn't it.

So, now I'm thinking that the ss chromuim oxide layer at this point is doing absolutely nothing.  When I etched my tubes with sandpaper, I had not increased the capacitance, but I had indeed increased the tube surface area, and hence the increase in current.

Why cant I discharge my cell instantly?  Well, quite simply, I cannot remove the ions off the electrodes in the water,  so shorting my cell does nothing!

Time to reconsider things, methinks.

[Farrah Day]

Having now realised how stupid I've been above, I think it's time to try to develop the oxide layer on my large test cell to see what can be gleamed from this.

I have placed the large test cell in tap water for this, otherwise I can't draw enough current from the PSU.  31 volts @ 768 mA when I left it.

So, at this point I must assume that the ss protective oxide layer on my test cells to be far too thin to realistically act as a dielectric or insulator of any kind, and hence it is effectively doing nothing in terms of capacitance.

I read that Ravi states that big bubbles were achieved once he had conditioned his electrodes.  He no longer had the small bubbles forming on the electrodes, just big bubbles occuring.  Now this would lead me to think that the oxide layer is now too thick for the ions in the water to easily exchange charges with the electrodes through this insulation, hence no small bubble formation. However every now and then, as the charges build up, a voltage spike will cause a weak area in the oxide layer to break down. The causes localised and rapid ionisation due to large charge build up occurs, and hence the big bubble formation.

Will periodically check my test cell to see what is happening and report back.  In the meantime I'm going to finish boxing my D14 cct.

[Spewing]

ravi said he used dish detergent.

[Farrah Day]

Curiosity got the better of me and I decided to make up yet another small test cell.

I managed to find some alloy tube of 1 and 3/4 inch outside diameter to use as an anode, and cut some ss tube to use as a cathode inside this.

The cell is 6 inches long, with the space between the outer alloy (primarily aluminium) electrode, and the inner ss electrode just over 1 mm.

Placed this test cell in normal tap water and ran it for about half hour under normal dc electrolysis, with 10 volts at around 1.5 A. 

Quite surprised then when I disconnected the PSU as in previous tests with my ss test cells to find that the voltage immediately drops to just a fraction of a volt (0.3v). Tried this a few times always with the same result; unlike ss cells, this electrode configuration holds next to no charge for any discernable period of time!

As I was using fresh tap water, and in case there was some unknown or unnoticed difference in my method, I immediately repeated the test with my small ss test cell.  After only a few minutes, my small ss test cell behaved just as it always did and held the charge as detailed in posts above.

So, what is this test telling me? 

Well, apart from not getting the results I expected, I'm not quite sure yet.

It could be due to the electrochemical difference between the ss and alloy electrodes, but nevertheless I had expected the protective aluminium oxide layer to act as some form of dielectric barrier, and hence hold a charge.

I need to try this test again now, but using alloy for both anode and cathode to eliminate the electrochemical potential difference, and so see if I get a different result.

Whatever the outcome of another alloy test, clearly ss at least, has the desirable property of holding a charge.