Meyer's WFC concept analysed

[Farrah Day]

My apologies TH

After reading your post asking where do the ?stripped off? electrons go, I realised that I may have confused you with an earlier post, which was not very clear (I was rushing at the time). I will amend that post.

Incidentally, I didn't respond to Eso's post as I did not understand what he was talking about. No offence intended Eso. Just to say that you are wrong to think that there are 'free' electrons mooching about in the solution.

There are no free electrons in the electrolyte (our water). In liquids (electrolytes), unlike solid conductors, electrons are not the charge carriers. In an electrolyte, ions carry the charge, not electrons. Ion?s being charged molecules or charged atoms.

Normal tap water is made to ionise when we apply a voltage potential across it above this ionisation threshold ? just a couple of volts.  It ionises into OH- and H+, with the H+ being attracted to the cathode and OH- to the anode.  Water does not ionise very readily, but we can help it along by applying more voltage.

To increase efficiency of normal electrolysis of water, a compound (such as sodium sulphate) is added. This compound makes for a good electrolytic solution because, unlike the water molecule it readily dissociates into charged ions within the water. Hence there are a lot more charges carriers, which so require far less voltage potential to do work. Basically we lower the resistance of the electrolyte. Now, although with a compound such as sodium sulphate the ion concentration is greatly increased, it does not itself take part in the reaction on the electrodes, but remains in the solution. We just still get our oxygen and hydrogen, but far more efficiently. This is due to reaction preferences that I?ll not go into here, but you can always research yourself.

Anyway, that is to say that electrons, themselves do not take part in carrying charges in ionic solutions but play their part by adding or reducing the charge of an atom or molecule so producing an ion.

I think that the water molecules will initially be aligned (polarised) by the electrostatic field.

If the cell is a capacitor with the oxide layer the dielectric, and the water simply an extension of the cathode, then I?m not sure plate spacing is important. This is because the negative plate is essentially the water and is spaced only microns from the anode (the thickness of the dielectric).  Usually, the simplest explanation (that makes scientific sense) is the one to go with, but there might well be more going on than I?ve considered, so don?t hold me to anything.

Now, I know I?m throwing a lot of ideas and theories about on this thread, but sometimes its good to ?air? your thoughts. I find that my thoughts don?t always make quite as much sense as they did in my head once that I?ve written them down and read through them.

One thing for sure though, if your going to assume or theorise on a subject, you do really need to be able to explain what reaction or reactions you think are actually occurring, and why.  It helps therefore if you have some scientific background or experience on which to ground your ideas and explanations. Many people seem to fall into this trap by simply assuming or accepting that something happens without questioning if it is actually possible or not. I try to work through all of my theories in order not to fall into this trap, but no doubt there are times when I get my science wrong.

In an attempt to research all avenues on this subject, I noted something of interest. Water, with a molecular weight of just 18, is actually lighter than many gases. Why is it not a gas?  Apparently it is because it ?clumps? together. Not as in covalent bonding, but simply the attraction between water molecules due to their bipolar nature. A little like how magnets would clump together when you throw them all in a bag.

This lead me to think that perhaps we are breaking the clumping bond at some point and actually getting water gas (not steam) produced.  Then I remembered reading something about Brown?s gas and how bubbles were emerging from between the plates (not on them).  There was also talk of ?expanded? water.  There could well be something important in this, though I?m not sure what the properties of water gas would be? would it ignite?  And what would stop it from immediately clumping?

More questions needing answers.

Now, I once read somewhere that Meyer was saying to use one type of ss over another, as most produced an oxide layer, which was unwelcome.  However, I?m fairly convinced we really do need that oxide dielectric layer ? so there?s another area of contention.  The high voltage across the dielectric will no doubt cause the water molecule to become polarised, but it is not until the dielectric oxide layer breaks down that the water becomes ionised and hence gas is produced.

Now Meyer suggested that the high electrostatic field on the plates broke the covalent bonds of the water molecule, but not into OH- and H+ as in electrolysis, but O-- and 2 x H+, hence, instead of one atomic ion and one molecular ion there are 3 atomic ions.  I?m no chemist and I?m not sure that this is even possible (there are laws to obey), but Meyer never seemed to let little things like that, spoil his conclusions.

Another thing I?m asking myself is what is the white deposit some people seem to be getting on their cathodes?  In hard water areas you can expect to get calcium carbonate or scale - essentially limestone - deposited on the elements of kettles, inside water carrying pipe work, on taps, etc. But on the cathode it would be a reduction reaction, which I think would be detrimental to the efficiency of our cells.

You say calcium hydroxide. I?m not sure at this point what it is (I need to research this further). However, I?m betting the ss is corroding in this way due to lack of oxygen, which is required to maintain the ?stainless?, in stainless steel.  It might be that this effect can be countered by swapping around the polarity of the electrodes from time to time ? that said, this would no doubt bring with it a whole new set of problems.  I?ll perhaps know more when I?ve got my cell up and running.

Sorry TH, I?m not on first name terms or anything like that with the moderator on the other forum, so can?t help you there. Just keep trying.

Farrah Day

[twohawks]

Thanks much, FD.  Your explanations appear very informative and useful... I am on the road right now so I need more time to digest, so no comments for now.

One thing, per 'contacting the admin' thing... I was only thinking that because you were a member, and members can always contact the admin at a bbs, that maybe it wouldn't be too much to ask you to simply ask him to check his registration submissions... or something? 

Cheers all,
HTH

[Garfield]

 
I've been following these posts which I find most interesting and would just like to help out with some important facts.
   I think one of the key items is the series resonant circuit itself. You can check the theory behind it by doing a search in google. But you will find that although the thing has zero impedance across the outer terminals, the voltage across the capacitor only can get exceedingly high.(this also  applies to the inductor only as well)
This effect becomes much greater with a high LC (high inductance and low capacitance) circuit.
The voltage is infinite and limited only by circuit resistance. The greater the current the greater the voltage.
    I think this is why Stan winds the resonant coils with resistance wire to limit the current.
He also uses a bunch of small cell electrodes rather than one large one. This is to take advantage of  the high LC effect.
  Some confusion also regarding the dc driving pulse. As the switching diode (fig.1) is wired to pass only the positive portion of the ac signal it still remains an ac signal which varies from ground potential to a high plus voltage. If there was a filtering network to smooth out these pulses and produce a pure dc voltage, then the resonant circuit would'nt work. You will notice in Fig.1 that one end of the resonant    coil goes to ground to ensure that the pulse never goes negative.
Also for what it's worth, in one of my electronic books there is a chart giving the dielectric constants of mant different materials. It says "Water (distilled) 34-78"
Hope all this helps and just don't add to the confusion.

Garfield

[Farrah Day]

Rav

if you're looking in can you please contact me on:

farrahday13@googlemail.com

Thanks
Farrah Day

[Farrah Day]

The problem with water.

Water does indeed have some very strange properties, and it would appear that though water is a very simple molecule, there are still things about it that science has not yet fathomed. Indeed there are things that scientists are at odds with and still can not agree on.

Now most people trust that Meyer is pulling water apart by creating a large electrostatic force between the electrodes and that no current conducts in the wfc itself. Simply assuming that the H2O molecules is 'pulled apart' into O and H, H. If this were the case, then designing a wfc around a glass jar, with the anode being aluminuim foil wrapped around the outside, and a cathode inside the jar immersed in water should work well.  It doesn't.

Then there is the 'pulling the H2O molecule apart' into its component parts of O and H, H.  Does this really make scientific sense?  If you managed to 'pull' one hydrogen atom off the oxygen atom, wouldn't the existing OH bond be now stronger due to the increased charge differential. The problem with this idea is that the charges are not taken into consideration... too much is assumed without proper scientific consideration.

No, the fact that both electrodes need to be in contact with the water, points to the same ionisation reaction you get from normal electrolysis. However, to become gases the ions MUST collect or dump a charge, which they can only get from the electrodes.

The interesting part is that even normal electrolysis is widely misunderstood, usually given as a simplistic equation for the reaction at the electrodes.  Now pure water contains very few OH- and H+ ions. That is why it makes for a very poor conductor; there are hardly any charge carriers.  Now the few ions that there are in pure water react at the electrodes in the simplistic way that is always shown, i.e. the H+ ions collect an electron at the cathode, bonds with another hydrogen atom which has just done the same and then is released as H2 gas. At the anode, the OH- ion drops off an electron, the OH splits and the O joins with another oxygen atom to be released as O2.

But, what happens to the hydrogen atom that has just been jilted at the anode by the oxygen atom?  You know, I'm not sure. I know it should have a neutral charge, i.e. be an atom rather than an ion, I also know its not given off at the anode as a gas, so doesn't bond with another hydrogen atom to become H2... so what happens to it?  This atom just seems to be forgotten about, but it must do something. Note to myself - look into that later.

Now, this is where things get a little more complicated. If we add a small amount of Sodium Sulphate to pure water, the Sodium Sulphate (Na2SO4), nearly all dissociates into ions, 2Na+ and SO4--. These ions carry charges to the electrodes once a small voltage is applied, but when they get there, they do not react. This is because these ions are less electrochemically reactive than the H2O water molecule itself. So instead of the 2Na+ ion collecting two electrons from the cathode, the favoured reaction is the water ionising into H+ and OH-, with the hydrogen reacting to be given off as H2.  At the anode, instead of the SO4-- reacting, again because the water molecule is more electrochemically reactive, it is the H2O that ionises into H+ and OH-, with the OH- breaking up to give oxygen when the ion deposits an electron.

So, that is what happens, but it?s still hard to get your head around why the Sodium Sulphate ions induce the water molecule to ionise, by being simply there and seemingly doing nothing other than provide competition. 

Now it occurs to me that if the sodium sulphate can induce the water molecule to ionise but not actually take part in the oxidation and reduction reaction, then there must be other ways of also inducing the water to ionise.

It would seem to me that once we have conditioned our wfc electrodes in order to form a better dielectric oxide layer, that we can do a very simple test with de-ionised water.

We know de-ionised water will hardly conduct with normal dc electrolysis, and very little, if any, gas will be seen to be rising from the electrodes.  Now using the dc voltage pulsing method, if we find that gas is given off to any great degree, we know that we must be ionising the water at the electrodes, without the help of any foreign compound ions.  This would then clearly indicate that, pulsing, high voltage or both would be instrumental in this process.  Also, comparing an unconditioned cell against a conditioned one should provide further insight.

All good stuff.

Would really appreciate some thoughtful feedback here.

Farrah Day