Basic Electrolysis

[Michael John Nunnerley]

Farrah Day

This I will only explain once:-

2HO that is atomic state of hydrogen and oxygen in a free mix, if you supply energy to this in the form of heat or spark you will combine the 2H to the O to form H2O "water" and this is an implosion reaction from a higher state to a lower state, "energy was needed to create this combination" from an external source, volume was changed from a high volume to it's resulting low volume H2O-energy.

Now H2+O2 note, none atomic, will explode to form H2O+energy, that extra energy came from the excess oxidiser O2 because the formula H2+O2 does not make H2O, we would have an O left over. H+H+O on the other hand is the exact molecular makeup of water H2O.

Now just to complicate matters further, if you start adding electrons to gases, for example O2, you have this:- O2+e gives you 2O+e, the oxygen has become atomic. Likwise if you add an electron to H2 you will get 2H+e. Now we have a totally different state of affairs, we have 2H+2O+shared electron, IF, they were produced in the same pot, so to say, as in a single electrolysis cell.

Mike

[Farrah Day]

Aaron

If you are simply an armchair theorist that never builds anything and has
no idea how things work in the real world, you might as well tell everyone.
But if you do have some hands on experience, then show it. It makes a
lot of difference and will give you some credibility to what you're talking
about.

You simply haven't gotten around much, Aaron, otherwise you'd know better than this.

Mike

This I will only explain once:-

2HO that is atomic state of hydrogen and oxygen in a free mix, if you supply energy to this in the form of heat or spark you will combine the 2H to the O to form H2O "water" and this is an implosion reaction from a higher state to a lower state, "energy was needed to create this combination" from an external source, volume was changed from a high volume to it's resulting low volume H2O-energy.

Now H2+O2 note, none atomic, will explode to form H2O+energy, that extra energy came from the excess oxidiser O2 because the formula H2+O2 does not make H2O, we would have an O left over. H+H+O on the other hand is the exact molecular makeup of water H2O.

Now just to complicate matters further, if you start adding electrons to gases, for example O2, you have this:- O2+e gives you 2O+e, the oxygen has become atomic. Likwise if you add an electron to H2 you will get 2H+e. Now we have a totally different state of affairs, we have 2H+2O+shared electron, IF, they were produced in the same pot, so to say, as in a single electrolysis cell.

Mike
Today 11:20 AM

Mike, with all due respect, I think your a little off the mark with your explanation here.

For a start you won't generally find atomic hydrogen or oxygen in a free mix as in these states they are unstable and more energetic than in molecular form. You don't need to add energy to these atoms to get them to form molecules, they will readily do this of their own accord and give off energy in the process.

If, for now, we forget any atomic species and just consider what happens when we ignite H2 in the presence of O2, this is the reaction process:

In molecular form, H2 and O2 need energy to dissociate into atoms. This can be - and usually is - provided by a spark, which results in a chain reaction. This part of the reaction, from molecules to atoms, is endothermic so gives out no excess energy, but instead absorbs energy, initially from the spark but also absorbs energy in the form of heat from the environment.

So we now have very unstable highly energetic atoms of hydrogen and oxygen, which (without the addition of any further energy) will now immediately endeavour to recombine to form the stable water molecule. This is the exothermic (explosive) part of the reaction, dissipating much energy in the form of heat. The resulting water molecules formed out of this reaction will absorb much of the heat produced and so maintain a gaseous state until they lose this excess energy to the environment once the residual heat has dissipated.

In it's gaseous state, the water molecule will take up half the volume of the two gases H2 and O2, but the residual heat from the exothermic reaction will have expanded the gas somewhat to compensate.

Not until the gaseous water molecules lose enough energy (cool) to form a liquid state will they contract to a fraction of the gaseous volume. And, by then in an ICE, the water molecule is somewhere on its way out of the tail pipe.

Hence we never see an implosion.

Farrah

[chasson321]

@ Faraday,

Nice explanation, but it will more than likely fall on deaf ears here, but nice anyways.

Tim

[Farrah Day]

Here's something that should be of interest to most people following this thread:

https://www1.eere.energy.gov/hydroge...fs/fcm03r0.pdf

Farrah

[Cherryman]

Maybe an unnessecary repost but, As you can clearly see:

- Volume before and after detonation
- Volume increase at explosion
- Volume decrease after explosion
- Explosion speed versus the "implosion" speed


( 12V 5amp water with soda, RVS+ Sinc-)

YouTube - YbborNetsrek's Channel

[Michael John Nunnerley]

grasping what I said

Aaron



You simply haven't gotten around much, Aaron, otherwise you'd know better than this.

Mike



Mike, with all due respect, I think your a little off the mark with your explanation here.

For a start you won't generally find atomic hydrogen or oxygen in a free mix as in these states they are unstable and more energetic than in molecular form. You don't need to add energy to these atoms to get them to form molecules, they will readily do this of their own accord and give off energy in the process.

If, for now, we forget any atomic species and just consider what happens when we ignite H2 in the presence of O2, this is the reaction process:

In molecular form, H2 and O2 need energy to dissociate into atoms. This can be - and usually is - provided by a spark, which results in a chain reaction. This part of the reaction, from molecules to atoms, is endothermic so gives out no excess energy, but instead absorbs energy, initially from the spark but also absorbs energy in the form of heat from the environment.

So we now have very unstable highly energetic atoms of hydrogen and oxygen, which (without the addition of any further energy) will now immediately endeavour to recombine to form the stable water molecule. This is the exothermic (explosive) part of the reaction, dissipating much energy in the form of heat. The resulting water molecules formed out of this reaction will absorb much of the heat produced and so maintain a gaseous state until they lose this excess energy to the environment once the residual heat has dissipated.

In it's gaseous state, the water molecule will take up half the volume of the two gases H2 and O2, but the residual heat from the exothermic reaction will have expanded the gas somewhat to compensate.

Not until the gaseous water molecules lose enough energy (cool) to form a liquid state will they contract to a fraction of the gaseous volume. And, by then in an ICE, the water molecule is somewhere on its way out of the tail pipe.

Hence we never see an implosion.

Farrah

You need to read carefully what I have said, I tried to use words that could be understood. The word free is in FREE, not combined or combining, and this should be used with the last parra: about added electrons and "sharing". The common duct electrolysis of water produces an abnormal gaseos situation, this can be seen in a video that Dr. Stiffler made on BURNING WATER, sorry I do not have the link.

He actually noticed a strange thing happening at the base of the flame. This is due to the way he was producing the electrolysis and added electrons from his sec exciter.

I agree that atomic H and O will not live together for very long without combining IF they are in there normal atomic states. I am just trying to explain that there is more than just H2 and O2 produced in electrolysis WHEN IN A COMMON DUCT, and you are talking about the reaction of atomic in an engine, or is that just for example?

There are four states of H and O produced in a common duct electrolysis, and these states depend on the frequency and or electrode configuration. The four states are molecular, atomic, vibrated and altered, the last two will be argued in the scientific world as one. These are the states in the instant of formation and will change in time depending on what state they are in.

You are on a sticky wicket with this subject, I am just trying to explain that not all is as seen by you, and others, and if you want to explain exactly what is happening, please get it right.

Just for comfort, your highschool teacher probably would not get it right either. And before you ask, yes I am qualified.

Mike

[Farrah Day]

Mike

There are four states of H and O produced in a common duct electrolysis, and these states depend on the frequency and or electrode configuration. The four states are molecular, atomic, vibrated and altered, the last two will be argued in the scientific world as one. These are the states in the instant of formation and will change in time depending on what state they are in.

Surely to-date this is only theoretical and has yet to be categorically substantiated. Vibrated and altered... what does that mean?

You are on a sticky wicket with this subject, I am just trying to explain that not all is as seen by you, and others, and if you want to explain exactly what is happening, please get it right.

I may not have covered all the common duct electrolyser gas angles in my post as I was simply detailing the reaction processes of combusting H2 in O2, but what I posted is correct... I have got it right!

As far as sticky wickets go, it may well depend on the point of view here Mike... as from where I'm sitting you're on a much stickier wicket than me!

Farrah

[Harvey]

Harvey

There are quite a few people on these forums that think that the ignition of Rhode's Gas simply causes an implosion (no explosion at all).

In fact the common myth was that such a gas would drive the enigine by sucking up the piston with this implosion, rather than pushing down the piston as in an explosion.

All sorts of rumours then came about whereby the timing would have to be set 180 degrees from normal to facilitate this perceived implosion issue.

Indeed, from what I see here, many people still seem to see implosion as a real issue in an ICE, completely disregarding any explosive element to the reaction.

All I've been trying to convey is that implosion is not an issue within an ICE, and that the gas still explodes.

Regards, Farrah

Surely you recognize from the empirical evidence that the net function of an HHO combustion cycle is a vacuum. In fact, that boys underwater test should be incontrovertible as long as others also get the same results - of which I am convinced.

I think the real problem here is that the experimenters are not using scrubbed gases - so they don't know precisely what the mixtures are and thus they get varied results.

Many years ago, Mercedes Benz did some hydrogen ICE tests on pure bottled hydrogen and the results were less than disappointing. The power to fuel volume ratio was reduced by such a large extent the project was found to be infeasible notwithstanding serious issues with excess heat.

More current tests have been done with some success:
http://avt.inel.gov/pdf/hydrogen/sae...tation2006.pdf

But the real consensus is that H2 ICE is not as effective as the fuel cell approach.

But therein lies the question - how different is the combustion of H2 compared to HHO? Experimental evidence strongly suggests that there are significant differences in how these two forms of Hydrogen gas burn. In order to prevent the inherent implosion effects the ICE must be operated at such an elevated temperature that the water produced is immediately formed in a vapor state and the problem with that is it takes energy away with it to which it would probably keep adding to the Global Warming complaints already rampant.

Therefore, I must lean toward the cold engine implosion approach to this combustion process - it just makes better sense all the way around. Why fight the natural characteristic of the process? Let the vacuum do the work and don't waste the energy to the atmosphere as heat. And toward that end, those proponents who have suggested a near 180° are correct in that suggestion. There is no doubt that some expansion must take place during the combustion, but the evidence would suggest that the force it produces is much less than that produced by the consequent vacuum.

I think that should be the focus of those wishing to employ the value of implosion - they should first identify those two pressures. Once identified, you will have the pre-stroke ratio needed for the fire timing. Just a wild guess on my part, I think it will be around 30° BBDC - this way you use the minor expansion to finish the downstroke and you take full advantage of the upstroke implosion. This requires an entirely new cam design for the valve timing as well as some means to evacuate the resulting liquid that results during the implosion. Trying to apply implosion technique to a gasoline internal combustion engine is only going to lead to frustration because the valve timing is wrong for it. Essentially, you want to fire just as the intake valves close, thus imploding on what is normally the compression cycle. This leaves you at TDC with water in the cylinder at a point where both valves are closed ready for what is the power stroke in the gasoline engine.

So what would be a way around this? Perhaps an inverted two cycle engine approach. Or at least some means to provide a vacuum on the exhaust system as the implosion cycle nears its end at TDC. A four cycle engine could be used if the power cycle were used as a purge. This would require a diverter on the exhaust system to allow atmosphere to enter the cylinder via the exhaust valve on the down stroke and remain open on the upstroke during the entire purge. Essentially a dead weight being pushed around, but it would help dry-out the water. Not too many engines allow the valves to be open at TDC, so the cam lobe would be a goofy thing on the water extraction portion of the implosion cycle. Or maybe forgo that entirely and hope the purge cycle is sufficient - intake atmosphere on the down stroke, switch the diverter valve to exhaust on the up stroke and hope we get most of the water out.

It would be interesting to see how we could apply the implosion to a pulse jet engine

[Aaron]

@Farrah Day

You simply haven't gotten around much, Aaron, otherwise you'd know better than this.

You know nothing about me. Your very consistent in evading questions.
What are you so afraid of by not posting your experiments? Are you ashamed
of your results or lack of? You have nothing to worry about if so because
you're not going to be judged based on that. All build attempts should
be respected in my opinion.

The picture you posted in the other thread - I'll just have to take your
word for it that it is yours and that you actually built it or had someone
build it for you. Could be something you found online somewhere for all
I know but I'll give you the benefit of the doubt that you're actually trying
to build something.

The reason I keep asking about experiments is because over the years
in all my "getting around" there are so many people that talk big and show
nothing. I've been contacted by more people than you can imagine that
have claimed to achieve all kinds of results or understanding of Stan
Meyer's technology, duplications, etc... and virtually every single one have
never been able to show one shred of evidence to back any of it.

Of course that is different than trying to innovate a new water technology,
but the fact is you have been debating myself and many members of this
forum that have a proven track record of being not only being innovative
in their thinking but have a proven track record as being builders that
have posted many experiments over the years all the while you have never
shown anything yourself - in this forum - and if you posted experiments in
other forums, please post the the links.

Smart little comments that say nothing or mean nothing are useless in
having anyone here understanding your experience (building).

[Farrah Day]

In order to prevent the inherent implosion effects the ICE must be operated at such an elevated temperature that the water produced is immediately formed in a vapor state and the problem with that is it takes energy away with it to which it would probably keep adding to the Global Warming complaints already rampant.

What elevated temperature? Water boils at just 100 deg. C, most engines operate at more than this when up and running.

I think you're missing the point, Harvey.

In the confines of an ICE cylinder, the heat from the explosion phase cannot easily dissipate. Due to this residual heat and the fact that the water molecule will readily absorb some of this heat, is what maintains the gaseous state of water. And don't forget vapour would be water's natural state anyway if it did not tend to cluster. The water produced would initially be in gaseous state until it lost energy and condensed on something cold... and there's not going to be anything cold in the combustion chamber after the first few explosions!

Therefore, I must lean toward the cold engine implosion approach to this combustion process - it just makes better sense all the way around. Why fight the natural characteristic of the process? Let the vacuum do the work and don't waste the energy to the atmosphere as heat. And toward that end, those proponents who have suggested a near 180° are correct in that suggestion. There is no doubt that some expansion must take place during the combustion, but the evidence would suggest that the force it produces is much less than that produced by the consequent vacuum.

You've got to also consider that we're never going to be just combusting Rhode's Gas or just stoichiometric volumes of H2 and O2, these gases are mixed with air in an ICE, so you will never get a total vacuum with just a tiny drop of liquid water in any condition.

Internal combustion engines can run on hydrogen, we know they can... and without setting the timing 180 deg out and relying on an implosion.

Take another look at this link Harvey and see if you still think implosion is the way to go:

https://www1.eere.energy.gov/hydroge...fs/fcm03r0.pdf

And there's me thinking I'd detailed it all so nicely... and Aaron telling me that I was imagining 'implosion' to be an issue.

Regards, Farrah

[Farrah Day]

The picture you posted in the other thread - I'll just have to take your
word for it that it is yours and that you actually built it or had someone
build it for you. Could be something you found online somewhere for all
I know but I'll give you the benefit of the doubt that you're actually trying
to build something

The benefit of the doubt... so kind of you. But hey, whatever... I doubt we're ever going to make each other happy, are we?

Farrah

[ashtweth]

SS passivation

Guys just worth a look, we already know how to do it as per Bob.
passivation.pdf - download now for free. File sharing. Software file sharing. Free file hosting. File upload. FileFactory.com


Ash

[HairBear]

Bob Boyce? I thought Aaron and Ravi discovered the passivation thing? I didn't see Bob's name anywhere in the PDF. I call passivation a perpetuated myth but Aaron and others will disagree with my opinion even though I have several hundred hours of research that shows me different. But then again are we talking normal electrolysis or something else?

The PDF is written by Dan Englebert. Who is Bob?

[ashtweth]

This is passivization for BRUTE FORCE, i am in the wrong thread i need a hybrid TAXI LOL

Bob =Bob Boyce . I tested a cell and found Aaron/Ravz theory of more gas via that white build up was true, did you test this?

Ash

[witsend]

Farrah Day - I'm still following this thread with bated breath - the more so as we're using a copper hot water cylinder in our first application. I'm inclined to think that there may well be some electrolytic effect that has not been brought into the equation.

Farrah - if I can impose on you - would the use of this material make a difference - considering we'll be applying those sharp voltage spikes to the system?

And guys - I think we're all on the same side here. Farrah is definitely an experimentalist Aaron. She's detailed the tests that she's applied to their own vehicle fleet and those tests, as I understand things, are promising. Not only that but I also understand that she's exploring different effects and we'll all be the happy beneficiaries of that knowledge. I don't think her expertise can be questioned. Certainly I won't.

EDIT. May I add, that nor would I question anyone else's expertise either. But if we're finding different results wouldn't it be more to the point to find out if there's some essential difference to the apparatus used? That's my impression - but I'm a newbie here.