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Gas production time. HighVoltage vs LowVoltage SameCurrent

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CrazyEwok:
50-80Kv per mm is the voltage range for the Dielectric breakdown of water. Google it as I am not at my home computer to list bookmarks. 80kv is the point for pure water, the more impurities the less the voltage required (lower dielectric strength). If you go over the dielectric breakdown point by about ~30% (that number is off the top of my head but there is a percentage range to achieve)  you reach the "avalanche effect". Highly unstable but is one of the only states of "confirmed over unity" (done on a microscopic level.


How to have KV running at ma. Look up "tesla coil" "van de graaf" "walton cockroft" these transformers/generators great massive potentials (MV in some cases) at uA and lower.


This is where my direction is taking me, you can read more on what i have done in my Projects section.


Back to your tests, I see that your amp flow increases, what sort of power supply are you using? Also your analogy for a linear resistance for the water cell I think is only 1/2 true, I would suggest you look at it more like a really leaky capacitor... Also from memory resistors work on a sliding scale in regards to V*A. In other words they are not wattage based,



--- Quote ---
Ohm's law[/size]The behavior of an ideal resistor is dictated by the relationship specified in Ohm's law:[/size][/size]Ohm's law states that the voltage (V) across a resistor is proportional to the current (I) passing through it, where the constant of proportionality is the resistance (R).
[/size]Equivalently, Ohm's law can be stated:[/size][/size]This formulation of Ohm's law states that, when a voltage (V) is present across a resistance (R), a current (I) will flow through the resistance. This is directly used in practical computations. For example, if a 300 ohm resistor is attached across the terminals of a 12 volt battery, then a current of 12 / 300 = 0.04 amperes (or 40 milliamperes) will flow through that resistor.
[/size]
--- End quote ---

[/size]As you can see contrary to your findings if your cell was a resistor your amp draw would change at the ratio above dependent on your voltage. Didn't (to some degree) we can place it that this not a simple resistor.
[/size]As you have stated that your not using pure water(if you do can you let me know where you got it!!!), we can then run on the fact that there is impurities in the water.- IMO -I would say that it is the conductive properties of your impurities that is allowing current to flow. I am envisioning the impurities as small "fuse" like connections. Once you overload these they "blow" and then reconnect into new "fuses". This limits your amp draw. - REMEMBER IMO!
[/size]Good Luck and let us know how you go!

[/size]

warj1990:
Lets start by saying we are all in the same boat, no overunity, no cars on water yet.

Now lets look into dielectric breakdown.

The breakdown of an insulator happens at a point – so you have a single pinhole dielectric failure in solid dielectrics.   
Even in air with lightning you are seeing a pinhole sized breakdown. The difference is air, and water, recover after this breakdown where solid dielectrics do not.

 There is no extra energy in the breakdown – as you simply have high resistance that dropped to low resistance – allowing the high voltage to convert to high amperage.

If I have a capacitor charged to 10 volts and discharge it through a 10 ohm resistor I will get a peak of 1 amp pulse.  (keep in mind we are not steadily recharging this cap)  For this example I will say the discharge lasts 2 seconds (without a size cap mentioned it is impossible to establish the time constant or duration of this event).

Now I charge this capacitor to 10 volts again and discharge it through a 1 ohm resistor.  With a peak of 10 amps.  The time of this pulse will be about 10 times faster than before, so 2 / 10 = 0.2 seconds.

Same energy discharged each time.
Now I charge this capacitor to 10 volts and it fails with a resistance of 0.001 ohm dielectric breakdown.
As per the example cross multiply and divide to calculate the time of this event.

2 seconds      x
10 ohms       0.001 ohms  = 0.0002 seconds.

Current would be 10000 amps. Remember this is for 0.0002 seconds.

The energy is the same, the gas production is going to be the same, xxx cc in 2 seconds or xxx cc in 0.0002 seconds
One benefit would be a faster rate of production as 0.0002 seconds could be repeated several times in 2 seconds, but keep in mind the power is going to be more as well.

(do keep in mind this is just a simple example and not intended to list all aspects of the current pulse event).

What I am working towards is lowering the resistance of water, google the lowest resistance of water.  Everyone seems to know the peak highest resistance of water, but no minimum resistance is listed.

Resistance is variable with: Size of cell, Spacing of cell, Electrolyte concentration, Water flow rate, Voltage applied.

Wait just a minute- Did I just name ¾ the variables Meyer mentioned?  Yes I left out frequency (as I have no chokes at present or pulse frequency.)
 
As a simple analogy you drive your car down the road and you see a sign “Bridge out ahead”  You ignore the sign and keep going, smack!  The car is now in the river – you made it out safely and get another car.
Again, you see a sign “Bridge out ahead”  Do you stop this time and find another route or keep going?

This analogy is what we are doing, Oh I didn’t get Meyers magic frequency to work…  The secret is in the cell size, that didn’t work.  The secret is in the spacing, that didn’t work.  The secret is in the electronics, that didn’t work.  The secret is listed in the patents, that didn’t help. The secret is in the VIC, did that help?

I have been there done that 5 years ago.

There have been so many varying levels of success in different areas in recent years.  Bingofuel, Plasma electrolysis, Putting a tesla coil output above water. All work to an extent.

Even Stan had problems.  Lets look into the computer industry for this example.  Way back when the computer came out – basically a massive calculator.  It was marketed, sold, etc… moved into the work force.  Then several updates happened, etc…. Now we have laptops that are more complex than most desktops.

Now compare to Stan and his water fuel cell.  He built a model, 9 tube demo cell.  He changed the model, and changed the model, and changed the model.

It never went to market – why not?  The capital for the future upgrades, or “better designs” could have come from the sales- as the computer industry has done (or any other business for that matter).
( Maybe it never worked right?)

My experiments are focused to lowing the resistance of water, allowing more current to flow at a lower voltage.  Is a variable resistance a form of dielectric breakdown? 

Maybe  - but my only interest is the resistance itself.  At this time further experiments need to be done in the water flow rate area VS. resistance of water.

Does pulsing the system make the water resistance lower?

Maybe - but I am not there yet with my work. 
(notice my work listed above)


I am looking for the lowest resistance in the simplest ways first - then moving into more complex areas, such as pulsing, chokes, etc...

Lektor:
just a few thoughts.
What is the dielectric breakdown in detail and what causes it. I could imagine that the strong electrical fields rip the water molecule apart, creating ions and increasing the conductivity of the water so that a current can flow. But we just want the first part, a field splitting the water but no current flow. So the thought of lowering the resistance of water isn't that bad.
What if we take two conductive plates for the capaciter and put an isolating material between them and the water with a higher breakthrough voltage than water. The field will act through the isolator and is able to dissociate the water. Since the breakthrough voltage of the isolator isn't reached there shouldn't be any current except from the ion and electron movement in the water.

What is wrong with this idea?
 

warj1990:
80 thousand volts didn't do anything with this setup, see photo.

Think of it this way.  You have a conductive wire.  You want this wire to be placed somewhere (in a field etc...) and completely destroy itself.

How can we do this without current flow from the source? 

By placing the wire across 130 volts dc we get the wire to burn up - destroy itself.  But the current flows from the source and causes this, not the voltage.

I have toyed with this idea for a long time and still don't have a good answer.

Perhaps in your example the choke and pulsing would prevent this high current surge and allow voltage to keep pulling on the molecules - however I am not that far yet.

Secondly I need to verify if the resistance is being lowered by water movement or the electron "orbitals" jumping states.

I have been at this a long time, in no hurry now to skip steps.  If you want to throw a choke into the water have at it. 

warj1990:
Why did my insulated system fail?
Theory:

The water split into H+  and OH- at about 3 volts.

Not having direct connection to the electrodes the water voltage/ field dropped to 0 - Just as a wire does not hold a charge but the charge is held in a capacitor.

Raising the voltage higher allowed the H+ again and O-

Skip a few steps...

Now I have 4 H+ (no electrons)   and 2 O- (8 electrons each).

Since O is happy with 8 electrons it will not link up to form O2 (gas).  It remains O (liquid)

The Hydrogen have no Electrons to share and remain H+ with no way to link up.

So it failed in 2 ways.
1, High voltage never truly existed (remained) across the water
2, No way to transport electrons from O side to H side.

There must be a way to transfer electrons to get the system to work.

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