Author Topic: Gas production time. HighVoltage vs LowVoltage SameCurrent  (Read 5587 times)

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Gas production time. HighVoltage vs LowVoltage SameCurrent
« on: July 18, 2011, 04:07:38 am »
I wanted to test gas production time for 2 setups:

This is using my 5.5 inch long tubes (lost 1/2 inch resealing the base area).
3/4 outside tube and 1/2 inch inside tube.

I marked a pop bottle at 120 cc from the top.  Cut the bottom off and slid this into my tube cell array.
I loosened the cap and allowed the bottle to fill with water.
As the gas is produced it displaces the water.  When the gas reached the line of 120cc I stopped the timer and test.

Setup 1 is natural water.
Rectified mains voltage, no filter cap.
Input was 130 volts @ 5.8 amps.
Gas production time: 1:44 seconds

Setup 2 is same natural water with KOH added.
Rectified mains voltage, no filter cap.
Input was 2.2 volts @ 5.85 amps.
Gas production time: 1:45 seconds.

This concludes current is still the key in gas production - regardless of voltage.

I have not added chokes and pulsing signals to these tests (as of yet).

Higher voltage is showing larger bubble production, the rate is still the same.

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Re: Gas production time. HighVoltage vs LowVoltage SameCurrent
« Reply #1 on: July 18, 2011, 06:40:07 am »
The use of voltage in your test is only valid if you are using it to move the current from one plate to the other, some things your have failed to mention is gap between the plates, water temp, water purity etc etc.
You will note that the following if you test them.
130V test
Temperature of water increased relatively rapidly
Gap between plates can be increased with no loss to gas production, bubbles get smaller.
If allowed to run for "X" period thermal runaway begins.


2.2V Test
Temperature in water increased slowly
Gap between plates can not be increased with no loss to gas production, bubbles get smaller till production stops.
Cells will reach thermal runaway after "X" period


Your explanation while holds some ideas for people doesn't really show things as conclusive as you think. Try putting approx 50kv per mm of cell gap into your cell. I suggest you use picoamps as sparking in the cell while the reaction takes place can be dangerous.

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Re: Gas production time. HighVoltage vs LowVoltage SameCurrent
« Reply #2 on: July 18, 2011, 08:04:05 am »

Water purity: drinkable, from kitchen sink to cell.  No smell of chlorine in the water, City water, processed.
Cell spacing is 1/8th inch.


For you 3.175 mm spacing.
130 volts / 3.175 mm = 40.945 volts / mm


At this spacing the cell is drawing 5.8 amps.


You suggest 50,000 volts / mm
Lets assume the water resistance remains constant at 33.3 ohms (which by my current tests it does not)
The voltage of 158,750 volts (50 kv * 3.175) needs to be applied to get the 50 kv / mm.


158,750 volts / 33.3 ohms = 4,767 amps.
I really didn't want to buy a power plant for this experiment.


How am I to get mili amps across the cell and keep the voltage high?  When the high Voltage pulse hits the water, 159 kv,  it will NEVER rise above the losses in the ampdraw. This high voltage at mili amps will read at the cell about 10 volts.
 
If I applied 158,750 volts to the cell at 5.85 amps  I would read 120 volts across the cell.


You are saying I can produce 120 cc of gas with a larger spacing in the same time? I have 4 tubes I will connect to each outside tube with 3/4 to 5/8 inch spacing.  Be right back with that experiment.

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Re: Gas production time. HighVoltage vs LowVoltage SameCurrent
« Reply #3 on: July 18, 2011, 08:27:25 am »
Ok did the next test.


Outside of both outer tubes, spacing about 3/4 inch.


130 volts @ 1.5 amps.
at 3 min 2 amps
at 4 min 2.16 amps


at 4:31 seconds I produced 120 cc of H2 and O2 combined gasses.


Gas production did diminish.




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Re: Gas production time. HighVoltage vs LowVoltage SameCurrent
« Reply #4 on: July 18, 2011, 22:05:55 pm »
Try putting approx 50kv per mm of cell gap into your cell. I suggest you use picoamps as sparking in the cell while the reaction takes place can be dangerous.
really interesting. could you give some more infos on that? where did you get these numbers from. I agree that the field strenght should be high via voltage and the current should be as low as possible. Do you have any setups or circuits to show ?

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Re: Gas production time. HighVoltage vs LowVoltage SameCurrent
« Reply #5 on: July 19, 2011, 06:49:56 am »
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]

[/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]
« Last Edit: July 19, 2011, 07:12:00 am by CrazyEwok »

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Re: Gas production time. HighVoltage vs LowVoltage SameCurrent
« Reply #6 on: July 19, 2011, 22:24:26 pm »
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...

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Re: Gas production time. HighVoltage vs LowVoltage SameCurrent
« Reply #7 on: July 19, 2011, 23:16:50 pm »
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?