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

[Farrah Day]

Did a couple of initial tests today. It was getting late so I really just wanted to test my transistor output stage to ascertain that it worked.

I'll write up some actual figures from the test when I do it again tomorrow, but I was quite surprised by something that occured so I thought I'd give it a quick mention beforehand.

I placed my ss test cell in de-ionised water and applied 10v olt straight dc. As expected there were no visible signs of gas and the current draw was just 1 milliamp.  I upped the voltage to 31 vdc and again, no visible signs of gas. Ammeter read 3 milliamps @ 31 vdc.

Used my newly built transistor output stage (which works fine) and pulsed at various voltages (unmodulated) between 8kHz and 28khz. Again no visible signs of gas given off, though current drawn seemed considerably higher at 10v than straight dc- will record results doing retest tomorrow.

So, everything as expected to this point.  Then I had a surprise. I switched off the power and saw that a voltage was maintained across my test cell. It dropped immediately from 10 volts to around 5.5volts, then steadily dropped a little at a time.  I stood there for a good five minutes while the test cell held the voltage, by this time it had dropped to around 2 volts, but as it was getting late I decided that would do for now and I'd retest properly next time. 

So it would appear that my test cell does indeed exhibit capacitance and is capable of holding a charge. Before I left I decided to short out the cell plates in oder to confirm that it was charges on the plates that were providing the voltage.  This I did, the voltmeter promptly dropped to 0 volts, however.... and this is the bit that surprised me, when I took away the shorting wire, I was gobsmacked to see the voltmeter reading a steady rise in voltage.  I looked on amazed as the voltage steadily rose from 0 v to just under the 2 volts it had been at before I shorted the plates. I then shorted out the plates a lot more times, each time the voltmeter zeroed, only to charge back up when I took the short away.  I could not seem to deplete the charge as would happen instantly with a normal capacitor.

Very interesting.

My test cell as you will have seen has a couple of floating plates, so tomorrow I'll short them out to the cathode and redo the test again to see if charges held on the floating plates were the reason, Though I did consider this at the time and attempted to short them all out, but still the voltage reappeared and rose.

Tomorrow I intend to log voltage, current, times on, off, how long to discharge, etc.

That's all for now.

[Kator01]

Hello Farrah Day,

your observation is based on the ohms-law. Using the figures of your measurement the formula
U = R * I  deliveres a value of 10,33 KiloOhm inner resistance of your de-ionisised water. Now you put into yout water-volume  93 millitwatt Power a second at a voltage level of 31 Volt.

Upon disconnecting the supply and waiting  until voltage dropped to 2 Volt you shorted the cell. But this time you have a discharge-current of 2Volt/10.33 Kilo-Ohm -> I = 193 *10 exp -6 ( myko-Ampere ).
This of course takes much more time to remove the stored charge, depending mainly of the time you have charged up your watercell

Indeed water can be charged up a lot as Dr. Stiffler has experience himself getting a heavy schock after charging 150 ml of water with 3 000 Volt.

Keep up your systematic work. I like your approach especially your scepticism concerning the claims of S.Meyer

Regards

Kator

[Farrah Day]

Hi Kator

I'm not sure I fully follow what you're saying, as though I understand that it will have took a time to charge up the plates of my cell with a very small current flow, surely when I short them, the current flow can surge largely unrestricted and deplete the charges within the blink of an eye.  This it did as my voltmeter immediately zeroed.

If I charge up a large electrolytic capacitor over a given time, by shorting the terminals it will discharge fast and completely with a very high current. Thats what I was expecting my cell to do.

For the plates to charge back up, yes surely this implies that the water holds a charge, but how can this be, it's deionised water... there are no ions in it, so no charge carriers.  And what would make charges in the water then attracted to uncharged plates again?  Always in the same polarity.

I find that 10 volts applied for 2 minutes drawing a current of 0.009 amps or 9 milliamps, leaves me with 0.5 of a volt across the cell after 5 minutes with the power off. After 12 minutes I still had 0.35 of a volt.  I disconnected the power supply and the voltmeter and left the cell.  About an hour later I put the voltmeter across the cell and found that it now read 2 volts.

* Furthermore, if I had my voltmeter probe connected to the central cathode and placed my probe directly in the water about an inch away from my cell I had a reading of 1.4 volts. If I now connected my positive probe to the outer tube of my cell (the anode) and placed my -ve probe in the water about an inch away from the anode, my meter read -1.2 volts (note the -ve polarity there).  Could the cell have been discharging through my voltmeter, maybe. I'll have to do the test again and this time not check with my meter until after a set time. *Correction, this measurement was taken with tap water, which obviously contains ions. Doing this test with de-ionised water gave me no voltage drop from cathode to water.

Need to give this all some thought as I would not have expected to get a voltage drop from the water itself without power.  Exactly how can de-ionised water hold a charge??

OK, to add further to my dilemma, once out of the water and dry, in the air, my cell has a reading of 1.5 volts across it.

It's funny how the simplest of little experiments already has me scratching my head.

I find this all extremely interesting though.

[Kator01]

Hey Farrah Day,

unfortunately I have to confine my comments to some basics because of a lack of time :

1 ) you understood that the inner resistance of your water-reservoir is 10 333 Ohm ?
2) If you short the plates the current flowing out of the cell  is limited by this inner resistance->
    I = U / R = 193 *10 exp -6 ( myko-Ampere ).
3) Any given amount of charge you have pumped in at 31 V discharges much slower a 2 Volt. In order to get a correct estimation :
   You first have to feed in a definite charge - lets say for 5 miniutes at 31 Volt and measuring at the same
    time the current. Then you have the total charge ( of 5 min ) in Watt-seconds
    Pin = u*I*time = 31 V * 3 milliamps * 300 sec  = 0.093 Watt * 300 sec = 27.9 Watt-sec = Joule

4) this very same calculation done at a level of 2 Volt and I = 193 *10 exp -6 ( myko-Ampere )
   gives you powerout per second of : 0.386 milliwatt/sec
5) 27.9 Wattsec pumped in devided by 0.386 milliwatt/sec = 72 279 sec / 3600 sec = 20.07 Hours
   Suprise, eh ?
   I am suprised myself because I myself have not done this calculation before.
  This means you would need 20 Hours to take all the charge out you have put in in 5 minutes.
  No surprise that the voltage builds up again after shorting a few seconds ( even minutes). And if you take
  out the cell an dry it a rest charge from the water-reservoir is transfered to the plates.

  Have you tried to discharge the dry cell-plates and measure voltage after that ?

6 ) But there is a mistake in this calc because the charge flowing out is much lower because of the
    fact that the voltage upon shorting the cell is almost Zero.The cell has a dynamic behaviour. So my calculation is the most optimistic case of a dicharge current at 2 Volt.

6) Now upon discharging the inner cell resistance will increase because conducting charge-carriers are
    removed and so the discharge process even goes much slower then.

7) Even if you use destilled water ( inner resistance is about 2 Meg-ohm ) water will hold charge. I have
    not done experiments on this but i will in the near future.

water is a complex thing and I really would like to introduce you into some scientific work of a
   german scientist ( biology) who discovered the dense water ( 1.3 to 1.5 gramm/cm exp3 specific
   weight)   which coexists with normal water at a certain percentage. But his work is only available in
   german language and I do not have the time to translate all this.
   Even in destilled water you have a fluctuation amount of H+ atoms which gives the ph-value ( about
   6 to 6.3 is for example is a normal low acidic value in fresh made reversed-osmosis-water )

So as you see this is a very complex system. But please do not get mentally blocked by the above statements. It is necessary to just go on and do systematic work. I just try to give you all of what I have learned from this scientist and from some good friend who is an electroic professional.

I even dare to say that de-ionisised water which ist pumped full of a certain amout of charge might be a good object for water-splitting studies because the charges remain and increase the voltage-level which aids to the necessary water-splitting energy.But this is just an unprooven idea.

Regards

Kator

[Farrah Day]

Thanks for that Kator, you're info is very thoughtful and I'm really finding this all very intriguing.

Water surely does seem to exhibit some strange properties.

Will be doing a few more tests again today.