### Author Topic: Retry N1001  (Read 39283 times)

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##### Re: Retry N1001
« Reply #256 on: January 07, 2019, 23:19:41 pm »
for a very small cell indeed and just dc going

now that i learned how to use molar quantity i feel more confident to understand what hell is happening in the water

something that surprises me is the heat not being so great

well 4 volts 8 amps are 32 watts of power so at least 9 grams of water should be  converted to hydrogen with this watts however with 8 amps im not sure if its going to happen

also if i´m applying almost 2,5v of over voltage where is this at least 20watts going?

i´m going to turn labview on and add a termometer into the cell

to split one mol of water we need 55 amps so 18 grams of water disappear... at 2volts this is 110watts at 1,23v 66watts

it would generate around 40 liters of gas in one hour one mole of hydrogen h2 and half mole of oxygen o2

Obs my power suppy is a variac 0-250v 500w connected to another similar variac working as a isolated step down transformer  that have a 110v output with thicker wire for max 250v input..

its being rectified by a full wave bridge rectifier that have a 100kohm resistor across the outputs and also 3x 10mf 70v capacitors in series.. each have a diode in parallel to prevent the capacitor to reverse the voltage..

for the readings im using my digital oscilloscope for the voltage and signal and the rms multimeter for the current
« Last Edit: January 08, 2019, 00:09:05 am by sebosfato »

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##### Re: Retry N1001
« Reply #257 on: January 07, 2019, 23:41:16 pm »
Ok  are you using the powder KOH?

Did you have the coil in series or parallel or none at all just the DC power?

With straight DC the current is the V / DC resistance across the anodes.
The closed you get the lower the resistance 3mm should be as close as you need with this or even 5mm.

Your cell it acting like a very leaky capacitor, that is normal with using KOH.

Series Resonance with the coil in series with the leaky cell is at resonance as you know is
XL = XC   then DC resistance of the coil and of the cell at resonance is all that the circuit will see.
With the coil in parallel then that now is a new ball game, it is still the same at resonance the DC resistance is then like two resistors in parallel.
Now if you have then a DC 50/50 pulse, when the pulse drops to zero some of the coil charge will discharge how much will depend on how quick the disconnection is.
Then the cell will discharge into the coil, until the voltage it is lower than that of the cell and so on until both are discharged.
If the set frequency is high enough at resonance so that the coil does not even get to the  T constant  it take 5 x T to reach full current.
You will still get the voltage across the cell, it still is the resistance in parallel but the coil discharge may all go in a small negative pulse so then you will have just the cell discharging the ringing is shorter.
You can then adjust the frequency and the resonant point to give you a high voltage the full say 12V with very little current and you then have the voltage doing the work.

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##### Re: Retry N1001
« Reply #258 on: January 07, 2019, 23:43:50 pm »
those caps in the power supply shoud be in parallel with the dc to reduce the ripple the circuit will see a DC supply with an AC pulces

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##### Re: Retry N1001
« Reply #259 on: January 08, 2019, 00:11:17 am »
those caps in the power supply shoud be in parallel with the dc to reduce the ripple the circuit will see a DC supply with an AC pulces

they are exactly there... i´m having a ac superimposed over a dc thats what it seems looking at the oscilloscope

there is ac current too although not reversing direction!

http://www.graylark.com/eve/ionic-resonance.html
http://www.electrochemsci.org/papers/vol8/80303731.pdf

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##### Re: Retry N1001
« Reply #260 on: January 08, 2019, 09:08:46 am »
capacitors are defined as 1faraday = 1 amp * 1 second  / 1 volt

meaning if we take one amp from a 1 faraday cap for 1 second it will loose 1 volt.

so a 1 milifaraday will loose 1000v in one second discharging at 1 amp rate

the current a capacito can  flow amps = faraday*volts *frequency   because frequency is = to seconds^-1

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##### temperature rise
« Reply #261 on: January 08, 2019, 09:37:34 am »
the temperature of our system is very important to determine how much power is becoming gas how much is becoming heat and how much is becoming excitation of the gas.

the temperature of the water running thru the cell change over time will give how much power its absorbing as heat

on my labview environment i´m doing a calculator that indicates directly the watts coming as heat from this type of calculations...

https://sciencing.com/how-8643971-convert-wattage-degrees.html

basically watts * seconds = joules

joules / mass / scpecific heat = temperature rise of the amount of substance...

an arduino could be used for this purpose too.. thought

i´m adding today a wall plug wattmeter to see how much power i´m wasting before become dc  too

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##### facts
« Reply #262 on: January 08, 2019, 15:05:57 pm »
Facts

for the electrons to want to take a trip on the external circuit is required to provide a potential and this potential is related to the force of attraction of the water ions or system to the electron so its not free to simply go around in the circuit and on the other hand on the negative electrode there is also a repelling force a negative potential or voltage that also dont want or dont allow the free insertion of electrons into the solution...

i believe here is where resonance comes into play

the resonance will allow the increase of potential of the ions to the point where it simply dont have the same attraction force it had before allowing the water to perhaps zero up the potential and leave only ohmic resistance...

still ohms is a pain in the *  but is more a controlable thing... will be about increase the surface area to increase efficiency and perhaps superconductors for the chokes..
if the cell had 100mohm to apply 10 amps will consume 10 watts and will be on the 100% efficiency range because 100mohm *10amps = 1v

therefore to increase the efficiency and power output this resistance must be in the mili to micro ohms range

seems to me that meyer was assuming that the voltage was producing an effect on water in the sense that it reduce the resistance the more voltage you apply.. there is a top limit however... the first is the efficiency... we dont want the cell to get over one volt of loss because this is around the 100% limit we want to get far from it..

this guy has nice videos

« Last Edit: January 08, 2019, 20:57:52 pm by sebosfato »