Amoeba Cell in diamond and horizontal positions compared.

[BoyntonStu]

http://www.youtube.com/watch?v=w4xkkd5hA-w

Amoeba Cell in diamond and horizontal positions compared.


15 plates, center is Positive, 13.7 Volts, 1/32” gap ~ 1.2mm.

BoyntonStu

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[Gary Diamond]

More questions, good video best video so far

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[BoyntonStu]

After a flush and continuous running for 3.5 hours The Amoeba Cell achieved these results:


Volts Amps ML Second Cells/side 3/26/2009
13.18 24.6 1000 30 7 1.88 Volts per cell!
Horizontal
Watts MMW ML/MIN 108*F
324.228 6.16 2000


Is 1.88V/cell a breakthough using 1/32" spacings?


BoyntonStu

[Painless]

Is 1.88V/cell a breakthough using 1/32" spacings?

Most of us have heard of the 1.24v ideal voltage per cell (or is it 1.23?). It stands to reason that a voltage closer to this ideal would be more efficient. "Anything over 2v is wasted" ... but also between 1.24 and 2.0 volts efficiency decreases.

Volts is purely a method for 'pushing' amps through a cell, if we can get the amount of amps that we want with a lower voltage, then a watts saving must be obtainable.

This is how I view it, anyhow.

Russ.

[BoyntonStu]

http://www.youtube.com/watch?v=LjKKOSqudeM

Amoeba Cell in horizontal position.

2 LPM and 104*F after running 3.75 hours in Florida sun.

Flushing the cell eliminated the gassing up problem.

1.88 Volts was a surprise.

Was this because of the 1/32” gasket?

BoyntonStu

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[Painless]

Thinner gasket equals less electrolyte width between plates equals less resistance. Less resistance equals less voltage potential to 'push' the amps between the plates?

[H2OPWR]

Thinner gasket equals less electrolyte width between plates equals less resistance. Less resistance equals less voltage potential to 'push' the amps between the plates?

From everything that I can find yes the electricity is 100% effecient at 1.24 volts per plate gap. That would require a zero electrical resistance cell (which does not exist). The problem is that there are just way too many variables and no easy way to calculate how much loss there is in your device. You could measure the resistance in the cell and calculate how much loss you were going to get at a given voltage and current draw. Then figure the voltage drop for that given set of conditions. Then build a cell to match exactly. But as the cell heated up or the amp draw changed you would then throw all the calculations out the window. It is chasing a moving target and would require a computer controlled cell capable of constantly changing the voltage to meet the current condition. It gives me a headache. I am comfortable with 2 volts per plate gap unless like Stu you tightened the gap up.

Larry