Someone please check me out

[dixiepc]

OK Gus here we go again,

If I build a build a unipolar cell in this configuration []+ - + - + [] + - + - + [] + - + - + [] + - + - + [] + - + - + [] + - + - + [] + - + - + [] with 9 square inches of emmisive surface area on one side of one plate running at 13.8 to 14.2 volts. And assuming that I am running full strength electrolyte. I should be making approximatly 9.03 liters of gas at 18 amps (assuming 14 amps per liter) and from around 1.97 to 2.03 volts per group.

I hope these numbers correct.

[myoldyourgold]

Sorry 14 amps per liter times 9 LPM is 126 amps not 18 amps.

[dixiepc]

So each group will be making 1.29 lpm at 18 amps for a reactor total of 9.03 lpm at 126 amps. Right?

[myoldyourgold]

What is the total amps you are planing to run? I most likely am misunderstanding what you are saying. If you take the total amps the reactor will draw and divide that by 14 amps you might be able to make that amount of HHO.

[BioFarmer93]

So each group will be making 1.29 lpm at 18 amps for a cell total of 9.03 lpm at 126 amps. Right?

YES, Your calculation is correct...
(but it's the REACTOR or ELECTROLYSER total, not cell total) We have to be careful with terms because this stuff is confusing enough anayway..

[dixiepc]

Sorry Gus, I edited my post to use the correct terms. Now that I have got that part straight in my brain I will move on to other questions later.

Monty

[dixiepc]

OK, my next question is about the number of bubblers. Gus, did you ever determine, for sure, that if the first and last plate of each group were the same polarity, that it would negate the need for separate bubblers?

[hhofox]

Thanks for the clarification.
I am wondering why you'd go that route when it uses so much energy and produces more HHO than is recommended for the fuel type and engine size. Unless you are using it on a tractor, steamroller, etc.