Lets do the math!

3 inch tube set, 1/8 inch gap, 5 gallons per hour, as per the control and driver patent

I know 1/16th inch gap is mentioned in other documents, but 5 gallons per hour is only mentioned with the 1/8th inch gap, as far as i know.

5 gallons per hour is 5.26 ml per second

1 US gallon = 3 785.41178 ml

((3 785.41178 x 5) / 60) / 60 = 5.25751636

5.26 ml of water is split per second

5.26 ml *1800 = 9460 ml of fuel gas per second (Hydrogen, Oxygen, Dissolved Air)

= 2.50 gallons of gas per second

What is the volume of the water gap?

V = A * L

L = 3 inch

A = Pi(1 inch)*2 - Pi(.5 inch)&2 = 2.356 inch^2

V = 2.356*3 = 7.0685 cubic inch

7.0685 (cubic inch) = 115.831962 ml

Basically, we have to have a flow rate equal (actually, GREATER than) to the gas production, which is 2.5 gallons at zero pressure, what pressure does this operate at? up to 50 psi?

Say at max pressure, 50 psi, and using the density of the gas as the same as air, you have 4.407 times the density, so at 50 psi we could have the flow rate 4.407 times less than 2.5 gallons per second, so that is 2.4/4.407 = 0.567 gallons per second, or ...

Water flow rate into tube set must be = 2147 ml per second or, 2.2 liter per second, or 129 liter per minute.

At pressures lower than 50 psi, you would need a higher flow rate than this... at 20 psi you would need... 4 liters per second, or 240 liters per minute

How fast is the water moving? (at 20 psi)

Flow rate = Area * Velocity

4000 ml = 0.0015199 square meters * Velocity

Velocity = 2.63 meters per second

That is not exceptionally fast...

so you would need to move about a gallon per second, to each tube set, now I assume the cell used in that patent was a single tube set

This cell Don is talking about with 11 tube sets (10 used) might have a 1/16th inch gap, i don't know. But I imagine he was not using a pump that big, and that he was not getting the same production.