Wanted to post a thought on this setup.

water is a voltage dependent resistor. The higher the voltage, driving the system, the lower the resistance. So the power input is about a squared function.

As an example

At 2.4 volts it draws 10.6 amps, Power is 25.44 watts. Resistance is (2.4/10.6) 0.2264 ohms.

At 4 volts it draws 37 amps, Power is 148 watts. Resistance is (4 / 37) 0.1081 ohms.

I know what you are thinking, been looking over this all day also. There are 2 things to look at here.

1 the water resistance went down - so the cell can pass more current and be more efficient.

2, connecting 4 tubes together in parallel will achieve the same thing, 2.4 volts @ 42.4 amps. This is still more efficient at 101.76 watts.

Now what IF...

You can achieve a high voltage pulse and deliver several amps low voltage? This would be kind of like a flash circuit where the hv pulse allows the electrons to flow. This is also similar to the plasma discharge spark plugs.

The big difference is how fast the water consumes the high voltage pulse.

The high voltage pulse could be 12 volts @ 1 amp, 12 watts, while the high current is 2 volt @ 20 amps, 40 watts.

Expected water resistance at 12 volts 0.055 ohms (need to verify this with more testing and data)

At this expected resistance the cell is able to handle (2 volts / 0.055 ohms) 36 amps.

So with 2 secondary windings on the transformer and a diode tied in from the high voltage to the low positive voltage we may be able to achieve greater than 100% efficiency. (based on current known methods).

The grounds on both windings would be tied together on one end and sent to the ground of the cell. The positive would be taken to the positive of the cell.

Current is going to take the path of least resistance, which is mostly the water with KOH added.