Electronic – Massive ultracapacitor banks, why don’t they work? (400V 3.38F)

As you say, capacitor plates experience attraction due to the opposite charge. However this typically cannot be seen from the outside. The "plates" press against the insulating dielectric that separates them.

In rare situations one can hear the mechanical effects of the rapid release of the charge from the capacitor: the capacitor makes an audible "clack" when the charge is released instantly - i.e. through an almost short circuit.

This is a common occurrence with for example high-power pulsed laser power supplies: Such a power supply stores high-voltage energy in a bank of large capacitors for a period of e.g. 1/10th of a second. All the stored energy is then released in about one billionth of a second. This is accompanied with a fairly noisy "clack" that originates from the release of mechanical stress in the storage capacitors. - With large laser systems the discharge noise is so loud that ear-protection has to be worn in the vicinity of the power supply.

Where to start?

First, your HV unit is not a transformer. It is a high-voltage module. It puts out a pulsing high voltage.

So, the high voltage transformer was made to make arcs, and as such does not allow ... a no-load situation.

You are correct. Your capacitor will store charge and will destroy your module. Furthermore, you have taken the input power specification, (and it would be nice if you would share with us where you got the numbers) of 18 watts at 4.5 volts, and then calculated an output power of 18 watts/20 kv, giving .9 mA. This is wrong. Input power does not equal output power.

You have also failed to notice the part of the description that reads

Work: Input voltage 1.5V ~ 3V can work for about 1 minute, the input voltage exceeds 3V continuous work does not allow more than 30 seconds

In other words, you will not be able to run your unit for more than 30 seconds at a time. And it doesn't say how long you need to let the unit recover from 30 seconds operation, either.

Having failed to understand your HV unit, you have equally failed to understand your water capacitor. To begin with, ultrapure water is does not have an infinite resistivity. It is, in fact, 18 Mohm-cm. For your described capacitor, this amounts to ~ 2 Mohm. At 20 kV, that will require 10 mA which you cannot provide. You mention a "coating on the plates" which you believe will prevent current flow, but you do not describe it, and I hope you will forgive my suspicion that it may not work as you plan.

You speak of your desire not to break down the dielectric (the water) and then talk about how pulsing or AC voltages will avoid this. You seem to be unaware that the breakdown potential for pure water is ~3 MV/m. Since the spacing for your capacitor is on the order of 12.5 mm, the breakdown voltage will be ~ 37 kV, or twice your voltage, and you have no need to worry.

Given the dielectric constant of water (80), the effective capacitance of your capacitor will be ~ 100 pF. Since your module has a diode in its output, you are correct in thinking that you will get just a DC output, since there is no major discharge path. You cannot discharge your capacitor from the input side. Any such discharge mechanism would have to be synchronized with the HV pulses, and would have to withstand 20 kV. Such a circuit will be neither simple nor cheap.

Your selection of series resistor is very strange. Your calculation for voltage drop make it clear that you have used the input figure of 4.5 amps, and this has absolutely no application to the output.

So, overall, I have no way give you advice, except to suggest that you do something else. I don't see any instance where you seem to actually understand how your proposed circuit would work, or why, and that is not a good starting point.