Hi all,
I was playing with modelling some asymetric grids, and lacking the FEM
software to do the job, settled on making then then immersing the test
grids in salt water, applying a few volts AC and measuring the potential at
varous points around the structure.
It turns out to be a very easy and quick way to explore the potential
gradients around various structures in 3D (More so if you have a CNC mill
so that you can mount the probe in the quill and drive the pan full of salt
solution around on the table!).
Anyway, while fiddling with this an interesting thought occured, if you have
an insulated conductor immersed in a conducting fluid (like my salt water
or a plasma) then in very short order the insulated section ceases to have
any effect on the local electrostatic field as the outer surface of the
insulator aquires charge to equal to the local potential.
Now apply this notion to a grid by extending the insulation from the stem to
cover the entire grid structure with the exception of a few points. You no
longer have a grid which acts as a giant attractor for deuterium, instead
only those points which are not insulated attract the deuterium.....
Obviously the heat load will severely limit input power, but it might be
something to explore.
Incidentally for the same reason an insulated stem will disapear in terms
of its electrostatic effects.
Just a thought.
Regards, Dan.
An observation about insulators in a conducting fluid
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longstreet
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Re: An observation about insulators in a conducting fluid
I'm not sure what you are trying to do. How can a grid act as an ion accelerator and at the same time not attract ions?
Re: An observation about insulators in a conducting fluid
The water tank method of mapping electric fields has a long and successful history in by gone days. It once was all that one had for the the analysis of complicated shapes. The results of these simulations tend to more accurately resemble actual results when the systems are both energized with low frequency AC.
Under DC excitation, resistivity and permitivity factor into the voltage distributions. In DC charged solid dielectrics, the temperature influenced resistivity is usually the determining factor with regard to how the voltages vary across pieces, since once the displacement current dies off (capacitance has fully charged) all that's left are the leakage currents from surface and bulk resistivity. When you are dealing with resistivities in the range of 1E+14 to 1E+18 ohm -cm for plastics, oils and such, and temperature effects that change these numbers by orders of magnitude for 10's of degrees C, the whole thing gets very, very condition dependent... everything practically becomes a special case, and very complicated to predict, except in the most general way.
In the water tank, the dielectric is actually making a significant disturbance to the equipotentials, since it places a high resistance in series with the low resistance of the salt water bath. This will locally create a very different potential distribution, which you should be able to see, by making two otherwise identical elelctrodes, one sheathed with plastic and one without.
It's a great learning experience... and thanks for sharing your observations!
Dave Cooper
Under DC excitation, resistivity and permitivity factor into the voltage distributions. In DC charged solid dielectrics, the temperature influenced resistivity is usually the determining factor with regard to how the voltages vary across pieces, since once the displacement current dies off (capacitance has fully charged) all that's left are the leakage currents from surface and bulk resistivity. When you are dealing with resistivities in the range of 1E+14 to 1E+18 ohm -cm for plastics, oils and such, and temperature effects that change these numbers by orders of magnitude for 10's of degrees C, the whole thing gets very, very condition dependent... everything practically becomes a special case, and very complicated to predict, except in the most general way.
In the water tank, the dielectric is actually making a significant disturbance to the equipotentials, since it places a high resistance in series with the low resistance of the salt water bath. This will locally create a very different potential distribution, which you should be able to see, by making two otherwise identical elelctrodes, one sheathed with plastic and one without.
It's a great learning experience... and thanks for sharing your observations!
Dave Cooper
Re: An observation about insulators in a conducting fluid
The point is that you probably DON'T benefit from the whole grid being an
attractor (and definately don't benefit from the stalk soaking up ions).
In the sort of geometry that is popular in these parts (small inner grid, large
outer) you ideally want a "grid" that appears from a distance to be a region
of negative charge, but that when observed close up degenerates into as
transparent a structure as possible.
For a totally uninsulated grid the capture cross section of the conductors
is going to be many times their actual area due to the electric field tending
to steer ions that would miss if not for the charge into collision.
If you insulate most of the grid, leaving say a cube of 8 points uninsulated
then the insulated elements will have a cross section determined by their
actual area (Or in a sufficiently far out of equalibrium device of less then
their actual area - as the outer surface of the insulation aquires positive
charge), thus your grid occludes less of the area.
Now you can take this to a logical conclusion and create a "grid" by
installling a series of insulated feedthrus with only the inner tip of the
conductors uninsulated, a nasty cooling problem, but not insurmountable.
Regards, Dan (Desparately looking for lab space in Liverpool).
attractor (and definately don't benefit from the stalk soaking up ions).
In the sort of geometry that is popular in these parts (small inner grid, large
outer) you ideally want a "grid" that appears from a distance to be a region
of negative charge, but that when observed close up degenerates into as
transparent a structure as possible.
For a totally uninsulated grid the capture cross section of the conductors
is going to be many times their actual area due to the electric field tending
to steer ions that would miss if not for the charge into collision.
If you insulate most of the grid, leaving say a cube of 8 points uninsulated
then the insulated elements will have a cross section determined by their
actual area (Or in a sufficiently far out of equalibrium device of less then
their actual area - as the outer surface of the insulation aquires positive
charge), thus your grid occludes less of the area.
Now you can take this to a logical conclusion and create a "grid" by
installling a series of insulated feedthrus with only the inner tip of the
conductors uninsulated, a nasty cooling problem, but not insurmountable.
Regards, Dan (Desparately looking for lab space in Liverpool).