Reply by A n g l e May 9, 20232023-05-09
Gggggggggg99999990 to hell and suck devil's cock...............


You are bastard..........


Good bye !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Have a nice day !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Reply by Lamont Cranston May 8, 20232023-05-08
On Sunday, May 7, 2023 at 10:08:07 PM UTC-5, Phil Allison wrote:
> Lamont Cranston wrote: > ------------------------------------------ > > I have the HV set up with a fullwave bridge. > > I pick off the AC at the input to the bridge and the DC out of the bridge. > > Without any load, (other that air), The DC voltage is 91% of the AC voltage, pretty close from 500V to 9000V.
> ** A standard DMM reads the *average* value of any wave on DC ranges. > On AC ranges, it reads the equivalent rms value of *sine waves* - by scaling their average rectified value up by a factor of 1.11. > So the average DC output from an ideal bridge rectifier will read 90% of AC rms voltage being supplied. > > I don't think it is the quality of the meter, I'm doubtful it is a crest factor problem at 60 Hz, unfiltered from a fullwave bridge. > > It seems to be more than diode loss, and also the 91% is from 500V to 9000V, straight line, > > diode loss would show more at the bottom end. > > Any Ideas? > ** There is no issue. > > > .... Phil
Thanks Phil, I hope they can run the DC vs AC experiment this week. Mikek
Reply by Lamont Cranston May 8, 20232023-05-08
On Saturday, May 6, 2023 at 9:44:33 PM UTC-5, Ricky wrote:
> On Saturday, May 6, 2023 at 10:30:55 AM UTC-4, Lamont Cranston wrote: > > On Saturday, May 6, 2023 at 7:35:23 AM UTC-5, Bob Engelhardt wrote: > > > On 5/5/2023 12:08 PM, Lamont Cranston wrote: > > > >[...] > > > > If one is run from AC out of the HV transformer and then the same AC is connected to a HV bridge, that is supplying HV DC to the other vessel. > > > > > > > > If, assuming no filter capacitor and ignoring and diode losses, > > > > Would the power delivered to each vessel be the same? > > > > > > > > [...] > > > > > > Only if the current is the same. Which is HIGHLY unlikely. > > I'm now going on the idea that there is no current flowing through solution* and it is just an electric field.
> Just to make sure, your electrodes are on the *outside* of the container, not in contact with the emulsion, right? Since there is no current through the glass, there's no current through the emulsion either.
At it is, glass tube is inserted into a an aluminum tune which is one electrode, the other electrode is a 12 gauge wire centered in the tube. Agreed, no current in emulsion.
> > So, it is just the electric field setup between the center electrode that is in the solution, and the electrode that is on the outside of the glass. > Oh! There is one electrode in the solution. Still, no current through the glass, other than by capacitive coupling. Is this 60 Hz? You might want to do a rough calculation of the capacitance of your set up. Is the outside electrode just a wire, or do you use foil around the container with a significant area? This could make a very significant capacitor!!!
Yes, aluminum tube (acting as one electrode) rather large, it is a holder for centrifuge tubes. The glass tube does fit snugly in the aluminum tube. The other electrode is centered in the tube and is in contact with the solution. I'll see if I can get a measurement of the capacitance.
> So the glass container inside the aluminum tube? Is the spacing fairly tight?
Yes, tight spacing.
> > We are spending the weekend tightening up the wiring to eliminate as much as we can.
> Not sure what you mean by "tightening".
Just shortening wires, doing what I think will eliminate corona currents.
> > -- > > Rick C. > > -- Get 1,000 miles of free Supercharging > -- Tesla referral code - https://ts.la/richard11209
Reply by Phil Allison May 8, 20232023-05-08
 Lamont Cranston wrote:
------------------------------------------
> I have the HV set up with a fullwave bridge. > I pick off the AC at the input to the bridge and the DC out of the bridge. > Without any load, (other that air), The DC voltage is 91% of the AC voltage, pretty close from 500V to 9000V.
** A standard DMM reads the *average* value of any wave on DC ranges. On AC ranges, it reads the equivalent rms value of *sine waves* - by scaling their average rectified value up by a factor of 1.11. So the average DC output from an ideal bridge rectifier will read 90% of AC rms voltage being supplied.
> I don't think it is the quality of the meter, I'm doubtful it is a crest factor problem at 60 Hz, unfiltered from a fullwave bridge. > It seems to be more than diode loss, and also the 91% is from 500V to 9000V, straight line, > diode loss would show more at the bottom end. > Any Ideas?
** There is no issue. .... Phil
Reply by Lamont Cranston May 7, 20232023-05-07
I have the HV set up with a fullwave bridge. I pick off the AC at the input to the bridge and the DC out of the bridge.
Without any load, (other that air), The DC voltage is 91% of the AC voltage, pretty close from 500V to 9000V.
 I don't think it is the quality of the meter, I'm doubtful it is a crest factor problem at 60 Hz, unfiltered from a fullwave bridge.
It seems to be more than diode loss, and also the 91% is from 500V to 9000V,  straight line, diode loss would show more at the bottom end.
                Any Ideas?
                                                Mikek
Reply by Ricky May 6, 20232023-05-06
On Friday, May 5, 2023 at 6:23:40 PM UTC-4, Lamont Cranston wrote:
> On Friday, May 5, 2023 at 12:16:46 PM UTC-5, Ricky wrote: > > > I'm surprised you are getting a result. An emulsion has droplets so tiny that they will not coalesce. But perhaps they can still be encouraged to coalesce. I don't think it is as simple as getting them to float to the top. They are so small, Brownian motion keeps them distributed. I'm not sure what keeps them from coalescing, perhaps surface tension? > It is not a electric field separation is not a recent, there are lots of papers on it.
Sorry, I don't understand your sentence.
> > It may be that the AC field moves them back and forth, causing collisions that overcome the reluctance to coalesce. I really don't remember much about emulsions. Maybe Bill has retained more. If the agitation of the AC field is causing them to coalesce, I would not expect much difference from the DC field, other than that the peak to peak field is half. If you can, try a smoothed DC field. If the effect goes away entirely, that would be a strong clue. But I realize 9000V caps are not so easy to come by. > The first test was with a electric field of 500V/cm, with the surprising result when compared to the control sample. > The only paper I have seen stopped at 1000V/cm, but the result was still improving. We did get inconsistent results with increasing voltage, > my need to revisit. One paper shows an improvement with squarewave, sinewave was not quite as good (it didn''t say 0.707% as good, but I wonder? :-) > and triangle was worse. The squarewave could have been DC, but I question it.
What does that mean, a DC square wave? Do you mean only one polarity, V+ and 0V?
> DC does work, but the paper says because of electrolytic corrosion it is out of favor. If there is no current through the glass, I didn't know how you get electrolytic corrosion, however there are systems that sink horizontal or vertical plates or mesh into the solution, so I can see corrosion in that method. Maybe they were just not complete with that idea.
I'm not following you. Do you mean they used TWO electrodes in the emulsion?
> I have about six 1uf 2000V caps I could string in series. > > > > Tell your son he is being pedantic. The water is pulled down harder than the oil is, so the oil floats to the top! > 6 of one 1/2 a dozen of the other :-) It happens,, no matter how it is described. > > > > You do find interesting things to occupy yourself with. > Do you mean from bean sprouts to oil/water separation, before that we troubleshot the air compressor motor and found a bad contact in the > centrifugal switch, and then it was the four pole double throw switch I should never have disassembled, but I did get it back together and working.
-- Rick C. -+ Get 1,000 miles of free Supercharging -+ Tesla referral code - https://ts.la/richard11209
Reply by Ricky May 6, 20232023-05-06
On Saturday, May 6, 2023 at 10:30:55 AM UTC-4, Lamont Cranston wrote:
> On Saturday, May 6, 2023 at 7:35:23 AM UTC-5, Bob Engelhardt wrote: > > On 5/5/2023 12:08 PM, Lamont Cranston wrote: > > >[...] > > > If one is run from AC out of the HV transformer and then the same AC is connected to a HV bridge, that is supplying HV DC to the other vessel. > > > > > > If, assuming no filter capacitor and ignoring and diode losses, > > > Would the power delivered to each vessel be the same? > > > > > > [...] > > > > Only if the current is the same. Which is HIGHLY unlikely. > I'm now going on the idea that there is no current flowing through solution* and it is just an electric field.
Just to make sure, your electrodes are on the *outside* of the container, not in contact with the emulsion, right? Since there is no current through the glass, there's no current through the emulsion either.
> So, it is just the electric field setup between the center electrode that is in the solution, and the electrode that is on the outside of the glass.
Oh! There is one electrode in the solution. Still, no current through the glass, other than by capacitive coupling. Is this 60 Hz? You might want to do a rough calculation of the capacitance of your set up. Is the outside electrode just a wire, or do you use foil around the container with a significant area? This could make a very significant capacitor!!!
> I'm assuming the RMS voltage will be about equal with AC or DC.
The RMS will be identical, as the polarity at any given instant doesn't matter after squaring. Root-mean-squared. The order is square the curve instant by instant, take the mean over a cycle, and take the square root. You are free to pick either polarity from the square root, but the positive number is usually more meaningful.
> * I'm pretty sure the current flow we saw was that spewing off the 8" x 1-3/4" aluminum tube to (everything else?)
So the glass container inside the aluminum tube? Is the spacing fairly tight?
> We are spending the weekend tightening up the wiring to eliminate as much as we can.
Not sure what you mean by "tightening". -- Rick C. -- Get 1,000 miles of free Supercharging -- Tesla referral code - https://ts.la/richard11209
Reply by Lamont Cranston May 6, 20232023-05-06
On Saturday, May 6, 2023 at 8:02:39 AM UTC-5, Martin Brown wrote:

> AT that sort of voltage you can get corona discharge off any > sharp edges of metal and leakage currents in ordinary water can be > large.
I'm now going on the idea that there is no current flowing through solution and it is just an electric field. So, it is just the electric field setup between the center electrode that is in the solution, and the electrode that is on the outside of the glass. But, yes plenty of corona discharge.
>We had to add the right combination of corrosion inhibitors to > our systems when operating water cooling systems floating up at 8kV. > > Backing pumps also required some clever protection too since although > air at STP is a pretty good insulator soft vacuum isn't at all. Hard > vacuum is an insulator again but the in between state is dodgy.
> >> What is the principle of your oil and water separation? Don't they > >> separate naturally? Or do you have an emulsion that does not > >> separate? What does the current do that separates them? > > > > I'm pretty clueless as to why, but I think it causes the oil to > > aggregate and the bigger droplets are more likely to float to the > > top. Yes it is an emulsion.
The quote from a paper, looks like I have it backwards, its the water that is that under goes coalescence. "Electrostatic coalescence was invented for the petroleum-related industries, and this method uses electric fields to promote the coalescence of small water droplets in water-in-oil emulsions. In electric fields, small water droplets move towards each other or collide with each other due to the electrostatic forces, such as dipole-dipole interaction and dielectrophoretic force, acting on them. As a result, the merging and coalescence of those droplets will increase the droplet size and eventually lead to settling and separation of water from the oil phase."
> I can't see this being generally applicable as a method of oil > separation since the currents flowing in normal water will be high! > > Dissolved salts make water not a particularly good insulator.
To reiterate, I think there is no current through the solution, but, there are system that have both electrode in solution. Mikek
Reply by Lamont Cranston May 6, 20232023-05-06
On Saturday, May 6, 2023 at 7:35:23 AM UTC-5, Bob Engelhardt wrote:
> On 5/5/2023 12:08 PM, Lamont Cranston wrote: > >[...] > > If one is run from AC out of the HV transformer and then the same AC is connected to a HV bridge, that is supplying HV DC to the other vessel. > > > > If, assuming no filter capacitor and ignoring and diode losses, > > Would the power delivered to each vessel be the same? > > > > [...] > > Only if the current is the same. Which is HIGHLY unlikely.
I'm now going on the idea that there is no current flowing through solution* and it is just an electric field. So, it is just the electric field setup between the center electrode that is in the solution, and the electrode that is on the outside of the glass. I'm assuming the RMS voltage will be about equal with AC or DC. * I'm pretty sure the current flow we saw was that spewing off the 8" x 1-3/4" aluminum tube to (everything else?) We are spending the weekend tightening up the wiring to eliminate as much as we can. Mikek
Reply by Martin Brown May 6, 20232023-05-06
On 05/05/2023 17:53, Lamont Cranston wrote:
> On Friday, May 5, 2023 at 11:34:44 AM UTC-5, Ricky wrote: > >> If your rectification is full wave, the power will be essentially >> the same. Of course, there will be some loss in the diode forward >> voltage, but in the context of "HV", this should be minute. > > >> You don't say how high your HV is. > > A couple of tests were done at 9000V, the minimum was 850V.
AT that sort of voltage you can get corona discharge off any sharp edges of metal and leakage currents in ordinary water can be large. We had to add the right combination of corrosion inhibitors to our systems when operating water cooling systems floating up at 8kV. Backing pumps also required some clever protection too since although air at STP is a pretty good insulator soft vacuum isn't at all. Hard vacuum is an insulator again but the in between state is dodgy.
>> What is the principle of your oil and water separation? Don't they >> separate naturally? Or do you have an emulsion that does not >> separate? What does the current do that separates them? > > I'm pretty clueless as to why, but I think it causes the oil to > aggregate and the bigger droplets are more likely to float to the > top, or as my son has corrected me, the water is pulled to the earth > by gravity. Yes it is an emulsion.
I can't see this being generally applicable as a method of oil separation since the currents flowing in normal water will be high! Dissolved salts make water not a particularly good insulator. -- Martin Brown