electrostatic current generation

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longstreet
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electrostatic current generation

Post by longstreet »

It's been a while since I have posted. I had to transfer schools last year so my fusor is on hold atm.

However, I had a thought about generating larger currents from electrostatic machines. It seems easy to use something like the van de graaff to generate extreamly large voltages but unusable current level for a fusor.

My idea though is to have a pair of independant capacitors that share a common free plate. One of the fixed capacitor plates is initially charged by a van de graaff to a high peak voltage. The other is at ground. The free plate is first grounded and brought to close proximity to the fixed charged plate which forms a capacitor and induces charge on the free plate. The free plate is then de-grounded and brought by force of a motor to the grounded plate which now forms another capacitor charged equal and opposite to the plate charged by the van de graaff, which can be discharged through a fusor. The current capability would only be limited by the power output of the motor driving the free plate between them. It acts as a physical pump moving the charge from low to high voltage.

I figure this could be placed on a rotating shaft powered by a standard industrial motor so that the free plate is rotated between the two potential plates. I was wondering if it might be cheaper creating kilowatts of power at such a high voltage (approaching million) mechanically than with solid state. What do you think?

Carter
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Re: electrostatic current generation

Post by Carl Willis »

Hi Carter,

On first blush it sounds like this is an electrostatic induction concept vaguely akin to the "Pelletron." As written I can't make sense of a few particulars, but I think I understand the concept well enough.

One of those particulars is that your method of charging the "free plate" relies on the very technology you have written off for insufficient current, the Van de Graaf generator. The Van de Graaf is a current source whose capabilities are limited in part by the charge density that the belt can carry. It seems to me that you want a voltage source as your charging power supply, something capable of handling the requisite high charging inrush currents when the free plates return.

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Re: electrostatic current generation

Post by longstreet »

Not quite. The grounded free plate along with the one connected to the van de graaff act as a capacitor. The plates build up a charge Q= CV, the capacitance depending on how well the plate is made. The van de graaff could be disconnected now with both plates having Q and -Q on them. Now both isolated they are pulled apart by a force applied by a motor. The free plate is brought to the other fixed plate, which is a ground potential, forming a second capacitory. It induces a Q on it and can be discharged at V = Q/C. The free plate can be regrounded, and brought back to the plate that still has it's full charge Q on it. The van de graaff does not have to replenish this. That Q now induces -Q on the free plate again and repeats the process.

This is not free energy. Work must be done by the motor to overcome the electrostatic attractions between the capacitor plates and move the charge. The van de graff is not doing this. It's only role is the initial charging of the plates.
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Re: electrostatic current generation

Post by DaveC »

I think quite a number of different permutations on the rotating capacitor method generating high voltage, were explored by Michael Faraday, and also Maxwell.

Within the limitations of the physical parameters, (and materials), permitivity, electric field intensity, permeability, and flux density, it is a matter of choice as to whether energy conversion is done with electric fields or magnetic fields, or as is the usual case, electromagnetic fields the typical combination of both.

Both E and B fields are conservative.. so what you put in, you can extract, minus some "handling" losses.

The electric utilities at one point in the past were interested in UHV Power Transmission at DC, and 50-60 Hz voltages of 1100 kV and above. Transformers were and are built for these voltages. Above 3 MV, the bushings become extremely long and difficult to manufacture. But the designs of innards of the transformer are probably not in trouble below about 5 - 7.5 MV.

Efficiency in power conversion, just depends on minimizing losses.

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Re: electrostatic current generation

Post by Carl Willis »

OK, I see what you're doing--kind of a mechanized "electrophorus."
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Richard Hull
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Re: electrostatic current generation

Post by Richard Hull »

I think that some calcs would be needed here. The surface area and speed would be in play and the 10ma needed for a fusor is continuous. Such systems as you envision would be hard pressed to hit 1 ma at full potential (what ever that design goal is) without ungainly size or super high speeds.

Again, I think nature will limit, through losses, such a concept without Herculean (read costly) efforts to limit them.

It intrigues me to consider the wall watts to the motor versus the watts out of the electrostatic system efficiency. Not that it is important in our application, but more from a curiousity standpoint. I would be rather stunned if it exceeded 10%. I think it would be pretty cool to get 100 watts of delivered continuous electrostatic energy using a 1000 watt (1.5 hp) motor. At 100kv this would be only 1ma!

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Re: electrostatic current generation

Post by tligon »

I looked up early accelerators when I wrote "World's Simplest" and found Van de Graaff's name right at the start ... his famous static generator was designed to run the earliest particle accelerators, and were easily capable of driving fusion reactions.

Miniscule numbers of fusion reactions.

But at voltage high enough to do fusion on almost everything under the sun. It was these early accelerators that discovered fusion, and generated much of the data available today on the subject.

As the others have said, this is an inefficient way to make high voltage, particularly higher currents at modest voltages (by accelerator standards) used on fusors. But it is worth a look for the history, and perhaps to understand ways in which natural phenomena such as static electricity can easily be expected to cause fusion reactions under the right circumstances.

Dr. Bussard had one of those little Edmund Van de Graaff generators, which we pulled out and hooked up just for fun and a dog and pony show or two. I was appalled that our 1000:1 high voltage probe (essentially a 10 gigohm resistor) was a sufficient load to pull its output nearly to zero.
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Re: electrostatic current generation

Post by Carl Willis »

I think there may be merit to sucn an a electrostatic-induction approach, even acknowledging that the concept has a major disadvantage of requiring moving parts at high potentials.

Some of the early x-ray equipment operated off huge multi-disc Wimshurst machines that could produce 100+ kV at lethal currents. The Felici generator appears to be a "modern" idea, good for 80-750 kV at currents of 0.2-14 mA: http://members.aol.com/lyonelb/felici.html

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Re: electrostatic current generation

Post by Richard Hull »

Therein lay the rub. Electrostatic systems are intrinsically very high impedance devices. Early X-ray devices loaded down what were often 4 foot diameter, multiple van degraff disks from 200kv plus to 20-30kv or a supplied a vastly varying intensity x-ray tube from 30-50kv. For x-ray purposes with the early orthochromatic films this wildly fluctuating x-ray blast was no real hassle. For the patient........well.......

It is tough to hold any electrostatic system to a current and voltage of choice.

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Re: electrostatic current generation

Post by DaveC »

In the late 60's and early 70's a company called Gianinni - Voltex made a modern version of the sector-less Wimshurst machine.

It had a rotating high resistivity glass cylinder on which charge was deposited at the bottom of the rotation, and then picked off 180 degrees away, after being dragged upward against the electrode field.

Their wrinkle was to use a special high resistivity inner glass sleeve to provide a nearly uniform electric field for the potential generation.

They were sealed units that operated at above atmospheric pressures in N, or possibly SF6. Not sure of the output, but I think it was in the 1-2 mA range... at several hundred kV. It was intended to power portable xray units.

On the Van deGraaff generators, the big Edmund Scientific units were actually built by an outfit in Indiana. Amazingly, they sold for almost the same price as you could buy direct. They were 400 kV units and could deliver a small fraction of a mA. The belt drive motor was only about 90 watts, so you can do the math as what power it could deliver. Other than leakage down the column and corona losses, motor output watts should be close to available electrostatic power. Not very large current, but properly operated, relatively efficient as these beasts go.

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Re: electrostatic current generation

Post by longstreet »

edit: I am sorry. Please disregard the first paragraph of this post it is wronge.

I have estimated the current for my example machine by this equation: I = (k*e_0*V*f*Pi*R^2)/(d). e_0 is the permitivity of free space, k the dielectric constant between the discs, V is voltage, f is rotational frequency, R is the radius of the discs, and d the seperation between them. Plug in say k = 1, V = 500kV, f = 1.67s-1 (about 100 rpm), R = 0.3m, d = 0.01m; I get a current of 0.2mA, or 100 watts. And this is from a single disc. Since they're flat you can stack a bunch together on a single axle. Also, you can make it much better by making it closer together, better dielectic material, faster etc... E.G. Say k = 2, V = 500kV, f = 8s-1 (about 500 rpm), R = 0.3m, d = 0.005m; I get a current of 4mA, or 2kw. Again, just a single plate.

The upper limit of rotational speed is determined by the charging rate of the system. The time constant would of course be t = RC. In my last example, C = 5e-10F. So as long as you don't have a very large resistance on the charging side, which you wouldn't anway, most mechanically feasable rotatinal speeds should be ok.

The biggest losses would be the contacts to the free plates, the heat generated by the motor itself, and whatever other resistance you have on there for some reason. Along with any stray arcs/ionization. The induction itself isn't lossy unless EM radiation is prominant for the rotating discs, I don't know.

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Re: electrostatic current generation

Post by longstreet »

I have thought about this more and am not yet sure how to solve it. What I realized is by bringing the free plate close to another grounded plate I am creating a capacitor which rapidly drops in voltage as I discharge it, since V = Q/C. However, what I want to do is destroy the capacitance of the system. As C drops the effective V rises and I can discharge my Q at the desired voltage. However, I am not sure how to calculate the capacitance with the free plate offset from the charged one. And it becomes even more complex if additional free/charging plates are added. Any thoughts?

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Re: electrostatic current generation

Post by AFW »

An ordinary school- lab wimshurst machine can give ~30 microamps at 20 - 30 kv. A van de graaf actually works on the continuous electrophorus principle . I wonder if its current capacity could be increased by having Al. segments on the belt, tho' this would obviously reduce its output voltage.

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Re: electrostatic current generation

Post by DaveC »

There have been many, many variations to both Wimshurst and Van deGraaff generators. There are commercial versions of Van deGraaffs that are designed to operate in SF6 or pressurized dry Nitrogen. Multimegavolt units are only a couple feet long and not much over a foot in diameter for the HV electrode. They are rather pricey, though.

A common charging method on the Van deGraaff is to have a HV DC corona charging system near the lower belt roller. With a grounded plate behind, such a technique increases the charging current by many times.

A well built electrophorus should be able to give a spark approaching at least half of its plate diameter. I have one built from a pizza pan, which is good for a few inches spark under ideal circumstances.

These simple machines will give you a lot of fun experimenting, and a good education in static electricity principles.

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Re: electrostatic current generation

Post by Richard Hull »

Current capacity is not mathematically related to the materials at all in any electrostatic machine.

Current is related to how much charge is transported per unit time to the output capacitance. The capacitance or top hat on a van degraff will ultimately determine break down or peak voltage attainable voltage, but sprayer voltage, if any, its impedance, belt width and velocity within the limits of coronal losses etc.,are the current determining factors.

Current is a matter of surface area to hold a charge and how fast youu can move, deliver, deposit and recharge that surface with minimal losses during the entire process. For the effective purposes of electrostatic uses, carbon plates would deliver the same current as pure gold.

So much in electrostatic machines is environmentally related that even doing good math during design is a joke unless every worldly parameter is rigidly fixed and controlled. This is why REAL van degraffs doing REAL work and supplying real currents are sealed, lay on their side and are under a gang of atmospheres of pressure in a dry gas of the designer's choice.

The book mentioned in the REFs forum on early accelerators is full of horror stories of open air van degraffs arcing 6 or more feet to supporting timbers and door jams.

I remain both impressed and stunned that any fusor user would even consider them as a source of viable electrical energy at the amateur level.

Richard Hull
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Fusion is the energy of the future....and it always will be
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Re: electrostatic current generation

Post by longstreet »

I think you still misunderstand what I was talking about. The van de graff was only a HV source to initially charge the inducing plate. It's not important to what I was talking about. For all my purposes you could replace the inducing plate with a dielectric which has been implanted with electrons by an electron gun. All the matters is the Q which is there to induce a charge on the free plates.

I am still trying to work out some math to model it but the geometries are not nice.

Carter
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Re: electrostatic current generation

Post by longstreet »

I think I have some numbers now. The machine again has the charged plate and the free plate. The free plate can rotate on half of an axle so that for half a rotation (call it stage A) it is parallel to the charged plate seperated by a distance S, forming a parallel plate capacitor. The area of each plate is Pi*R^2/2 (half a circle). The other half of the rotation (call it stage B) the plates are parallel but adjacent to each other seperating by a distance 2*D. D is the distance from each plate to the axis. There is a third stationary plate parallel to the axis but perpendicular to the charged and free plate. This third plate is grounded to now isolate the free plate from the charged plate, and has a capacitance with the free plate.

In stage A the free plate is grounded and charge flows onto it induced by the charge plate. It is isolated and rotated into stage B where it is connected to the HV output. The output voltage of free plate is related to the relative capacitances of stages A and B.

I think I have a decent estimate for the capacitance of B now which was my problem, which is weaker than stage A which forces the voltage higher. As Q drains from the free plate the voltage drops but can still be higher than your desired voltage.

The Q you extract is as much as you can discharge from the free plate before the voltage drops below the input voltage.

Q_out = C_i*V_i - C_f*V_i = (C_i - C_f)*V_i

The current capability is just the Q_out times the frequency at which the free plate is rotated back and fourth.

My example machine has R = 0.3m, S = 0.005, D = 0.05, frequency = 8Hz (480rpm), input voltage = 500kV. I got a capacitance at stage A at C_i = 0.24nF, and at stage B at C_f = 10pF. This gives a output current of 0.92mA. Again, this is a very basic ouput with the simplest setup. I think it could be improved many times over with added stages etc...
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Re: electrostatic current generation

Post by Richard Hull »

As always, the best of luck on the endeavor. Electrostatic stuff is always fun and interesting to work with.

Richard Hull
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Fusion is the energy of the future....and it always will be
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Re: electrostatic current generation

Post by Steven Sesselmann »

Carter,

I think I understand what you are trying to do. You want to use the strong electrostatic field of the van DeGraaf generator without actually depositing any charge from it, in much the same way as a permanent magnet functions in a normal dynamo or generator.

You will need the equivalent of a commutator to switch your plate arrangement. In theory this should work, but you would have to sort out the charge exchanges, without too much sparking.

Maybe part of the devise can run in transformer oil.

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Re: electrostatic current generation

Post by longstreet »

Yes I have been thinking about a how to do a HV commutator.

Also, one of the advantages I have thought about is that the machine could also be driven by an electrostatic motor. So, theoretically fusion product ions could drive it since they come out at high voltage. And assuming breakeven is ever acheived it could drive a dynamo connected to the shaft. but that is neither here nor there atm... lol
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