TallDave wrote:Yeah... rough Faraday cage.
...
Of course a dodec will be less rough than a truncube. I thinik we expect a reactor to be a dodecahedron.
Yup, a bow sided rectified dodecahedron (aka icosadodecahedron); though a bow sided rectified cube (aka cuboctahedron) may do almost as well and would probably be much easier to build.
The space-ship variety may want the theorized improved performance of the former, a land based would go for peak value, not peak performance.
Again, I don't think the ions are getting outside the WB boundary. The ions are heavier, but very slow at that point, and for them the electrostatic force and the magnetic field gradient are pointing the same way. For the electrons, the magnetic field gradient is opposed to the direction of the electrostatic force.
And if you have most of the ions slow and colliding in an area where they're picking up transverse momentum from the B field it's going to be hard to get them to fuse again.
And the plasma isn't losing equal number of ions and electons. Losses are not ambipolar; they can't be when we're pumping in excess electrons.
Again, I don't think the ions are getting outside the WB boundary. The ions are heavier, but very slow at that point, and for them the electrostatic force and the magnetic field gradient are pointing the same way. For the electrons, the magnetic field gradient is opposed to the direction of the electrostatic force.
And if you have most of the ions slow and colliding in an area where they're picking up transverse momentum from the B field it's going to be hard to get them to fuse again.
And the plasma isn't losing equal number of ions and electons. Losses are not ambipolar; they can't be when we're pumping in excess electrons.
To maintain the charge balance (at what ever potential) the number of ions leaving must equal the number of electrons entering once equilibrium is reached.
Engineering is the art of making what you want from what you can get at a profit.
Again, I don't think the ions are getting outside the WB boundary. The ions are heavier, but very slow at that point, and for them the electrostatic force and the magnetic field gradient are pointing the same way. For the electrons, the magnetic field gradient is opposed to the direction of the electrostatic force.
And if you have most of the ions slow and colliding in an area where they're picking up transverse momentum from the B field it's going to be hard to get them to fuse again.
And the plasma isn't losing equal number of ions and electons. Losses are not ambipolar; they can't be when we're pumping in excess electrons.
To maintain the charge balance (at what ever potential) the number of ions leaving must equal the number of electrons entering once equilibrium is reached.
Thats one straight forward way to look at it. Another way is to say the number of electrons injected must equal the number of electrons lost within the limits imposed by the allowable ratio of ions/ electrons. This a more liberal attitued towards the dynamics of the electrons- the knobs that can be adjusted, not only to maintain the potential well, but also the shape of the potential well, allowences for electron thermalization time, percentage of recirculation input compared to injection input. So there are two limiting factors-:the input- output ratios nessisary to maintain the space charge, and the input-output of electrons to modify things within the space charge limits ( if you have any wiggle room past breateven conditions).
MSimon wrote: To maintain the charge balance (at what ever potential) the number of ions leaving must equal the number of electrons entering once equilibrium is reached.
But the ions leaving are (it is to be hoped) only the fusion PRODUCT ions, not the fusion FUEL ions.
To maintain the charge balance (at what ever potential) the number of ions leaving must equal the number of electrons entering once equilibrium is reached.
I think you have that mixed up? If an electrons enters (-1) and an ion leaves(-1), that's a -2. I'll assume you meant one of each must leave.
That's true unless you're adding more electrons than ions because you're losing more electrons. My understanding is we're adding a LOT more electrons than ions. If we're pumping in 10MW of electrons and nearly all the losses are electron losses...
I'm not sure what the ion current is. I guess I've always assumed it was a lot lower than that.
Has someone done the math on how many ions we're losing to fusion? I'm lazy and hungry at the moment.
TallDave wrote:Again, I don't think the ions are getting outside the WB boundary. The ions are heavier, but very slow at that point, and for them the electrostatic force and the magnetic field gradient are pointing the same way. For the electrons, the magnetic field gradient is opposed to the direction of the electrostatic force.
A hole is a hole... and bigger for the ions than the electrons, as Art explained.
TallDave wrote:And if you have most of the ions slow and colliding in an area where they're picking up transverse momentum from the B field it's going to be hard to get them to fuse again.
TallDave wrote:Again, I don't think the ions are getting outside the WB boundary. The ions are heavier, but very slow at that point, and for them the electrostatic force and the magnetic field gradient are pointing the same way. For the electrons, the magnetic field gradient is opposed to the direction of the electrostatic force.
A hole is a hole... and bigger for the ions than the electrons, as Art explained.
TallDave wrote:And if you have most of the ions slow and colliding in an area where they're picking up transverse momentum from the B field it's going to be hard to get them to fuse again.
Go on...
A magnetic hole is indeed larger for an ion. That is why the mantra is that the electrons are magneticallly contained (in excess) and this sets up the conditions that allows the ions to be electrostatically contained.
As far as the magnetic interface adding transverse motion, I'm not sure that the half spiral that an ion makes when turning on the Wiffleball border introduces any significant lateral motion because each magnetic turning may contribute only a tiny lateral motion. Also, because of the convex surface of each lobe, the field will turn the ion slightly away from the center, or if it hits the other half of the lobe, it will turn slightly towards the center, eg, on the left side of the lobe the magnetic turning might be 179 degrees relative to the Wiffleball center. On the right side the ion may turn at 181 degrees (both ions starting from the Wiffleball center and hitting an equal distance from the center of the lobe). The net effect may be zero. Also, the ions suposedly stays out of the magnetic domain, which means it is reversed by the electrostatic field, and only upscattered ions would see the magnetic field . Finally, this periferal region of slow ions is advertised to be the reigon where annealing occurs.
Dan Tibbets
Last edited by D Tibbets on Sun Nov 15, 2009 1:46 pm, edited 2 times in total.
To maintain the charge balance (at what ever potential) the number of ions leaving must equal the number of electrons entering once equilibrium is reached.
I think you have that mixed up? If an electrons enters (-1) and an ion leaves(-1), that's a -2. I'll assume you meant one of each must leave.
That's true unless you're adding more electrons than ions because you're losing more electrons. My understanding is we're adding a LOT more electrons than ions. If we're pumping in 10MW of electrons and nearly all the losses are electron losses...
I'm not sure what the ion current is. I guess I've always assumed it was a lot lower than that.
Has someone done the math on how many ions we're losing to fusion? I'm lazy and hungry at the moment.
Using some assumptions and hopefully acurate unit conversions
It takes ~ 10^20 fusions (D-D) to generate ~ 100 MW of fusion energy. That would consume 2 X 10^20 deuteruim ions.
1 Coulomb= 1 Amp sec. and 1 Coulomb = ~ 6 X 10^18 single charged ions.
So, a current of ~ 30 A of singly charged deuterium ions would be required. This would be ~ 0.0003 moles of deuterium atoms/ions, or ~ 0.6 milligrams of deuterium per second.
Assuming a 10 million fold in fusion energy yield versus burning the deuterium with oxygen, it would take ~ 6 kilograms per second to get the same amount of energy through the chemical process. That would be ~ 20,000 kg per hr and ~ 500,000 kg per day. Pretending that liquid deuterium has a similar chemical energy per unit weight/ volume of coal (which it doesn't) that would amount to ~ 12 railroad cars per day. This is probably a little low, but I will blame it on the different chemical energy of hydrogen compared to coal. I'm assuming that a 500 MW coal plant will burn through ~ 100 railroad cars of coal per day (~ 40 metric tons per car).
alexjrgreen wrote:3. Between the holes and the magrid: As they leave the negatively charged wiffleball, the ions on the outside of the jet are decelerating and the electrons in the middle are accelerating, so that the jet becomes electron rich as it moves away.
No one seems to have picked up that this potentially places a limit on the size of a polywell.
Maintaining this structure in a larger device is going to be interesting.
TallDave wrote:My understanding is we're adding a LOT more electrons than ions.
Until the ion gun was installed this year, no ions were added at all. Gas was puffed in and electrons ionised the neutrals.
Well, I was thinking of a reactor (hence the 10MW of electrons), but yes even for the pulsed machine the losses should be overwhelmingly electron losses, at least at the beta~1 areas of interest.
A hole is a hole... and bigger for the ions than the electrons, as Art explained.
But the ions also have a MUCH harder time getting to and through that hole. For the electrons, it's a hole (or downward-leaning tunnel) near the bottom of the well. For the ions, it's a hole (upward-leaning tunnel) near the top.
TallDave wrote:
And if you have most of the ions slow and colliding in an area where they're picking up transverse momentum from the B field it's going to be hard to get them to fuse again.
Go on...
And then you don't get net power and your technology is worthless? I'm not sure where you thought I was going.
TallDave wrote:
And if you have most of the ions slow and colliding in an area where they're picking up transverse momentum from the B field it's going to be hard to get them to fuse again.
Go on...
And then you don't get net power and your technology is worthless? I'm not sure where you thought I was going.
I was hoping you might expand on your reasoning. The holes are (almost) symmetrical, so most processes contributing transverse momentum will cancel out.