Net energy question

[bcglorf]

OK, I've just concluded that the alpha's must come into contact with the positively charged plates and rob electrons of them to produce DC. Am I right?

As MSimon said, it's more like a linear accelerator in reverse. Instead of doing work to push alphas, the alphas push and you get work(DC) out.

[Stefan]

That's correct rogerlam, you get power in the form of an electron current from the MA-Grid ("only" point where the electrons can escape) to the collector plates.

[Keegan]

pop quiz !

The highest energy reaction product for p+B11 is a 3.76Mev alpha particle.

Does this mean the positively charged grid would have to be near +3.76 million volts to successfully extract its energy ?

If so, this would definitely interfere with other facets of the machine.

[JohnP]

Does this mean the positively charged grid would have to be near +3.76 million volts to successfully extract its energy ?

If so, this would definitely interfere with other facets of the machine.

I was just thinking the same thing. Tom Ligon writes about this somewhat in "World's Easiest Fusion Reactor" but how do you have a 3 megavolt field in your chamber that doesn't interfere with the fusion fuel & electrons?

[Keegan]

Skimming through the literature i did find one reference to direct power conversion (Bussard-FusionTechnology-Mar1991.pdf). More a concept than a technical drawing, it mentions a grid with a 1Mev positive bias. Its gonna reak havoc with electron injection/recirculation. One could place the grid far enough away so its effect would be negligible, but would make the machine excessively large and would put more stress on the vacuum system.

*Scratchy SSB transmission* Er, Houston.... We have a problem.

[JohnP]

t mentions a grid with a 1Mev positive bias

Since the alpha has a charge of +2 you "only" need a 1.5 MV bias, not 3 MV.
This diagram you have, looks like the 1 MV field is a continuation of the electric field that's already there. But you need a + center to accelerate the electrons, right?

[Keegan]

Hi John,

p + B11 -> He 2.46Mev + He 2.46Mev + He 3.76Mev

and yes a positive center is need to accelerate the electrons. Some fantastic modeling can be found here courtesy of Indrek

[Stefan]

My guess is that you can use the same setup you'd use anyway, but
around the faraday cage you put the extraction grid.
There'd need to be quiet some distance between them or there'd be paschen discharge.
On the inside of the faraday cage the E-fields wouldn't be affected.

The faraday cage is usually biased to ground.
What's important is, that it's slightly below electron gun voltage, so the electrons (which are supposed to be recirculating) can't escape to the vacuum chamber wall or in this case to the extraction grid (this would mean huge energy losses).

Because of their speed most alphas should be able to pass the cage although they are attracted by it.

[JohnP]

Because of their speed most alphas should be able to pass the cage although they are attracted by it.

sound familiar?

What's going to happen to the cage when you're running a 100 MW reactor. The cage will disintegrate.

[MSimon]

The way to handle this is:

Outer shell - grounded
Decelerator grid - +1.7 MV (roughly)
Potential "equalization" grid - grounded
Accelerator (mag shielded grid) - +50 to +200 KV (depending)

Electron guns will be located in the "equalization" grid area.

Good question Keegan. Stumped me too for a while.

Getting all this to work of course is going to be tricky. Lots of tough engineering problems.

Simon

[MSimon]

pop quiz !

The highest energy reaction product for p+B11 is a 3.76Mev alpha particle.

Does this mean the positively charged grid would have to be near +3.76 million volts to successfully extract its energy ?

If so, this would definitely interfere with other facets of the machine.

Alphas are +2. So divide by 2 to get the grid voltage (minus a little to make up for actual particle speed distribution. Around 1.9 MV nominal. Probably 1.6 to 1.85 MV in a real machine.

[Roger]

break even device would be 1.5-2m for 100MW output.

IIRC,

P-B11 100MW net power=3m
P-B11 break even =2m
D_D 100MW net power=2m
D-D break even=1.5m

[JohnP]

Outer shell - grounded
Decelerator grid - +1.7 MV (roughly)
Potential "equalization" grid - grounded
Accelerator (mag shielded grid) - +50 to +200 KV (depending)

I'm not seeing it. What I see is some fraction of the 3 MeV alphas hitting the equalization grid and frying it.

Maybe I'm thick here, but isn't this the same particles-hit-grid problem that EMC2 solved for electrons, come back to haunt us in the guise of alphas?

[MSimon]

This is a inverse linear accelerator.

The decelerator grids can be shielded by the shadow of the magnetic grids.

There is still the problem of where to put the e-guns.

If you look at the e-guns for a color cathode ray tube it becomes more obvious. The accelerator/focusing grids are cylinders.

[drmike]

Outer shell - grounded
Decelerator grid - +1.7 MV (roughly)
Potential "equalization" grid - grounded
Accelerator (mag shielded grid) - +50 to +200 KV (depending)

I'm not seeing it. What I see is some fraction of the 3 MeV alphas hitting the equalization grid and frying it.

Maybe I'm thick here, but isn't this the same particles-hit-grid problem that EMC2 solved for electrons, come back to haunt us in the guise of alphas?

Yes and no. Yes some fraction will hit and no it is not a problem. With electrons, the total number of particles is many orders of magnitude higher, so the same fraction becomes a mechanical heat load.

It's like my kids example of thermo - what has more energy, a bathtub full of luke warm water or a tea cup with boiling water? If you stick your finger in one nothing happens, but it burns off in the other. But because the bathtub is so much larger, it has more total energy.

Same thing with electron current vs ion current. Yes, each ion has a lot more energy, but the total number is a lot less. Sucking the energy out before any damage happens is an engineering problem.