Ports for ion guns

For the design and construction details of ion guns, necessary for more advanced designs and lower vacuums.
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sharks63128
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Re: Ports for ion guns

Post by sharks63128 »

Thank you all very much for the help. Below is a sketch of my plan. Will the location of the ports work with future ion guns?
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Richard Hull
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Re: Ports for ion guns

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Yes, this looks oK if two guns are your goal.

Thanks to Brian for catching the error on the calcs. I was still locked in D-D mode.......As are we all.

D-T makes a big difference don't it!

Richard Hull
Progress may have been a good thing once, but it just went on too long. - Yogi Berra
Fusion is the energy of the future....and it always will be
The more complex the idea put forward by the poor amateur, the more likely it will never see embodiment
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Brian McDermott
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Re: Ports for ion guns

Post by Brian McDermott »

Specifically, what trouble did he run into?

Perhaps this should be in a new thread?
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Richard Hull
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Re: Ports for ion guns

Post by Richard Hull »

Joe found that the filament and ionizer system really performed too well and the main fusor acceleration current was seen to suddenly spike. The ionizer system current had a more far reaching effect on accelerator current to the point that the ionizer system had to be run at vey low levels that were tough to control. All of this made the pressure much more critical to control, as well. Lots of variables. perhaps too many for manual control. The only way to evaluate is to actually do it oneself, I figure.

I sort of wished Joe had reported on this personally, but he was not much for online interaction then.
Richard Hull
Progress may have been a good thing once, but it just went on too long. - Yogi Berra
Fusion is the energy of the future....and it always will be
The more complex the idea put forward by the poor amateur, the more likely it will never see embodiment
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Brian McDermott
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Re: Ports for ion guns

Post by Brian McDermott »

Just to correct those calculations a bit:

I recall you saying that Hirsch's 10^12 machine ran on D-T fuel. If that is the case, every neutron signifies one fusion, and each fusion releases 17MeV. (The original calcs assumed D-D fusion.)


So the calculation ought to be:

10e12 (total fusions=the number of neutrons) X ~1.7X10e7 (average eV/fusion) X 1.6X10e-19 (joules/eV) = 27.2 joules or 27.2 watt-second

When he achieved these results, what were the voltage and current being used? Assuming the total input power was between 1 and 10 kilowatts, this is (relatively) much closer to breakeven than the typical 7-10 orders of magnitude.

I also posted this in the construction/operation forum, since this topic is no longer pertinent to the original thread.
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