In this space, visitors are invited to post any comments, questions, or skeptical observations about Philo T. Farnsworth's contributions to the field of Nuclear Fusion research.Subject: Re: Fusor Operation - a general question
Date: Jul 06, 2:59 am
Poster: Dave CooperOn Jul 06, 2:59 am, Dave Cooper wrote:
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>I have not done the calorimetry, but I would imagine grid losses, while significant, can't be even 2% of the total losses. I find all of the real loss occurs due to electrons slamming into the outer shell.
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>After about 10 minutes of operation at 10ma - 22kv, the outer SS shell can't be touched at about 180 degrees F! I think all of that 220 watts goes into heating the outer shell. It is just part of the simple fusor's natural loss mechanism.
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> Finding a way to turn them or utilize their kinetic energy would be a positive step. Some have suggested multiple decelerator grids and other schemes. Some parties (pros) are working on this now. We amateurs can't respond to these issue quickly due to budgetary and time restrictions. Most of us do this out of pocket and after we leave our real jobs.
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>I'll be curious to see who and if we amateurs will address this issue physically.
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>Richard Hull
...an interesting point Richard... I have been assuming that the electron or ion guns have a sufficient electrostatic presence (effect) to create a more or less uniform potential shell at the gun potential.. that was why I said the ions which arrived back at the ground plane would be at relatively low energy... i.e.: just those that leaked through.
You experimental results indicate otherwise.. that the electrostatic shell is rather incomplete in its coverage. Allowing most of the electrons streaming from the center to collide full force at the fusor wall.
In order to produce a reasonably good electrostatic screen at the gun potential the mesh size must be on the order of the mesh wire diameter...maybe just a couple of times larger.
I have a good program to model this.. will check out numbers.
a fine mesh spherical screen shouldn't be too costly..but it should greatly reduce input currents, by trapping more of the injected charge.
In fact by reducing input currents, at the same kV, this is a sign that the system is becoming more efficient. Ultimately it should charge up like a capacitor... and the "leakage current " will be an indication of either fusion or losses.
Dave Cooper