Fusion Message Board

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 Deuterium Consumption
Date: Oct 25, 09:36 am
Poster: Richard Hull

On Oct 25, 09:36 am, Richard Hull wrote:

Robert has done a great job of showing just how little D2 is used up in the system. I believe I have noted this in the past. There are countless trillions of D2 molecules in an ideal fusor chamber at 1 micron. Burning off a trifling 100,000 to 1,000,000 per second would leave one with years of head room.

I prefer the continuous flowing method as my current vac system, at present, relies on a pure mechanical pump which operates at its less than ideal mode of operation in the molecular flow regime. This flow assists in the purging of the fusor under what is a terrible pumping situation. As a matter of fact, all who attempt to use only a mechanical pump will be virtually forced to use this method. It is somewhat wasteful of D2, but at the leak rate, a small cylinder will last for years. With an ultra deep pumping (deeper than 10e-10 torr) and a super bakeout and glow cleaning, the fusor could be sealed off and used as is for months, if not years once simply backfilled with D2.

This is not a useful experimental situation though as you could not modify, alter or change many parameters on the fly.

I would suspect that for 90% of those on this list, the calculations presented below are hardly insulting, but instead, are incredibly enlightening. It is the mechanics of the answer to the original question layed open for all to see.

For the rest of us, it is always good to see the order of magnitudes of D2 in the chamber on a regular basis as shown by this calculational effort. It is often easy for "old hands" to lose sight of some of the basics. Thanks Robert!


>If you mean the rate at which deuterium is consumed due to fusion, it is negligable. Here is why (I hope that this is not insulting, but it is an interesting calculation anyway):
>operating pressure (sealed),P = 10 mTorr (1.33 Pa)
>gas temperature,T = 300 K
>fusion rate: 10^6 neutrons/second
>chamber radius: 1 ft (0.3 m) (Volume=0.113 m^3)
>k (boltzmann's constant) = 1.38E-23 J/K
>molecule density=n=P/(kT)=3.2E20/m^3
># of deuteriums=2*n*0.133m^3=7E19 deuterium atoms
>time to
>deuterium~(10^19 deuteriums)/(10^6 fusions/sec)
> ~10^13 seconds~10^5 years
>In short, you will be waiting a long, long time before you burn up all of the deuterium in your chamber.
>Hope this answers your question,