## Fusion Rate Calculation Questions

It may be difficult to separate "theory" from "application," but let''s see if this helps facilitate the discussion.
Liam David
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### Fusion Rate Calculation Questions

Since I hope to be fusing within a couple months, I decided to try and calculate my approximate reaction rate using the outline in the fusortip.doc. However, I have repeatedly tried and failed to get any reasonable or realistic reaction rate. The typical numbers I get are 10^25 fusions a second. Could someone please outline, in a detailed way, how to calculate the reaction rate without implementing calculus?

Numbers I used are:
grid diameter= .25in=.635cm
tungsten wire diameter= .25mm- 2 loops make the grid
accelerating voltage= 50kV
current= 10mA

Thanks,
-Liam David

Richard Hull
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### Re: Fusion Rate Calculation Questions

I don't like to throw cold water on your efforts, but any calculation you do will be wrong.....I mean way off wrong.

There are too many variables that cannot and will never be fully controlled. Your fusion rate is operator driven assuming all of many variables are held fairly tightly aligned during a fusion run attempt. You will get better at running the fusor with experience and your realized rate will steadily improve up and until you reach the construction and operational limits of your system......whatever they may be.

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
Retired now...Doing only what I want and not what I should...every day is a saturday.

Dan Tibbets
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### Re: Fusion Rate Calculation Questions

This is a very imprecise approach, but...

Calculate the number of Coulomb collisions you would expect (based on the Mean Free Path) per second at that temperature and density and volume. Then divide by 10,000.This is a perhaps reasonable estimate of Coulomb collisions to fusion collision ratio at that temperature Beam- cold neutral collision fusion rates probably dominate, either that or beam- wall collisions resulting in fusion. Beam beam fusion collisions are probably rare in a glow discharge fusor. Then divide by an additional amount to account for things like cold neutrals, actual lower average temperature due to heat losses, and what ever else you can think of. A number of 100 may be a very optimistic number.
This would give you a fusion rate of ~ 1/ 1,000,000 that of the Coulomb collision rate. Compare the result to reported fusor outputs (1 million fusions per second may be an optimistic target). Adjust you calculation accordingly,then add a range of perhaps +/- factor of 10X for operational variables as mentioned, though a range of +10/-1000X might be more appropriate.

Dan Tibbets

Liam David
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### Re: Fusion Rate Calculation Questions

Well, I met with a physics professor at my school today and talked with him about my fusor. In order to buy deuterium through the school and possibly move the fusor to the building, I have to provide the expected fusion rate, x-ray and neutron energies, and all that stuff. In other words, I need to calculate the fusion rate, however off it will be.
-Liam David

Richard Hull
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### Re: Fusion Rate Calculation Questions

If you report a number like 10e25 fusions per second you'll never get it in your school. Dan probably gave the best advice. You could derive the whole thing using actual reported values and tweak with a magic K factor and call that the empirically derived, high pressure operational fusion factor.

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
Retired now...Doing only what I want and not what I should...every day is a saturday.

Tom McCarthy
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### Re: Fusion Rate Calculation Questions

Had a Doctor of plasma physics who works with the Max Planck Tokamak in Germany for about half of his working year check over some numbers I had come up with and provide some calculations. Maybe it'll help - results pasted below.

Tom

The max. 60kV operating voltage puts an upper limit on the X-ray energy of 60keV. The NIST graph and table of photon mass attenuation coefficients for iron (close enough to steel) at:
http://physics.nist.gov/PhysRefData/Xra ... b/z26.html
shows that for E < 60 keV, the mass attenuation coefficient is > 1.2 cm^2/gm. It is 8.2 cm^2/gm for 30 keV and rises steeply to 170 cm^2/gm for 10 keV.

The calculation of penetrating flux follows de Beer's law of exponential attenuation :

I(x) = Io exp(-m x) where the thickness x is expressed as mass per unit area (gm/cm^2) and m is the mass attn. coefficient. For your chamber, x = rho_m * t where rho_m, the mass density is 8.0 gm.cm^2 for steel and t is the thickness in cm. My own plasma chamber is 4mm stainless steel, so taking that as an example, one gets
x = 8 x .4 = 3.2 gm/cm^2 and hence the attenuation of 60 keV X-rays is exp(- 1.2 x 3.2) = 0.021, i.e. only 2.1% of 60 keV X-days will penetrate. For 30 keV this is much smaller: exp(-8.2 x 3.2) = 4 E-12, already negligible.

The results are obviously high dependent on your chamber thickness. They are even more highly dependent on the
X-ray energy. Actually, since the inner shell binding energy of electrons in iron is (from recall) > 5 keV this already makes a big (positive) difference since from the above table, m would be about 1.55 giving 0.7% instead of 2% (for a thickness of 4mm).

The publication Kwon et al. Journal of the Korean Nuclear Society, Vol.12 (1980) p.171 et seq. http://www.kns.org/jknsfile/v12/A048032 ... f91bb5a89f
gives convenient factors converting neutron flux rates to neutron dose rates. For 2.5 MeV neutrons, the calculation is: dose rate (REM/hour) = 1.24 x 10-4 x flux (neutrons per cm2 per sec). For your numbers, this gives, for a maximum flux of 7 neutrons/cm2/s , a maximum dose rate of 0.875 mrem/hour which is very close to your quoted does rate!

Thus I can verify that your calculation is correct. You could also quote it in terms of the radius from the centre of the vessel, when (replacing (1/1.5 m) 2 by 1/r2) it becomes (based on your 0.833 mrem/hour): maximum dose rate = 1.874/ r2 mrem/hour (r in metres). The Environmental Protection Agency states that the average radiation dose per year to Irish people is 4037 microSieverts: http://www.epa.ie/radiation/radexp/expo ... EbFT0voF38

A Sievert is 100 Rem hence this average annual dose corresponds to 403.7 mrem/yr. I can’t find a recommended maximum annual dose on the epa site, but I would guess that the Office of Radiation Protection
would be concerned if the likely additional dose from the fusor were comparable to or in excess of the average dose. Thus it would be quite important to state how many hours you envisage operating the fusor, and hence
the expected annual dose. At 1 mrem/hour (to use a round number), you would equal the average Irish annual dose in 400 hours – admittedly a lot of operation time, and I’m sure you wouldn’t be planning to run it for anything like as long as this cumulative period over a year. You should, however, specify how long you do intend to run it (and I am assuming that you will be present when it is running).

Richard Hull
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### Re: Fusion Rate Calculation Questions

Remember, any figures you come up with related to what is produced in the fusor has to be degraded based on any shielding you might use. I work my fusor mostly in the high 30kv decade range to the low 40kv decade and use no shielding whatsoever. I typically sit about 6 feet from the device while operating it. The actual run time in an operating session at those voltages is usually a tiny fraction of my time spent in the session.

(A session being from the time I turn on the first pump to the time I turn off the last pump.)

In a typical fusor session I might spend 3 hours around the device at a distance of 6 feet. Of this time, I might only spend 20 minutes at or above 30kv in several timed, data collection runs. I might have a maximum of 10 such sessions each year. Looking at my log book, I now have had 8 operating sessions in 2014.

Even using no shielding, you will have to degrade the figures you compute based on the inverse square law related to your operational distance during the time of maximum operational voltage.

I realize you have to jump through these hoops, but it is much ado about nothing for an operator at only a moderate distance from the device.

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
Retired now...Doing only what I want and not what I should...every day is a saturday.