New Scientist on ITER

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Steven Sesselmann
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New Scientist on ITER

Post by Steven Sesselmann » Sun Oct 18, 2009 4:13 am

Hi Guys,

Just read the NS article in the 10. october issue on ITER.

I found it fascinating that they expect the neutron flux to be so large that the stainless steel walls of the reactor will become warm and soft....600-700 C˚?? I quote the article...

"The main reactor wall, known as the blanket, will be made from 440 stainless steel blocks nearly half a metre thick and riddled with high-pressure water pipes. This steel blanket should absorb most of the neutrons, which will heat the blanket from within. Near the inner wall, the water pipes can be no more than 2.5 centimetres apart, otherwise the steel between them would become dangerously warm and soft."

So it sounds as if 50 cm of stainless steel will stop a 2.5 mev Neutron...

Does anyone have any data on what the half thickness of stainless steel is for fusion neutrons?

Steven

Reference:
http://www.newscientist.com/article/mg2 ... ?full=true
http://www.gammaspectacular.com - Gamma Spectrometry Systems
https://www.researchgate.net/profile/Steven_Sesselmann - Various papers and patents on RG

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Carl Willis
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Re: New Scientist on ITER

Post by Carl Willis » Sun Oct 18, 2009 5:24 am

Hi Steven,

The total cross-section of Fe-56 (the predominant constituent of steel) at 14 MeV, typical of a DT neutron, is about 2.6 barns. 1.2 b elastic scattering, ~1 b inelastic scattering with gamma emission, a little bit of (n,2n), a little bit of (n,p).

http://www.nndc.bnl.gov/sigma/index.jsp ... .0&nsub=10

To calculate the mean free path of a 14-MeV neutron in iron (standing in as an example for a steel composition we don't really know), you take the inverse of the product of atom number density and the cross-section.

Atom density of iron (N): ~8.5E+22 / cc
Cross-section (sigma): 2.6E-24 cm^2

The mean free path is thus about 4.5 cm.

The unscattered flux is exponentially attenuated in a uniform material:

F(x) = F(0) * exp [-N * sigma * x]

The half-value thickness x_hvt for the unscattered flux corresponds to F(x_hvt)/F(0) = 1/2, so x_hvt = ln(1/2) / (-N * sigma) = ~3.1 cm. You can calculate the unscattered flux at any thickness, and obviously it doesn't ever fall to zero. So the 50-cm wall will scatter or absorb 99.999% of the incident neutrons and thus most of their kinetic energy. Some will get through unscattered, and quite a few will get out after one or more interactions in the steel. All this neglects the substantial volume of cooling water, which is highly effective at slowing and stopping neutrons, and the smaller amounts of other elements that make up a stainless steel alloy. But it gives you an idea.

-Carl
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Steven Sesselmann
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Re: New Scientist on ITER

Post by Steven Sesselmann » Sun Oct 18, 2009 8:07 am

Carl,

Thanks for the complete answer, it took me a while to work through it, but I get it now, and it will come in useful, it feels good when you learn something new

It shows that the neutron attenuation in stainless steel is not insignificant, and that anyone building a fusion devise in a thick steel block will have to evaluate the neutron absorption.

The cathode I was using in my STAR experiments had 0.5 cm thick walls, so the flux would have been affected by around...10%

ln(0.9) / ((-8.5e22) * 2.6e-24) = 0.476744415

Add to that a further 7 cm of dielectric oil, which we already calculated in an old post, and the absorption was most likely in the order of 20-30%

I see the possibility of machining a fusor chamber with thick walls and a thin beam port, so that the fast neutron flux mainly comes out in one beam.

Food for thought only, at this stage...

Steven
http://www.gammaspectacular.com - Gamma Spectrometry Systems
https://www.researchgate.net/profile/Steven_Sesselmann - Various papers and patents on RG

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Re: New Scientist on ITER

Post by Carl Willis » Sun Oct 18, 2009 9:54 am

Hi Steven,

Jon and I discussed the impact of the steel shell on an earlier thread, where I actually did a calculation in MCNP and I believe Jon made measurements with a BTI detector:

viewtopic.php?f=13&t=5803#p34247

Your star reactor differs from ITER in that yours is DD, ITER is DT. So your 2.5-MeV cross-section is different than what I Iooked up for the 14-MeV neutrons from DT. But you can find the cross-sections for any energies or materials of interest at that ENDF portal on NNDC.

-Carl
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Re: New Scientist on ITER

Post by Quantum » Sun Oct 18, 2009 10:18 am

Carl, I don't think your maths is correct here.

You can't just use the figure for Iron.

Stainless steel works because the gaps between the iron atoms are filled with other atoms.

Simplistically, it is a mix of body centred cubic and face centred cubic, this is why it's density is roughly 1.5 times that of Iron.(this is the reason it is used in these applications)

The density of nuclei is roughly 1.5 times that of Iron. I'll leave the maths to someone else.

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Re: New Scientist on ITER

Post by Carl Willis » Sun Oct 18, 2009 10:23 am

>Carl, I don't think your maths is correct here.

Your thinking has been noted.

>I'll leave the maths to someone else.

Hilarious.
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Re: New Scientist on ITER

Post by Quantum » Sun Oct 18, 2009 10:46 am

Ok, Carl, Your maths may be correct, but your figures are not.

It is your theory that is incorrect.

I humbly apologise for my slip.

You can't just use the figures for Iron here. The crystal lattice structure is entirely different in stainless steel. (this is why it it used in these applications)

Thankyou for 'picking me up' on that minor detail.

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Re: New Scientist on ITER

Post by Carl Willis » Sun Oct 18, 2009 11:02 am

>It is your theory that is incorrect.

Thanks for the honors, but I must humbly admit that it is not my theory.

>The crystal lattice structure is entirely different in stainless steel.

So, in your own theory, how does the crystal lattice structure figure into the neutron attenuation estimate? How do your results differ from the ones I showed?

-Carl
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Re: New Scientist on ITER

Post by Steven Sesselmann » Sun Oct 18, 2009 11:05 am

Carl,

Not sure if I am using that cross section calculator correctly...

is it right that the cross section in barns, can vary that much over a small range of energies ?
(see chart below)

I would have expected a smooth curve...

steven
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Re: New Scientist on ITER

Post by Carl Willis » Sun Oct 18, 2009 11:18 am

Hi Steven,

The structure comes from quantized energy levels in the nucleus. At lower energies the elastic scattering cross-section does indeed become smooth, the inelastic scattering cross-section goes to zero, and the radiative capture cross-section dominates (often with many resonances arising from the quantized energy levels in the nucleus).

You can also print out tables that make reading the data easier. You probably figured that function out.

-Carl
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