Fusor Forums

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

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

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

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

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.

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

Your thinking has been noted.

>I'll leave the maths to someone else.

Hilarious.

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.

>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

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

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

The figures I qouted may not be entirely correct regarding density.

It's not MY theory, Carl.

I'm just saying the reason stainless steel is used for these applications is due to it's properies, which are due to it's structure, and theredore the figures you quote for Iron are irrelevant in this instance.

I'm not a mathematician, I'm just saying your theory is incorrect in this instance

You are effectively comparing chalk with cheese. Both contain calcium, but both have different properties.

>The figures I qouted may not be entirely correct regarding density.

Don't worry, that's understood...

>I'm just saying the reason stainless steel is used for these applications is due to it's properies

Indeed. Because of the properties.

>I'm not a mathematician, I'm just saying your theory is incorrect in this instance

I heard you the first time (and made the point that it wasn't my theory). But you must have an alternative that you feel IS correct?

>You are effectively comparing chalk with cheese. Both contain calcium

Doing this same calculation for stainless steel is not conceptually any more difficult than with iron, and as far as my education tells me, makes use of exactly the same "theory." If you know the alloy in use at ITER and can do the work of looking up the 14-MeV neutron cross-sections for the constituents, we will compare "cheese to cheese." Only if you want to, though.

-Carl

I neither claim to know the figures for 440 stainless steel, or where to look them up.

I'm just saying you can't just assume the figures for Iron can be substituted.

And my knowledge relates to (specifically) 316 stainless. But it is reasonable to assume that 440 stainless will also differ from Iron.

Thanks again Carl,

Yes I did figure that out, I was just puzzled about the variation in the crosss section, which you have explained.

For DD fusion neutrons 2.45 mev, the cross section appears to be around 3.3 barns.

So around 86% of the neuts would make it through 53 mm of Fe56

ln(0.86) / ((-8.5e22) * 3.3e-24) = 0.53

So back to my relatively thick walled STAR cathode (5.0 mm 306 SS), the absorbtion of the cathode wall plus the dielectric oil could have been quite a bit more.

I think I need to go back and revisit this experiment, there are a number of things I can improve next time.

Thanks for teaching me the method, ...

Steven

So to summarize, you think the theory I use is wrong. You can't explain how or why it's wrong, you don't offer an alternative, you aren't a "mathematician," you don't know the figures or where to look them up, you do have some ideas about what it's "reasonable to assume." Also, cheese and chalk.

Thanks for playing, Ash.

-Carl

No, Carl, I'm saying you can't use the figure for Iron, and assume it's the same for 440 stainless.

Simple really.

Whoa!

Did I use a figure for iron and assume it was the same for some kind of stainless? Or is that a manufactured controversy--a wee bit of trolling? To answer, we need to read my posts. (That CAN be a herculean task, especially if one has the eyesight and / or reading comprehension of a barnacle.)

>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)

Hmmm....standing in as an example?

And 440 stainless? Wherever did you come to the conclusion that this is the actual alloy in use?

-Carl

I believe 440 was mentioned earlier in this thread, however, 316 is usually used for nuclear applications.

FYI (and FWIW), just looking at older publications on ITER, all they say is "a ferritic/martensitic steel", implying a custom alloy "tbd", by my reading of it.

Oak Ridge National Laboratory (under the U.S. ITER Project) has been developing their own casting steel for the specific purpose for ITER and they have a candidate alloy under test at the moment, as far as I can gather.

maybe of interest:

http://www.ornl.gov/~webworks/cppr/y2001/rpt/121054.pdf

Reading between the lines on other documents, I suspect they're looking at a low chromium (9Cr-1Mo) martensitic steel and/or nanostructured steel like 12YWT 14YWT, MA957 or 14WT with dispersed Y2O3, Y2Ti2O7 and/or YTiO3 strengthening particles. Some combination thereof, I presume...

edit: just came across this whilst looking under the DoE contract #DE-AC05-00OR22725 that may spill a few more beans on the subject;

http://www.ms.ornl.gov/programs/fusionm ... 012-24.pdf

Ferriic/martensitic sounds like 400 series to me. Certainly not 300 series (Austenitic).

nice post

I never chimed in on this posting. This was a better post to learn from than to comment on. Thanks Carl.

Regardless of grain structure or constituents iron is the overwhelming content of any SS and thus iron is the best material to work on in base level, simple, basic calcs.

Richard Hull