New Scientist on ITER
Re: New Scientist on ITER
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.
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.
- Carl Willis
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Re: New Scientist on ITER
>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
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
Re: New Scientist on ITER
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.
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.
- Steven Sesselmann
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Re: New Scientist on ITER
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
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
http://www.gammaspectacular.com - Gamma Spectrometry Systems
https://www.researchgate.net/profile/Steven_Sesselmann - Various papers and patents on RG
https://www.researchgate.net/profile/Steven_Sesselmann - Various papers and patents on RG
- Carl Willis
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Re: New Scientist on ITER
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
Thanks for playing, Ash.
-Carl
Re: New Scientist on ITER
No, Carl, I'm saying you can't use the figure for Iron, and assume it's the same for 440 stainless.
Simple really.
Simple really.
- Carl Willis
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Re: New Scientist on ITER
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
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
Re: New Scientist on ITER
I believe 440 was mentioned earlier in this thread, however, 316 is usually used for nuclear applications.
- Chris Bradley
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Re: New Scientist on ITER
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
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
Re: New Scientist on ITER
Ferriic/martensitic sounds like 400 series to me. Certainly not 300 series (Austenitic).
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Re: New Scientist on ITER
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
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
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
The more complex the idea put forward by the poor amateur, the more likely it will never see embodiment
Fusion is the energy of the future....and it always will be
The more complex the idea put forward by the poor amateur, the more likely it will never see embodiment