It is concerning FISPACT and water activation. 16N and 17N are determined to be the most important activation products in the coolant water. In the attachment you see input file and spectrum, which is employed. Water isotopic composition is in FUEL card I get the results without N (the same using MASS). What I am doing wrong? For information, I'm using FISPACT-2010 with EAF-2010. As far as I know 16N and 17N should be produced from reactions 16O(n,p)16N and 17O(n,p)17N reactions. Another path is by 16O(n,γ)17O17(n,p)17N. But non of them actually appears in FISPACT-2010 run.
Thanks for your support!
Regards,
Gediminas
water activation problem using FISPACT-2010
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water activation problem using FISPACT-2010
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Re: water activation problem using FISPACT-2010
Dear Gediminas,
Thank you for contacting us through the forum.
Legacy FISPACT-2010 (and previous) software is no longer supported by the UKAEA. We support, maintain and develop FISPACT-II, which is available through the OECD-NEA Data Bank, RSICC or directly through the UKAEA.
I've run your simulation using FISPACT-II and the TENDl-2015 neutron data after converting to 709 group with a 10 meV cutoff. The simulation shows the reaction channels that you are looking for:
Note that reactions above 30 MeV are treated through a generic production channel (mt=5) rather than individual channels, so these appear as (n,O) in the pathways analyses.
I hope this helps - the full simulations are attached below.
Best regards,
Michael
Thank you for contacting us through the forum.
Legacy FISPACT-2010 (and previous) software is no longer supported by the UKAEA. We support, maintain and develop FISPACT-II, which is available through the OECD-NEA Data Bank, RSICC or directly through the UKAEA.
I've run your simulation using FISPACT-II and the TENDl-2015 neutron data after converting to 709 group with a 10 meV cutoff. The simulation shows the reaction channels that you are looking for:
Code: Select all
Target nuclide N 17 99.999% of inventory given by 1 path
--------------------
path 1 99.999% O 17 ---(R)--- N 17 ---(S)---
99.98%(n,p)
0.02%(n,O)
Target nuclide N 16 99.972% of inventory given by 1 path
--------------------
path 1 99.972% O 16 ---(R)--- N 16 ---(S)---
97.19%(n,p)
2.81%(n,O)
I hope this helps - the full simulations are attached below.
Best regards,
Michael
Michael Fleming, PhD
D3/1.56 Culham Science Centre
UK Atomic Energy Authority
Abingdon, OX14 3DB
+44 (0) 1235 466884
michael.fleming@ukaea.uk
D3/1.56 Culham Science Centre
UK Atomic Energy Authority
Abingdon, OX14 3DB
+44 (0) 1235 466884
michael.fleming@ukaea.uk
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