From: Alberto Fasso' (fasso@SLAC.Stanford.EDU)
Date: Mon Feb 25 2008 - 03:18:03 CET
To: fluka-discuss@fluka.org
cc: Buthaina Abdalla Suleiman Adam <buthaina@aims.ac.za>
Subject: Re: Flux
In-Reply-To: <200802242108.m1OL8Y6q029377@smtp1.mi.infn.it>
Message-ID: <Pine.LNX.4.64.0802241645390.21318@noric07.slac.stanford.edu>
References: <200802242108.m1OL8Y6q029377@smtp1.mi.infn.it>
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Dear Buthaina,
you have an incident beam of protons with a momentum of 0.2 GeV,
which means about 21 MeV kinetic energy. Protons of 21 MeV energy
have a very short range: about 2.8 mm in beryllium.
And your energy cutoff for protons is the default, equal to 10 MeV.
So, as soon as each proton has lost half of its energy, it is ranged
out without any nuclear interaction. At 10 MeV, the residual range
is about 0.8 mm, so each of your protons has only a short distance
available (2.8 - 0.8 = 2 mm) to have a nuclear interaction.
The proton nuclear cross section between 10 and 21 MeV is anyway
very small. And I must add that this energy range is
at the limit of the FLUKA preequilibrium model, so any result must
be taken with some caution.
(If your intention was to simulate 200 MeV protons, then you should
have input -0.2 and not 0.2).
If you look at the end of your .out output, you will see that:
- on average, 99% of the proton energy is dissipated as dE/dx
- there are only very few inelastic interactions (stars): about 1600
for 1.E5 protons
- only a very few neutrons are produced: about 1100 for 1.E5 protons
I would like to add another comment about your input: from a test I
did, it seems to work, but it would be better to avoid building
contiguous regions by means of bodies having a common surface (in
your case, regions regBE4 and regST5 have a common face at z=1.96
and regST5 and regFe6 at z=201.96).
You can obtain that by making some of the bodies a little longer
and cutting them out. Instead of:
............
RPP body3 -1.0 +1.0 -1.0 +1.0 0.0 +1.96
RCC body5 0.0 0.0 +1.96 0.0 0.0 200.0
1.5
RPP body6 -2.375.0 +2.375 -2.375 +2.375 +201.96 +401.96
.............
* Be target 2nd half
regBE4 5 +body3 -body4
* beam tube
regST5 5 +body5
* Iron Collimator
regFe6 5 +body6
you could have
............ make body3 0.04 cm longer, and body5 1 cm longer
RPP body3 -1.0 +1.0 -1.0 +1.0 0.0 +2.00
RCC body5 0.0 0.0 +1.96 0.0 0.0 201.0
1.5
RPP body6 -2.375.0 +2.375 -2.375 +2.375 +201.96 +401.96
...................
* Be target 2nd half -- subtract body5 to make it end at z=1.96
regBE4 5 +body3 -body4 -body5
* beam tube -- subtract body6 to make it end at z=201.96
regST5 5 +body5 -body6
* Iron Collimator
regFe6 5 +body6
This latter definition is more likely to work without problems.
Best regards,
Alberto
On Sun, 24 Feb 2008, Buthaina Abdalla Suleiman Adam wrote:
> Hi
> I want to get the flux of neutron at 400cm away from the target outside
> Steel tube 200cm (regST5) and iron tube200cm (regFe6). My problem is the
> flux value is zero, even when I made the iron tube shorter than 200cm
> (10cm, 100cm). Also I made regFe6 just as vacuum to see if I can get any
> flux but still the same. I don't is this problem due to the number of
> regions (there is a flux outside regST5)? Please any help, I attached the
> input file
>
> Thank You in Advance,
> Buthaina Adam
>
-- Alberto Fasso` SLAC-RP, MS 48, 2575 Sand Hill Road, Menlo Park CA 94025 Phone: (1 650) 926 4762 Fax: (1 650) 926 3569 fasso@slac.stanford.edu
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