Thank you very much!
In this case, the entire radiator is exposed to the scintillator, there is no collimation, which should create a 1/r effect, compared to a 1/r^2 effect, naively, so I think I need to simulate the entire cylinder as being an activated source.
Will this still accomplish that?
> On Aug 11, 2016, at 11:56 AM, Joachim Vollaire <joachim.vollaire_at_cern.ch> wrote:
>
> Hi James
>
> Looking at your input a few things.
>
> The geometry errors:
>
> You define your blakchole external and internal surface with the same sphere radius (use a big number for the blakchole)
>
> SPH blkbody 0.0 0.0 0.0 50.0
> * Air sphere
> SPH air 0.0 0.0 0.0 50.0
>
> To define the blackbody replace :
>
> BLKBODY 5 +blkbody -(air+Plastic)
> By (“Plastic” is already contained inside air…)
> BLKBODY 5 +blkbody -air
>
> The error problem should be solved with the above.
>
> For scoring, you can use a region based USRBIN Scoring for DOSE (GeV/g) in the region describing your scintillator.
>
> For accuracy of the results, did you consider simulating the collimation system of the irradiator and a point source and isotropic source inside ? I have no experience with irradiator but I am not sure how you can derive from the source activity the photon fluence out of the collimator. I guess that using only solid angle does not work due to scattering effect ?
>
> For the source description, note that instead of using mono-energetic 667 keV photon you can use an isotope as the source and benefit from the decay data in FLUKA (of interest for more complicated cases and to get accurate branching ratio which for example is not 100 % for decay of Cs-137 in Ba-137m emitting the 661.6 keV gamma…).
> To do so look at the course material related to definition of sources slide 21 in the following presentation (don’t forget to activate the decay of the particle by calling the RADDECAY card:
>
> https://indico.cern.ch/event/442634/contributions/1096523/attachments/1186817/1721027/18_Activation_2015.pdf <https://indico.cern.ch/event/442634/contributions/1096523/attachments/1186817/1721027/18_Activation_2015.pdf>
>
> For normalization, the results (deposited energy) will be by isotope decay so you just need to multiply the results by the source activity ( in Bq)
>
> Cheers
> Joachim
>
>
>
>
> From: James Wetzel [mailto:jwwetzel_at_icloud.com <mailto:jwwetzel_at_icloud.com>]
> Sent: 11 August 2016 17:26
> To: Joachim Vollaire <joachim.vollaire_at_cern.ch <mailto:joachim.vollaire_at_cern.ch>>
> Cc: Giuseppe Battistoni <giuseppe.battistoni_at_mi.infn.it <mailto:giuseppe.battistoni_at_mi.infn.it>>; Paola Sala <Paola.Sala_at_cern.ch <mailto:Paola.Sala_at_cern.ch>>
> Subject: Re: FLUKA
>
> Thank you very much!
>
> Attached is my .inp file I am working with.
>
> I have a cylindrical gamma radiator of cesium 137, with dimensions Diameter: 2.54 cm and Length: 30.48 cm, and 6 Ci activity.
>
> I have a sample of plastic scintillator with dimensions 10cm x 10cm x 0.1cm, which is placed a distance 5 cm from the radiator as in the picture below, where the 10x10cm scintillator plane is parallel with the cylinder, and the center of the plastic is aligned with the center of the cylinder, as below:
>
> Radiator: ===============
> ▲
> 5cm
> ▼
> Plastic Scint: ———
>
>
> What I need at the end of the day is the total dose absorbed by the plastic scintillator, in Gy or Mrad.
>
> I am getting Geometry errors (I think), and I’m not sure how to get the dose out properly with scoring.
>
> Thank you very much for your help!
>
>
> <CS173.inp>
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Received on Thu Aug 11 2016 - 21:59:20 CEST