# Re: [fluka-discuss]: USR Track in FLUKA

From: Mikhail Polkovnikov <pmk_at_ihep.ru>
Date: Fri, 2 Sep 2016 15:45:43 +0300

On 01.09.2016 22:37, Trinh Ngoc-Duy wrote:
> Dear Mikhail,
>
> Thank you so much for your answer. I know also that the ratio should be 1.0. But it's not clear for me how can i obtain this value from the value of 0.61 that Usrtrack give me ? I do not know if it is possible to calculate the production rate in thick target using this technique or should i just put some detector around the target from 0 to 360 deg.
>
> Thank you.
> Sincerely.
> ND Trinh.
>
> ________________________________________
> De : Mikhail Polkovnikov <Михаил Полковников> [pmk_at_ihep.ru]
> Envoyé : jeudi 1 septembre 2016 18:03
> À : Trinh Ngoc-Duy; fluka-discuss_at_fluka.org
> Objet : Re: [fluka-discuss]: USR Track in FLUKA
>
> On 01.09.2016 13:12, Mikhail Polkovnikov <Михаил Полковников> wrote:
> On 31.08.2016 17:14, Trinh Ngoc-Duy wrote:
> Dear Fluka users,
>
> I'm using FLUKA to calculate the production rate of particle by interaction of a primary beam with a thick target. I'm using USR Track to obtain the spectre of particle produced in the target. I see that USR TRACK give results with units of particle /(cm2/GeV/primary beam).
>
> So if I send a beam into a target. The reaction is happen inside the target, and i score the spectre of produced proton for example. I obtain the total flux of proton :
>
> 8 proton /(cm2/GeV/primary beam). with a energy bin of [0:0.1] GeV.
>
> The target is a box with each side of 1 cm. So the area of a face is : 1*1=1cm2.
>
> To obtain the total production rate of proton, i multiple 8*0.1*1=0.8 (proton per primary beam).
>
> I'm not sure about this value because when I simulate a isotropic beam inside a target, and score the total flux with the same method. Normally i should obtain the value of 1 because my target cover all the beam. But FLUKA gives me 6.1 (proton/cm2/GeV/primary beam) = 6.1*1*0.1=0.61 (proton per primary beam) (with a energy bin of [0:0.1] GeV).
>
> I put my input and my results in attachment. Please give me an idea.
>
> Thank you very much.
>
> ND Trinh, GANIL, France.
>
>
>
>
> ________________________________
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>
> Dear ND Trinh,
>
>
> The track length of each proton produced inside of the thick target is different! Even in you example with isotropic source, the track length within the target varies from minimum (half of the edge) to maximum (half of the diagonal). As it shown in *sum.lis file:
>
>
> Tot. response (p/cmq/pr) 0.6107748 +/- 2.8535640E-02 %
> ( --> Track l. (cm/pr) 0.6107748 +/- 2.8535640E-02 % )
>
>
> Best regards,
>
> Mikhail
>
> Dear ND Trinh,
>
>
> Forgot to mention!
>
> In your example, the ratio (Total fluence / track-length) is exactly 1.0 with units [number of particles per primary per cm^3].
>
>
> Best regards,
>
> Mikhail
>
>
> ________________________________
> Préservons notre environnement, n’imprimez ce mail que si nécessaire.
> Preserve our environment, print this email only if necessary.

Dear ND Trinh,

If it is acceptable, you can estimate number of particles from TARGET to
VOID by means of USRYIELD card:

*...+....1....+....2....+....3....+....4....+....5....+....6....+....7....+....
USRYIELD PROTON 21. TARGET VOID 1.yield
USRYIELD 0.1 0.0 1. 0.1 0.0 3. &

Integration over IE and IA gives you a number of particles per primary.

According to the scoring lecture material
<https://indico.cern.ch/event/442634/contributions/1096540/attachments/1184625/1716758/07_Scoring_2015.pdf>
(at the end of third slide), to calculate reaction rate inside the
region, you have to know: cross section, fluence, region volume.

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Received on Fri Sep 02 2016 - 16:36:35 CEST

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