RE: deposit_fraction

From: Santana, Mario <>
Date: Thu, 21 Apr 2011 14:56:43 -0700


Your formula does not lead to the right dimension
([Energy^2/(Volume*time)], but I see what you mean.

Do the following:
1) Run FLUKA using USRBIN with what(2)=3DENERGY to get the TOTAL energy
deposited in your region. In your case I'd score per region (unless you
want to see the energy deposition pattern or peak), WHAT(1)=8.0
2) If you score per region and you haven't specified otherwise the
region volume is 1 cm^3, so your USRBIN result will actually be
expressed in GeV/primary, for example, it could be something like 1E-2
3) In order to obtain the fraction of your beam power deposited in that
given volume now just divide by the energy of your primary. For a
monoenergetic beam that would be what you wrote in WHAT(1) of the beam
card (if you used WHAT(1)<0). For example, if your beam energy was 5
GeV, just do the following: 1E-2 [GeV/primary] / 5 [GeV/primary] 2E-3
> 0.2 % is your deposited fraction


-----Original Message-----
From: [
] On Behalf Of Ertan Arikan
Sent: Thursday, April 21, 2011 4:53 AM
Subject: deposit_fraction

Dear fluka experts,

I want to calculate the beam power deposition=3D2E I know that the formula
of beam power deposition is:

Power Deposition [kW] =3D Eb [GeV] * Deposit-Fraction (GeV/cm^3) * 1.6022
[10^-19 (J/eV)] * Np * f [Hz]
    {Eb ->beam energy, Np -> number of particles}

question 1....
Is this formula correct?
if it is the wrong formula, what is the correct formula?

question 2....
if it is the correct formula, how can I obtain the Deposit-fraction value?

Thank you for your helps

Ertan Ar=FDkan
Received on Fri Apr 22 2011 - 09:43:33 CEST

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