NOrmalizing Tally to get Flux value

[King]

Hye everybody,

i have performed a calculation for IAEA benchmark reactor and i need to get flux in various direction like axial profile and radial profiles. The tally i used in tally.xml is shown below.

 <mesh id="2" type="rectangular">
    <dimension>1 1 30</dimension>
    <lower_left>-0.5 -0.5 -35</lower_left>
    <upper_right>0.5 0.5 35</upper_right>
  </mesh>

  <tally id="2">
    <filter type="mesh" bins="2"/>
    <filter type="energy" bins="0. 0.625e-6 5.531e-3 20.0"/>
    <scores>flux</scores>
  </tally>

for example if i get a value of 0.225861832063,0.036718304271,0.053172569348 for thermal, epitehrmal and fast flux for above mesh setting. Can anyone guide me how to normalize these values to get flux values? Thanks to everyone and sorry if this answer to this question is too obvious. 

[Adam Nelson]

The tally results are integrals over the entire phase space - that includes the volume of the mesh. So if you want a value of flux in [n/(cm^2-s)] then you will need to divide by the volume of each mesh cell. Is that what you were looking for?
Adam

[King]

More or less. In MCNP we multiply F4 tally with power and nu and divide them by 1.6022E-13 x 193 x keff. This gives neutron per cm2 per sec. But here how do we get units of flux balanced

[Adam Nelson]

Maybe I'm not understanding correctly what you want, but to my knowledge the OpenMC flux score outputs the same units as the MCNP F4 tally. So if what you do for MCNP suffices, then it should here as well.

Is my understanding correct: you want to take the OpenMC flux score and take it from the normalized values that OpenMC outputs in to the mesh element averaged values of the flux that should be expected when operating at power P, right?

[King]

thank you nelson for answering. Maybe i am confused by your answer in another post on the forum. there you wrote and i quote 

"Hey Anthony,
The tally responses are integrated over volume, and so are [# cm^3]....." 

so if the units of tally are #-cm^3 then multiplying by power and nu and dividing them by 1.6022E-13x 193 x keff x volume of mesh, it would give #/sec i guess. so my question is do we have to divide the calculation done above by area or not to get the required units of (#/cm^2-sec)?

[Paul Romano]

The tally is not normalized by volume, so the units are something like [neutrons-cm/source]. This can be understood quite easily: in a collision estimator tally, OpenMC accumulates 1/Sigma_t which has units of cm, and similarly for a tracklength tally, OpenMC accumulates the tracklength in cm. When you divide by volume, you get [neutrons/cm^2-source]. The normalization factor P*nu/(Q*k) gives you [J/sec*neutrons/fission/(J/fission*neutrons/source)] = [source/sec] so that when you multiply the flux by the normalization factor, you get [neutrons/cm^2-sec].

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[King]

thank you very much Paul :-)

[yangch...@gmail.com]

Thank you,Paul. But how can I  get P with a  unit of (J/sec) ?  I also notice that we can get a unit of fission/sec by P/Q, is this what we get by fission tally ?

在 2014年8月27日星期三 UTC+8上午1:49:12，Paul Romano写道：

[Paul Romano]

P is the reactor power and is generally known a priori. For example, if you have a 1000 MWe reactor with a thermal efficiency of 33%, the thermal power is 3000 MW = 3e9 J/s. P/Q would tell you how many fission reactions per second are necessary to sustain the given power. Once you account for nu (neutrons/fission), then you have the number of neutrons needed to sustain a given power level. All tallies in OpenMC are 'per source neutron', so you need to determine how many neutrons/sec there are in your system of interest using a combination of these factors.

[Javier Gonzalez]

Hi Paul,

I am interested in expressing the flux in “neutrons/cm²s”. I was reading this post where it is explained how to do it. I would like to know if I understood the normalization factor that you presented [P*nu/(Q*k)]:

  P – reactor power [J/s]

 Q = 200 MeV for U-235 or 3.2x10^-11 [J/fission]

nu – score ‘nu-fission’ [neutrons/fission]

  k – eigenvalue [neutrons/source]

Is my interpretation okay?

Thanks,

Javier

On Tuesday, August 26, 2014 at 1:49:12 PM UTC-4, Paul Romano wrote:

The tally is not normalized by volume, so the units are something like [neutrons-cm/source]. This can be understood quite easily: in a collision estimator tally, OpenMC accumulates 1/Sigma_t which has units of cm, and similarly for a tracklength tally, OpenMC accumulates the tracklength in cm. When you divide by volume, you get [neutrons/cm^2-source]. The normalization factor P*nu/(Q*k) gives you [J/sec*neutrons/fission/(J/fission*neutrons/source)] = [source/sec] so that when you multiply the flux by the normalization factor, you get [neutrons/cm^2-sec].

On Tue, Aug 26, 2014 at 1:23 PM, King <khurru...@gmail.com> wrote:

thank you nelson for answering. Maybe i am confused by your answer in another post on the forum. there you wrote and i quote 

"Hey Anthony,
The tally responses are integrated over volume, and so are [# cm^3]....." 

so if the units of tally are #-cm^3 then multiplying by power and nu and dividing them by 1.6022E-13x 193 x keff x volume of mesh, it would give #/sec i guess. so my question is do we have to divide the calculation done above by area or not to get the required units of (#/cm^2-sec)?

On Tuesday, 26 August 2014 17:47:07 UTC+5, Adam Nelson wrote:

Maybe I'm not understanding correctly what you want, but to my knowledge the OpenMC flux score outputs the same units as the MCNP F4 tally. So if what you do for MCNP suffices, then it should here as well.

Is my understanding correct: you want to take the OpenMC flux score and take it from the normalized values that OpenMC outputs in to the mesh element averaged values of the flux that should be expected when operating at power P, right?

On Monday, August 25, 2014 11:58:06 AM UTC-4, King wrote:

More or less. In MCNP we multiply F4 tally with power and nu and divide them by 1.6022E-13 x 193 x keff. This gives neutron per cm2 per sec. But here how do we get units of flux balanced

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[Paul Romano]

All that is correct except for nu. While the units are indeed [neutrons/fission], the 'nu-fission' score doesn't quite give you what you need. 'nu-fission' gives you the total fission neutron production, so you need to divide by the fission reaction rate in order to get the average number of neutrons per fission.

Best,

Paul

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[Javier Gonzalez]

Thanks Paul.

Javier

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[Javier Gonzalez]

Paul, 

Continuing with this post, a new question comes to my mind. 

If I am modeling one fuel element, in the normalization factor, is "P" the total reactor power or the total reactor power divided by the number of fuel elements?

Thanks,

Javier 

On Friday, December 6, 2019 at 8:17:37 AM UTC-5, Paul Romano wrote:

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[Paul Romano]

If modeling a single fuel element, then you would use the power of just one element as opposed to the power for the whole core.

Best,

Paul

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