Query regarding 'score' card and activation of fission phenomena only

Dear experts,

  1. In my simulation, I wanted to get the information regarding fission density. So, I used a SCORE card. But in the output, I got zero result, although I have resnucle output. I have attached here the output file. Can you please guide?

  2. Also, if I want to get activity of radionuclides produced due to fission only, (i.e., excluding the contribution coming from the decay of another fission product, then I have to disable/omit RADDECAY, DECYTIMES and DCYSCORE, right?

  3. What does NEU-BALA refer to? In the manual, it is written as net neutron production in each region. Is it same as scoring neutrons using USRBIN? Can we get information regarding how many fission neutrons are produced?

Mo_01001.out (3.1 MB)
neutron_spectrum.txt (3.6 MB)

Mo.inp (2.3 KB)

Regards,
Riya

1 Like

Dear @riya,

  1. Noticing that you are using the pointwise treatment of neutrons, please not that unfortunately the scoring for fission processes are not yet fully implemented, and thus, you would be required to customize a user routine (e.g. mgdraw) in order to flag the fission neutrons or use other work-arounds. Would you be interested in just counting them or also to score more information?

As a temporary and simpler solution, you could revert to the group wise treatment of neutrons in FLUKA and user the SCORE card as you already do.

  1. Indeed, if you do not wish to have radioactive decays, you should not use the RADDECAY, DECYTIMES and DCYSCORE cards.

  2. The NEW-BALA according to the user manual is the Neutron balance (algebraic sum of outgoing neutrons minus incoming neutrons for all interactions). As such, it is not the same as scoring neutrons using USRBIN since the USRBIN would score all neutrons, and NEW-BALA would score only the difference (either positive or negative) between outgoing/incomig neutrons. Note that neutrons could be produced not just from fission phenomena (neutron hitting a larger target), but also due to other inelastic interactions.

Best,
Daniel

2 Likes

Thank you @dprelipc,

Using the group wise treatment, I am able to get output in SCORE.
For e.g. considering 0.1 eV neutron and pure U235 target, the SCORE output gives 2.07 fissions/primary neutron.

Is it possible to get number of neutrons/fission as well from SCORE card? In the .out file, it is not showing any value.
1

Or for this mgdraw is required? If so, can you please mention the variable names for fission neutron?

Regards,
Riya

Dear @riya,

In the output file, you get the following information as well:

1Region # name     volume         FISSIONS Star Density   HE-FISS  Star Density   LE-FISS  Star Density            Star Density
                in cubic cm       Stars/cm**3             Stars/cm**3             Stars/cm**3             Stars/cm**3
                                  /one beam particle      /one beam particle      /one beam particle      /one beam particle

      1 BLKBODY  1.000000000D+00         0.000000000D+00         0.000000000D+00         0.000000000D+00         0.000000000D+00
      2 VOID     1.000000000D+00         0.000000000D+00         0.000000000D+00         0.000000000D+00         0.000000000D+00
      3 VOID1    1.000000000D+00         0.000000000D+00         0.000000000D+00         0.000000000D+00         0.000000000D+00
      4 TARGET   1.000000000D+00         8.700000000D-01         0.000000000D+00         8.700000000D-01         0.000000000D+00

 Total (integrated over volume):         8.700000000D-01         0.000000000D+00         8.700000000D-01         0.000000000D+00

The above has been obtained for 0.1 eV neutron and pure U235 target. Isn’t this what you are looking for already?

Best,
Daniel

1 Like

Dear @dprelipc ,

Is fission star density per beam particle same as the number of neutrons per fission ? My intention was to get this value of nu and total number of fissions:

For a very large number of fissions of U-235 by thermal neutrons it is known that 2.7% of the fissions give no neutrons, 15.8% give one neutron, 33.9% give two, and so on. The average number of neutrons released per fission is denoted by v and has a value of 2.43 for thermal fission in U-235.

Regards,
Riya

Dear @riya,

The star density denotes the number of fission events per primary. If you are interested in the number of neutrons per fissions, you can find in the output file a table as follows:

 Number of secondaries created by low energy neutron per beam particle:
 Prompt radiation      Radioactive decays
   6.7700E+00 (100.%)   0.0000E+00 (100.%)
   2.5000E-01 ( 3.7%)   0.0000E+00 ( 0.0%) elastic NEUTRON
   2.2100E+00 (32.6%)   0.0000E+00 ( 0.0%) inelastic NEUTRON
   0.0000E+00 ( 0.0%)   0.0000E+00 ( 0.0%) fission NEUTRON
   0.0000E+00 ( 0.0%)   0.0000E+00 ( 0.0%) PROTON
   6.5200E+00 (96.3%)   0.0000E+00 ( 0.0%) PHOTON
   0.0000E+00 ( 0.0%)   0.0000E+00 ( 0.0%) DEUTERON
   0.0000E+00 ( 0.0%)   0.0000E+00 ( 0.0%) TRITON
   0.0000E+00 ( 0.0%)   0.0000E+00 ( 0.0%) 3-HELIUM
   0.0000E+00 ( 0.0%)   0.0000E+00 ( 0.0%) 4-HELIUM

where for inelastic neutron you get \nu \approx 2.21 (for a simulation of 10 000 primaries). Would you be interested to obtain the number of neutrons for each fission event?

Best,
Daniel

1 Like

Thank you @dprelipc for the explanation. I expected that number of fission neutrons will be printed on ‘fission Neutron’ , which is printed as zero here. But as you mentioned nu is printed in inelastic neutron part. Also, as you asked, I think number of neutrons in each fission event can be obtained using = no of inelastic neutrons printed in .out / number of fission obtained from SCORE card. Is it right ?

Regards,

Riya

Dear @riya,

You may have channels other than fission producing secondary neutrons that would also be counted, so your procedure may not be 100% correct.

My question is whether you be interested to obtain the number of neutrons for each fission event, which is not possible with existing/default scorings and a special user routine would be required.

Best,
Daniel

Thank you @dprelipc for the clarification. Yes i would like to get the value of number of neutrons per fission. Do you suggest to use mgdraw?

regards,
Riya

Dear @riya,

First of all, let me apologize for the delayed answer. After looking carefully at this use case, it is unfortunately not possible to retrieve the number of neutrons per each fission event, but let me add more information.

a. Currently, there is no flag in FLUKA that identifies fission events to the user, but one can work around this by looking at the number of heavy products from an interaction. We know that for a fission event, there should always be NPHEAVY=2 products in a fission event, therefore one could indeed use an mgdraw.f (11.4 KB) routine for this, and you can see the results plotted here.

Screenshot from 2023-04-18 10-13-05

FLUKA outputs only 2 or 3 neutrons (respecting the ~2.43 mean for 235Ur), instead of a Poisson distribution with events also at 0, 1 or 4+ neutrons per fission.

Inside FLUKA there are several approaches for sampling the fission neutrons.

  1. The default is the binomial sampling between 2-3 such a way to reflect the database neutron multiplicity.
  2. There is Poissonian way of sampling with an average of \nu.
  3. Applying Bayesian Distortion to the fission yield distribution such as to the distribution of the fission fragments to reflect the library.

Since all distributions in the neutronic libraries are inclusive there is no easy way to reproduce all distributions when we are sampling the final states. For the fission reaction it was chosen the 1st approach, since it was the one that was minimizing the distortion of the other distributions. The Poisson sampling was altering the second fission fragment distribution, so we do not offer it to the user, hence the behaviour above.

b. Also, please note that there is a table printed in the output file mentioned above. In your reply, you were using the neutron group-wise treatment, as initially suggested. However, if you use the point-wise neutron treatment, then you get more sensible results, e.g. as:

Number of secondaries created by low energy neutron per beam particle:
 Prompt radiation      Radioactive decays
   3.3584E+00 (100.%)   0.0000E+00 (100.%)
   1.2870E-01 ( 3.8%)   0.0000E+00 ( 0.0%) elastic NEUTRON
   3.3100E-02 ( 1.0%)   0.0000E+00 ( 0.0%) inelastic NEUTRON
   2.0891E+00 (62.2%)   0.0000E+00 ( 0.0%) fission NEUTRON
   0.0000E+00 ( 0.0%)   0.0000E+00 ( 0.0%) PROTON
   1.1406E+00 (34.0%)   0.0000E+00 ( 0.0%) PHOTON
   0.0000E+00 ( 0.0%)   0.0000E+00 ( 0.0%) DEUTERON
   0.0000E+00 ( 0.0%)   0.0000E+00 ( 0.0%) TRITON
   0.0000E+00 ( 0.0%)   0.0000E+00 ( 0.0%) 3-HELIUM
   0.0000E+00 ( 0.0%)   0.0000E+00 ( 0.0%) 4-HELIUM

Unfortunately, these are not fully correct, since they are per BEAM particle, and not per FISSION event. One could have more fission events per beam (consider the case of a secondary neutron generating a fission event and so on in a chain reaction), or beam neutrons not interacting.

Please let me know if there is something else I could help with!

Thank you,
Daniel

1 Like

Thank you @dprelipc for the suggestion and useful discussion.

However, I have not understood fully how you have plotted the figure. The output of mgdraw will give total number of PRODNEU, i.e. whenever there is a heavy ion production, it will add the counts of neutron in PRODNEU variable. Hence, it will be helpful if you can explain which variables from mgdraw you have plotted as x and y in the figure.

Regards,
Riya

Dear @riya,

In the plot I attached above, the blue histogram from FLUKA is the output of the mgdraw, filtered on NPHEAVY=2, while the orange one is the “theoretical one”:

“In each fission event the number of neutrons emitted must be an integer. For a very large number of fissions of U-235 by thermal neutrons it is known that 2.7% of the fissions give no neutrons, 15.8% give one neutron, 33.9% give two, and so on. The average number of neutrons released per fission is denoted by v and has a value of 2.43 for thermal fission in U-235.” From: Fission Event - an overview | ScienceDirect Topics

Best,
Daniel

Thank you @dprelipc for the clarification.

Regards,
Riya