I’ve tried to calculate the material arrangement of three concentric circles on a two-dimensional plane, which are fuel, monolith and reflector respectively from inside to outside. The boundary condition is the white boundary. And I use 50,0000 particles for 300 batches (100 inactive batches).
However, the k-eff shown in RESULTS is 1.86732 +/- 0.00009, and the k-eff calculated from fiss_rate / abs_rate (since the leak is zero) is 1.720066 +/- 0.000238.
I don’t know how this happens. Please excuse my lack of knowledge on OpenMC as I am practically new to it and still on a learning curve.
@Mio welcome to the community! k-effective is defined as production / loss, and the fission rate alone doesn’t tell you the production; for that you need the “nu-fission” tally score, which accounts for the number of neutrons produced per fission (nu). Once you account for that, you should get a number closer to what is reported for k-effective.
Hi, Paul. I still can’t know what’s going wrong in my code. Could you please help me?
Since I’m a new user and I can’t upload my .py file, I’ll paste all the codes below.
If you look at the result for the absorption score, you’ll see that it is about 1.08. In a problem without leakage, you normally expect to see a value very close to 1 because the units are “per source neutron” and without leakage, each neutron that is born eventually has to be absorbed somewhere. In your problem, you actually have an appreciable amount of (n,2n) reactions occurring, which increases the population. If you add the (n,2n) score to one of your tallies, you’ll see that it is about ~0.08. So, to recreate the k-effective value from these tallies, you can do nu-fission/(absorption - (n,2n)).
There is actually no universal agreement on how to define k-effective when (n,xn) reactions are present. See this 2009 report by Red Cullen that explains how two codes (MCNP and TART) handle it differently. The definition that is used in OpenMC is consistent with MCNP and Serpent and several other codes.
Hi, many thanks for your reply. Your answer has perfectly solved my problem!
But here comes another question. If I want to use OpenMC to generate cross-sections, how to define settings in the tally regarding to this (n,2n) reactions.
Maybe we should make this clearer: for a problem listed above, we have been trying to use OpenMC to generate multi-group cross sections for deterministic calculations. However, the multi-group cross sections seemed to incur significant error when we compare deterministic results with openMC results. Using the generated multi-group cross-sections and multi-group neutron flux do not reproduce exactly the OpenMC k-effective. As you mentioned, the discrepancy was caused by our omission of (n,xn) reaction rate. In this case, how do we take into account the contribution of (n,xn) reaction when generating multi-group cross-sections?
We tried activating
‘nu=True’ in mgxs.ScatterXS(), and
‘nu=True’ in mgxs.TransportXS()
but it doesn’t help improving the deterministic results. What else should we do to fix this problem?
I’m probably not the best person to address questions about downstream use for deterministic codes (perhaps @mkreher13 or @agnelson can chime in) but I think to get a matching answer, you would need to replace your absorption cross section with a “removal” cross section that accounts for the extra production of neutrons from (n,2n), i.e., removal = absorption - (n,2n). If you want to get an (n,2n) MGXS, you can do so with the ArbitraryXS class.
Thank you very much for your timely response. Given your hints, I believe we should try removing the contribution of (n,2n) in the total XS.
By the way, I feel that this could be a underlying trouble for researchers who use OpenMC to generate multi-group XSs. The special treatment of (n,xn) is automatically considered in deterministic lattice codes like Helios and others. And we just found out that this problem was also noticed by SERPENT users:
Yes, that’s correct, but I believe we also need to account for it in absorption as well in order for the total cross section to be reproduced accordingly. That is, (n,xn) production is added to scattering and subtracted from absorption so that absorption + scatter = total.