## Analog vs. Implicit keff

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mathieu.hursin
Posts: 35
Joined: Thu Sep 22, 2011 5:27 pm

### Analog vs. Implicit keff

Dear all,

I m a new user to SERPENT. I have been running some simple examples (bwr 1 from the lattice validation package). Looking at the output, 2 values for k-eff are available, one implicit and one analog...I haven't been able to figure out the difference between the two, difference which can be sometime bigger than a standard deviation.

Can someone help me out with the meaning of the various definition of keff?

Thank you,

Regards,

Mathieu

Jaakko Leppänen
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### Re: Analog vs. Implicit keff

Analog k-eff is the ratio of new source neutrons produced in two consecutive generations. Implicit k-eff is calculated using reaction rate estimators for source and loss terms (more precisely imp_keff = nsf/(fiss + abs + leak - n2n). The values should be equivalent within statistics. What kind of differences do you see?
- Jaakko

mathieu.hursin
Posts: 35
Joined: Thu Sep 22, 2011 5:27 pm

### Re: Analog vs. Implicit keff

Jaakko,

for the BWR 1 case, after the 500 cycles here is what I obtain:

k-eff (analog) = 1.07512 +/- 0.00115 [1.07286 1.07738]
k-eff (implicit) = 1.07705 +/- 0.00060 [1.07587 1.07823]

I m also puzzled by the fact that the standard deviation for the implicit k-eff is much smaller that the analog one even though the implicit keff comes from the combination of various reaction rates with standard deviation that should be in the range of the analog k-eff...

Also talking about the different k-eff in the output, I m guessing that col_keff is the equivalent of k(collision) in MCNP and abs_keff the equivalent of (k absorption). Is that correct?

Mathieu

Jaakko Leppänen
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### Re: Analog vs. Implicit keff

Well, the value of the implicit estimator lies within the 95% confidence interval of the analog estimator, so statistically the difference isn't that significant.

The larger statistical error for the analog estimator comes from the fact that there is more variation in the number of new source points per generation than there is in the source and loss rates.

The naming convention is not the same as in MCNP, and the terms are used somewhat loosely. Collision estimate of k-eff is similar to the MCNP track-length estimate, with the difference that the scores are based on collisions instead of track lengths. The absorption estimator, on the other hand, is similar to the collision estimator in MCNP. Notice also that ABS_KEFF and IMP_KEFF are the same parameter.

The naming of the parameters is something I should have payed more attention to in the early stages of code development. The reason I got the terminology confused is that the code was designed more for group constant generation than criticality calculations.
- Jaakko

emilf
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### Re: Analog vs. Implicit keff

Jaakko,

Should it be:
imp_keff = nsf/(abs + leak - n2n)

imp_keff = nsf/(fiss + abs + leak - n2n) ?

Jaakko Leppänen
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### Re: Analog vs. Implicit keff

You are right. By absorption I meant capture. In your formula abs = fiss + capt. so imp_keff = nsf/(abs + leak - n2n) = nsf/(fiss + capt + leak - n2n)
- Jaakko

mathieu.hursin
Posts: 35
Joined: Thu Sep 22, 2011 5:27 pm

### Re: Analog vs. Implicit keff

Dear all,

I'm studying Molten Salt Reactor type lattice with Thorium/MOX/F,Li,Be fuel. There is an unrealistic discrepancy between k-analog and k-implicit for this type of fuel (k_analog = 0.9 and k_implicit = 1.1)....when replaced with regular UO2 fuel, there is no discrepancy anymore...

Have you come across such behavior? Is that a user error (I couldn't attached the input, so I copied my input below), if so what am I doing wrong?

Mathieu

==========================================================================
% --- PWR MOX/UOX lattice (SCALE-style input format) ---------

% --- Problem title:

set title "MSR lattice"

% --- Cross section library file path:

set acelib "/afs/psi.ch/project/fast_lrs/workspace/COD/SERP/xsdata/endfb7/sss_endfb7u.xsdata"

% ------------------------------------------------------------

% --- Material definitions ("comp block"):

% --- Molten Salt Fuel

mat fuel 8.762e-2
90232.03c 3.778e-3
94238.03c 6.359e-3
94239.03c 3.906e-5
94240.03c 1.513e-5
94241.03c 5.018e-6
94242.03c 3.177e-6
3007.03c 2.260e-2
4009.03c 5.037e-3
9019.03c 4.978e-2

% --- VVER fuel
mat fuel2 -10.45700
92235.03c -0.03173
92238.03c -0.84977
8016.03c -0.11850

% --- Water with Grid plates:

mat graphite 9.226e-2
6000.03c 9.226e-2

% ------------------------------------------------------------

% --- Parameters ("param block"):

% --- p-table for unresolved resonance region

set ures 1

% --- Periodic boundary condition:

set bc 3

% --- Neutron population and criticality cycles:

set pop 2000 50 1

% ------------------------------------------------------------

% --- Geometry ("geom block"):

% --- Core lattice ("global unit 0"):

surf 1 cyl 0.0 0.0 6.64
surf 2 hexyc 0.0 0.0 11.55

cell 100 0 fuel -1
cell 101 0 graphite 1 -2
cell 102 0 outside 2

% -----------------------------------------------------------

orca.blu
Posts: 59
Joined: Wed Apr 20, 2011 1:39 pm

### Re: Analog vs. Implicit keff

Ciao Mathieu!

I don't have serpent on this computer so I can not really help you... sorry.
I have just few considerations, looking at your input:

- can you post the statistical error of k (both) ?

- in mat "fuel" you have a huge amount of Pu-238 (94238.03c),
it is probably U-238 (92238.03c)

- you put in "set pop" almost no skip cycles,
I believe that in this reactor the distribution of fission source
(near the center of the fuel channel or closer to the graphite)
is quite important

anyway these considerations can not explain such a discrepancy between k-analog and k-implicit...

Ciao

Manu
Manuele Aufiero
LPSC/IN2P3/CNRS Grenoble

mathieu.hursin
Posts: 35
Joined: Thu Sep 22, 2011 5:27 pm

### Re: Analog vs. Implicit keff

Hey Manu,

The isotope ID was good (Pu-238; I m looking at a Th cycle, so no U-238) but the amount was wrong...it should have been 6.359e-7.

Changing the material description fixed the issue, ie here is the final k-eff:
k-eff (analog) = 0.78810 +/- 0.00133 [0.78549 0.79072]
k-eff (implicit) = 0.78710 +/- 0.00062 [0.78589 0.78831]

I m still wondering why the k-eff analog was so different from the implicit one with a big amount of Pu-238...it is not obvious to me.

Thanks for your help, though !

M

Jaakko Leppänen