Unstable 3D pin-cell burnup problem
In certain burnup problems such as long, axially symmetric 3D assemblies or fuel rods or large symmetric core geometries traditional Monte Carlo burnup schemes may run into instabilities[1][2]. This page describes the simulation of one such case first using the traditional burnup schemes (to showcase the instability) and then using a stable burnup scheme.
Contents
Problem description
Base input for unstable 3D pin-cell problem
/***************** ** Run options ** *****************/ % --- Neutron population to be used set pop 10000 500 200 % --- 200 W/cm linear power set power 60000 /************************* ** Geometry definition ** *************************/ % --- Fuel Pin definitions: pin p1 fuel 0.47 void 0.48 clad 0.54 cool % --- Lattice lat l1 1 0.0 0.0 1 1 1.5 p1 % --- Surrounding surfaces: % Boundary of geometry: surf 3 cuboid -0.75 0.75 -0.75 0.75 -160 160 % Lower boundary of fuel surf 4 pz -150 % Upper boundary of fuel surf 5 pz 150 % --- Cell definitions: % Active fuel pin cell 3 0 fill l1 -3 4 -5 % Coolant below active fuel (bottom reflector) cell 4 0 cool -3 -4 % Coolant above active fuel (top reflector) cell 5 0 cool -3 5 % outside world cell 99 0 outside 3 % Outside world % --- Reflective boundary conditions in XY, black in Z: set bc 3 3 1 /************************** ** Material definitions ** **************************/ % --- Fuel material (4.85 % enrichment): mat fuel -10.283 vol 208.19 rgb 200 200 125 92235.09c 0.016166667 92238.09c 0.317166667 8016.09c 0.666666667 % --- Cladding (Zr-4) mat clad -6.56000E+00 rgb 180 180 180 8016.06c -1.19276E-03 8017.06c -4.82878E-07 24050.06c -4.16117E-05 24052.06c -8.34483E-04 24053.06c -9.64457E-05 24054.06c -2.44600E-05 26054.06c -1.12572E-04 26056.06c -1.83252E-03 26057.06c -4.30778E-05 26058.06c -5.83334E-06 40090.06c -4.97862E-01 40091.06c -1.09780E-01 40092.06c -1.69646E-01 40094.06c -1.75665E-01 40096.06c -2.89038E-02 50112.06c -1.27604E-04 50114.06c -8.83732E-05 50115.06c -4.59255E-05 50116.06c -1.98105E-03 50117.06c -1.05543E-03 50118.06c -3.35688E-03 50119.06c -1.20069E-03 50120.06c -4.59220E-03 50122.06c -6.63497E-04 50124.06c -8.43355E-04 % --- Coolant: mat cool -0.75 moder lwtr 1001 rgb 50 50 255 1001.06c 0.666666667 8016.06c 0.333333333 % --- Thermal scattering data for light water: therm lwtr lwj3.11t
The base input for the problem is given above. The input describes a 300 cm long fuel rod in infinite lattice geometry. Axially the fuel rod is reflected from top and bottom with 10 cm water layers after which a black boundary condition is applied. The radial geometry is shown here:
Let's say that we want to calculate the axial power distribution and flux distribution using 100 axial bins over the active fuel length at burnups between 0 MWd/kgU and 20 Mwd/kgU. In order to capture the axial burnup distribution we will divide the fuel material axially into a rather small number of division zones (10) using the div card.
Unstable solution using explicit Euler's scheme
We'll first run the calculation using the explicit Euler's scheme for the discretization of the Bateman equations. This scheme is equal to using the constant values for the beginning-of-step (BOS) flux and cross sections to burn the materials through each of the burnup steps. Explicit Euler's scheme can be chosen with
set pcc ce
Where "ce" refers to "constant extrapolation".
The input used for this calculation is given below:
Input for explicit Euler's scheme (constant extrapolation) burnup calculation
/*****************
** Run options **
*****************/
% --- Neutron population to be used
set pop 10000 500 200
% --- 200 W/cm linear power
set power 60000
/******************
** Burn options **
******************/
% --- Divisor for the fuel material
div fuel subz 10 -150 150
% --- Use constant extrapolation burnup scheme
% (explicit Euler's method)
set pcc ce
% --- Depletion history
dep butot 0.01 0.25 0.5 1.0 2.0 3.0 5.0 7.5 10.0 12.5 15.0 17.5 20.0
% --- Fission yield and decay data libraries
set nfylib "sss_jeff311.nfy"
set declib "sss_jeff311.dec"
% --- Depletion output for divided materials
set depout 1
% --- Nuclides included in depletion output
set inventory
531350 % Iodine 135
541350 % Xenon 135
621490 % Samarium 149
922350 % Uranium 235
942390 % Plutonium 239
/***************
** Detectors **
***************/
% --- Fission heat deposition detector
det power dr -6 void dz -150 150 10
% --- Neutron flux detector
det flux dz -150 150 10
/*************************
** Geometry definition **
*************************/
% --- Fuel Pin definitions:
pin p1
fuel 0.47
void 0.48
clad 0.54
cool
% --- Lattice
lat l1 1 0.0 0.0 1 1 1.5
p1
% --- Surrounding surfaces:
% Boundary of geometry:
surf 3 cuboid -0.75 0.75 -0.75 0.75 -160 160
% Lower boundary of fuel
surf 4 pz -150
% Upper boundary of fuel
surf 5 pz 150
% --- Cell definitions:
% Active fuel pin
cell 3 0 fill l1 -3 4 -5
% Coolant below active fuel (bottom reflector)
cell 4 0 cool -3 -4
% Coolant above active fuel (top reflector)
cell 5 0 cool -3 5
% outside world
cell 99 0 outside 3 % Outside world
% --- Reflective boundary conditions in XY, black in Z:
set bc 3 3 1
/**************************
** Material definitions **
**************************/
% --- Fuel material (4.85 % enrichment):
mat fuel -10.283 burn 1 vol 208.19 rgb 220 220 115
92235.09c 0.016166667
92238.09c 0.317166667
8016.09c 0.666666667
% --- Cladding (Pure zirconium)
mat clad -6.56000E+00 rgb 180 180 180
40090.06c -3.32125E+00
40091.06c -7.32348E-01
40092.06c -1.13172E+00
40094.06c -1.17187E+00
40096.06c -1.92818E-01
% --- Coolant:
mat cool -0.75 moder lwtr 1001 rgb 50 50 255
1001.06c 0.666666667
8016.06c 0.333333333
% --- Thermal scattering data for light water:
therm lwtr lwj3.11t
mesh 2 300 900
Running the calculation will produce one yz-meshplot for each of the burnup steps (animation shown on right) and already based on these meshplots it is easy to see that the solution is neither symmetric or stable.
In any case, we can plot the axial power distribution and (e.g.) the axial 135Xe distribution for some of the steps:
These plots show that the solution really is quite unstable. Even a small asymmetry in the fission power produces an asymmetry in the 135Xe distribution, which in turn affects the power distribution. Setting the equilibrium xenon calculation on (set xenon 1) typically offers some relief, but similar asymmetries in other nuclide densities can destabilize the solution.
Unstable solution using predictor-corrector scheme
Stable solution using SIE burnup scheme
Stable solution using a symmetry boundary
References
- ^ Dufek, J. and Hoogenboom, E. "Numerical Stability of Existing Monte Carlo Burnup Codes in Cycle Calculations of Critical Reactors", Nucl. Sci. Eng., 162 (2009) 307-311
- ^ Dufek, J. et al. "Numerical stability of the predictor–corrector method in Monte Carlo burnup calculations of critical reactors", Ann. Nucl. Energy, 56 (2013) 34-38
