# Difference between revisions of "Unstable 3D pin-cell burnup problem"

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− | These plots show that the solution really is quite unstable. Even a small asymmetry in the fission power produces an asymmetry in the <sup>135</sup>Xe distribution, which in turn affects the power distribution. Setting the equilibrium xenon calculation on ( | + | These plots show that the solution really is quite unstable. Even a small asymmetry in the fission power produces an asymmetry in the <sup>135</sup>Xe 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 == | == Unstable solution using predictor-corrector scheme == |

## Revision as of 13:28, 29 September 2017

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 ^{135}Xe 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 ^{135}Xe 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