Examples of Complete XSProc Input Data

Infinite homogeneous medium input data

Examples of XSProc input data for infinite homogeneous media problems are given below. In these cases the cross section library name “fine_n” indicates that the latest recommended fine-group SCALE library will used in the calculations.

EXAMPLE 1. Default cell definition.

Consider a cylindrical billet of 20 wt % enriched UO2, having a density of 10.85 g/cm3 that is 26 cm in diameter and 26 cm tall.

The average mean-free path in the uranium dioxide is on the order of 2.5 cm. Because only a small fraction of the billet is within a mean-free path of the surface, the material can be treated as an infinite homogeneous medium; therefore the CELL DATA block can be omitted. The XSProc data follows:

20%  ENRICHED  UO2  BILLET fine_n READ COMP UO2  1  0.99  293  92235  20  92238  80  END END  COMP

The volume fraction used for the UO2, 0.99, is calculated by dividing the actual density by the theoretical density obtained from the Isotopes in standard composition library table in the STDCMP chapter, (10.85/10.96). Since the enrichment was specified as 20%, it is assumed that the remainder is 238U.

An alternative input data description follows:

20% ENRICHED UO2 BILLET fine_n READ COMP UO2 1 DEN=10.85  1  293  92235  20  92238  80  END END COMP

EXAMPLE 2. Specify the cell definition.

Consider a 5-liter Plexiglas bottle with an inner radius of 9.525 cm and inner height of 17.78 cm that is filled with highly enriched uranyl nitrate solution at 415 g/L and 0.39 mg of excess nitrate per gram of solution. The uranium isotopic content of the nitrate solution is 92.6 wt % 235U, 5.9 wt % 238U, 1.0 wt % 234U, and 0.5 wt % 236U. Solution density will be calculated from the given data.

The size of the nitrate solution is on the order of 16 to 20 cm in diameter and height. The average mean-free path in the nitrate solution is on the order of 0.5 cm. Therefore, infinite homogeneous medium is an appropriate choice for this problem. By default BONAMI is used for self-shielding the infinite medium of Plexiglas, while CENTRM is used to shield the infinite medium fissile solution.

SET UP 5 LITER URANYL NITRATE SOLUTION IN A PLEXIGLAS CONTAINER fine_n READ COMP PLEXIGLAS   1  END SOLUTION   MIX=2   RHO[UO2(NO3)2]=415            92235   92.6   92238   5.9   92236   0.5 END SOLUTION END  COMP READ CELLDATA INFHOMMEDIUM  2  END END CELLDATA

LATTICECELL input data

Examples of XSProc input data for LATTICECELL problems are given below.

EXAMPLE 1. SQUAREPITCH ARRAY.

Consider an infinite planar array (infinite in X and Y and one layer in Z) of 20 wt % enriched U metal rods with a 1-cm pitch. Each fuel rod is bare uranium metal, 0.75 cm OD × 30.0 cm long. The rods are submerged in water.

Because the diameter of the fuel rod, 0.75 cm, is only slightly larger than the average mean-free path in the uranium metal, approximately 0.5, and because the configuration is a regular array, LATTICECELL is the appropriate choice for proper cross-section processing. The parm field is not provided, so the default CENTRM/PMC self-shielding method is used. XSProc data follows:

INFINITE  PLANAR  ARRAY  OF  20%  U  METAL  RODS fine_n READ COMP URANIUM  1  1  293  92235  20  92238  80  END H2O      2  END END  COMP READ CELLDATA LATTICECELL  SQUAREPITCH   PITCH=1.0  2  FUELD=0.75  1  END END CELLDATA

Since the MORE DATA and CENTRM DATA blocks were omitted, default options will be used in the self-shielding calculations. The default CENTRM/PMC computation options for a square pitch lattice cell are the method-of-characteristics (MoC) method with P0 scatter in CENTRM calculations.

EXAMPLE 2. SQUAREPITCH PWR LATTICE.

Consider an infinite, uniform planar array (infinite in X and Y and one layer in Z) of PWR-like fuel pins of 2.35% enriched UO2 clad with zirconium. The density of the UO2 is 9.21 g/cm3. The fuel in each pin is 0.823 cm in diameter, the clad is 0.9627 cm in diameter, and the length of each pin is 366 cm. The fuel pins are separated by 0.3124 cm of water in the horizontal plane.

LATTICECELL is the appropriate choice for cross-section processing. Assume that all defaults are appropriate; thus the CENTRM/PMC methodology is used, and the MORE DATA and CELL DATA blocks are not entered. The input cross section library named “broad_n” indicates that the recommended broad group SCALE library will be used. In this case CENTRM uses the 2D MoC transport solver. The XSProc data follows:

PWR-LIKE FUEL BUNDLE; uniform infinite array model. broad_n READ COMP UO2   1  .84  293.  92235  2.35  92238  97.65  END ZR    2  1  END H2O   3  1  END END  COMP READ CELLDATA LATTICECELL  SQUAREPITCH  PITCH=1.2751  3  FUELD=0.823  1  CLADD=0.9627  2  END END CELLDATA

EXAMPLE 3. SQUAREPITCH PWR LATTICE, with non-uniform Dancoff.

This example is a single PWR assembly of fuel pins of the type described above, contained in a water pool. The interior pins in the assembly can be self-shielded using the same uniform, infinite lattice model in previous example. However self-shielding of the outer boundary-edge pins will be modified to account for being adjacent to a water reflector, rather than surrounded on all sides by similar pins. This requires that the MCDancoff module be executed previously to obtain non-uniform Dancoff factors for the edge pins. The average edge-pin value of 0.61 is used to represent Dancoff factors of all boundary pins. The default CENTRM MoC transport solver is used for both cells, but the original pitch of 1.2751 cm for the second cell (i.e., boundary pin) is modified to a new pitch corresponding to a Dancoff value of 0.61.

PWR-LIKE FUEL BUNDLE, with boundary-pin corrections broad_n READ COMP ' mixtures for interior pins UO2   1  .84  293.  92235  2.35  92238  97.65  END ZR    2  1  END H2O   3  1  END ' mixtures for boundary pins UO2   4  .84  293.  92235  2.35  92238  97.65  END ZR    5  1  END H2O   6  1  END END  COMP READ CELLDATA LATTICECELL  SQUAREPITCH PITCH=1.2751 3 FUELD=0.823 1 CLADD=0.9627 2  END LATTICECELL  SQUAREPITCH PITCH=1.2751 6 FUELD=0.823 4 CLADD=0.9627 5  END   CENTRM DATA  DAN2PITCH=0.61    END CENTRM END CELLDATA

EXAMPLE 6. SPHTRIANGP ARRAY.

Consider an infinite array of spherical pellets of 2.67% enriched UO2 with a density of 10.3 g/cm3 and a diameter of 1.0724 cm arranged in a “triangular” pitch, flooded with borated water at 4350 ppm. The boron is natural boron; the borated water is created by adding boric acid, H3BO3, and has a density of 1.0078 g/cm3. The temperature is 15ºC and the pitch is 1.1440 cm. The standard composition data for the borated water are given in Example 2 of Combinations of basic and user-defined standard compositions to define a mixture.

Because the diameter of the fuel pellet, 1.0724 cm, is smaller than the average mean-free path in the UO2, approximately 1.5 cm, and because the configuration is a regular array, LATTICECELL is the appropriate choice for proper cross-section processing.

The density fraction for the UO2 is the ratio of actual to theoretical density (10.3/10.96 = 0.9398). Assume that the U is all 235U and 238U. See Combinations of basic and user-defined standard compositions to define a mixture for how to define borated water.

The XSProc data follows:

SPHERICAL  PELLETS  IN  BORATED  WATER fine_n READ COMP UO2   1  .9398  288  92235  2.67  92238  97.33  END ATOMH3BO3  2  0.025066  3  5000  1  1001  3  8016  3        1.0  288  END H2O   2  0.984507  288  END END  COMP READ CELLDATA LATTICECELL  SPHTRIANGP  PITCH=1.1440  2  FUELD=1.0724  1  END END CELLDATA

MULTIREGION input data

Examples of XSProc input data for MULTIREGION problems are given below.

EXAMPLE 1. SPHERICAL.

Consider a small highly enriched uranium sphere supported by a Plexiglas collar in a tank of water. The uranium metal sphere has a diameter of 13.1075 cm, is 97.67% enriched, and has a density of 18.794 g/cm3. The cylindrical Plexiglas collar has a 4.1275-cm-radius central hole, extends to a radius of 12.7 cm and is 2.54 cm thick. The water filled tank is 60 cm in diameter.

The density fraction of the uranium metal is the ratio of actual to theoretical density, where the theoretical density is obtained from the Isotopes in standard composition library table in section 7.2.1. Thus, the density multiplier is 18.794/19.05 = 0.9866. The abundance of uranium is not stated beyond 97.67% enriched, so it is reasonable to assume the remainder is 238U. The Plexiglas collar is not significantly different from water and does not surround the fuel, so it can be ignored. If it is ignored, the problem becomes a 1-D geometry that can be defined using the MULTIREGION type of calculation, and the eigenvalue of the system can be obtained without additional data by executing CSAS1. However, the Plexiglas has been included in this data so it can be passed to a code such as KENO V.a which can describe the geometry rigorously. The XSProc data follow:

SMALL  WATER  REFLECTED  SPHERE  ON  PLEXIGLAS  COLLAR fine_n READ COMP URANIUM    1  .9866  293.  92235  97.67  92238  2.33  END PLEXIGLAS  2  END H2O        3  END END  COMP READ CELLDATA MULTIREGION SPHERICAL RIGHT_BDY=VACUUM END 1 6.55375 3 30.0 END ZONE END CELLDATA

EXAMPLE 2. BUCKLEDSLAB.

This example features a 93.2% enriched uranyl-fluoride solution inside a rectangular Plexiglas container immersed in water. The fissile solution contains 578.7 g of UO2F2 per liter and has no excess acid. The critical thickness of the fuel is 5.384 cm. The finite height of the fuel slab is 147.32 cm, and the depth is 71.58 cm. The Plexiglas container is 1.905 cm thick and is reflected by 20.32 cm of water.

The half thickness of the fuel (2.692) will be used with a reflected left boundary and a vacuum right boundary (default). The XSProc data follow:

CRITICAL SLAB EXPERIMENT USING URANYL-FLUORIDE SOLUTION fine_n READ COMP SOLUTION  MIX=1  RHO[UO2F2]=578.7           92235  93.2  92238  6.8  TEMP=300 END SOLUTION PLEXIGLAS  2  END H2O        3  END END  COMP READ CELLDATA MULTIREGION  BUCKLEDSLAB  LEFT_BDY=REFLECTED DY=71.58 DZ=147.32  END  1  2.692  2  4.597  3  24.917  END ZONE END CELLDATA

DOUBLEHET input data

EXAMPLE 1: A doubly-heterogeneous spherical fuel element with 15,000 UO2 particles in a graphite matrix.

Grain fuel radius is 0.025 cm. Grain contains one coating layer that is 0.009-cm-thick. Pebbles are in a triangular pitch on a 6.4-cm-pitch. Fuel pebble fuel zone is 2.5‑cm in radius and contains a 0.5-cm-thick graphite clad that contains small amounts of 10B. Pebbles are surrounded by 4He. In this case we designated the homogenized mixture as mixture 10. If we have a KENO V.a or KENO-VI input section, we would use mixture 10 in that section. Note that the keyword “FUELR=” is followed by the fuel dimension only, i.e., no mixture number. That is because the fuel mixture number is specified with “FUELMIX=” and therefore need not be repeated.

INFINITE ARRAY OF UO2-FUELLED PEBBLES fine_n READ COMP ' UO2 FUEL KERNEL U-235  1 0 1.92585E-3 293.6 END O      1 0 4.64272E-2 293.6 END ' FIRST COATING C      2 0 5.26449E-2 293.6 END ' GRAPHITE MATRIX C      6 0 8.77414E-2 293.6 END ' CARBON PEBBLE OUTER COATING C      7 0 8.77414E-2 293.6 END B-10   7 0 9.64977E-9 293.6 END HE-4   8 0 2.65156E-5 293.6 END END COMP READ CELLDATA DOUBLEHET  RIGHT_BDY=WHITE FUELMIX=10 END  GFR=0.025  1 COATT=0.009 2 MATRIX=6 NUMPAR=15000 END GRAIN PEBBLE SPHTRIANGP RIGHT_BDY=WHITE HPITCH=3.2 8 FUELR=2.5 CLADR=3.0 7  END END CELLDATA

EXAMPLE 2: A doubly-heterogeneous spherical fuel element with 10,000 UO2 particles and 5,000 PuO2 particles in a graphite matrix.

Grain fuel radii for UO2 and PuO2 particles are 0.025 cm and 0.012 cm, respectively. UO2 grains contain one coating layer that is 0.009‑cm-thick. PuO2 grains contain one coating layer that is 0.0095-cm-thick. Pebbles are in a triangular pitch on a 6.4-cm-pitch. Fuel pebble fuel zone is 2.5-cm in radius and contains a 0.5-cm-thick graphite clad that contains small amounts of 10B. Pebbles are surrounded by 4He. Since number of particles is entered, the total volume fraction and the pitch can be calculated by the code.

INFINITE ARRAY OF UO2- AND PUO2-FUELLED PEBBLES fine_n READ COMP ' UO2 FUEL KERNEL U-235  1 0 1.92585E-3 293.6 END O      1 0 4.64272E-2 293.6 END ' FIRST COATING C      2 0 5.26449E-2 293.6 END ' GRAPHITE MATRIX C      6 0 8.77414E-2 293.6 END ' CARBON PEBBLE OUTER COATING C      7 0 8.77414E-2 293.6 END B-10   7 0 9.64977E-9 293.6 END HE-4   8 0 2.65156E-5 293.6 END ' PUO2 FUEL KERNEL PU-239  11 0 1.24470E-02 293.6 END O       11 0 4.60983E-02 293.6 END ' FIRST COATING C      12 0 5.26449E-2 293.6 END ' GRAPHITE MATRIX C      16 0 8.77414E-2 293.6 END END COMP READ CELLDATA DOUBLEHET  RIGHT_BDY=WHITE FUELMIX=10 END  GFR=0.025  1 COATT=0.009 2 MATRIX=6 NUMPAR=10000 END GRAIN  GFR=0.012 11 COATT=0.0095 12 MATRIX=16 NUMPAR=5000 END GRAIN PEBBLE SPHTRIANGP RIGHT_BDY=WHITE HPITCH=3.2 8 FUELR=2.5 CLADR=3.0 7 END END CELLDATA

EXAMPLE 3: A doubly-heterogeneous slab fuel element with flibe salt coolant

Grain fuel radii for UO2 particles are 0.025 cm. The UO2 grains contain four coating layers with thicknesses of 0.01, 0.0035, 0.003, and 0.004 cm, respectively. The fuel grains are embedded in a carbon matrix material to form the fuel compact. The x-dimension of fuel plate consists of a 0.5 cm (half-thickness) fuel compact region, a carbon clad with outer dimension of 1.27, followed by the flibe coolant with an outer reflected dimension of 1.62 cm. The width (y-dimension) of the slab plate is 22.5 cm and the height (z-dimension) is 500 cm. The y and z dimensions are only used to define volumes for the fuel plate.

slab doublehet sample problem: double-het for slab v7.1-252n read comp ' fuel kernel u-238  1 0 2.12877e-2 293.6 end u-235  1 0 1.92585e-3 293.6 end o      1 0 4.64272e-2 293.6 end b-10   1 0 1.14694e-7 293.6 end b-11   1 0 4.64570e-7 293.6 end ' first coating c      2 0 5.26449e-2 293.6 end ' inner pyro carbon c      3 0 9.52621e-2 293.6 end ' silicon carbide c      4 0 4.77240e-2 293.6 end si     4 0 4.77240e-2 293.6 end ' outer pyro carbon c      5 0 9.52621e-2 293.6 end ' graphite matrix c      6 0 8.77414e-2 293.6 end b-10   6 0 9.64977e-9 293.6 end b-11   6 0 3.90864e-8 293.6 end ' carbon slab outer coating c      7 0 8.77414e-2 293.6 end b-10   7 0 9.64977e-9 293.6 end b-11   7 0 3.90864e-8 293.6 end Li-6         8    0   1.38344E-06   948.15  end Li-7         8    0   2.37205E-02   948.15  end Be           8    0   1.18609E-02   948.15  end F            8    0   4.74437E-02   948.15  end end comp read celldata   doublehet  fuelmix=10 end     gfr=0.02135   1     coatt=0.01    2     coatt=0.0035  3     coatt=0.003   4     coatt=0.004   5     vf=0.4     matrix=6     end grain   slab symmslabcell     hpitch=1.62   8     cladr=1.27    7     fuelr=0.5     fuelh=500     fuelw=22.500   end   centrm data ixprt=1 isn=8 end centrm end celldata

Two methods of specifying a fissile solution

The standard composition specification data offer flexibility in the choice of input data. This section illustrates two methods of specifying the same fissile solution.

Create a mixture 3 that is aqueous uranyl nitrate solution:

UO2(NO3)2, solution density = 1.555 g cm3/

0.2669 g U/g-soln., 0.415 g U/ cm3; excess nitrate = 0.39 mg/g-soln

Uranium isotopic content: 92.6 wt % U-235 5.9 wt % U-238

1.0 wt % U-234 and 0.5 wt % U-236

The SCALE atomic weights used in this problem are listed as follows:

H 1.0078

O 15.999

N 14.0067

U-234 234.041

U-235 235.0439

U-236 236.0456

U-238 238.0508

Two methods of describing the uranyl nitrate solution will be demonstrated. Method 1 is more rigorous, and method 2 is easier and as accurate.

METHOD 1:

This method involves breaking the solution into its component parts [(HNO3, UO2(NO3)2, and H2O)] and entering the basic standard composition specifications for each.

1. Calculate the density of the HNO3 0.39 × 10−3 g NO3/g soln × [(62.997 g HNO3/mole HNO3)/(61.990 g NO3/mole NO3)] × 1.555 g soln/ cm3soln = 6.16 × 10−4 g HNO3/cc soln.

2. Calculate the density fraction of HNO3 (actual density/theoretical density). In the Standard Composition Library the theoretical density of HNO3 is 1.0. 6.16 × 10−4/1.0 = 6.16 × 10−4.

3. Calculate the molecular weight of the uranium

The number of atoms in a mole of uranium is the sum of the number of atoms of each isotope in the mole of uranium.

Let AU = the average molecular weight of uranium, g U/mole U

GU = the density of uranium in g/cm3.

Then the number of atoms in a mol of uranium =

(6.023 × 10+23 * 10−24 * GU)/AU

or 0.6023 * GU/AU.

The weight fraction of each isotope is the weight % * 100.

Therefore, F235 = 0.926, the weight fraction of U-235 in the U

F238 = 0.059, the weight fraction of U-238 in the U

F236 = 0.005, the weight fraction of U-236 in the U

F234 = 0.010, the weight fraction of U-234 in the U

A235 = 235.0442, the molecular weight of U-235

A238 = 238.0510, the molecular weight of U-238

A236 = 236.0458, the molecular weight of U-236

A234 = 234.0406, the molecular weight of U-234.

Then the number of atoms of isotopes in a mol of uranium =

6.023 × 10+23 * 10−24 * ( (GU*F235/A235) + (GU*F238/A238) +

GU*F236/A236) + (GU*F234/A234) )

or

0.6023*GU * ( 0.926/235.0442 + 0.059/238.0510 +

0.005/236.0458 + 0.010/234.0406 ).

Because the number of atoms of uranium equals the sum of the atoms of isotopes,

0.6023 * GU/AU = 0.6023 * GU *( 0.926/235.0442 + 0.059/238.0510 +

0.005/236.0458 + 0.010/234.0406 )

1/AU = 0.926/235.0442 + 0.059/238.0510 + 0.005/236.0458 + 0.010/234.0406

AU = 235.2144.

1. Calculate the molecular weight of the UO2(NO3)2.

235.2144 + (8 × 15.9954) + (2 × 14.0033) = 391.184 g UO2(NO3)2/mole

1. Calculate the density of UO2(NO3)2

0.415 g U/cc × [(391.184 g UO2(NO3)2/mol)/(235.2144 g U/mole)] =

0.69018 g UO2(NO3)2/ cm3.soln.

Calculate the density fraction (actual density/theoretical density) of UO2(NO3)2.

[In the Standard Composition Library the theoretical density of UO2(NO3)2 is given as 2.2030 g/cm3.]

The density fraction is 0.69018/2.2030 = 0.31329.

1. Calculate the amount of water in the solution

1.555 g soln/ cm3. soln − 6.16 × 10−4 g HNO3/cm3 soln − 0.69018 g UO2(NO3)2LL/ cm3. soln = 0.8642 g H2O/cc soln.

1. Calculate the density fraction (actual density/theoretical density) of water.

HNO3       3   6.16-4   293  END UO2(NO3)2  3  .31329  293  92235  92.6  92238  5.9  92234  1.0                    92236   0.5  END H2O        3    .86575  293  END

METHOD 2:

This method utilizes the solution option available in the standard composition specification data. Because the density is specified in the input data, this method should yield correct number densities that should agree with method 1 except for calculational round-off.

1. Calculate the fuel density

0.415 g U/cc is 415 g U/L.

1. The molecular weight of nitrate NO3 is 61.9895.

2. Calculate the molarity of the solution.

0.39 mg nitrate/g soln × 1000 cm3soln/L soln × 1 g/1000 mg × 1.555 g soln/ cm3soln = 0.60645 g excess nitrate/L soln.

A 1-molar solution is 1 mole of acid/L of solution:

(For nitric acid 1 molar is 1 normal because there is only one atom of hydrogen per molecule of acid in HNO3.)

(0.60645 g nitrate/L soln)/(61.9895 g NO3/mole NO3) = 9.783 × 10−3 mole nitrate/L is identical to mole of acid/L, which is identical to molarity.

1. The density fraction of the solution is 1.0. Do not try to use the density of the solution divided by the theoretical density of UO2(NO3)2 from the Standard Composition Library for your density multiplier. The UO2(NO3)2 listed there is the solid, not the solution.

The solution specification data follow:

SOLUTION      MIX=1   RHO[UO2(NO3)2] = 415    92235   92.6    92238   5.9                       92234    1.0    92236   0.5               MOLAR [HNO3] = 9.783-3               TEMP = 293      DENSITY = 1.555 END SOLUTION

Comparison of number densities from the two methods

The number densities of methods 1 and 2 should agree within the limits of the input data. The density multipliers in method 1 are 5 digits and the density multipliers in method 2 are 4 digits. Therefore, the number densities calculated by the two methods should agree to 4 or 5 digits.