.. _7-1c: Examples of Complete XSProc Input Data ====================================== .. _7-1c-1: 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 UO\ :sub:`2`, having a density of 10.85 g/cm\ :sup:`3` 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: .. highlight:: scale :: 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 UO\ :sub:`2`, 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 :sup:`238`\ U. 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 % :sup:`235`\ U, 5.9 wt % :sup:`238`\ U, 1.0 wt % :sup:`234`\ U, and 0.5 wt % :sup:`236`\ U. 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 .. _7-1c-2: 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 :math:`\times` 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 UO\ :sub:`2` clad with zirconium. The density of the UO\ :sub:`2` is 9.21 g/cm\ :sup:`3`. 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 UO\ :sub:`2` with a density of 10.3 g/cm\ :sup:`3` 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, H\ :sub:`3`\ BO\ :sub:`3`, and has a density of 1.0078 g/cm\ :sup:`3`. The temperature is 15\ :math:`^{\circ}`C and the pitch is 1.1440 cm. The standard composition data for the borated water are given in Example 2 of :numref:`7-1a-9`. Because the diameter of the fuel pellet, 1.0724 cm, is smaller than the average mean-free path in the UO\ :sub:`2`, 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 UO\ :sub:`2` is the ratio of actual to theoretical density (10.3/10.96 = 0.9398). Assume that the U is all :sup:`235`\ U and :sup:`238`\ U. See :numref:`7-1a-9` 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 .. _7-1c-3: 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/cm\ :sup:`3`. 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 :sup:`238`\ U. 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 UO\ :sub:`2`\ F\ :sub:`2` 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 .. _7-1c-4: DOUBLEHET input data -------------------- EXAMPLE 1: A doubly-heterogeneous spherical fuel element with 15,000 UO\ :sub:`2` 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 :sup:`10`\ B. Pebbles are surrounded by :sup:`4`\ He. 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 UO\ :sub:`2` particles and 5,000 PuO\ :sub:`2` particles in a graphite matrix. Grain fuel radii for UO\ :sub:`2` and PuO\ :sub:`2` particles are 0.025 cm and 0.012 cm, respectively. UO\ :sub:`2` grains contain one coating layer that is 0.009 cm thick. PuO\ :sub:`2` 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 :sup:`10`\ B. Pebbles are surrounded by :sup:`4`\ He. 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 UO\ :sub:`2` particles are 0.025 cm. The UO\ :sub:`2` 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 EXAMPLE 4: A doubly-heterogeneous triangular-pitch fuel element with 1,302 UO\ :sub:`2` particles in a graphite matrix with the DAN2PITCH option for grain. Grain fuel radius for UO2 particles are 0.02125 cm. The UO2 grains contain four coating layers with radii of 0.03125, 0.03525, 0.03875, and 0.04275 cm, respectively. The fuel grains are embedded in a carbon matrix material to form the fuel compact. Fuel compact is in a triangular pitch on a 1.8796-cm-pitch. Fuel zone is 0.6225-cm in radius and there is a 0.0125 cm gap between fuel and graphite moderator. Since number of particles is entered, the total volume fraction and the pitch can be calculated by the code. Dancoff factor of 0.6552 is inputted for a grain to consider neutron leakage effect. :: DH_dan2pitch_nonuniform v7.1-252 read composition u-235 1 0 3.6676E-03 600.0 end u-238 1 0 1.9742E-02 600.0 end o-16 1 0 3.5114E-02 600.0 end c 1 0 1.1705E-02 600.0 end c 2 0 5.2646E-02 600.0 end c 3 0 9.5263E-02 600.0 end si-28 4 0 4.4159E-02 600.0 end si-29 4 0 2.2433E-03 600.0 end si-30 4 0 1.4805E-03 600.0 end c 4 0 4.7883E-02 600.0 end c 5 0 9.5263E-02 600.0 end c-graphite 6 0 7.2701E-02 600.0 end he 7 0 2.4006E-05 600.0 end c-graphite 8 0 9.2756E-02 600.0 end end composition read celldata doublehet fuelmix=9 end gfr=0.02125 1 coatr=0.03125 2 coatr=0.03525 3 coatr=0.03875 4 coatr=0.04275 5 numpar=1302 matrix=6 end grain centrm data alump=0.0 dan2pitch=0.6562 end centrm rod triangpitch fuelr=0.6225 gapr=0.635 7 hpitch=0.9398 8 fuelh=1.000 right_bdy=white left_bdy=reflected end centrm data iup=12 isn=16 alump=0.0 end centrm end celldata .. _7-1c-5: 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: UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`2`, solution density = 1.555 g cm\ :sup:`3`/ 0.2669 g U/g-soln., 0.415 g U/ cm\ :sup:`3`; 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. .. centered:: METHOD 1: This method involves breaking the solution into its component parts [(HNO\ :sub:`3`, UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`2`, and H\ :sub:`2`\ O)] and entering the basic standard composition specifications for each. 1. Calculate the density of the HNO\ :sub:`3` 0.39 :math:`\times` 10\ :sup:`-3` g NO\ :sub:`3`/g soln :math:`\times` [(62.997 g HNO\ :sub:`3`/mole HNO\ :sub:`3`)/(61.990 g NO\ :sub:`3`/mole NO\ :sub:`3`)] :math:`\times` 1.555 g soln/ cm\ :sup:`3`\ soln = 6.16 :math:`\times` 10\ :sup:`-4` g HNO\ :sub:`3`/cc soln. 2. Calculate the density fraction of HNO\ :sub:`3` (actual density/theoretical density). In the Standard Composition Library the theoretical density of HNO\ :sub:`3` is 1.0. 6.16 :math:`\times` 10\ :sup:`-4`/1.0 = 6.16 :math:`\times` 10\ :sup:`-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/cm\ :sup:`3`. Then the number of atoms in a mol of uranium = (6.023 :math:`\times` 10\ :sup:`+23` \* 10\ :sup:`-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 :math:`\times` 10\ :sup:`+23` \* 10\ :sup:`-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. 4. Calculate the molecular weight of the UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`2`. .. 235.2144 + (8 :math:`\times` 15.9954) + (2 :math:`\times` 14.0033) = 391.184 g UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`2`/mole 5. Calculate the density of UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`2` .. 0.415 g U/cc :math:`\times` [(391.184 g UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`2`/mol)/(235.2144 g U/mole)] = 0.69018 g UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`2`/ cm\ :sup:`3`.soln. Calculate the density fraction (actual density/theoretical density) of UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`2`. [In the Standard Composition Library the theoretical density of UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`2` is given as 2.2030 g/cm\ :sup:`3`.] The density fraction is 0.69018/2.2030 = 0.31329. 6. Calculate the amount of water in the solution .. 1.555 g soln/ cm\ :sup:`3`. soln - 6.16 :math:`\times` 10\ :sup:`-4` g HNO\ :sub:`3`/cm\ :sup:`3` soln - 0.69018 g UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`2`\ LL/ cm\ :sup:`3`. soln = 0.8642 g H\ :sub:`2`\ O/cc soln. 7. 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 .. centered:: 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. 2. The molecular weight of nitrate NO\ :sub:`3` is 61.9895. 3. Calculate the molarity of the solution. .. 0.39 mg nitrate/g soln :math:`\times` 1000 cm\ :sup:`3`\ soln/L soln :math:`\times` 1 g/1000 mg :math:`\times` 1.555 g soln/ cm\ :sup:`3`\ soln = 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 HNO\ :sub:`3`.) (0.60645 g nitrate/L soln)/(61.9895 g NO\ :sub:`3`/mole NO\ :sub:`3`) = 9.783 :math:`\times` 10\ :sup:`-3` mole nitrate/L is identical to mole of acid/L, which is identical to molarity. 4. 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 UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`2` from the Standard Composition Library for your density multiplier. The UO\ :sub:`2`\ (NO\ :sub:`3`)\ :sub:`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 .. centered:: 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. +----------------+--------------+--------------+ | | Method 1 | Method 2 | +----------------+--------------+--------------+ | Nuclide number | Atom density | Atom density | +----------------+--------------+--------------+ | 92235 | 9.84603E-04 | 9.84603E-04 | +----------------+--------------+--------------+ | 92238 | 6.19415E-05 | 6.19415E-05 | +----------------+--------------+--------------+ | 92234 | 1.06784E-05 | 1.06784E-05 | +----------------+--------------+--------------+ | 92236 | 5.29387E-06 | 5.29387E-06 | +----------------+--------------+--------------+ | 07014 | 2.13092E-03 | 2.13092E-03 | +----------------+--------------+--------------+ | 08016 | 3.74135E-02 | 3.7410E-02 | +----------------+--------------+--------------+ | 01001 | 5.77973E-02 | 5.77983E-02 | +----------------+--------------+--------------+ .. _7-1c-6: Multiple unit cells in a single problem --------------------------------------- Consider a problem that involves three different UO\ :sub:`2` fuel assemblies: a 1.98%-enriched assembly, a 2.64%-enriched assembly, and a 2.96%-enriched assembly. All fuel rods are UO\ :sub:`2` at 10.138 g/cm\ :sup:`3` and are 0.94 cm in diameter. The Zircaloy-4 clad has an inside radius of 0.4875 cm and an outside radius of 0.545 cm. The rod pitch is 1.44 cm. Each fuel assembly is a 15 :math:`\times` 15 array of fuel pins with water holes, instrumentation holes, and burnable poison rods. For cross-section processing, the presence of the water holes, instrumentation holes, and burnable poison rods in the assemblies are ignored. The following XSProc input use the CENTRM/PMC method for self-shielding three latticecells with different fuel enrichments. The remaining mixture (SS-304), not specified in a unit cell, is processed as an infinite homogeneous medium using the BONAMI method. Each mixture can appear only in a single zone of one unit cell. For square pitch latticecells the default CENTRM transport solver is MoC with P0 scatter; however in this input, the solver for the 3\ :sup:`rd` cell is modified through CENTRM DATA to use the two-region approximation for the CE calculation [npxs=5], and discrete S\ :sub:`N` transport calculation with P1 anisotropic scatteringfor the MG solutions in the fast and thermal energy ranges [nfst=0, nthr=0]. :: DEMONSTRATION PROBLEM WITH MULTIPLE RESONANCE CORRECTIONS REQUIRED broad_n READ COMP UO2 1 .925 300 92235 1.98 92238 98.02 END UO2 2 .925 300 92235 2.64 92238 97.36 END UO2 3 .925 300 92235 2.96 92238 97.04 END ZIRC4 4 1.0 300 END H2O 5 1.0 300 END ZIRC4 6 1.0 300 END H2O 7 1.0 300 END ZIRC4 8 1.0 300 END H2O 9 1.0 300 END SS304 10 1.0 300 END END COMP READ CELLDATA LATTICECELL SQUAREPITCH PITCH=1.44 5 FUELD=0.94 1 CLADD=1.09 4 GAPD=0.975 0 END LATTICECELL SQUAREPITCH PITCH=1.44 7 FUELD=0.94 2 CLADD=1.09 6 GAPD=0.975 0 END LATTICECELL SQUAREPITCH PITCH=1.44 9 FUELD=0.94 3 CLADD=1.09 8 GAPD=0.975 0 END CENTRM DATA npxs=5 nthr=0 nfst=0 isct=1 END CENTRM DATA END CELLDATA .. _7-1c-7: Multiple fissile mixtures in a single unit cell ----------------------------------------------- The following problem involves large units having the bulk of their fissile material more than one mean-free path away from the surface of the unit. The interaction between the units that occurs in the resonance range is a very small fraction of the total interaction because an overwhelming percentage of the interaction occurs deep within each unit. Therefore, the resonance range interaction between the units can be ignored, and the default infinite homogeneous medium cross-section processing in the resonance range can be considered adequate for this particular application. Consider a problem that consists of four 20.96-kg 93.2%-enriched uranium metal cylinders, density 18.76 g/cm\ :sup:`3`, and four 5-liters Plexiglas bottles filled with highly enriched uranyl nitrate solution at 415 g/L, a specific gravity of 1.555, and 0.39 mg of excess nitrate per gram of solution. The isotopic content of the uranium metal is 93.2 wt % :sup:`235`\ U, 5.6 wt % :sup:`238`\ U, 1.0 wt % :sup:`234`\ U, and 0.2 wt % :sup:`236`\ U. The uranium isotopic content of the nitrate solution is 92.6 wt % :sup:`235`\ U, 5.9 wt % :sup:`238`\ U, 1.0 wt % :sup:`234`\ U and 0.5 wt % :sup:`236`\ U. The size of the metal cylinders is between 10 and 12 cm in diameter and height, and the size of the nitrate solution is on the order of 16 and 20 cm in diameter and height. The average mean-free path in the uranium metal is on the order of 1.5 cm, and 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 and the use of CENTRM/PMC is valid. See Examples 1–4 of :numref:`7-1a-2` for data input details for the Plexiglas and uranium metal. See Example 1 of :numref:`7-1a-5` for data input details for the uranyl nitrate solution. The XSProc data for this problem follow: :: SET UP 4 AQUEOUS 4 METAL fine_n READ COMP URANIUM 1 0.985 293 92235 93.2 92238 5.6 92234 1.0 92236 0.2 END SOLUTION 2 RHO[UO2(NO3)2]=415 92235 92.6 92238 5.9 92234 1.0 92236 0.5 MOLAR[HNO3]=9.783-3 DENSITY=1.555 TEMPERATURE=293 END SOLUTION PLEXIGLAS 3 END END COMP Consider the same materials above except rearrange them so that a 10 cm diameter uranium metal sphere sits inside a 50 cm diameter spherical tank of uranyl nitrate solution having a 1-cm thick Plexiglas wall. This problem can be modeled in SCALE but only CENTRM/PMC will treat the resonance processing correctly. This problem is modeled below. :: SET UP 4 AQUEOUS 4 METAL fine_n READ COMP URANIUM 1 0.985 293 92235 93.2 92238 5.6 92234 1.0 92236 0.2 END SOLUTION 2 RHO[UO2(NO3)2]=415 92235 92.6 92238 5.9 92234 1.0 92236 0.5 MOLAR[HNO3]=9.783-3 DENSITY=1.555 TEMPERATURE=293 END SOLUTION PLEXIGLAS 3 END END COMP READ CELLDATA MULTIREGION SPHERICAL END 1 5.0 2 25.0 3 26.0 END ZONE END CELLDATA .. _7-1c-8: Cell weighting an infinite homogeneous problem ---------------------------------------------- Cell weighting an infinite homogeneous medium has no effect on the cross sections because there is only one zone and one set of cross sections. However, a cell-weighted mixture number can still be supplied using the keyword **CELLMIX**\ = followed by an unique mixture number. This cell-weighted mixture number can be used in subsequent codes and will produce results similar to the cross sections of the original mixture. EXAMPLE 1 This problem would probably be run with CSAS1 to provide the k-infinity of 20%-enriched UO\ :sub:`2`. :: 20% ENRICHED UO2 BILLET fine_n READ COMP UO2 1 0.99 293 92235 20 92238 80 END END COMP READ CELLDATA INFHOMMEDIUM 1 CELLMIX=100 END END CELLDATA .. _7-1c-9: Cell weighting a LATTICECELL problem ------------------------------------ Cell weighting used with a **LATTICECELL** problem creates cell-weighted homogeneous cross sections that represent the characteristics of the heterogeneous unit cell. This cell-weighted mixture can then be used in a subsequent code for the overall volume where the cells are located without having to mock up the actual 3-D heterogeneous array of cells. This cell-weighted homogeneous mixture is designated by the user with the keyword **CELLMIX**\ = immediately followed by an unused mixture number. This needs to follow immediately after the cell description. Note that the mixtures used in the unit cell data cannot be used in a subsequent code because they have been flux weighted to create the user specified mixture. Therefore, if a mixture used in the unit cell description is also to be used in a subsequent code, another mixture must be created that is identical except for the mixture number. Every mixture that is to be used in a subsequent code except zero (i.e., void) must be defined in the standard composition data. A byproduct of the cell-weighting calculation is the eigenvalue (k-effective) of an infinite array of the cell described as the unit cell. EXAMPLE 1 Consider a cylindrical stainless steel tank filled with spherical pellets of 2.67%-enriched UO\ :sub:`2` arranged in a close-packed "triangular" pitch, flooded with borated water at 4350 ppm. The cylindrical stainless tank is sitting in a larger tank filled with borated water at 4350 ppm. The data for the UO\ :sub:`2` and borated water were developed in detail in Example 3 of :numref:`7-1c-2`. The stainless steel must be defined, and mixture 3 was chosen because mixture 1 was the UO\ :sub:`2` and mixture 2 was the borated water. Because the borated water will be used as a reflector for the stainless steel tank and has been used in the unit cell data, it must be repeated with a different mixture number (in this case, as mixture 4). In the subsequent calculation, user specified cell mixture 100 will be used to represent the UO\ :sub:`2` pellets in the borated water, mixture 3 will represent the stainless steel tank, and mixture 4 will represent the borated water reflector around the stainless-steel tank. The XSProc data for creating the cell-weighted cross sections on mixture 100 follow: :: SPHERICAL PELLETS IN BORATED WATER fine_n READ COMP UO2 1 .9398 293. 92235 2.67 92238 97.33 END ATOMH3BO3 2 0.025066 3 5000 1 1001 3 8016 3 1.0 293 END H2O 2 0.984507 293 END SS304 3 1.0 293 END ATOMH3BO3 4 0.025066 3 5000 1 1001 3 8016 3 1.0 293 END H2O 4 0.984507 293 END END COMP READ CELLDATA LATTICECELL SPHTRIANGP PITCH 1.0724 2 FUELD 1.0724 1 CELLMIX=100 END END CELLDATA .. _7-1c-10: Cell weighting a MULTIREGION problem ------------------------------------ A **MULTIREGION** problem is cell weighted primarily to obtain a cell-weighted homogeneous cross section that represents the characteristics of the heterogeneous unit cell. The eigenvalue obtained for a **MULTIREGION** problem with cylindrical or spherical geometry having a white boundary condition specified on the right boundary approximates an infinite array of the cells. A vacuum boundary condition would represent a single cell. A slab with reflected boundary conditions for both boundaries represents an infinite array of slab cells. The cell-weighted cross sections for spherical or cylindrical geometries with a white right boundary condition do not use a Dancoff correction and thus may not be accurate for representing a large array of the specified units. EXAMPLE 1 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/cm\ :sup:`3`. 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 Plexiglas collar is not significantly different from water and does not surround the fuel, so it will be ignored. Because this makes the problem a 1-D geometry, it can be defined using the **MULTIREGION** type of calculation and the eigenvalue of the system can be obtained without additional data by executing CSAS1 with CENTRM/PMC, if PARM=CENTRM is specified on the command line. The abundance of uranium is not stated beyond 97.67% enriched, so assume the remainder is :sup:`238`\ U. The XSProc data follow: :: =CSAS5 SMALL WATER REFLECTED SPHERE ON PLEXIGLAS COLLAR fine_n READ COMP URANIUM 1 DEN=18.794 1 293. 92235 97.67 92238 2.33 END H2O 2 END END COMP READ CELLDATA MULTI SPHERICAL CELLMIX=100 END 1 6.5537 2 30.0 END ZONE END CELLDATA • • • KENO DATA THAT USES MIX=100 FOR A HOMOGENEOUS SPHERE OF 30-CM RADIUS GOES HERE. • • END