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.. _7-1a:

XSProc: Standard Composition Examples
=====================================

.. _7-1a-1:

Standard composition fundamentals
---------------------------------

The standard composition specification data are used to define mixtures
using standardized engineering data entered in a free-form format. The
XSProc uses the standard composition specification data and information
from the Standard Composition Library to provide number densities for
each nuclide of every defined mixture according to :eq:`eq7-1a-1`.

.. math::
  :label: eq7-1a-1

  NO = \frac{RHO \times AVN \times C}{AWT} ,

where

   NO is the number density of the nuclide in atoms/b-cm,

   RHO is the actual density of the nuclide in g/cm\ :sup:`3`,

   AVN is Avogadro’s number, 6.02214199 × 10\ :sup:`23`, in atoms/mol,

   C is a constant, 10\ :sup:`−24` cm\ :sup:`2`/b,

   AWT is the atomic or molecular weight of the nuclide in g/mol.


The actual density, RHO, is defined by

.. math::
  :label: eq7-1a-2

  RHO = ROTH \times VF \times WGTF ,

where

   RHO is the actual density of the standard composition in
   g/cm\ :sup:`3`,

   ROTH is either the specified density of the standard composition or
   the theoretical density of the standard composition in
   g/cm\ :sup:`3`,

   VF is a density multiplier compatible with ROTH as defined by Eq. ,

   WGTF is the weight fraction of the nuclide in the standard
   composition. This value is automatically obtained by the code from
   the Standard Composition Library. WGTF is 1.0 for a single nuclide
   standard composition.

.. math::
  :label: eq7-1a-3

  VF = DFRAC \times VFRAC ,

where

   VF is the density multiplier,

   DFRAC is the density fraction,

   VFRAC is the volume fraction.

To illustrate the interaction between ROTH and VF, consider an Inconel
having a density of 8.5 g/cm\ :sup:`3`. It is 7.0% by weight iron, 15.5%
chromium, and 77.5% nickel. The Inconel occupies a volume of
4 cm\ :sup:`3`.

**Method 1**:


To describe the iron, enter 8.5 for ROTH and 0.07 for VF.

To describe the chromium, enter 8.5 for ROTH and 0.155 for VF.

To describe the nickel, enter 8.5 for ROTH and 0.775 for VF.

**Method 2**:


Do not enter the density, and by default the theoretical density of each
component will be used for ROTH. DFRAC will be the ratio of the
specified density to the theoretical density. The specified density of
each component is the density of the Inconel × the weight fraction of
that component.

Thus, the density of the iron is 8.5 × 0.07   = 0.595 g/cm\ :sup:`3`

                         chromium is 8.5 × 0.155 = 1.318 g/cm\ :sup:`3`

                         nickel is 8.5 × 0.775 = 6.588 g/cm\ :sup:`3`.

To calculate DFRAC, the theoretical density of each material must be
obtained from the table *Elements and special nuclide symbols* in the
STDCMP chapter. These values are

7.86 g/cm\ :sup:`3` for iron

8.90 g/cm\ :sup:`3` for nickel

7.20 g/cm\ :sup:`3` for chromium

The DFRAC entered for the iron is 0.595/7.86 = 0.0757

                  for the nickel is 1.318/8.90 = 0.1481

                  for the chromium is 6.588/7.20 = 0.9163.

Since there are no volumetric corrections, VFRAC is 1.0 and the values of DFRAC are entered for VF.

**Method 3**:


Assume the Inconel, which occupies 4 cm\ :sup:`3`, is to be spread over
a volume of 5 cm\ :sup:`3`. Then the volume fraction, VFRAC, is
4 cm\ :sup:`3`/5 cm\ :sup:`3` = 0.8 and can be combined with the density
fraction, DFRAC, to obtain the density multiplier, VF.

To describe the iron, enter 8.5 for ROTH and  0.07 × 0.8 = 0.056 for VF

            or chromium, 	enter 8.5 for ROTH and 0.155 × 0.8 = 0.124 for VF

            for nickel, 	enter 8.5 for ROTH and 0.775 × 0.8 = 0.620 for VF.


Alternatively, the volume fraction can be applied to the density before
it is entered. Then the ROTH can be entered as 8.5 g/cm\ :sup:`3` × 0.8
= 6.8 g/cm\ :sup:`3`, and DFRAC is entered for the density multiplier,
VF.

To describe the iron, enter 6.8 for ROTH and 0.07 for VF

                for chromium, enter 6.8 for ROTH and 0.155 for VF

                for nickel, enter 6.8 for ROTH and 0.775 for VF.

**Method 4**:


Assume the Inconel, which occupies 4 cm\ :sup:`3`, is to be spread over
a volume of 5 cm\ :sup:`3`. Then the volume fraction, VFRAC, is
4 cm\ :sup:`3`/5 cm\ :sup:`3` = 0.8. Do not enter the density, and by
default the theoretical density of each component will be used for ROTH.

VF is then entered as the product of VFRAC and DFRAC according to Eq. .
The specified density of each component is the density of the Inconel ×
the weight fraction of that component.

Thus, the density of the 	iron is 8.5 × 	0.07   = 	0.595 g/cm\ :sup:`3`

                          chromium is 8.5 × 	0.155 = 	1.318 g/cm\ :sup:`3`

                          nickel is 8.5 × 	0.775 = 	6.588 g/cm\ :sup:`3`.

To calculate DFRAC, the theoretical density of each material must be obtained from :numref:`tab7-2-3`.  These values are

  7.86 g/cm\ :sup:`3` for iron
  8.90 g/cm\ :sup:`3` for nickel
  7.20 g/cm\ :sup:`3` for chromium.

Then DFRAC 	for the iron is 0.595/7.86 = 0.0756

            for nickel is 1.318/8.90 = 0.1481

  	        for chromium is 6.588/7.20 = 0.9150.


Then VF is DFRAC × VFRAC

VF 	for the iron is 0.0757 × 0.8 = 0.0606

    for nickel is 0.1481 × 0.8 = 0.1185

    for chromium is 0.9150 × .8 = 0.7320.


.. _7-1a-2:

Basic standard composition specifications
-----------------------------------------

EXAMPLE 1. Material name is given. Create a mixture 3 that is Plexiglas.

   Since no other information is given, the information on the Standard
   Composition Library can be assumed to be adequate. Therefore, the
   only data to be entered are the standard composition name and the
   mixture number

.. highlight:: scale

::

  PLEXIGLAS  3  END

EXAMPLE 2. Material name and density (g/cm\ :sup:`3`) are given.

  Create a mixture 3 that is Plexiglas at a density of
  1.15 g/cm\ :sup:`3`. Since no other data are specified, the defaults
  from the Standard Composition Library will be used. Therefore, the
  only data to be entered are the standard composition name, the
  mixture number, and the density.

::

  PLEXIGLAS  3  DEN=1.15  END

EXAMPLE 3. Material name and number density (atoms/b-cm) are given. Create a mixture 2 that is aluminum having a number density of 0.060244.

  ::

    AL  2  0  .060244  END

EXAMPLE 4. Material name, density (g/cm\ :sup:`3`) and isotopic abundance are given.

  Create a mixture 1 that is uranium metal at 18.76 g/cm\ :sup:`3` whose
  isotopic composition is 93.2 wt % :sup:`235`\ U, 5.6 wt % :sup:`238`\ U,
  and 1.0 wt % :sup:`234`\ U, and 0.2 wt % :sup:`236`\ U. This example
  uses the DEN= keyword to enter the density and define the standard
  composition. Example 5 demonstrates another method of defining the
  standard composition.

::

  URANIUM   1  DEN=18.76 1 300  92235  93.2  92238  5.6  92234  1.0  92236  0.2  END

EXAMPLE 5. Material name, density (g/cm\ :sup:`3`) and isotopic abundance are given.

   Create a mixture 7 defining B\ :sub:`4`\ C with a density of
   2.45 g/cm\ :sup:`3`. The boron is 40 wt % :sup:`10`\ B and 60 wt %
   :sup:`11`\ B. This example utilizes the **DEN**\ = keyword. Example 6
   illustrates an alternative description.

::

  B4C 7  DEN=2.45  1.0 300  5010  40.0  5011  60.0  END

EXAMPLE 6. Material name, density (g/cm\ :sup:`3`) and isotopic abundance are given.

   Create a mixture 7 defining B\ :sub:`4`\ C with a density of
   2.45 g/cm\ :sup:`3`. The boron is 40 wt % :sup:`10`\ B and 60 wt %
   :sup:`11`\ B. This example incorporates the known density into the
   density multiplier, *vf*, rather than using the **DEN**\ = keyword.
   The default density for B\ :sub:`4`\ C given in the COMPOUNDS table
   in the SCL section 7.2 is equal to 2.52 g/cm\ :sup:`3`.

::

  B4C  7  0.9722 300  5010  40.0  5011  60.0  END

.. note:: In the above examples, the actual density is input for
  materials containing enriched multi-isotope nuclides (uranium in
  Examples 4 and 5 and boron in Examples 6 and 7). The default density
  should never be used for enriched materials, especially low atomic mass
  neutron absorbers such as boron and lithium. The default density is a
  fixed value for nominal conditions and naturally occurring distributions
  of isotopes. Use of the default density for enriched materials will
  likely result in incorrect number densities

.. _7-1a-3:

User-defined (arbitrary) chemical compound specifications
---------------------------------------------------------

The user-defined compound option allows the user to specify materials
that are not found in the Standard Composition Library and can be
specified by the number of atoms of each element or isotope that are
contained in the molecule. To define a user-defined compound, the first
four characters of the standard composition component name must be
**ATOM**. The remaining characters of the standard composition component
name are chosen by the user. The maximum length of the standard
composition name is 16 characters. All the information that would
normally be found in the Standard Composition Library must be entered in
the user-defined compound specification. :ref:`7-1-3-3` contains data
input details for arbitrary compounds.

EXAMPLE 1. Density and chemical equation are given.

  Create a mixture 3 that is a hydraulic fluid,
  C\ :sub:`2`\ H\ :sub:`6`\ SiO, with a density of 0.97 g/cm\ :sup:`3`.
  The input data for this user-defined compound are given below:

::

  ATOM     3  0.97  4 6000 2 1001 6 14000 1 8000 1 END

EXAMPLE 2. Density and chemical equation are given. Create a mixture 7,
TBP, also known as phosphoric acid tributyl ester or tributylphosphate,
(C\ :sub:`4`\ H\ :sub:`9`\ O)\ :sub:`3`\ PO, having a density of 0.973
g/cm\ :sup:`3`.

::

  ATOMtbp         7  0.973  4 1001 27 6000 12 8016 4 15031 1 end

.. _7-1a-4:

User-defined (arbitrary) mixture/alloy specifications
-----------------------------------------------------

The user-defined compound or alloy option allows the user to specify
materials that are not found in the Standard Composition Library and are
defined by specifying the weight percent of each element or isotope
contained in the material. To define a user-defined weight percent
mixture, the first four characters of the standard composition component
name must be *wtpt*. The remaining characters of the standard
composition component name are chosen by the user. The maximum length of
the standard composition name is 16 characters. All the information that
would normally be found in the Standard Composition Library must be
entered in the arbitrary mixture/alloy specification. :ref:`7-1-3-3`
contains data input details for user-defined compounds.

EXAMPLE 1. Density and weight percents are given.

   Create a mixture 5 that defines a borated aluminum that is 2.5 wt %
   natural boron. The density of the borated aluminum is
   2.65 g/cm\ :sup:`3`.

::

  SOLUTION MIX=2 RHO[UO2(NO3)2]=415 92235 92.6 92238 5.9 92234 1 92236
  0.5 MASSFRAC[HNO3]=6.339-6 TEMPERATURE=293 END SOLUTION

EXAMPLE 2. Density, weight percents, and isotopic abundance are given.

   Create a mixture 5 that defines a borated aluminum that is 2.5 wt %
   boron. The boron is 90 wt % :sup:`10`\ B and 10 wt % :sup:`11`\ B.
   The density of the borated aluminum is 2.65 g/cm\ :sup:`3`. The
   minimum generic input specification for this arbitrary material is

::

  WTPTBAL  5  2.65  2  5000 2.5  13027 97.5  1  293  5010 90.  5011 10. END

.. _7-1a-5:

Fissile solution specifications
-------------------------------

Solutions of fissile materials are available in the XSProc. A list of
the available solution salts and acids is given in the table *Available
fissile solution components* in :ref:`7-2-3`. When the XSProc processes
a solution, it breaks the solution into its component parts (basic
standard composition specifications) and uses the solution density to
calculate the volume fractions.

EXAMPLE 1. Fuel density, excess acid and isotopic abundance are given.

   Create a mixture 2 that is a highly enriched uranyl nitrate solution
   with 415 g/L and 0.39 mg of excess nitrate per gram of solution. The
   uranium isotopic content 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 temperature is 293 Kelvin.

::

  SOLUTION MIX=2 RHO[UO2(NO3)2]=415 92235 92.6 92238 5.9 92234 1 92236
  0.5 MASSFRAC[HNO3]=6.339-6 TEMPERATURE=293 END SOLUTION

where

  The molecular weight of NO\ :sub:`3` is 62.0049 g/mole, of H is
  1.0078 g/mole, so the grams of excess H per gram of solution is
  1.0078 / 62.0049 × (0.39 mg/g) × (1 g/1000 mg) = 6.339 ×
  10\ :sup:`-6`.

.. _7-1a-6:

Combinations of standard composition materials to define a mixture
------------------------------------------------------------------

Frequently more than one standard composition is required to define a
mixture. This section contains such examples.

EXAMPLE 1. Boral from B\ :sub:`4`\ C and Aluminum.

   Create a mixture 6 that is Boral, 15 wt % B\ :sub:`4`\ C and 85 wt %
   Al, having a density of 2.64 g/cm\ :sup:`3`. Natural boron is used in
   the B\ :sub:`4`\ C. Note that Example 2 demonstrates the use of the
   keyword **DEN**\ = to enter the density of the mixture and avoid
   having to look up the theoretical density from the table *Isotopes in standard
   composition library,* in the section 7.2.2, and calculate the density
   multiplier (VF)

::

  B4C  6  0.1571  END
   AL  6  0.8305  END

EXAMPLE 2. Boral from B\ :sub:`4`\ C and Aluminum.

   This is the same problem as Example 1 using a different method of
   specifying the input data. Create a mixture 6 that is Boral, 15 wt %
   B\ :sub:`4`\ C and 85 wt % Al, having a density of
   2.64 g/cm\ :sup:`3`. Natural boron is used in the B\ :sub:`4`\ C.

::

  B4C  6  DEN=2.64  0.15  END
   AL  6  DEN=2.64  0.85  END

EXAMPLE 3. Boral from Boron, Carbon, and Aluminum.

    If neither Boral nor B\ :sub:`4`\ C were available in the Standard
    Composition Library, Boral could be described as follows:

    Create a mixture 2 that is Boral composed of 35 wt % B\ :sub:`4`\ C and
    65 wt % aluminum with an overall density of 2.64 g/cm\ :sup:`3`. The
    boron is natural boron.

     *vf* is the density multiplier. (The density multiplier is the ratio
     of actual to theoretical density.) From the Standard Composition
     Library chapter, table *Isotopes in standard composition library*,
     the theoretical density of aluminum is 2.702 g/cm\ :sup:`3`; boron is
     2.37 g/cm\ :sup:`3`; and carbon is 2.1 g/cm\ :sup:`3`. The density
     multiplier, *vf*, for Al is (0.65)(2.64)/2.702 = 0.63509. The
     isotopic abundances in natural boron are known to have some
     variability. Here it is assumed that natural boron is 18.4309 wt %
     :sup:`10`\ B at 10.0129 amu and 81.5691 wt % :sup:`11`\ B at
     11.0096 amu. C is 12.000 amu.

    Convert the weight percents to atom percents for the natural boron where
    *w* denotes weight fraction, *a* denotes atom fraction, and *M* denotes
    atomic mass:

.. math::

  w_{B10} = 0.184309 \equiv \frac{a_{B10}M_{B10}}{a_{B10}M_{B10} + a_{B11}M_{B11}} = \frac{a_{B10}(10.0129)}{a_{B10}(10.0129) + (1-a_{B10}))(11.0093)}

Solving for :math:`a_{B10}` gives:

.. math::

  [{{\text{a}}_{\text{B10}}}\text{=0.184309}\ \ \text{=}\ \ \frac{\text{(0.184309)}\ \text{(11.0093)}}{\ \text{(0.184309)}\ \text{(11.0093)-(0.184309)}\ \text{(10.0129)+(10.0129)}}\quad \text{=}\ \ \text{19.900}

Therefore the atom percent of :sup:`11`\ B is, *a\ B*\ :sub:`11` = 80.1
a%.

Similarly, the mass of the B\ :sub:`4`\ C molecule is

   [(0.199 × 4 × 10.0129) + (0.801 × 4 × 11.0093) + (12.000)] =
   55.24407 amu.


The mass of the boron is (55.24407 − 12.000) = 43.24407 amu.

The *vf* of boron would be :math:`\left( \frac{43.24407}{55.24407} \right)\left( \frac{(0.35)(2.64)}{2.37} \right)` = 0.30519

The *vf* of C would be

.. math::

  \left( \frac{12.0000}{55.24407} \right)\left( \frac{(0.35)(2.64)}{2.1} \right) = 0.09558


.. math::

  \left(\frac{12.000}{55.25045}\right)\left[\frac{(0.35)(2.64)}{2.30}\right] = 0.08725

The standard composition input data for the Boral follows:

::

  AL	2	0.63509	END
  BORON	2	0.30519	END
  C	2	0.09558	END

EXAMPLE 4. Boral from :sup:`10`\ B, :sup:`11`\ B, Carbon, and Aluminum.

  Create a mixture 2 that is Boral composed of 35 wt % B\ :sub:`4`\ C
  and 65 wt % aluminum. The Boral density is 2.64 g/cm\ :sup:`3`. The
  boron is natural boron.

  *vf* is the density multiplier. Use 0.63581 for AL and 0.08725 for C
  as explained in Example 3 above. From the Standard Composition
  Library chapter, *Isotopes in standard composition library* table,
  the theoretical density of :sup:`10`\ B is 1.00 g/cm\ :sup:`3` and
  :sup:`11`\ B is 1.00 g/cm\ :sup:`3`. As computed in Example 3, the
  mass of the B\ :sub:`4`\ C molecule is 55.25045 amu, and the boron is
  19.764 atom % :sup:`10`\ B and 80.236 atom % :sup:`11`\ B. The mass
  of :sup:`10`\ B is 10.0129 amu and the :sup:`11`\ B is 11.0096. Thus,
  the *vf* of :sup:`10`\ B is

  .. math::

    \left( \frac{(4)(0.199)(10.0129)}{55.24407} \right)\left( \frac{(0.35)(2.64)}{1.0} \right)\ \ =\ \ 0.13331\ .

  The *vf* of :sup:`11`\ B is

  .. math::

    \left( \frac{(4)(0.801)(11.0093)}{55.24407} \right)\left( \frac{(0.35)(2.64)}{1.0} \right)\ \ =\ \ 0.58998\ .

The standard composition input data for the Boral are given as

::

  AL	2	0.63509	END
  B-10	2	0.13331	END
  B-11	2	0.58998	END
  C	2	0.09558	END


EXAMPLE 5. Specify all of the number densities in a mixture.

  Create a mixture 1 that is vermiculite, defined as

    hydrogen at a number density of 6.8614−4 atoms/b-cm

    oxygen at a number density of 2.0566−3 atoms/b-cm

    magnesium at a number density of 3.5780−4 atoms/b-cm

    aluminum at a number density of 1.9816−4 atoms/b-cm

    silicon at a number density of 4.4580−4 atoms/b-cm

    potassium at a number density of 1.0207−4 atoms/b-cm

    iron at a number density of 7.7416−5 atoms/b-cm

  In this example we use the 2\ :sup:`nd` syntax option described in
  :ref:`7-1-3-3`, in which the 3rd entry must be 0. The standard
  composition input data for the vermiculite are given below:

  ::

    H	1	  0	  6.8614-4	  END
    O	1	  0	  2.0566-3	  END
    MG	1	  0	  3.5780-4	  END
    AL	1	  0	  1.9816-4	  END
    SI	1	  0	  4.4580-4	  END
    K	1	  0	  1.0207-4	  END
    FE	1	  0	  7.7416-5	  END

.. _7-1a-7:

Combinations of user-defined compound and user-defined mixture/alloy to define a mixture
----------------------------------------------------------------------------------------

Mixtures can usually be created using only basic standard composition
specifications. Occasionally, it is convenient to create two or more
user-defined materials for a given mixture. This procedure is
demonstrated in the following example.

EXAMPLE 1. Specify Boral using a user-defined compound and user-defined mixture/alloy.

   Create a mixture 6 that is Boral, 15 wt % B\ :sub:`4`\ C and 85 wt %
   Al, having a density of 2.64 g/cm\ :sup:`3`. Natural boron is used in
   the B\ :sub:`4`\ C. Boral can be described in several ways.
   For demonstration purposes, it will be described as a combination of
   a user-defined compound and user-defined mixture/alloy. This is not
   necessary, because both B\ :sub:`4`\ C and Al are available as
   standard compositions. A method of describing the Boral without using
   user-defined compounds or user-defined mixtures/alloys is given in
   Examples 1 and 2 of :ref:`7-1a-6`. The minimum generic input
   specifications for this user-defined compound and alloy are

   ::

     ATOM-B4C	6	2.64 	2	5000		4	6012	1	0.15	END
     WTPT-AL	6	2.64	1	13027		100.0		0.85	END

.. _7-1a-8:

Combinations of solutions to define a mixture
---------------------------------------------

This section demonstrates the use of more than one solution definition
to describe a single mixture. The assumptions used in processing the
cross sections are likely to be inadequate for solutions of mixed oxides
of uranium and plutonium. Therefore, this section is given purely for
demonstration purposes.

EXAMPLE 1. Solution of uranyl nitrate and plutonium nitrate.

  Note that the assumptions used in processing the cross sections are
  likely to only be adequate for CENTRM/PMC calculations of mixed-oxide
  solutions. This example is given purely for demonstration purposes.
  Create a mixture 1 consisting of a mixture of plutonium nitrate
  solution and uranyl nitrate solution. The specific gravity of the
  mixed solution is 1.4828. The solution contains 325.89 g (U + Pu)/L
  soln. The acid molarity of the solution is 0.53. In this solution
  77.22 wt % of the U+Pu is uranium. The isotopic abundance of the
  uranium is 0.008% :sup:`234`\ U, 0.7% :sup:`235`\ U, 0.052%
  :sup:`236`\ U, and 99.24% :sup:`238`\ U. The isotopic abundance of
  the plutonium is 0.028% :sup:`238`\ Pu, 91.114% :sup:`239`\ Pu, 8.34%
  :sup:`240`\ Pu, 0.426% :sup:`241`\ Pu, and 0.092% :sup:`242`\ Pu.
  Note that a single quote in the first column indicates a comment line
  in SCALE input.

  ::

    '   Uranium density of 77.22% of 325.89 g/L
    SOLUTION  MIX=1  RHO[UO2(NO3)2]=251.65  92234 .008 92235 .700 92236 .052
                                            92238 99.240
    '   Plutonium density if 22.78% of 325.89 g/L
                     RHO[PU(NO3)4]=74.24  94238 .028 94239 91.114 94240 8.34
                                          94241 .426 94242 .092
    '   Acid molarity is 0.53 M
                     MOLAR[HNO3]=0.53
    '   Specifying the density over specifies the problem, which means the solution may
    '   not be in thermodynamic equilibrium.  The specification below adds about 0.3%
    '   extra hydrogen to the problem
                     DENSITY=1.4828
    END SOLUTION

.. _7-1a-9:

Combinations of basic and user-defined standard compositions to define a mixture
--------------------------------------------------------------------------------

EXAMPLE 1. Burnable poison from B\ :sub:`4`\ C and Al\ :sub:`2`\ O\ :sub:`3`.

   Create a mixture 6 that is a burnable poison with a density of
   3.7 g/cm\ :sup:`3` and composed of Al\ :sub:`2`\ O\ :sub:`3` and
   B\ :sub:`4`\ C. The material is 1.395 wt % B\ :sub:`4`\ C. The boron
   is natural boron. This material can be easily specified using a
   combination of user-defined material to describe the
   Al\ :sub:`2`\ O\ :sub:`3` and a simple standard composition to define
   the B\ :sub:`4`\ C. The minimum generic input specification for this
   user-defined material and the standard composition are

   The density multiplier of the B\ :sub:`4`\ C is the density of the
   material times the weight percent, divided by the theoretical density
   of B\ :sub:`4`\ C [(3.7 × 0.01395)/2.52] or 0.02048; the density
   multiplier of the Al\ :sub:`2`\ O\ :sub:`3` is 1.0 – 0.01395 or
   0.98605 (the theoretical density of B\ :sub:`4`\ C was obtained from
   *Isotopes in standard composition library* table in the STDCMP
   chapter).

   The input data for the burnable poison are given below:

   ::

     ATOM-AL2O3  6  3.70  2  13027  2  8016  3  0.98605  END
     B4C  6  2.048-2  END

   The B\ :sub:`4`\ C input can be specified using the **DEN**\ = parameter
   as shown below:

   ::

     ATOM-AL2O3  6  3.70  2  13027  2  8016  3  0.98605  END
     B4C  6  DEN=3.7  0.01395  END

   The fraction of B\ :sub:`4`\ C in the mixture is ((3.7 × 0.01395)/2.52)
   = 0.02048. The fraction of Al\ :sub:`2`\ O\ :sub:`3` in the mixture is
   1.0 – 0.02048 = 0.979518. The density of the Al\ :sub:`2`\ O\ :sub:`3`
   can be calculated as shown below.

   .. image:: figs/XSProcAppA/math1.png
    :align: center
    :width: 500

   Input data using the density of Al\ :sub:`2`\ O\ :sub:`3` are given
   below:

   ::

     ATOM-AL2O3  6  3.72467  2  13027  2  8016  3  END
     B4C  6  2.048-2  END

EXAMPLE 2. Borated water from H\ :sub:`3`\ BO\ :sub:`3` and water.

  Create a mixture 2 that is borated water at 4350 parts per million
  (ppm) by weight, resulting from the addition of boric acid,
  H\ :sub:`3`\ BO\ :sub:`3` to water. The density of the borated water
  is 1.0078 g/cm\ :sup:`3` (see “Specific Gravity of Boric Acid Solutions,” Handbook of Chemistry, 1162, Compiled and Edited by Norbert A. Lange, Ph.D, 1956.). The solution temperature
  is 15ºC and the boron is natural boron.

An easy way to describe this mixture is to use a combination of a
user-defined compound to describe the boric acid, and a basic
composition to describe the water.

STEP 1.  INPUT DATA TO DESCRIBE THE USER-DEFINED COMPOUNDThe generic input data
for the boric acid are given below.  The actual input data are derived in steps 2 through 5.

::

  ATOMH3BO3  2  0.025066  3  5000  1  1001  3  8016  3  1.0  288.15  END

STEP 2.  AUXILIARY CALCULATIONS FOR THE USER-DEFINED COMPOUND INPUT DATA

In calculating the molecular weights, use the atomic weights from SCALE,
which are available in the table *Isotopes in standard composition
library* in :ref:`7-2-2` of the SCALE manual. The atomic weights used
in SCALE may differ from some periodic tables. The SCALE atomic weights
used in this problem are listed below:

  H (1001) 1.0078

  O (8016) 15.9949

  :sup:`10`\ B 10.0129

  :sup:`11`\ B 11.0093

The natural boron abundance, in weight percent, is defined to be:

  :sup:`10`\ B 18.4309

  :sup:`11`\ B 81.5691

The molecular weight of natural boron is given by

  DEN nat B/AWT nat B = DEN :sup:`10`\ B/AWT :sup:`10`\ B + DEN :sup:`11`\ B/AWT :sup:`11`\ B

                        DEN :sup:`10`\ B = WTF :sup:`10`\ B × DEN nat B

                        DEN :sup:`11`\ B = WTF :sup:`11`\ B × DEN nat B

where:

  DEN is density in g/cm\ :sup:`3`,

  AWT is the atomic weight in g/mol,

  WTF is the weight fraction of the isotope.

Substituting,

  DEN nat B/AWT nat B = DEN nat B × ((WTF :sup:`10`\ B/AWT
  :sup:`10`\ B) + (WTF :sup:`11`\ B/AWT :sup:`11`\ B))

Solving for AWT nat B yields:

  AWT nat B = 1/((WTF :sup:`10`\ B/AWT :sup:`10`\ B) + (WTF
  :sup:`11`\ B/AWT :sup:`11`\ B))

The atomic weight of natural boron is thus

  1.0/((0.184309 g :sup:`10`\ B/g nat B/10.0129 g :sup:`10`\ B/mol
  :sup:`10`\ B) +
  (0.815691 g :sup:`11`\ B/g nat B/11.0093 g /mol :sup:`11`\ B)) =
  10.81103 g nat B/mol nat B

The molecular weight of the boric acid, H\ :sub:`3`\ BO\ :sub:`3` is
given by:

   (3 × 1.0078) + 10.81103 + (3 × 15.9949) = 61.8191

Calculate the grams of boric acid in a gram of solution:

   Boric acid, H\ :sub:`3`\ BO\ :sub:`3` is 61.8222 g/mol

   Natural boron is 10.81261 g/mol

   (4350 × 10\ :sup:`–6` g B/g soln) × (1 mol/10.81261 g B) × (61.8191 g
   boric acid/mol) =

   0.024874 g boric acid/g soln (2.4874 wt %)

Interpolating from the referenced page from Lange's Handbook of Chemistry, the specific gravity of the boric acid
solution at 2.4872 weight percent is 1.0087. This value is based on
water at 15ºC. The density of pure air free water at 15°C is
0.99913 g/cm\ :sup:`3`. Therefore, the density of the boric acid
solution is 1.0087 × 0.99913 g/cm\ :sup:`3` = 1.0078 g
soln/cm\ :sup:`3`.

Calculate ROTH, the theoretical density of the boric acid.

   1.0078 g soln/cm\ :sup:`3` × 0.024874 g boric acid/g soln =
   0.025068 g boric acid/cm\ :sup:`3`

STEP 3. DESCRIBE THE BASIC STANDARD COMPOSITION INPUT DATA

::

  H2O     2  0.984507  288.15  END

where the volume fraction =0.984506 (see step 4 auxiliary calculations below)

STEP 4. AUXILIARY CALCULATIONS FOR THE BASIC STANDARD COMPOSITION INPUT
DATA

Calculate the volume fraction of the water in the solution, assuming
0.9982 is the theoretical density of water from :numref:`tab7-2-4`. Each gram
of solution contains 0.024872 g of boric acid, so there is 0.975128 g of
water in each gram of solution. The volume fraction of water is then
given by:

  (1.0078 g soln/cm\ :sup:`3` × 0.975128 g water/g soln)/0.9982 g
  water/cm\ :sup:`3` = 0.984506

STEP 5.  CREATE THE MIXTURE FOR BORATED WATER

::

  ATOMH3BO3  2  0.025068  3  5000  1  1001  3  8016  3  1.0  288.15  END
  H2O                   2  0.984506  288.15  END

.. _7-1a-10:

Combinations of basic and solution standard compositions to define a mixture
----------------------------------------------------------------------------

The solution specification is the easiest way of specifying the
solutions listed in the *Available fissile solution components* table in
:ref:`7-2-3`. A combination of solution and basic standard compositions
can be used to describe a mixture that contains more than just a
solution as demonstrated in the following example.

EXAMPLE 1. Uranyl nitrate solution containing gadolinium.

   Create a 4.306% enriched uranyl nitrate solution containing 0.184 g
   gadolinium per liter. The uranium in the nitrate is 95.65%
   :sup:`238`\ U, 0.022% :sup:`236`\ U, 4.306% :sup:`235`\ U, and 0.022%
   :sup:`234`\ U. The uranium concentration is 195.8 g U/L and the
   specific gravity of the uranyl nitrate is 1.254. There is no excess
   acid in the solution. The presence of the gadolinium is assumed to
   produce no significant change in the solution density. The solution
   is defined to be mixture 3.

::

  SOLUTION  MIX=3
    RHO[UO2(NO3)2]=195.8 92238 95.65 92236 0.022 92235 4.306 92234 0.022
    VOL_FRAC=0.99985
    DENSITY=1.254
  END SOLUTION
  GD  3  0.000184  293  END

.. _7-1a-11:

Combinations of user-defined compound and solution to define a mixture
----------------------------------------------------------------------

The solution specification is the easiest way of specifying the
solutions listed in the *Available fissile solution components* table in
:ref:`7-2-3` of the SCALE manual. A solution specification and
user-defined compound specification can be used to describe a mixture
that contains more than just a solution as demonstrated in the following
example.

EXAMPLE 1. Uranyl nitrate solution with gadolinium nitrate.

   Create a 4.306% enriched uranyl nitrate solution containing
   gadolinium in the form of Gd(NO\ :sub:`3`)\ :sub:`3`. The uranium in
   the nitrate is 95.65% :sup:`238`\ U, 0.022% :sup:`236`\ U, 4.306%
   :sup:`235`\ U, and 0.022% :sup:`234`\ U. The uranium concentration is
   195.8 g U/L and the density of the uranyl nitrate is 1.254. There is
   no excess acid in the solution. The concentration of the gadolinium
   is 0.184 g/L. The volume fraction of the mixture that is uranyl
   nitrate (0.99985 = 1.254/ (1.254 + 0.000184)). The solution is
   defined to be mixture 3.

::

  SOLUTION  MIX=3
    RHO[UO2(NO3)2]=195.8 92238 95.65 92236 0.022 92235 4.306 92234 0.022
    VOL_FRAC=0.99985
    DENSITY=1.254
  END SOLUTION

The density of the gadolinium is given as 0.184 g/L. To describe the
user-defined compound, the density of the Gd(NO\ :sub:`3`)\ :sub:`3` is
needed. The atomic weights from the Standard Composition Library are:

   Gd 157.25

   N 14.0067

   O 15.999

Therefore, the density of the Gd(NO\ :sub:`3`)\ :sub:`3` = 0.000184 g
Gd/cm\ :sup:`3` × (157.25 + 3(14.0067 + 3(15.999))/157.25) =
0.0004017 g/cm\ :sup:`3`.

The input data for this user-defined compound are given below:

::

  ATOMGD(NO3)3  3  .0004017  3  64000  1  7014  3  8016  9  1.0  300  END

The complete input data for the mixture of uranyl nitrate and gadolinium nitrate are given as:

::

  SOLUTION  MIX=3
    RHO[UO2(NO3)2]=195.8 92238 95.65 92236 0.022 92235 4.306 92234 0.022
    VOL_FRAC=0.99985
    DENSITY=1.254
  END SOLUTION
  ATOMGD(NO3)3  3  .0004017  3  64000  1  7014  3  8016  9  1.0  300  END

.. note:: Since the default temperature (300 K) is to be used, it can be
  omitted from the user-defined compound standard composition. The
  temperature must be entered if the standard composition contains a
  multiple-isotope nuclide whose isotopic abundance is to be specified.






..