.. _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.
..