How to Calculate Atomic Density for Borated Water (750 ppm) at 300 K?

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Hello everyone,

I am trying to calculate the atomic density (in atom/barn-cm) for borated water (750 ppm) at 300 K in the active core of a reactor. Could anyone guide me on the correct approach or provide a reference formula for this calculation?

Thanks in advance!
Hello everyone,

I am trying to calculate the atomic density (in atom/barn-cm) for borated water (750 ppm) at 300 K in the active core of a reactor. Could anyone guide me on the correct approach or provide a reference formula for this calculation?
I came across the following atomic densities (in atom/barn-cm) for borated water:
O-16.03c 3.323848e-01
O-17.03c 1.266139e-04
O-18.03c 6.830486e-04
H-1.03c 6.663122e-01
H-2.03c 7.663472e-05
B-10.03c 8.293027e-05
B-11.03c 3.338047e-04
I would like to understand how these atomic densities were calculated. Could someone provide a step-by-step explanation of the process?
Specifically, I am interested in: The formula used to compute atomic density.How the density of borated water (at a given temperature) is determined and converted to atomic density.The role of isotopic fractions in these calculations.
If possible, a numerical example demonstrating the calculations would be very helpful.

Thanks in advance!
 
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  • #2
To calculate an atomic density, one takes the mass density, divide mass density by the mass/g-mole and multiply by Avogadro's number, 6.02214076×10²³ mol⁻¹. That would yield atoms/unit volume, e.g, g/cm3 or kg/m3. Some codes like MCNP use cgs.

One can multiply atoms/cm3 by 1 x 10-24 cm2/barn, and the result is atoms/barn-cm.

Think about how one calculates the concentration of an acid in terms of molarity (moles of solute per liter of solution, or moles per 1000 cm3. It is convenient for some codes to use atoms/barn-cm since the microscopic cross-sections are given in barns, and one calculates the macroscopic cross-section (cm-1) from atoms/barn-cm * barn.

Take density of water at 1 g/cm3. To find density in atoms/barn-cm,


This yields, 3.3456 E-2 atoms/barn-cm.

The density giving 3.323848 E-2 atoms/barn-cm would be about 0.9935 g/cm3.
One can address isotopic vectors by using the isotopic abundances to adjust the atomic/molecular mass.

Also, it is important with boric acid to know if the concentration in ppm applies to B or to the molecule H3BO3. The difference in concentration between 750 ppm B and 750 ppm H3BO3 is substantial.

The values B-10.03c 8.293027e-05 and B-11.03c 3.338047e-04 indicate ~0.2 B-10, 0.8 B-11, which indicates natural boron. One may use enriched boron to lower the boric acid concentration.

Also, it is customary to buffer boric acid with LiOH to achieve a pH of 6.9 to 7.4, and ideally in the upper half of that range, and in the case of PWR cores, the Li is something like 0.999 7Li.

https://multimedia.3m.com/mws/media/950625O/3m-7li-enriched-lithium-hydroxide.pdf
 
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Astronuc said:
To calculate an atomic density, one takes the mass density, divide mass density by the mass/g-mole and multiply by Avogadro's number, 6.02214076×10²³ mol⁻¹. That would yield atoms/unit volume, e.g, g/cm3 or kg/m3. Some codes like MCNP use cgs.

One can multiply atoms/cm3 by 1 x 10-24 cm2/barn, and the result is atoms/barn-cm.

Think about how one calculates the concentration of an acid in terms of molarity (moles of solute per liter of solution, or moles per 1000 cm3. It is convenient for some codes to use atoms/barn-cm since the microscopic cross-sections are given in barns, and one calculates the macroscopic cross-section (cm-1) from atoms/barn-cm * barn.

Take density of water at 1 g/cm3. To find density in atoms/barn-cm,


This yields, 3.3456 E-2 atoms/barn-cm.

The density giving 3.323848 E-2 atoms/barn-cm would be about 0.9935 g/[cmsup]3[/sup].
One can address isotopic vectors by using the isotopic abundances to adjust the atomic/molecular mass.

Also, it is important with boric acid to know if the concentration in ppm applies to B or to the molecule H3BO3. The difference in concentration between 750 ppm B and 750 ppm H3BO3 is substantial.

The values B-10.03c 8.293027e-05 and B-11.03c 3.338047e-04 indicate ~0.2 B-10, 0.8 B-11, which indicates natural boron. One may use enriched boron to lower the boric acid concentration.

Also, it is customary to buffer boric acid with LiOH to achieve a pH of 6.9 to 7.4, and ideally in the upper half of that range, and in the case of PWR cores, the Li is something like 0.999 7Li.

https://multimedia.3m.com/mws/media/950625O/3m-7li-enriched-lithium-hydroxide.pdf
Dear expert, I did not understand how the calculations for Boron isotopes were made here. The values B-10.03c 8.293027e-05 and B-11.03c 3.338047e-04 How to calculate. Can you do the calculations you did for water for Boron?
 
  • #4
emilmammadzada said:
Can you do the calculations you did for water for Boron?
One can take the numbers for boron (0.2 10B and 0.811B), which would give a mean atomic mass of 10.8 g/mol and do some calculations as I did for water, or one can do similar calculations for each isotope.

The density of white crystalline boric acid is 1.435 g/cm3.

From B-10.03c 8.293027e-05 and B-11.03c 3.338047e-04, one can determine the moles of each B isotope present, and compare to the theoretical density to see how much is present.

750 ppm simply means 750 micrograms of boric acid (or boron) to 1 gram of water.

Consider this a homework assignment. We expect students to demonstrate effort and show their work.

Edit/update: I did a correction to the latex expression, where I changed atoms/g-mole to molecules/g-mole for water. When dealing with molecular mass, one needs to consider molecules rather than atoms. One molecule of water has 3 atoms, 2 of H and 1 of O. Also, normally when atomic masses are given in the periodic table, the isotopic abundance is considered. For diatomic hydrogen, one might find H2, HD, HT, DT, D2, T2 in the environment. Probably one would more likely find HD and HT in the environment, since there is much more H than D or T.
 
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Serpent Data Calculate Data
O-16.03c 3.32E-01 O-16.03c 3.31E-02
O-17.03c 1.27E-04 O-17.03c 1.28E-05
O-18.03c 6.83E-04 O-18.03c 6.64E-05
H-1.03c 6.66E-01 H-1.03c 6.64E-02
H-2.03c 7.66E-05 H-2.03c 9.96E-06
B-10.03c 8.29E-05 B-10.03c 8.19E-06
B-11.03c 3.34E-04 B-11.03c 3.35E-05
Dear expert, I did the calculation and when I compare the results, there is a difference. I got this data from the NuScale SMR example in serpent.I am attaching the excell file I made the calculation below. How do I solve my error and problem?
 

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  • #6
emilmammadzada said:
I am attaching the excell file I made the calculation below. How do I solve my error and problem?
One's isotopic abundances look reasonable. One appears to be off by about an order of magnitude.

In one's initial post, is the number reported for O-16.03c given as 3.323848e-01, or 0.33223284e-01, which = 3.32384e-02? Notice the decimal place. That might be the reason one's numbers appear to be off by an order of magnitude or a factor of 10.

In my water example, the result was 3.3456 E-2 atoms/barn-cm, which is 0.33456 e-01, which is close to 0.33223284e-01.
 
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Astronuc said:
One's isotopic abundances look reasonable. One appears to be off by about an order of magnitude.

In one's initial post, is the number reported for O-16.03c given as 3.323848e-01, or 0.33223284e-01, which = 3.32384e-02? Notice the decimal place. That might be the reason one's numbers appear to be off by an order of magnitude or a factor of 10.

In my water example, the result was 3.3456 E-2 atoms/barn-cm, which is 0.33456 e-01, which is close to 0.33223284e-01.
I have attached the original ppm file below. I think there is no shift here. I don't know where the calculation error is coming from.
 

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  • #8
emilmammadzada said:
I don't know where the calculation error is coming from.
Note that the input deck has " boron 750 ppm ", which is different than 750 ppm of boric acid!

So 750 ppm B * 61.83/10.8 = 4293.75 ppm boric acid.
 
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Astronuc said:
Note that the input deck has " boron 750 ppm ", which is different than 750 ppm of boric acid!

So 750 ppm B * 61.83/10.8 = 4293.75 ppm boric acid.
How do I do the calculation in this case, what steps do I follow to calculate atomic densities?
 
  • #10
emilmammadzada said:
How do I do the calculation in this case, what steps do I follow to calculate atomic densities?
I will look at the calculations later today, but usually one takes the molecular density, and multiplies by the ratio of atoms to molecule, and in the case of each isotope, one would multiply the atom density by the isotopic abundance.

For example, find the molecular density of H3BO3, then multiply by 1 atom B/molecule of H3BO3, and 3 atoms of H per molecule and 3 atoms of O/molecule, and then multiply by the appropriate isotopic abundance to get the atomic density of each species, e.g., 0.99986 for H, 0.00014 for D, and 0.9942 for O16, 0.00384 O17, and 0.002 for O18.
 
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Astronuc said:
I will look at the calculations later today, but usually one takes the molecular density, and multiplies by the ratio of atoms to molecule, and in the case of each isotope, one would multiply the atom density by the isotopic abundance.

For example, find the molecular density of H3BO3, then multiply by 1 atom B/molecule of H3BO3, and 3 atoms of H per molecule and 3 atoms of O/molecule, and then multiply by the appropriate isotopic abundance to get the atomic density of each species, e.g., 0.99986 for H, 0.00014 for D, and 0.9942 for O16, 0.00384 O17, and 0.002 for O18.
I performed the atomic density calculations as you suggested, but the results were ten times different from the data in the Serpent sample file. I hope your calculation of these values will also be insightful for me. Thank you again!
 
  • #12
emilmammadzada said:
I performed the atomic density calculations as you suggested, but the results were ten times different from the data in the Serpent sample file.
I check several sources for calculations of atom density. The references cited in the OP are off by an order of magnitude, or a factor of 10 too great.
emilmammadzada said:
O-16.03c 3.323848e-01
O-17.03c 1.266139e-04
O-18.03c 6.830486e-04
H-1.03c 6.663122e-01
H-2.03c 7.663472e-05
B-10.03c 8.293027e-05
B-11.03c 3.338047e-04

According to one source, the atomic density of O in water is 0.033428 atoms/barn-cm, and for H it is 0.066853 atoms/barn-cm. Using the isotopic abundance of O-16 at 0.9942, the atomic density is 0.033234 atoms/barn-cm.

One can multiply those numbers by the respective isotopic abundance to get the atoms/b-cm, for each isotope.

Taking Boron (B) with an atomic mass of 10.81 amu, and water at 18 amu,

0.00075 g B / (10.81 g/g-mole) = 6.938 e-5 moles of B
1 g H2O / (18 g/g-mole) = 5.556 e-2 moles of H20,

or 1.24884 e-3 atoms of B/molecule of H2O, and since there is one atom of O in H2O, one can simply multiply the stomic density of O in the H2O at 1 g/cm3, or 0.033428 atoms/b-cm * 1.24884 e-3 to obtain 4.17464 e-5. And since for B, there is 1 atom of B10 for 4 atoms of B11 in natural boron,
the atomic density of B10 ~ 0.2 * 4.17464 e-5 = 8.349 e-6 atoms/b-cm and the atomic density of B11 is 4* that of B10, or 3.334 e-5 atoms/b-cm.

Number will vary according to assumptions about density, isotopic abundance, and how the stoichiometry of solutions. For example 750 ppm B can be taken as 0.00075 g B in 1 g H2O, or more precisely, 0.00075 g B mixed with 0.99925 g H2O, or 1 gram of solution. Assumptions about isotopic abundance may vary, and the density is affected by pressure and temperature (and compressibility of a liquid).
 
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Astronuc said:
I check several sources for calculations of atom density. The references cited in the OP are off by an order of magnitude, or a factor of 10 too great.


According to one source, the atomic density of O in water is 0.033428 atoms/barn-cm, and for H it is 0.066853 atoms/barn-cm. Using the isotopic abundance of O-16 at 0.9942, the atomic density is 0.033234 atoms/barn-cm.

One can multiply those numbers by the respective isotopic abundance to get the atoms/b-cm, for each isotope.

Taking Boron (B) with an atomic mass of 10.81 amu, and water at 18 amu,

0.00075 g B / (10.81 g/g-mole) = 6.938 e-5 moles of B
1 g H2O / (18 g/g-mole) = 5.556 e-2 moles of H20,

or 1.24884 e-3 atoms of B/molecule of H2O, and since there is one atom of O in H2O, one can simply multiply the stomic density of O in the H2O at 1 g/cm3, or 0.033428 atoms/b-cm * 1.24884 e-3 to obtain 4.17464 e-5. And since for B, there is 1 atom of B10 for 4 atoms of B11 in natural boron,
the atomic density of B10 ~ 0.2 * 4.17464 e-5 = 8.349 e-6 atoms/b-cm and the atomic density of B11 is 4* that of B10, or 3.334 e-5 atoms/b-cm.

Number will vary according to assumptions about density, isotopic abundance, and how the stoichiometry of solutions. For example 750 ppm B can be taken as 0.00075 g B in 1 g H2O, or more precisely, 0.00075 g B mixed with 0.99925 g H2O, or 1 gram of solution. Assumptions about isotopic abundance may vary, and the density is affected by pressure and temperature (and compressibility of a liquid).
This seems to be the definition of material in serpent. I calculated it by applying what was written in this section at first. I could not find the reason why x10 was different.

"Material name is used to identify the material in cell cards (see Sec. 3.3.1 on page 25). Thenuclide names correspond to the identifier determined in the directory file. These identifiersdefine the cross section data used in the calculation. Densities and fractions can be givenas atomic or mass values. Positive entries refer to atomic densities (in units of 1024/cm3)and atomic fractions, respectively, and negative entries to mass densities (in units of g/cm3)and mass fractions. Isotopic compositions are normalized before the calculation and mixedentries are not allowed."
 
  • #14
emilmammadzada said:
Dear expert, I did the calculation and when I compare the results, there is a difference. I got this data from the NuScale SMR example in serpent.I am attaching the excell file I made the calculation below. How do I solve my error and problem?
From what document did one obtain this information? Does one have a link?
 
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  • #16
emilmammadzada said:
OK, I found it. I downloaded the gz file and found waters_czp_750ppm.inc.

I will have to study it and do an independent calculation. However, at first glance, it does not look correct.
 
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  • #17
emilmammadzada said:
TL;DR Summary: Hello everyone,

I am trying to calculate the atomic density (in atom/barn-cm) for borated water (750 ppm) at 300 K in the active core of a reactor. Could anyone guide me on the correct approach or provide a reference formula for this calculation?

Thanks in advance!

I came across the following atomic densities (in atom/barn-cm) for borated water:
O-16.03c 3.323848e-01
O-17.03c 1.266139e-04
O-18.03c 6.830486e-04
H-1.03c 6.663122e-01
H-2.03c 7.663472e-05
B-10.03c 8.293027e-05
B-11.03c 3.338047e-04

In checking another official source, I noticed that one table provides atom fractions of nuclides in one column and atomic densities (atom/b-cm) in an adjacent column. They are about an order of magnitude different in the case of water (with or without dilute solutions).

So, I wondered, and I checked the seven cited numbers above, and they sum to 1.000000 e+00, so it appears that the numbers given in the Serpent example are atomic fractions (dimensionless) for each nuclide, and not the atomic densities (atom/barn-cm) of the borated water (with 750 ppm B).

So the calculated atomic densities provided are probably correct as reasonable approximations with reasonable uncertainties.

The lesson here is to be careful about units, and a responsible scientist/engineer must understand the numbers (and units) he or she is observing. Formatting is absolutely crucial (critical) in writing code and input decks to codes (in modeling and simulation). Small errors could lead to significant errors and adverse quality (and consequences).


I just now checked the Serpent Input Syntax Manual
http://serpent.vtt.fi/mediawiki/index.php/Input_syntax_manual#mat_.28material_definition.29

Where is shows FRACn : fraction of n-th nuclide in composition (positive value = atomic fractions/density, it seems to imply one can specify atomic fraction or atomic density. So one has to be careful in using input decks and mixing/misinterpreting units.

mat (material definition)​


See Chapter 4 of [1].

mat NAME DENS [ tmp TEMP ]
[ tms TEMP ]
[ tft TMIN TMAX ]
[ rgb R G B ]
[ vol VOL ]
[ mass MASS ]
[ burn NR ]
[ fix ID TEMP ]
[ moder THNAME ZA ]
NUC1 FRAC1
[ NUC2 FRAC2 ]
[ ... ]


Material definition. The mandatory parameters are:


NAME : name of the material
DENS : density of the material (positive value = atomic density [in b-1cm-1], negative value = mass density [in g/cm3])
NUCn : Identifier of n-th nuclide in composition
FRACn : fraction of n-th nuclide in composition (positive value = atomic fractions/density, negative values = mass fractions/density)

The remaining parameters are defined by separate key words followed by the input values, being optional.

Notes:


  • There is a special entry for the DENSparameter:
    • "sum": to calculate the density from given nuclide fractions
  • The nuclide identifier for nuclides with associated cross-sections corresponds to ZZAAA.ID and, for nuclides without associated cross-sections, e.g., decay nuclides, to ZZAAAI.
    • The identifiers include Z, the atomic number; A, the mass number of the nuclide; I, the isomeric state (0 = ground state, 1 = metastable state); and ID, the library identifier.
  • For more information, see the detailed description on Definitions.

. . . .
 
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