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Re: Phreeqc

Hello Dr. Parkhurst,

My name is Barb Petrunic and I am a student at the University of New
working under the supervision of Kerry MacQuarrie and Tom Al.

I am trying to work through example problem 15 that is provided in the
program however, I am having trouble understanding the rates data block for

the Biomass in the problem.  I was wondering if you could answer a couple
questions with regards to this example problem:

1) why does the "moles" of biomass have to be negative?

The signs get a little confusing, but positive numbers indicate addition to
solution and negative numbers mean removal from solution. Still "positive
number" is a product of three factors, the "moles" saved in the basic
program (-3 in the following example), the coefficient in the -formula of
KINETICS (positive or negative, -1 in the following example), and possibly
a stoichiometric coefficient in the reactant part of -formula (always
positive, 1 for C and O and 2 for H in the following example). So a "saved"
moles of -3, of the -formula CH2O -1, would result in 3 mol C, 6 mol H, and
3 mol O being added to solution. If saved moles is 3 and -formula CH2O -1,
then the result would remove C, H, and O from solution. Similarly, if saved
moles is -3 and -formula CH2O 1, the result would remove moles from
solution. Finally, a saved moles of 3, and -formula CH2O 1, would add the
elements to solution.

Now the formula for Biomass is 0.0 H, so nothing is added or removed from
solution. However, a cumulative moles of kinetic reactant is calculated as
given by the "M" Basic function, and initialized with -M in KINETICS, is
calculated. The quantity of biomass is needed to calculate the rate
reactions because the rates of substrate usage depend on the amount of
biomass present. By convention, if the saved reaction moles is negative, M
is incremented; if the saved reaction moles is positive, M is decremented.
Note, that this is independent of the stoichiometry given in -formula. The
intention was that if something comes out of solution, then the moles of
the kinetic reactant increased, and if it goes into solution, the moles of
the kinetic reactant is decreased. In this case, nothing really comes out
of solution, but the convention is used to cause the biomass to increase,
and to increase the biomass (M), the saved moles must be negative.

One other point is that M can not be negative. If M is decremented to zero,
then a kinetic reaction that consumes the kinetic reactant will stop
because it has run out of reactant.

2) why is the "if" statement required in line 60 in the rates data block?

It is probably not needed, but it does point out that you can not have less
than 0 biomass. I think the program would automatically limit the removal
of reactant so that M can not be less than 0, but I'm not positive. Line 60
may help with the numerical integration of the rate equation.


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