0405mix

MIXING CALCULATIONS (Faure, Chpt 18).

        The history of the Universe and of the Earth can be looked at in terms of continuous compositional fractionation - from hydrogen to helium, from helium to the periodic elements, to galactic systems, to solar systems, to planetary systems, to Earth reservoirs, etc. In the case of the Earth, subsequent to the formation of the Earth's core, the most important fractionation process involves unmixing in response to decreasing pressure and/or temperature as hot mantle material convects to levels of lower gravitational potential. Further unmixing fractionation takes place with the crystallization of olivine and other mafic phases as the magma temperature drops, and the denser material separates from the magma under the influence of gravity.
      Nevertheless, many geological processes involve the mixing of phases with similar physical properties. Mafic and felsic magmas may mix to form magmas of intermediate composition, and magmatic gases, e.g. Mount Pinatubo, inevitably mix into the atmosphere. Mafic magmas also commonly assimilate the lower temperature melting crustal material into which they intrude, and liquids produced by unmixing in subduction zones may remix with the overlying mantle causing it to melt and rise towards the surface.
        Hydrous fluids of different composition mix under many different circumstances: mineral rich hydrothermal fluids rise out of the oceanic crust to mix with sea water, oceanic layers of different composition are mixed by convective overturn, meteoric waters mix with brines, and rivers inevitably mix at their confluence.
       Eroded material undergoes physical mixing during riverine and glacial transportation and deposition.
       Metamorphic rocks may also mix as a result of diffusive reactions.

Given the importance of mixing in geological processes, it is therefore appropriate that we develop some appreciation of the arithmetic of mixing.
References to Faure in the following text are to the book 'Principles and Applications of Geochemistry', First Edition, by Gunter Faure, Prentice Hall.

Binary mixtures

        Since the problems set by Faure concern the mixing of K and Mg, the latter 
elements are used as components in the following mixing equations.

        The potassium content, Km, of a material formed by the mixing of a fraction 'fa' 
of material A containing KA concentration units (e.g. %; kg per litre) of K and a 
fraction (1-fa) of material B containing KB concentration units of K, is given by the equation:

1)                              Km = fa*KA + (1-fa)*KB

2)                  and the proportion of KA = fa = (Km-KB)/(KA-KB)
                    and the proportion of KB = fb = (Km-KA)/(KB-KA);
                 (Note (Km-KA) and (KB-KA) are negative terms in this case.)

        and the Mgm content of a material composed of a fraction 'fa' of material A with MgA 
        of Mg and a fraction (1-fa) of material B with MgB of Mg  is:

3)                 Mgm = fa*MgA + (1-fa)*MgB  (Problem 1, 8, 10, 12, Chpt 18, Faure)

        Solving for 'fa' in 1) and 3), then:

4)                  fa = (Km-KB)/(KA-KB) = (Mgm-MgB)/(MgA-MgB)

         (Problem 3, 4c, Chpt 18, Faure)

        By cross multiplication and re-ordering of the terms in 4), we get:

        Km(MgA-MgB) - KB(MgA-MgB) = Mgm(KA-KB) - MgB(KA-KB)
        Km(MgA-MgB) = Mgm(KA-KB) - MgB(KA-KB)+ KB(MgA-MgB) 
        Km          = Mgm(KA-KB)/(MgA-MgB) - (MgB(KA-KB)- KB(MgA-MgB)) / (MgA-MgB)Km
                    = Mgm(KA-KB)/(MgA-MgB) - (MgBKA-MgBKB - MgAKB+MgBKB)/(MgA-MgB)

5)      Km        = Mgm(KA-KB)/(MgA-MgB)  +  (KB.MGA - KA.MgB)/(MgA-MgB)
                (Problem 2, 5, 6, 11, Chpt 18, Faure)

        This equation represents a straight line in coordinates of Km and Mgm. What 
is the slope and intercept?

        If the mixing is represented by a set of samples, the equation of the line can be 
easily established using least squares regression analysis. It is important to note however 
that the equation does not establish the composition of the end members.  We only know 
that the end members lie on the mixing line.
        Notice that if K and Mg are absent from component B, that is KB and MgB = 
0, then :

6)                              fa=Km/KA = Mgm/MgA, and

7)                              Km = Mgm. (KA/MgA) + 0 = Mgm.(KA/MGA)

(See problems 4a, 4b, 7, Chpt 18, Faure)
        Equation 7) represents a dilution (concentration) line along which all points 
have the same ratio of K to Mg. Consequently, the K/Mg ratio at any point is also the 
ratio of K to Mg in the primary undiluted solution.

        Ternary Mixtures

        Where mixing involves three end members, the intermediate members of the 
mixing series occupy an area on a bipolar graph. If one of the components contains none 
of the elements being plotted and serves only to dilute the other two components, the 
graph takes on a special form. (see problems 4, 7, Chpt 18, Faure):

        Mixing ratios

        In many cases, e.g. Rb, Sr and their isotopes, it is convenient to represent 
chemical variation in terms of element ratios rather than absolute abundances. 
Nevertheless, it is still necessary to know the absolute abundance of one of the ratioed 
elements in each end member, or of three ratios. Ratios cannot cannot be treated as if 
they are simple single element variables. In other words, the Rb/Sr of a mixture 
cannot be calculated knowing only the Rb/Sr ratio of the end members. 

8a)     RbM = RbA*fa + RbB*(1-fa)
8b)     SrM = SrA*fa + SrB*(1-fa)

        if RM = RbM/SrM = [Rb/Sr]M, the Rb/Sr ratio in the mixture, then in terms 
of RbA, RbB, SrA and SrB:

8c)     RM = (RbA*fa + RbB*(1-fa))/(SrA*fa + SrB*(1-fa))
        and
9)      fa = (RbB-RM*SrB)/(RM*(SrA-SrB)-(RbA-RbB))

        In terms however of RbA, RbB, and the ratio terms [Rb/Sr]A, and [Rb/Sr]B, 

10)     RM = (fa*RbA + (1-fa)*RbB)/(fa*RbA/[Rb/Sr]A+(1-fa)*RbB/[Rb/Sr]B)

        It is also possible to rewrite this equation in terms of the THREE ratios 
[Rb/Sr]A, [Rb/Sr]B, and [SrB/SrA]

11)     RM = (fa*[Rb/Sr]A+(1-fa)*[Rb/Sr]B*[SrB/SrA])/(fa+(1-fa)*[SrB/SrA])

        Proof:
        RM = (fa*RbA        + (1-fa)*RbB)             / (fa*SrA+(1-fa)*SrB)
           = (fa*RbA/SrA + (1-fa)*RbB/SrA)            / (fa         +(1-fa)*SrB/SrA)
           = (fa*RbA/SrA + (1-fa)*RbB/SrA*(SrB/SrB))  / (fa         +(1-fa)*SrB/SrA)
           = (fa*RbA/SrA + (1-fa)*RbB/SrB*(SrB/SrA))  / (fa         +(1-fa)*SrB/SrA) 
           = (fa*[Rb/Sr]A + (1-fa)*[Rb/Sr]B*[SrB/SrA])/ (fa         +(1-fa)*[SrB/SrA])

        ISOTOPIC RATIOS

        Many branches of geology involve a study of isotope ratios, in particular the 
isotopes of Rb and Sr where 87Sr is the daughter product of 87Rb. They are particularly 
valuable in scenarios involving fractionation or unmixing because we can assume that 
the original material and the fractionation products will all have the same daughter 
product isotopic ratio at the time of differentiation, irrespective of the amount of 
fractionation. For this reason we are able to calculate the age of rocks. However, 
variation in daughter product isotopic ratio can be produced by mixing two end member 
materials with different daughter product isotopic ratios, as in the case of the mixing of 
sediments or waters derived from sources with different Sr/Sr ratios. For this reason it is 
important to test whether or not the variation in any given set of isotopic data could be 
the result of mixing.

        Mixing lines of 87Sr/86Sr are hyperbolic and are described by the following 
equations, where 'g' refers to granite, 'b' to basalt, 't' to total Sr, and X to the mixing 
fraction.

12)     [87Sr/86Sr]MIX = (X * [87Sr/86Sr]g * Srgt / (X * Srgt + (1 - X)Srbt)
                                 + ((1 - X) * [87Sr/86Sr]b * Srbt / (X * Srgt + (1 - X) * Srbt),
the equation (18.20) given by Faure, Chpt 18, p. 379.

        An alternate formula involving the variables  [87Sr/86Sr]g, [87Sr/86Sr]b, and 
Srbt/Srgt is due to Anderson:

13)     [87Sr/86Sr]MIX = (X * [87Sr/86Sr]g + (1-X) * [87Sr/86Sr]b * Srbt/Srgt)
                                 / (X + (1 - X) * Srbt/Srgt),

(Compare [Rb/Sr]M   = (fa*[Rb/Sr]A + (1-fa)*[Rb/Sr]B*[SrB/SrA]) / (fa +(1-fa)
                                *[SrB/SrA]) in 12) above.)

(see problem 8, 12, Chpt 18, Faure)

    Anderson's equation is derived as follows:

[Sr87/Sr86]mix = ( X * 87Srg        + (1 - X)*87Srb       / ( X * 86Srg+(1-X)*86Srb)

        Divide the terms in the dividend and divisor by 86Srg

= (X*87Srg/86Srg + (1- X)*87Srb/86Srg)               / (X*86Srg/86Srg + (1-X)*86Srb/86Srg)

= (X*87Srg/86Srg + (1- X)*87Srb/86Srg*(86Srb/86Srb))   / (X + (1- X)*86Srb/86Srg)

= (X*87Srg/86Srg + (1- X)*87Srb/86Srb*(86Srb/86Srg))   / (X + (1- X)*86Srb/86Srg) 

= (X*[87Sr/86Sr]g + (1- X)*[87Sr/86Sr ]b*[86Srb/86Srg])/(X + (1- X)*[86Srb/86Srg])

        and since [86Srb / 86Srg]) = [Srbt/Srgt]

= (X*[87Sr/86Sr]g + (1 - X)*[87Sr/86Sr ]b*[Srbt/Srgt])/(X + (1 - X)*[Srbt/Srgt])

=       A               +               B          *[Srbt/Srgt]/        C

        Andersons equation can be converted to that of Faure by dividing terms A, B, and C by 
the term C, where: 

C = (X+(1-X)* [Srbt/Srgt]) = (X+(1-X)*[Srbt/Srgt])*Srgt/Srgt = (X*Srgt+(1-X)*Srbt)/Srgt,

        Term C is thus reduced to unity to give equation 12):

[87Sr/86Sr]MIX = (X*[87Sr/86Sr]g * Srgt / (X * Srgt + (1 - X) * Srbt)  +  ((1 - 
X)*[87Sr/86Sr]b *Srbt / (X * Srgt + (1 - X)*Srbt


        If the ratio 87Sr/86Sr is the same in the end member mixing components then 
dividing through by 87Sr/86Sr, equation 13) [87Sr/86Sr]MIX = (X*[87Sr/86Sr]g* 
Srgt/(X*Srgt+(1-X)Srbt)+((1-X)*[87Sr/86Sr]b*Srbt/(X*Srgt+(1-X)*Srbt), reduces to:

14a)    1 = X*Srgt/X*Srgt+(1-x)*Srbt)+(1-X)*Srbt/(X*Srgt+(1-x)*Srbt), 
and

14b)    Total Sr in the mixture = X*Srgt+(1-X)*Srbt, the equation (18.20) given by 
Faure, Chpt 18, p. 379.

********************************

        If the Sr concentration of the end member components (Srgt, Srbt) are equal, 
then :

        Srgt/(X*Srgt+(1-x)*Srbt) = Srbt/(X*Srgt+(1-x)*Srbt) = 1, and

15)     [87Sr/86Sr]MIX = (X*[87Sr/86Sr]g) + ((1-X)*[87Sr/86Sr]b), equation 
(18.21) given by Faure, Chpt 18, p. 379.

        This equation is usually employed when examining the mixing relationships of 
isotopes of oxygen, hydrogen, or carbon, when it can be assumed that the end members 
of the mixture have similar concentration of these elements. Note that this may not be 
assumed to be the case for the mixing of brine and meteoric water.

**********************************

        The radioactive isotope of Rb, 87Rb, breaks down to 87Sr by the conversion of 
one neutron to one proton. The change in concentration of 87Sr due to the breakdown of 
87Rb over a period of time T is given by the relationship:

16)     87Sr/86Sr (Total, now) = 87Sr/86Sr(Original) + 87Rb/86Sr(now) [e^LT - 1], 
where L is the decay constant for 87Rb - 87Sr

        Consequently the 87Sr/86Sr ratio of rocks may vary simply as a consequence of 
the variation in concentration of 87Rb, no mixing being implied whatsoever.
        Whether the variation is the result of mixing can be tested however, by 
combining equation 12) with the simple mixing relationship for total strontium:

17)                     SrMIX = X*Srgt+(1-X)*Srbt and X = (SrMIX-Srbt)/Srgt-Srbt)

        By substituting for X in equation 12), we obtain the relationship :

        [87Sr/86Sr]MIX = {Srgt*Srbt*([87Sr/86Sr]b-[87Sr/86Sr]g)/(SrMIX*(Srgt-Srbt))}
                                +{(Srgt *       [87Sr/86Sr]g-Srbt*[87Sr/86Sr]b)/(Srgt-Srbt)}

        This equation has the form of a straight line:

18)                     [87Sr/86Sr]MIX  = a/SrMIX+b

        A plot of 87Sr/86Sr against the reciprocal of SrMIX should generate a straight 
line if the variation of 87Sr/86Sr is the result of mixing, and the value of a and b in 
hyperbolic equation 18) can be easily determined by regression of the data set. Faure 
provides an interesting example of this procedure in the case of brine mixing in the 
Clinton sandstones of Ohio.



        Solving equations by Iteration

        The equation for calculating the Rb/Sr ratio of a mixture is:

        RBSRMIX(R) = (X*RBG + (1-X)*RBB) / (X*RBG/RBSRG+(1-X)*RBB/RBSRB)

        To solve for X the equation can be rearranged into the form:

        X = (RBB-RBSrMIX*RBB/RBSrB)/(RBSrMIX*(RBG/RBSrG-RBB/RBSrB)-(RBG-RBB))

        However the use of a computer allows another approach called iteration, 
involving a simpler rearrangment of the equation. If all the terms in equation 16) are 
balanced, then by simple cross multiplication:

        (RBSrMIX * (X*RBG/RBSrG+(1-X)*RBB/RBSrB)) - (X*RBG + (1-X)*RBB) = 0

        In the case of equation 13) where one knows the value of [87Sr/86Sr]MIX and 
wishes to determine the value of the mixing fraction, the equation would be rearranged 
in the form:

        [87Sr/86Sr]MIX - (87Sr/86SrG*X*SrGt/(SrGt*X+SrBt*(1-X))+87Sr
        /86SrB*(1-X)*SrBt/(SrGt*X+SrBt*(1-X))) = 0

        Now we can ask the computer to carry out the calculation in the equation in a 
loop for values of X varying from 0 to 1 in incremental steps, say of .1. We also tell the 
computer to stop calculating when the value calculated is less than 0. The value of X at 
this point approximates the solution to the problem. In a spreadsheet the iterative 
prodedure is carried out by a procedure called ‘GOAL SEEK’. . Alternatively, a 
GWBASIC program segment would look like this:

2000 for r=0 to 1 step .1
3000 [87Sr/86Sr]MIX-(87Sr/86SrG*X*SrGt/(SrGt*X+SrBt*(1-X))+87Sr/86SrB
        *(1-X)*SrBt/(SrGt*X+SrBt*(1-X)))=ans
4000 if ans <0 go to 6000
5000 next r
6000 print X
        After examining the result we can narrow the range of values of 'r' and decrease 
the increment. In this way we can obtain a more accurate estimate of X.

FIGURES

Structural Provinces of North America.

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