University:
Massachusetts Institute of Technology
Course:
12.520 | Geodynamics
Academic year:

2023
Views:

125

Pages:

9
Author:

atapitfbn

Gibbs method - takes advantage of constraints from thermo. equilib. & mass balance. equilib. constraints Σᵢ Yᵢₘ = 0 i=1,...,s one for every stoichiometric equation written among the phase components of an equilibrium assemblage. only necessary to write a linearly independent subset of all the possible reactions. of linearly indep. = total # of reactions # of system components in all phases stoichiometric constraints Σᵢ xᵢₖ = 1 i=1,...,s mass balance constraint one for each independent sys comp. specifies the bulk comp. of the rock by describing the amt. of each comp. present nph - nm² = E one for each sys comp. in all phases K - # of sys comp. mᵢ - the number of moles of system component i is generally known because it defines the bulk composition. The variable nᵢₖ can be expressed in terms of xᵢₖ - the mole fraction of component i in phase k - and mₖ - the # of moles of each phase component (i,k) the mole fraction of phase component j in the mineral or melt... Xⱼₖ or nᵢₖ = mₖ Xⱼₖ and the full mass balance becomes mᵢ = ∑ₖ mₖ nᵢₖ = ∑ₖ mₖ Xⱼₖ intensive variables in the Gibbs minimization of free energy are T, P, alk the number of independent variables is given by the Gibbs phase rule F = nsc + 2 - nph For each phase i in the system, there is a Gibbs-Duhem equation that can be used to specify relations at T, P, μᵢ 0 = dT - VdP + Σᵢ nᵢdμᵢ one for each phase Number of reactions For any stoichiometric reaction among the phases in the system Σᵢ νᵢᵢ = 0 and there are nrx = npt - nsc independent reactions npt = # of phases nsc = # of system components For system MgO-FeO-SiO₂ 2 linearly independent reactions: (20) 2MgSiO₃ = Mg₂SiO₄ + SiO₂ (21) 2MgSiO₃ + Fe₂SiO₄ = 2FeSiO₃ + Mg₂SiO₄ conditions of equilibrium: (20) -RT ln K²₀ = ΔH°₂₀ + ∑ᵢ Sᵢ^(0) dT + dV_P - T(∑ᵢ Sᵢ^(0) + dV_P) (21) -RT ln K²₁ = ΔH°²₁ + ∑ᵢ Sᵢ^(0) dT + dV_P - T(∑ᵢ Sᵢ^(0) + dV_P) Stoichiometric constraints: XSiO₂ + XFeO + XMgO = 1 of intensive variables = 7 TP, XFeO, XMgO, XSiO₂, En, Fs, Fo also constraints: mass balance = 5 2 degrees of freedom the total # of constraints = 5 thermo + stoichiometric + mass balance Total # of variables = 10 T, P, XFeO, XMgO, XSiO₂, XFe₂O₃, XMg₂SiO₄, XFeSiO₃, XFe₂SiO₄, XMg₂SiO₄ For a variance of 2 - so - any 2 variables can be taken as independent now - to do with all of this information = to choose variables as independent specify values for each independent variable solve for dependent variables we have been doing this graphically throughout showing a T-X diagram and fixing P₀ choosing T,x as independent variables Solving equations to use Newton-Raphson - need to take total derivative of each equation with respect to the independent variables: T, P, Xᵢ so for each M in an stoichiometric constraint [d\mu_{i} = \left(\frac{\partial \mu_{i}}{\partial T}\right){P,X{j}} dT + \left(\frac{\partial \mu_{i}}{\partial P}\right){T,X{j}} dP + \sum_{j=1}^{n} \left(\frac{\partial \mu_{i}}{\partial X_{j}}\right){T,P} dX{j}] since the nonideal term RT ln Xᵢ was in the phase - all the other Xᵢs of the composition of the phase change its value [\frac{\partial \mu_{i}}{\partial X_{j}} = RT \left(\frac{\partial \ln X_{i}}{\partial X_{j}}\right){T,P} + \frac{\partial \mu{i}^{ex}}{\partial X_{j}}] plug these constraints into the original 2 stoic. coefficients -ΔS dT + ΔV dP + ∑ᵢ=1^m Vᵢ RT ln Xᵢ + ∑ᵢ=1^m ∑ⱼ=1^nct Yᵢⱼ RT ln Xⱼ|i,T,P,vₖ dXᵢ Stoic. constraints ∑ᵢ=1^nct dXᵢₖ = 0 mass balance constraints for a closed system ∑ₖ=1^nph mₖ ∑ᵢ=1^nct nᵢₖ Xᵢₖ + ∑ₖ=1^nph dmk ∑ᵢ=1^nct nᵢₖ Xᵢₖ solve using Newton-Raphson find f(x) = 0 where f(x) = f(x₀) + (∂f/∂x)|x₀ Δx