Page 134 - Fundamentals of Computational Geoscience Numerical Methods and Algorithms

6.1 Problems between Fluids Mixing, Heat Transfer and Redox Chemical Reactions 123

kinds of equations: One is the same as the first category of mass transport equation
without any reaction term; while another remains the same as the third category of
reactive transport equation with a strong reaction term. Since the solution of a reac-
tive transport equation with a chemical reaction term is computationally much more
expensive than that of a mass transport equation without a chemical reaction term,
any reduction in the total number of reactive transport equations can significantly
save computer time in a numerical computation. Based on this idea, a decoupling
numerical procedure, which can be used to remove the coupling between redox
types of reactive transport equations, is proposed to solve coupled problems between
fluids mixing, heat transfer and redox chemical reactions in fluid-saturated porous
rocks. This allows the interaction between the solute molecular diffusion/dispersion,
advection and chemical kinetics to be investigated within and around faults and
cracks in the upper crust of the Earth.

6.1 Statement of Coupled Problems between Fluids Mixing, Heat
Transfer and Redox Chemical Reactions

For pore-fluid saturated porous rocks, Darcy’s law can be used to describe pore-
fluid flow and the Oberbeck-Boussinesq approximation is employed to describe a
change in pore-fluid density due to a change in the pore-fluid temperature. Fourier’s
law and Fick’s law can be used to describe the heat transfer and mass transport
phenomena respectively. If the pore-fluid is assumed to be incompressible, the
governing equations of the coupled problem between fluids mixing, heat transfer
and redox chemical reactions in fluid-saturated porous rocks can be expressed as
follows:

∂u ∂ν
+ = 0, (6.1)
∂x ∂y

K x ∂ P
u = − , (6.2)
μ ∂x

K y ∂ P
ν = − + ρ f g , (6.3)
μ ∂y
2
2
* ∂T ∂T ∂T ∂ T ∂ T
)
+ ρ f c u + ν ,
φρ f c pf + (1 − φ)ρ s c ps pf = λ ex + λ ey
∂t ∂x ∂y ∂x 2 ∂y 2
(6.4)
2 2

∂C i ∂C i ∂C i ∂ C i ∂ C i
φ + u + ν = D ex + D ey +φR i (i = 1, 2, ...,N),
∂t ∂x ∂y ∂x 2 ∂y 2
(6.5)
ρ f = ρ f 0 [1 − β T (T − T 0 )], (6.6)