Page 166 - Formation Damage during Improved Oil Recovery Fundamentals and Applications

144 Thomas Russell et al.

Eq. (3.149) shows that:
S a 5 0:
½Š (3.152)
Thus, the mass balance condition indicates that:
C ½Š 5 α C ½Š: (3.153)

To satisfy the condition that α can be less that one, this becomes:
C 5 0:
½Š (3.154)
Thus, the suspended concentration is continuous, which differs from
the solution presented in Section 3.5 for equilibrium particle detachment.
The solution of C 5 0 for all X . T follows in the same manner as
presented previously. As the suspended concentration is zero ahead of the
concentration front, and given that C is continuous across this front, the
suspended concentration must also be zero behind it. This gives the initial
condition for the second region:

T 5 X:C 5 0: (3.155)
Substituting the equation for the straining rate (Eq. (3.95)) and the
solution for the attached concentration (Eq. (3.149)) into the dimension-
less mass balance equation presents a first-order hyperbolic differential
equation in C that can be solved using the initial condition (Eq. (3.155)):

@C @C @S a
1 α 52 αΛC 2 : (3.156)
@T @X @T
Between the particle and salinity fronts (αT , X , T), the PDE
(Eq. (3.156)) reduces to an ordinary differential equation:
dC @S a
52 αΛC 2 ; (3.157)
dT @T
along parametric curves given by,
dX
5 α: (3.158)
dT
From the solution for the attached concentration, it follows that for all
X , T:

@S a ð S a1 2 S a0 Þ X2T
52 e ε : (3.159)
@T ε