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Here J stands for the mean fluid flux density in the porous channels. In our case the
Onsager coefficients are L 12 D L 21 D C, L 22 D , and L 11 D k= where k is
the rock permeability and is the viscosity of underground fluid.
The laboratory tests on zeta potential measurement are usually based on mea-
surements of streaming potential coefficient and using Eq. ( 8.8 ) or other analogous
dependences in order to extract the value of zeta potential from these measurements
(e.g., see Wiese et al. 1971 ; Ishido and Mizutani 1981 ; Hunter 1981 ; Jouniaux and
Pozzi 1999 ; Jouniaux and Bordes 2012 ; Luong and Sprik 2013 ). The experiments
with granite, quartz, and other water-saturated rocks have shown that the zeta
potential is usually negative in sign. Typically the zeta potential varies from several
unities to several tens mV and these values can increase with the enhancement
of pH or temperature of electrolyte. The observed values & D 100-120 mV at
the temperature T D 300 K (Ishido and Mizutani 1981 ) seem to have been
overestimated because the maximal value of the zeta potential corresponds to the
condition e& max k B T whence it follows that & max 50 mV.
To estimate the streaming potential coefficient we choose such groundwater
parameters as " D 80, D 10 4 Pa s, f D 1 15 S/m and & D 10 2 -10 1 V.
Substituting these parameters into Eq. ( 8.6 ) we obtain the numerical estimate C
0.01-1 V/Pa which is close to the abovementioned values measured for granites,
sandstones, and other porous rocks.
8.1.2
Electrokinetic Effect in Homogeneous Media
The low-frequency electric field is derivable from a potential function dž through
E Dr dž. Substituting this relation into Eq. ( 8.8 ) and taking the notice of
continuity equation for the electric current; that is r j D 0, we come to the following
equation:
r . r dž C C r P/ D 0:
(8.11)
Here P stands for the excess fluid pressure in pores with respect to hydrostatic fluid
pressure. Assuming that the ground is a homogeneous conductor with constant
and C, one can reduce Eq. ( 8.11 ) to the following:
2 D 0;
r
(8.12)
where ‰ D dž C CP.Let z be the upward-directed axis perpendicular to the
ground surface z D 0. As the atmosphere is considered as an insulator, the vertical
component of the total current has to be equal to zero at z D 0. Equation ( 8.12 )
under the boundary conditions j z D @ z D 0 and ‰ D 0 has the trivial solution
D 0 everywhere. Whence we get
dž D CP:
(8.13)
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