Page 32 - Physical Principles of Sedimentary Basin Analysis

14 Properties of porous media

h 2

h 1
h

k=?

Δx
Figure 2.10. See Exercise 2.10.

Exercise 2.10 Fluid is driven through a porous core (or plug) by the weight of the fluid in
a cylinder, see Figure 2.10. Show that the permeability of the plug is
μ xln (h 2 /h 1 )
k = (2.33)
f gt
when the time t is needed for the fluid to drop from height h 2 to h 1 . Assume that the area
of the cross-section is the same for the cylinder as for the core.

2.5 Potential flow and gravity
We have seen that a fluid pressure gradient drives fluid through a porous medium. But, in
the vertical direction we have to subtract the effect of gravity, and Darcy’s law becomes
k ∂ p f
v D =− − f g (2.34)
μ ∂z
where p f is the fluid pressure and f is the fluid density. The z-axis is pointing downwards,
which implies that the weight of a fluid column is increasing with increasing z-coordinate.
We see that a fluid pressure equal to the weight of the fluid column, p f = f gz, leads
to zero Darcy flux. This explains why we had to subtract f g in the vertical direction in
Darcy’s law. In three dimensions Darcy’s law can be written as

k
v D =− ∇ p f − f gn z (2.35)
μ
where n z is the unit vector in the vertical direction. It is often convenient to express Darcy’s
law in terms of a pseudo-potential defined as the fluid pressure minus a hydrostatic
pressure p h,0 relative to a reference level as

= p f − p h,0 (2.36)