Page 48 - Physical Principles of Sedimentary Basin Analysis

30 Properties of porous media

(b) The average permeability can be written as

k f w f
k av = k 0 1 + (2.94)
k 0 w 0
which shows that the average permeability is much larger than the rock permeability k 0
when k f w f
k 0 w 0 .
3
(c) The condition for fracture-dominated average permeability becomes w /12
k 0 w 0
f
or w f
(12k 0 w 0 ) 1/3 .
(d) w
10 −5 m.
Exercise 2.20 Let a thin horizontal sheet of thickness h 1 be inserted into a layer of rock
with thickness h 0 . The sheet and the layer have the permeabilities k 1 and k 0 , respectively.
The sheet has a thickness that is much less than the layer (h 1
h 0 ).
(a) Show that the average vertical permeability of the composite layer is

k 0 , h 1 /h 0
k 1 /k 0 , rock-dominated
k av = (2.95)
k 1 (h 0 /h 1 ), h 1 /h 0
k 1 /k 0 , sheet-dominated.
There are two regimes, one where the sheet does not alter the permeability of the block sig-
nificantly (rock-dominated), and another regime where the sheet controls the permeability
of the block (sheet-dominated).
(b) Assume that there is a pressure difference p across the layer in the vertical direction.
Show that this pressure difference is almost entirely across the sheet when the average
permeability is sheet-dominated.
Solution:
(a) We have
h 0 + h 1 k 0
k av = ≈ . (2.96)
(h 0 /k 0 ) + (h 1 /k 1 ) k 0 h 1
1 +
k 1 h 0
(b) The vertical Darcy flow through the layer is
k av p k 1 p
u = ≈ (2.97)
μ (h 0 + h 1 ) μ h 1
when h 0
h 1 .

Exercise 2.21 A rectangular block is composed of two parts – one part of permeability
k 1 and thickness h 1 and another part of permeability k 2 and thickness h 2 , as shown by
Figure 2.19. The permeabilities parallel and normal to the layer are

k 1 h 1
k 2 h 2
Figure 2.19. A block is composed of two layers – one with permeability k 1 and thickness h 1 and
another of permeability k 2 and thickness h 2 . See Exercise 2.21.