Page 41 - Physical Principles of Sedimentary Basin Analysis

2.9 Anisotropic permeability 23

(a) Check that the unit vectors are orthonormal.
(b) Find the rotation matrix (2.57).
(c) Check that the transpose of the rotation matrix is its inverse.

Exercise 2.15
(a) Show that the rotation of the gradient operator is the same as the rotation of a vector.
(b) Show that the Laplace operator is rotation invariant.
Solution: (a) Equation (2.53) is a linear transformation of a coordinate position in the
unprimed to the primed reference system. Let f be a function of x and z. The chain rule of
differentiation then gives
∂ f ∂ f ∂x ∂ f ∂z ∂ f ∂ f
= + = cos θ − sin θ (2.63)
∂x ∂x ∂x ∂z ∂x ∂x ∂z

∂ f ∂ f ∂x ∂ f ∂z ∂ f ∂ f
= + = sin θ + cos θ (2.64)

∂z ∂x ∂z ∂z ∂z ∂x ∂z

which is the same as

∇ f = R −T ∇ f (2.65)
or

∇ = R∇. (2.66)
(b) We have that
T
2 T T T T 2
∇ = (∇ ) ∇ = R∇ R∇ =∇ R R∇= ∇ ∇= ∇ . (2.67)

2.9 Anisotropic permeability
The permeability of rocks is rarely isotropic. Especially sedimentary rocks often have a
layered structure caused by the deposition process. The permeability normal to the layers
is often much less than the permeability parallel to the layers because of low permeable
sheets. It is therefore necessary to distinguish between the two perpendicular directions.
Assume that a rock has a layered structure in the z-direction, and that the permeability in
the xy-plane is isotropic. Darcy’s law for both perpendicular directions at the same time
can be written as the vector
⎛ ⎞

k x ∂
v x − 1 k x 0
v D = = ⎝ μ ∂x ⎠ =− ∇ (2.68)
v z k z ∂ μ 0 k z
−
μ ∂z
where k x is the permeability in the bedding plane and k z is the permeability normal to the
bedding plane. The anisotropic permeability is a diagonal matrix

k x 0
K = . (2.69)
0 k z