12.520 Lecture Notes 27
Flow in Porous Media
Problem of great economic importance (also scientific)
• hydrology (ground water migration, toxic waste)
• oil migration
• soil stability, fault mechanics (pore pressure)
• melt migration in mantle
• geysers and hot springs
Porous medium ⇒ voids ⇒ porosity φ
φ ≡ volume fraction of voids
For example,
Sand: φ ∼ 40%
Pumice: φ ∼ 70%
Oil shales: φ ∼ 10−20%
If pore connected ⇒ permeable
Pressure gradient ⇒ flow
k
Darcy’s law ⇒ v = − ∇p
η
v ≡ volumetric flow rate
k ≡ permeability
We can use Poiseuille flow for simple geometries. For example, cubical matrix, circular
tubes or pipes.
b
δ
Figure 27.1. An idealized model of a porous medium.
Circular tubes of diameter δ form a cubical matrix with
dimensions b.
Figure 27.1
Figure by MIT OCW.
⎛δ ⎞
1
12 ⋅ ⋅ π ⋅ ⎜ ⎟ ⋅ b
4
⎝2⎠
2
φ=
b3
=
3π δ 2
4 b2
dp
(one direction only)
dx
δ 2 dp
[Poiseuille flow]
In each pipe (along x), u = −
32η dx
Consider
⎛δ ⎞
1
4⋅ ⋅u ⋅ π ⋅⎜ ⎟
4
⎝2⎠
2
Darcy velocity: v =
b
2
v=−
b2φ 2 dp
72πη dx
⇒
k=
1 2 2
bφ
72π
=
πδ 2
4b
2
u=
φ
3
u
Large b ⇒ large v?
b2 =
Large φ ⇒ large v?
k=
3π δ 2
4 φ
π δ4
128 b2
Compare to cubes separated along faces (channel flow)
b
δ
Figure 27.2
Figure by MIT OCW.
1
6 ⋅ ⋅ δ b2
δ
φ= 23
=3
b
b
Again,
dp
directed along one edge
dx
u=
(
1 dp 2
Z − (δ / 2)2
2η dx
)
δ/2
1 dp ⎛ Z 3 δ 2 Z ⎞
5δ 2 dp
u=
=−
⎜ −
⎟
2ηδ dx ⎝ 3
2 ⎠ −δ / 2
24η dx
bδ
5 δ 3 dp
5 b2φ 3 dp
Darcy velocity: v = 2 2 u = −
=−
12 bη dx
324 η dx
b
k=
5b2φ 3
324
k is different depending on φ .
k=
135 δ 3
324 b
Clearly, porosity distribution is important.
Figure 27.3
Figure by MIT OCW.
Also -- more easily measured than figured out theoretically – more complicated
geometries → numerical simulation.
Consider “Lawn Sprinkler” example – flow in unconfined aquifer.
y
"Phreatic surface"
Land surface
h(x)
u(x)
x
Figure 27.4
Figure by MIT OCW.
h ≡ “hydraulic head”
u → Darcy velocity
Dupuit approximation:
dp
∂h
= ρg
dx
∂x
∂h
= 1 flow is one-dimensional.
∂x
k ρ g ∂h
Darcy’s law: u = −
η ∂x
For
Conservation of mass: Assume no input
Flux Q = u(x)h(x) = −
k ρ g dh
h = const.
η dx
⇒ phreatic surface is a parabola
For h = h0 at x = 0
⎛
2Qη x ⎞
h = ⎜ h0 2 −
⎟
kρ g ⎠
⎝
1/ 2
Suppose we have a porous dam of width w. The relation between Q, h0 and h1 is:
Q=
kρg 2
h0 − h12
2η w
Q=
kρ g ⎡
h0 − h1 h0 + h1 ⎤⎦
⎣
2η w
(
)
or
(
)(
)
y
w
Phreatic Surface
A
Dupuit Parabola
h0
B
Seepage face
C
h1
x
E
Impermeable Layer
D
Figure 27.5. Unconfined flow through a porous dam. The Dupuit parabola AC is the solution
if (h0-h1)/h0<<1. The actual phreatic surface AB lies above the Dupuit parabola resulting in a
seepage face BC.
Figure 27.5
Figure by MIT OCW.