12.005 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 by MIT OCW.
2
1
⎛δ ⎞
12 ⋅ ⋅ π ⋅ ⎜ ⎟ ⋅ b
3π δ 2
4
⎝2⎠
φ=
=
b3
4 b2
dp
(one direction only)
dx
δ 2 dp
[Poiseuille flow]
In each pipe (along x), u = −
32η dx
Consider
2
1
⎛δ ⎞
4 ⋅ ⋅ u ⋅π ⋅ ⎜ ⎟
2
4
⎝ 2 ⎠ = πδ u = φ u
Darcy velocity: v =
b2
4b 2
3
v=−
b 2φ 2 dp
72πη dx
⇒
k=
1 2 2
bφ
72π
Large b ⇒ large v?
b2 =
Large φ ⇒ large v?
k=
3π δ 2
4 φ
π δ4
128 b 2
Compare to cubes separated along faces (channel flow)
b
δ
Figure 27.2
Figure by MIT OCW.
1
6 ⋅ ⋅ δ b2
δ
=3
φ= 23
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
Darcy velocity: v = 2
5 δ 3 dp
5 b 2φ 3 dp
bδ
=
−
=
−
u
12 bη dx
324 η dx
b2
5b 2φ 3
k=
324
k is different depending on φ .
135 δ 3
k=
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:
For
∂h
∂x
dp
∂h
= ρg
∂x
dx
1 flow is one-dimensional.
Darcy’s law: u = −
k ρ g ∂h
η ∂x
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
1/ 2
⎛
2Qη x ⎞
h = ⎜ h0 2 −
⎟
kρg ⎠
⎝
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=
kρg
⎡( h0 − h1 )( h0 + h1 ) ⎤⎦
2η w ⎣
or
y
w
A
Phreatic Surface
Dupuit Parabola
h0
B
Seepage face
C
E
Impermeable Layer
h1
x
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 by MIT OCW.