Geodynamics - Flow in Porous Media

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 ⎛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 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 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 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