PART 4 Flow in Porous Media

Read: Ch 2.1 and 2.2 PART 4 Flow in porous media

Darcy’s law

Imagine a point (A) in a column of water (figure below); the point has following characteristics: (1) elevation z (2) pressure p (3) velocity v (4) density  A  h , v, p Point A has some energy z which can be described as datum, z=0, p=p0 a sum of potential, kinetic (p0 = atmospheric) and elastic energies:

Potential energy = mgz (1)

1 2 Kinetic energy= ---mv (2) 2

p dp m Elastic energy= m -----  ---- pp– (3)    0 p0

[the near equality in the ‘elastic energy’ term is true under assumption of incompressible fluid].

1 2 m Total energy= mgz ++---mv ---- pp– 2  0

Hydrogeology, 431/531 - University of Arizona - Fall 2019 Dr. Marek Zreda Flow in porous media 18

Energy per unit mass is called fluid potential:

1 2 pp–  = gz ++---v ------0 2 

Because velocity (v) is low in porous media, the kinetic energy term is small, that is, mv20, and we can write

pp– 0   gz + ------

Pressure at point A is p = g + p0 where  iswater column above A and p0 is the atmospheric pressure. We now have g p0 –+ p0  ===gz +gzghz------+gzghgz– –+ 

 = gh Fluid potential

Fluid at point A has potential  = gh. Fluid will flow from point of higher potential to point of lower potential.

Dividing by g (which can be assumed constant), we get hydraulic head:

h =  + z Hydraulic head

Hydraulic head has two components:

 = pressure head (due to pressure of water above point A) z = elevation head (due to elevation of point A above the datum)

Hydrogeology, 431/531 - University of Arizona - Fall 2019 Dr. Marek Zreda Flow in porous media 19

Darcy’s experiment

A = cross-sectional area

z1, p1 yy

h1 yydL Q datum: z = 0 = z2

z , p Henry (Henri) Darcy yyy2 2 Darcy observed that flow Q is: (1) proportional to head difference dh (2) inversely proportional to column length dL (3) proportional to cross-sectional area A

Add a proportionality constant (K), which depends on properties of fluid and properties of soil (porous medium), to get Darcy’s law:

dh QKA==– ------–KAh Darcy’s law dL

h = hydraulic head [L] A = column cross-sectional area [L2] L = column length [L] K = hydraulic conductivity [LT-1] Q = flow rate [L3T-1] = hydraulic gradient = -dh/dL [-]

Darcy’s law per unit area:

Q dh q ==---- –K------=–Kh q = specific discharge [LT-1] A dL

Three questions to ponder: • What is q per unit dh/dL? • If dh → ∞, is q → ∞ as well? Why or why not? • Is travel time t = distance/q? Why or why not?

Hydrogeology, 431/531 - University of Arizona - Fall 2019 Dr. Marek Zreda Flow in porous media 20

Seepage (linear) velocity and travel time:

Flow velocities are faster than the specific discharge because flow occurs in pores only. There- fore, we have seepage (linear) velocity, v:

q K dh K -1 v ==--- –------=–---- h v = seepage velocity [LT ] n n dL n

This is average macroscopic velocity of water.

Travel time, t, is distance divided by velocity:

d n dL t ==--- –d------t = travel time [T] v K dh

Components of hydraulic head

h =  + z

total head = pressure head + elevation head

Go back to our Darcy’s experiment. Plot zh as a function of l:

h, , z

l1 l2 h1 h h h1 h2 z1 z z1 z2 z 2, h2  h - z h - z  1 1 2 2 1 Procedure: determine h, determine z, calculate z 2 l  l1 l2

Hydrogeology, 431/531 - University of Arizona - Fall 2019 Dr. Marek Zreda Flow in porous media 21

Validity of Darcy’s law

We assumed the following: (1) uniform media (2) incompressible fluid (3) water only (single phase flow) (4) isothermal fluid (5) constant cross-sectional area (6) steady flow (7) laminar flow

Upper limit of Darcy’s law

Look at laminar flow assumption (determined using the Reynolds number Re):

inertial forces forces due to acceleration R ==------e viscous forces forces due to friction inertia = mass  acceleration acceleration = velocity / time inertia = mass  velocity / time [M L T-2] mass = density  length3 viscous forces = viscosity  velocity  length [M L T-2]

density length2 velocity density length velocity R = ------= ------e time dynamic viscosity velocity length dynamic viscosity

Lv R = ------e  dimensionless number

L = some characteristic length of pore size, such as d50 (median grain size), d10 (10-th percentile), or k, and v = water velocity.

Rules: q Re  1 laminar flow; Darcy’s law applies 1 < Re  10 transition zone; Darcy’s law is questionable Re  10 “turbulent” flow; Darcy’s law does not apply

tan a = K a dh/dl

Re = 1 Re = 10

Hydrogeology, 431/531 - University of Arizona - Fall 2019 Dr. Marek Zreda Optional reading: J. Bear, 1972, Dynamics of fluids in Flow in porous media porous media, Ch. 5.10; G. de Marsily, 1986, Ch. 3. 22

Lower limit of Darcy’s law

In fine soils, such as clays, flow does not occur below some mini- q mum gradient.

Explanations: (1) small pores; water molecules strongly influenced by electri- cal charge on solid particles; leads to increased “effective viscosity.” dh/dl

(2) streaming potential; water carries cations which are attracted minimum gradient to solid surfaces and slow the movement of water. (3) non-Newtonian fluid; only in capillary spaces. (4) electrostatic counterflow.

Hydrogeology, 431/531 - University of Arizona - Fall 2019 Dr. Marek Zreda Flow in porous media 23

Navier-Stokes equation for flow in a tube

A(r, u, x) p1 p2

r a x u

Tube along x Radius a Radial (cylindrical) coordinates: radial distance r, angle u, (axial distance x)

Assume flow parallel to x and steady state; no r and ucomponents

Navier-Stokes equations (Bear, 19721; de Marsily, 19862) reduce to:

p 1  V – ----- +0--- r------x = x r rr

Multiply both sides by r to get:

p  V –----- r +0 r------x = x rr 1-D, steady state flow

BC1: velocity is zero at the wall of the tube (due to viscosity and adhesive water) Vx = 0 @ r = a

BC2: velocity is maximum in the center of the tube (due to symmetry of Vx in the tube) Vx/r = 0 @ r = 0 Note: some information has been removed so that you would have to do your own derivations. Note: some information has been removed so

1. Bear, J., 1972, Dynamics of fluids in porous media, Elsevier (reprinted by Dover). 2. de Marsily, G., 1986, Quantitative hydrogeology, Academic Press.

Hydrogeology, 431/531 - University of Arizona - Fall 2019 Dr. Marek Zreda Flow in porous media 24

First integration gives:

1 2p V –---r ----- + r------x = C 2 x r 1

Use BC2: 0 + 0 = C1 then C1 = 0

Plug C1 in and divide by r, and the equation becomes:

1 p V –---r----- +0------x = 2 x r

Second integration gives:

1 2p –---r ----- + V = C 4 x x 2

Use BC1:

1 2p –---a ----- = C 4 x 2

substitute for C2:

1 2p 1 2p –---r ----- + V = –---a ----- 4 x x 4 x

and the solution becomes:

1 2 2 p V = ------r – a ----- radial distribution of velocity x 4 x

Results:

(1) Velocity is zero @ r = a (tube wall)

(2) Velocity is maximum @ r = 0 (tube axis)

(3) Velocity depends on: r, 1/, and dp/dx

Hydrogeology, 431/531 - University of Arizona - Fall 2019 Dr. Marek Zreda Flow in porous media 25

The discharge through the tube is calculated by integrating the velocity over the tube area:

a a a 2 QV== x Ad Vxdr =Vx2rrd 0 0 0

a a 1 2 2 p  p 3 2 = 2r------r – a ----- rd = ------r – a r rd  4 x 2x 0 0

a  p 1 4 1 2 2 = ----------- ---r – ---a r 2 x 4 2 0

 p 1 4 1 4  p 1 4 = ------------a – ---a –= ------------a 2 x 4 2 2 x 4

– 4 p Total discharge through Q = ------a ----- 8 x a tube of radius a

Hydrogeology, 431/531 - University of Arizona - Fall 2019 Dr. Marek Zreda Flow in porous media 26

We can calculate discharge per unit area, or specific discharge, q. It is often used instead of total discharge Q because it is independent of the size of the analyzed system and thus, it is useful in comparison of different flow systems (for example different aquifers or different laboratory col- umns).

Q Q q = ---- = ------A a2

–a2 p q = ------ ----- specific discharge 8 x

Significance of this result: (1) Flow depends on a2, 1/ and px (2) Velocity is zero at tube wall (3) Velocity is maximum at tube center

Exercise:

Compare the above expression for q with that from Darcy’s law (p. 19). Look at each term (do term-by-term comparison). What is the hydraulic conductivity (K) in the solution of the Navier- Stokes equation? What are components of hydraulic conductivity?

Hint: write pressure (p) in terms of hydraulic head (h) to obtain comparable expressions for q from Darcy’s law and from the Navier-Stokes equation.

Flow in fractures

g Nb3 K = ------------ 12

N = number of fractures per unit distance;

b = aperture [L].

3 dh g Nb dh qK– ------== –------------------dx  12 dx

Hydrogeology, 431/531 - University of Arizona - Fall 2019 Dr. Marek Zreda Flow in porous media 27

Factors controlling flow in porous media

Back to solution of Navier-Stokes equation in its form integrated over cross-sectional area:

–a4 p Q = ------ ----- 8 x

Now, make a medium from m parallel tubes in cube with side length b. 2 b Side area:B=b Area of tubes:A = ma2 Define porosity:n = void space / total space b n = A/B = ma2/B a Then:m = nB/a2 2 Total flux:QT = mQ = nB/a Q

nB – 4 p QT = ------ ------a ----- a2 8 x

na2 p Q = –B ------T 8 x

Observe that flow is proportional to: (1) total area of the medium (2) porosity (3) pore size (4) external gradient (5) inverse of viscosity (6) 1/8, which is a constant valid for circular tube only. Other pore geometries will have different constants.

Hydrogeology, 431/531 - University of Arizona - Fall 2019 Dr. Marek Zreda