No Slide Title · Whitaker (1986) and Bear (1988) (see website course) Transport in Permeable Media...

Transport in Permeable Media

Surface tensions

Curved surface

Pressure difference

Unsaturated

Saturated

Surface tensions

Curved surface

Pressure difference

wnc rp γ2

Liquid ‘fast’ Vapour ‘slow’

Same macroscopic pressure: suction

Transport in porous media

Saturated Un-saturared

soil-groundwater soil-plants

oil/water drying

bio buildings

chemical reactors

Most famous publication

Henry Darcy “As much as possible, one should favor the free drawing of water because it is necessary for public health. A city that cares for the interest of the poor class should not limit their water, just as daytime and light are not limited.”

Darcy’s experiment reported in Appendix D of Les Fontaines Publiques…

Flow tube: Hagen-Pousseullie flow

Laminair Re<<2000

∆∆

−=µ

∆∆

−==µπ 8

Average velocity

For porous medium flux: mean volume per area

(speed m/s)

∆∆

−=µ

‘porous medium’ with single pore

∆∆

−==µ

single pore

v1 ≠ v2 µ8

)( 4ilo

lppQ −

pore I with radius ri

total volume rate:

ilotot

∑−=

Rewrite in average pores size

µ8)( 4 ><−

lppQ lo

Multiple pores

Tortuosity (kronkeligheid)

Flow paths increase in length. Instead of proceeding straight through a volume of porous medium, the flow must avoid all the empty pores, making the path more tortuous

Common empirical expression: L = straight-line distance Le = actual (effective) path T ≈ 0.7 for sand ?

Multiple pores

v1 ≠ v2

Correction tortuosity total

volume rate:

µ8)( 4 ><−

lTAppq

porous

Flux of porous medium

µ8)( 4 ><−

lTppQ lo

ArNn ><

><><−

lppq lo

tot µ

Multiple pores

v1 ≠ v2

Flux porous medium:

∂∂

−=µ

k: intrinsic permeability

Law of Darcy

><><−

lppq lo

tot µ π T

=⋅××

dp/dxqk µ

dxdpkq

µ−=

Darcy’s law

Darcy flux: volume flux

Permeability

Darcy : 9.8692 10-13 m2

Phenomemological Laws

∂∂

−=µ

dxdTQ λ−=

∂∂

Darcy’s law : porous flow

Fourier’s law : heat flow

Fick’s law : mass flow

∂∂

−= σ Ohm’s law : electrical current

Actual flow length, L e

"Darcian" flow length, L

"Darcian" velcoity q = flow/area

Darcy: macro law

Darcy’s law can be derived from volume averaging the momentum equation by making the following assumptions (Bear and Verruijt, 1987; Sahimi, 1995)

• inertial effects are negligible • the internal friction inside the fluid is negligible

• I.E. viscous force and gravity dominate the flow of groundwater in porous media, which are called the driving forces

Theoretical derivation of Darcy’s law can be found, among others, Whitaker (1986) and Bear (1988) (see website course)

Bernoulli's Equation

Elevation Head, m

Fluid Pressure Head, m

Velocity Head, m

water travels very slowly through soil as opposed to channel flow

Total suction, m

Total Energy = velocity energy + potential energy + pressure energy

rewrite

µρgkK = hydraulic conductivity

K is a function of soil and fluid properties.

Darcy 1D

Pressure in meters

∂∂

−=µ

𝑝 = 𝜌𝜌𝜌

𝑞 = −𝑘 𝜌𝜌𝜇

𝜕𝜌𝜕𝜕

𝑞 = −𝐾𝜕𝜌𝜕𝜕

Darcy = three-dimensional motion equation

q h= − • ∇K

, ,x y zh h hq K q K q Kx y z

∂ ∂ ∂= − = − = −

∂ ∂ ∂

– For homogeneous and isotropic medium, K = constant

– For heterogeneous and isotropic medium, K = K(x,y,z)

( ) ( ) ( ), , , , , , , ,x y zh h hq K x y z q K x y z q K x y zx y z

∂ ∂ ∂= − = − = −

∂ ∂ ∂

– For homogeneous and anisotropic medium

x xx xy xz

y yx yy yz

z zx zy zz

h h hq K K Kx y zh h hq K K Kx y zh h hq K K Kx y z

∂ ∂ ∂= − − −

∂ ∂ ∂∂ ∂ ∂

= − − −∂ ∂ ∂∂ ∂ ∂

= − − −∂ ∂ ∂

VLKAth

• Constant-head permeameter : for noncohesive sediments such as sand and rocks

(After Fetter, 1994)

( )A Bh hQ KAJ KA

= = −

( )( )

elevation head only

pressure head onlyA

( )L h LQ KA t

L− −

⇒ = −

/Qt V KAht L⇒ = =

V = volume of water discharging in time t (L3)

Constant head

For cohesive sediments with low conductivities, such as

silt and clay

(After Fetter, 1994)

Rate of water flowing from the tube into the chamber : in bdhq Adt

Rate of water flowing out of the chamber : out chq KAL

KA hdhAdt L

⇒ − =

A LKdt dhA h

⇒ = −

A LK dt dhA h

⇒ = −∫ ∫0lnb

A L hKA t h

⇒ = Ab = cross-sectional area of the tube

Ac = cross-sectional area of the chamber

Falling-head permeameter

Testing loose sands and other materials

Shelby Tube Permeameter

Field Methods for Determining Permeability

Double Ring Infiltrometer

K (m/s)

Clay Silt

Sand Gravel

Concrete 2 10-13 Clay 10-11 – 10-8

Silt 10-8 – 10-6

Fired clay brick 2.5 10-6

Silty Sand 10-7 – 10-5

Sands 10-5 – 10-3

Gravel 10-4 – 1-2

The Kozeny-Carman predictive Model for Ksat

The Kozeny-Carman approach provides unsatisfactory estimates of Ks in many soils, due to the assumption of uniform pore radii.

Works well for sands and other materials with uniform pore size distribution.

Apparent K as a function of hydraulic gradient

1.E-09 1.E-08 1.E-07 1.E-06 1.E-05 1.E-04 1.E-03Hydraulic Gradient

ity (m

0.001 0.01 0.1 1 10 100 1000Approximate Reynolds Number

Darcy-Forchheimer Equation

τ = 1 mdhq K

dl = −

turbulent

Real Reason: due to forces in acceleration of fluids passing particles at the microscopic level being as large as viscous forces: increased resistance to flow, so flux responds less to applied pressure gradients

N-decane

N-heptane

Iso-propanol

µρgkK =

scaling

Example permeability-porosity relationship

From Tiab and Donaldson

Porosity Permeability

Ability to hold water Ability to transmit water Size, Shape, Interconnectedness

Porosity Permeability

Hydraulic Conductivity

Unit mD

– 1 darcy (or d) is defined as the permeability that will lead to a specific discharge of 1 cm/s for a fluid with a viscosity of 1 cp under a hydraulic gradient that makes ρgdh/dl to 1 atm/cm (Freeze and Cherry, 1979)

k dhq gdl

cm / s cpdarcyatm / cm

cm / s Pa s. Pa / cm

×⇒ =

× ⋅=

×= ×

1 1 1 1

1 10 1013 10

987 10

1 md = 10-3 d

Charbeneau, 2000.

• the top of the saturated zone of groundwater • the level to which water will rise in a hole • the level to which water will rise in an unconfined aquifer

Cross Section of Unconfined and Confined Aquifers

A GUSHER OF WATER from a "true" artesian well at Slough, Bucks, which gave an initial supply of 25,000 gallons and a final delivery of 100,000 gallons an hour. This was the first example of a "true" artesian well sunk in the neighbourhood of London

http://www.engrailhistory.info/e021.html

Artesian well in Iași, Romania

Mountains

Groundwater can discharges from the

wall (natural springs).

Groundwater flow An aquafer sandwiched between two aquacludes forms a confined aquafer.

Layer 30 m ,

and 5 km width

Pressure 5 m, over 1 km

The solution • Cross-Sectional area=

30(5.1000) = 15 x 104 m2

• Hydraulic gradient = (55-50)/1000 = 5 x 10-3

• Sand 5 10-4~ 50 m/day

• Darcy Velocity: V = 50 5 x 10-3 = 0.25 m/day

• Volume flux Q = 15 x 104 0.25 = 37,500 m3/day

∂∂

−=ψ

• Darcy’s law gives an apparent velocity, but strictly speaking this is a discharge per unit cross-sectional area of aquifer

• But if we are interested in the velocity of the water molecules themselves, or a solute, we have to consider the porosity of the aquifer (which is the cross-sectional area of the pores)

Average Linear Velocity

• Darcy velocity is the flow per unit cross- sectional area of the aquifer

• Much of the cross-sectional area is “blocked” by particles

Volume porosity =

Surface porosity

• Seepage Velocity: Vs = V/n = (0.25) / (0.2) = 1.25 m/day

• Time to travel 1 km downstream: T = (1000m)/(1.25m/day)

800 days or ~2 years

• This example shows that water moves very slowly underground.

Human Activities that Can Contaminate Groundwater

∂∂

qin qout

(Mass in per time unit) – (Mass loss per time unit) = increase in weight per time unit

Mass balance

t∂∂θ

Continuity (1D)

∆x x

xqqqout ∆

∂∂

+=qqin =

∆∂∂

∆∂∂θ

qxt ∂

∂∂∂θ

0=+ qxt ∂

∂∂∂θ

t∂∂θ

Continuity qout

∫∫∫∫∫ −=∂∂

dsnqvdt

∫∫∫∫∫ ∇=VS

vdqdsnq ..

∂∂

∫∫∫V

vdqtθ

Any volume

0. =∇+∂∂ q

Saturated medium: n=θ=constant

0. =∇ q

ψ∇−= Kq

0=∇∇ ψK

02 =∇ ψ Laplace equation

0)(2 =−−∇ ∑ mm xxQK δψ Point sources/ sinks

Flow under dam: seepage

river lake

FLOW???

Homework: pde toolbox

Fluid flow porous medium

• porous medium

• fluid potential

• flux q

• hydraulic conductivity

Electricity conducting medium

• conducting medium

• current i

• electric conductivity

“Analog computer”

Homework: pde toolbox

Electric analog model of Long Island, New York

Streamlines Y and Equip. lines φ are ⊥. Streamlines Y are parallel to no flow boundaries. Grids are curvilinear squares, where diagonals cross at right angles. Each stream tube carries the same flow.

FLOWNET

• Flow through a channel between equipotential lines φ1 and φ2 per unit width is: ∆q = K(dm x 1)(∆h1/dl)

Contour map of the piezometric surface near Savannah, Georgia, 1957, showing closed contours resulting from heavy local groundwater pumping

Regional Aquifer Flows are affected by Pump Centers

Geology & Pumping Impacts

Flow Modeling

Predicting heads (and flows) and Approximating parameters

Solutions to the flow equations Most ground water flow models are solutions of some form of the ground water flow equation

“e.g., unidirectional, steady-state flow within a confined aquifer

The partial differential equation needs to be solved to calculate head as a function of position and time, i.e., h=f(x,y,z,t)

h(x,y,z,t)?

Kxqhxh −= 0)(

Kxqhhdx

dxdh xh

h−=−⇒−=⇒−= ∫∫ 000

Darcy’s Law Integrated

How much water has to be pumped to

keep excavation dry?

Impermeable Base L

Dupuit Assumptions

• Hydraulic gradient = slope of water table.

• For small hydraulic gradients, flow lines are horizontal and equipotential lines are vertical (no vertical flow)

• The water table or free surface is only slightly inclined

• Free surface p=0 > h

Observation made by Dupuit (1863): the slope of the phreatic surface is very small (range of 1 in 1000 to 10 in 100). sinθ = tanθ

X = 0 X = L X

Building side

Sand K=5 m/day

h1=5 m

h2=2 m

L =10 m

Equilibrium water table profile

• Strip of aquifer, 1 meter wide, parallel to the page

X = 0 X = L X

∂∂

X = 0 X = L X

Khdhqdx −=

∫∫ −=2

hdhKdxq

integral

xhhKq x

2)( 22

1 −−=

Kqxhhx

221 −=

X = 0 X = L X

1 −−=

Kqxhhx

221 −=

q= constant

Lhhxhhx

12 −

Dupuit parabola

X = 0 X = L X

Building side

Sand K=5 m/day

h1=5 m

h2=2 m

L =10 m

q=5(52-22)/2 10

q=5.25 m2/day

Dupuit

• simple analytic expression

• in -out flow

• approximation, but Q is exact expression

• distances 1.5/2 times height of domain, Dupuit sufficiently accurate for practical purposes

During the Wet Season…

During the Dry Season…

The Water Table

• Downward infiltration from surface • Percolation from influent streams • Pore space completely filled = saturation • Zone of saturation below water table.

Unconfined aquifer – upper boundary of saturated zone is a water table atmospheric pressure and connected directly to the atmosphere

X = 0 X = L X

Stream Channel

(P – E)

Hill 1000 m wide K = 0.5 m/day average rainfall = 15 cm/yr evaporation = 10 cm/yr

Example

Evaporation estimation

4 ft (1.22 m) 10” (25.4 cm)

At equilibrium:

( )dxdhKhxEP −=−

( ) xdxK

EPhdh −−=

X = 0 X = L X

Stream Channel

(P – E)

Equilibrium water table profile

P=precipitation (rain)

E= evaporation

∫∫

−= xdxK

−=22

Boundary condition: at x = xL h = hs

EPh Ls +

−= 22

22Ls x

+=X = 0 X = L X

Stream Channel

(P – E)

[ ]2222 xxK

EPhh Ls −

[ ]222 xxK

EPhh Ls −

The result is a convex-upward water table, the height and steepness of which depends on the height of the stream surface and the ratio of recharge to conductivity.

X = 0 X = L X

Stream Channel

(P – E)

X = 0 X = L X

Stream Channel

(P – E)

Hill 1000 m wide K = 0.5 m/day average rainfall = 15 cm/yr evaporation = 10 cm/yr

Example

(P-E) = 5 cm/yr = 1.369 x 10-4 m/day

[ ]224

2 010005.0

104.15 −

h=17 m

More complicated?

Domestic well

Overpumping can produce a large cone of depression causing shallow wells to go dry.

In the San Joaquin Valley of California, over-pumping has not only depleted aquafers but caused the ground to subside drastically.

Figure from Hornberger et al. 1998

unconfined aquifer

confined aquifer

A well penetrates an unconfined aquifer. Prior to pumping the water leavel is ho=25m. After a long period of pumping at a constant rate of 0.05 m3/s, the drawdowns at the distances of 50 and 150 m from the well were observed to be 3 and 1.2 m. Compute the hydraulic conductivity of the aquifer?

( ) ( )rvrhrQ π2=

( ) ( ) ( )rqrQrdrdhKrhrQ ππ 22 =

rhhrKQ

∂∂

−= π2

∫∫ ∂−=∂ 2

r∂−=

∂ π2

212ln hhK

πThiem’s equation (parabolic)

Well: Darcy radial Dupuit

h1=25-3=22 m h2=25-1.3=23.8 m

𝐾 =𝑄

𝜋 (𝜌22 − 𝜌12)ln

𝑟2𝑟1

𝐾 =4320

𝜋 (23.82 − 222)ln

= 18.3 𝑚/𝑑𝑑𝑑

Q=0.05 m3/s= 4320 m3/day

• Inverse Modeling: Aquifer Characterization

– The Thiem Equation can also be solved for K

– Pump Test: This inverse model allows measurement of K using a steady state pump test

• A pumping well is pumped at a constant rate of Q until heads come to steady state, i.e.,

• The steady-state heads, h1 and h2, are measured in two observation wells at different radial distances from the pumping well r1 and r2

• The values are “plugged into” the inverse model to calculate K (a bulk measure of K over the area stressed by pumping)

)(tfh ≠

ln)( r

Measuring Permeability

Laboratory • Constant head test • Falling head test • Other Field • Pumping tests • Borehole infiltration

How good is the

sample? Need to know soil

profile (incl. WT) and boundary

conditions

Conceptual model

Analytical solution Numerical solution

Verification of solution

Prediction

Solution is correct ? No

Revision

Modeling Chronology

1960’s Flow models are great! 1970’s Contaminant transport models are great!

1975 What about uncertainty of flow models?

1980s Contaminant transport models don’t work. (because of failure to account for heterogeneity)

1990s Are models reliable?

Problem

• variations soil properties • K, suction + how to measure

• grid of measurement

• precipitation, evaporation distribution

Perfect land surface model

Garbage in

Garbage out

No Slide Title · Whitaker (1986) and Bear (1988) (see website course) Transport in Permeable Media...

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