No Slide Title · Whitaker (1986) and Bear (1988) (see website course) Transport in Permeable Media...
Transcript of No Slide Title · Whitaker (1986) and Bear (1988) (see website course) Transport in Permeable Media...
Transport in Permeable Media
TPM
Surface tensions
Curved surface
Pressure difference
Unsaturated
Saturated
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Surface tensions
Curved surface
Pressure difference
wnc rp γ2
−=
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Liquid ‘fast’ Vapour ‘slow’
Same macroscopic pressure: suction
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Transport in porous media
Saturated Un-saturared
soil-groundwater soil-plants
oil/water drying
bio buildings
chemical reactors
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Most famous publication
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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.”
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Darcy’s experiment reported in Appendix D of Les Fontaines Publiques…
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Flow tube: Hagen-Pousseullie flow
Laminair Re<<2000
xprQ
∆∆
−=µ
π8
4
xpr
rQv
∆∆
−==µπ 8
2
2
Average velocity
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For porous medium flux: mean volume per area
(speed m/s)
xprQ
∆∆
−=µ
π8
4
‘porous medium’ with single pore
xp
Ar
AQq
∆∆
−==µ
π8
4
single pore
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v1 v2
v1 ≠ v2 µ8
)( 4ilo
ir
lppQ −
=
pore I with radius ri
total volume rate:
µ8)(
4i
ilotot
r
lppQ
∑−=
Rewrite in average pores size
µ8)( 4 ><−
=rN
lppQ lo
tot
Multiple pores
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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 ?
2
=
eLLT
Multiple pores
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v1 v2
v1 ≠ v2
Correction tortuosity total
volume rate:
µ8)( 4 ><−
=rN
lTAppq
porous
lotot
Flux of porous medium
µ8)( 4 ><−
=rN
lTppQ lo
tot
=n
ArNn ><
=2
><><−
= 2
4
81)(
rrn
lppq lo
tot µ
Multiple pores
π
π T
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v1 v2
v1 ≠ v2
Flux porous medium:
xpkq
∂∂
−=µ
k: intrinsic permeability
Law of Darcy
><><−
= 2
4
81)(
rrn
lppq lo
tot µ π T
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=
sm
m/sm2
3
A
x
=⋅××
= 2m
mPa
sPasm
dp/dxqk µ
dxdpkq
µ−=
Darcy’s law
Darcy flux: volume flux
Permeability
Darcy : 9.8692 10-13 m2
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Phenomemological Laws
xpkq
∂∂
−=µ
dxdTQ λ−=
xCDJ
∂∂
−=
Darcy’s law : porous flow
Fourier’s law : heat flow
Fick’s law : mass flow
xVi
∂∂
−= σ Ohm’s law : electrical current
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Actual flow length, L e
"Darcian" flow length, L
"Darcian" velcoity q = flow/area
q
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)
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Bernoulli's Equation
2gv
gpzh
2
++=ρ
Elevation Head, m
Fluid Pressure Head, m
Velocity Head, m
water travels very slowly through soil as opposed to channel flow
0
Total suction, m
Total Energy = velocity energy + potential energy + pressure energy
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rewrite
µρgkK = hydraulic conductivity
K is a function of soil and fluid properties.
Darcy 1D
Pressure in meters
xpkq
∂∂
−=µ
𝑝 = 𝜌𝜌𝜌
𝑞 = −𝑘 𝜌𝜌𝜇
𝜕𝜌𝜕𝜕
𝑞 = −𝐾𝜕𝜌𝜕𝜕
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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
∂ ∂ ∂= − − −
∂ ∂ ∂∂ ∂ ∂
= − − −∂ ∂ ∂∂ ∂ ∂
= − − −∂ ∂ ∂
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VLKAth
⇒ =
• Constant-head permeameter : for noncohesive sediments such as sand and rocks
(After Fetter, 1994)
( )A Bh hQ KAJ KA
L−
= = −
A
B
( )( )
elevation head only
pressure head onlyA
B
h L
h L h
=
= +
( )L h LQ KA t
L− −
⇒ = −
/Qt V KAht L⇒ = =
V = volume of water discharging in time t (L3)
Constant head
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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
=
cb
KA hdhAdt L
⇒ − =
b
c
A LKdt dhA h
⇒ = −
00
t h
b
ch
A LK dt dhA h
⇒ = −∫ ∫0lnb
c
A L hKA t h
⇒ = Ab = cross-sectional area of the tube
Ac = cross-sectional area of the chamber
Falling-head permeameter
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Testing loose sands and other materials
Shelby Tube Permeameter
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Field Methods for Determining Permeability
Double Ring Infiltrometer
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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
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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.
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Apparent K as a function of hydraulic gradient
0
5
10
15
20
25
30
35
40
1.E-09 1.E-08 1.E-07 1.E-06 1.E-05 1.E-04 1.E-03Hydraulic Gradient
Hyd
raul
ic C
ondu
ctiv
ity (m
s-1)
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
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N-decane
water
N-heptane
Iso-propanol
µρgkK =
scaling
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Example permeability-porosity relationship
From Tiab and Donaldson
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=
Porosity Permeability
VS
Ability to hold water Ability to transmit water Size, Shape, Interconnectedness
Porosity Permeability
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Hydraulic Conductivity
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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
. cm
ρµ
−
−
= −
×⇒ =
× ⋅=
×= ×
3
5
9 2
1 1 1 1
1 10 1013 10
987 10
1 md = 10-3 d
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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
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Cross Section of Unconfined and Confined Aquifers
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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
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Mountains
Groundwater can discharges from the
wall (natural springs).
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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
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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
xKq
∂∂
−=ψ
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• 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)
sm
msm
AQ
==−
2
13
sm
msm
AnQ
==−
2
13
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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
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Volume porosity =
Surface porosity
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• 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.
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Human Activities that Can Contaminate Groundwater
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tm
∂∂
qin qout
(Mass in per time unit) – (Mass loss per time unit) = increase in weight per time unit
Mass balance
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t∂∂θ
qin
Continuity (1D)
qout
∆x x
xqqqout ∆
∂∂
+=qqin =
xxq
∆∂∂
xt
∆∂∂θ
qxt ∂
∂∂∂θ
−=
0=+ qxt ∂
∂∂∂θ
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t∂∂θ
qin
Continuity qout
∫∫∫∫∫ −=∂∂
SV
dsnqvdt
.θ
∫∫∫∫∫ ∇=VS
vdqdsnq ..
0. =
∇+
∂∂
∫∫∫V
vdqtθ
Any volume
0. =∇+∂∂ q
tθ
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Saturated medium: n=θ=constant
0. =∇ q
ψ∇−= Kq
0=∇∇ ψK
02 =∇ ψ Laplace equation
0)(2 =−−∇ ∑ mm xxQK δψ Point sources/ sinks
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Flow under dam: seepage
river lake
FLOW???
Homework: pde toolbox
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Fluid flow porous medium
• porous medium
• fluid potential
• flux q
• hydraulic conductivity
Electricity conducting medium
• conducting medium
• V
• current i
• electric conductivity
“Analog computer”
Homework: pde toolbox
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Electric analog model of Long Island, New York
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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
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dm
∆h1
dl
Φ1
Φ3
∆q
Φ2
∆h2
∆q
n
m
• Flow through a channel between equipotential lines φ1 and φ2 per unit width is: ∆q = K(dm x 1)(∆h1/dl)
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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
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Geology & Pumping Impacts
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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)?
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Kxqhxh −= 0)(
x
x x
ho
x 0
h(x)
x
K q
Kxqhhdx
Kqdh
Kq
dxdh xh
h−=−⇒−=⇒−= ∫∫ 000
Darcy’s Law Integrated
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How much water has to be pumped to
keep excavation dry?
h1 h2
Impermeable Base L
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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
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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
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X = 0 X = L X
h1
h2
Building side
Sand K=5 m/day
h1=5 m
h2=2 m
L =10 m
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Equilibrium water table profile
• Strip of aquifer, 1 meter wide, parallel to the page
X = 0 X = L X
h1
h2
xhKhq
∂∂
−=
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X = 0 X = L X
h1
h2
Khdhqdx −=
∫∫ −=2
10
h
h
x
hdhKdxq
integral
xhhKq x
2)( 22
1 −−=
Kqxhhx
221 −=
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X = 0 X = L X
h1
h2
LhhKq
2)( 2
22
1 −−=
Kqxhhx
221 −=
q= constant
or
Lhhxhhx
)( 21
222
12 −
−=
Dupuit parabola
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X = 0 X = L X
h1
h2
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
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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
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During the Wet Season…
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During the Dry Season…
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The Water Table
• Downward infiltration from surface • Percolation from influent streams • Pore space completely filled = saturation • Zone of saturation below water table.
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Unconfined aquifer – upper boundary of saturated zone is a water table atmospheric pressure and connected directly to the atmosphere
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X = 0 X = L X
h hS
Stream Channel
(P – E)
(P – E)
Hill 1000 m wide K = 0.5 m/day average rainfall = 15 cm/yr evaporation = 10 cm/yr
Example
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Evaporation estimation
4 ft (1.22 m) 10” (25.4 cm)
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At equilibrium:
( )dxdhKhxEP −=−
( ) xdxK
EPhdh −−=
X = 0 X = L X
h hS
Stream Channel
(P – E)
(P – E)
Equilibrium water table profile
P=precipitation (rain)
E= evaporation
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∫∫
−
−= xdxK
EPhdh
CxK
EPh+
−
−=22
22
Boundary condition: at x = xL h = hs
CxK
EPh Ls +
−
−= 22
22Ls x
KEPhC
−
+=X = 0 X = L X
h hS
Stream Channel
(P – E)
(P – E)
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[ ]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
h hS
Stream Channel
(P – E)
(P – E)
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X = 0 X = L X
h hS
Stream Channel
(P – E)
(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
h=17 m
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More complicated?
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Domestic well
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Overpumping can produce a large cone of depression causing shallow wells to go dry.
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In the San Joaquin Valley of California, over-pumping has not only depleted aquafers but caused the ground to subside drastically.
1925
1955
1977
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Figure from Hornberger et al. 1998
unconfined aquifer
confined aquifer
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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?
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( ) ( )rvrhrQ π2=
( ) ( ) ( )rqrQrdrdhKrhrQ ππ 22 =
−=
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rhhrKQ
∂∂
−= π2
∫∫ ∂−=∂ 2
1
2
1
2h
h
r
r
hKhQr
r π
hKhQr
r∂−=
∂ π2
( )22
21
1
2
212ln hhK
Qrr
−=
π
=−
1
222
21 ln
rr
KQhh
πThiem’s equation (parabolic)
Well: Darcy radial Dupuit
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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
15050
= 18.3 𝑚/𝑑𝑑𝑑
Q=0.05 m3/s= 4320 m3/day
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• 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 ≠
=−
1
222
21 ln
rr
KQhh
π
−=
1
22
22
1
ln)( r
rhh
QKπ
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Measuring Permeability
Laboratory • Constant head test • Falling head test • Other Field • Pumping tests • Borehole infiltration
tests
How good is the
sample? Need to know soil
profile (incl. WT) and boundary
conditions
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Conceptual model
Analytical solution Numerical solution
Verification of solution
Prediction
Solution is correct ? No
Yes
Revision
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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?
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Problem
• variations soil properties • K, suction + how to measure
• grid of measurement
• precipitation, evaporation distribution
Perfect land surface model
Garbage in
Garbage out