Post on 04-Jun-2018
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Movement of groundwaterdepends on rock and sediment properties
and the groundwaters flow potential. Porosity, permeability, specific
yield and specific retention are important components of hydraulic
conductivity.
HYDRAULIC CONDUCTIVITY = K (or P)
units = length/time (m/day)
Ability of a particular material to allow water to pass through
it
The definition of hydraulic conductivity (denoted "K" or "P" inhydrology formulas) is the rate at which water moves through material.
Internal friction and the various paths water takes are factors affecting
hydraulic conductivity. Hydraulic conductivity is generally expressed
in meters per day.
Groundwater Movement
S. Hughes, 2003
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WELL SORTED
Coarse (sand-gravel)
POORLY SORTED
Coarse - Fine
WELL SORTED
Fine (silt-clay)
Permeability and Hydraulic Conductivity
High Low
Sorting of material affects groundwater movement. Poorly sorted (well
graded) material is less porous than well-sorted material.
S. Hughes, 2003
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Groundwater Movement
Table 10.6 in textbook (Keller, 2000)
Porosity and hydraulic conductivity of selected earth materials
HydraulicPorosity Conductivity
Material (%) (m/day)
Unconsolidated
Clay 45 0.041Sand 35 32.8
Gravel 25 205.0
Gravel and sand 20 82.0
RockSandstone 15 28.7
Dense limestone or shale 5 0.041
Granite 1 0.0041
S. Hughes, 2003
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Groundwater Movement
The tortuous path of groundwater moleculesthrough an aquifer
affects the hydraulic conductivity. How do the following properties
contribute to the rate of water movement?
Clay content and
adsorptive properties
Packing density
Friction
Surface tension
Preferred orientation
of grains
Shape (angularity orroundness) of grains
Grain size
Hydraulic gradient
S. Hughes, 2003
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Water table contour linesare similar to topographic lines on a map.
They essentially represent "elevations" in the subsurface. These
elevations are the hydraulic head mentioned above.
Water table contour lines can be used to determine the direction
groundwater will flowin a given region. Many wells are drilled and
hydraulic head is measured in each one. Water table contours (called
equipotential lines) are constructed to join areas of equal head.Groundwater flow lines, which represent the paths of groundwater
downslope, are drawn perpendicularto the contour lines.
A map of groundwater contour lines with groundwater flow lines is
called a flow net.
Remember:groundwater always moves from an area of higher
hydraulic head to an area of lower hydraulic head, and perpendicular
to equipotential lines.
Groundwater Flow Nets
S. Hughes, 2003
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6
Flow nets
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Groundwater Flow Nets
Aquitard (granite)
Qal100 50
Qal
WT
A simple flow net
Cross-profile view
well
Aquitard
Qal
Effect of a
producing well
Notice theapproximate diameter
of the cone of
depression
S. Hughes, 2003
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Groundwater Flow Nets
70
70
80
80
90
90
100
100
Aquitard
Aquitard
Qal
Qal
Water table contours
Water is flowing from Qal togranite
Water is flowing from graniteto Qal
Distorted contours may occur
due to anisotropic conditions
(changes in aquifer
properties).
Area of high permeability (high conductivity)
S. Hughes, 2003
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DRAINAGE
BASIN
NWTcontours
Flow lines
Groundwater Flow NetsWater table contours in
drainage basinsroughly
follow the surface topography,but depend greatly on the
properties of rock and soil that
compose the aquifer:
Variations in mineralogy andtexture
Fractures and cavities
Impervious layers
Climate
Drainage basins are often used to collect clean,
unpolluted water for domestic consumption.
S. Hughes, 2003
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Groundwater Flow Net
400
402
404
406
408
410
412414
N
Water Table Contours
Water Flow Lines
Well
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Flow Nets
Graphical representation of the steady-state velocity
potential and stream function. Used to determine flow velocities, flow paths, and travel
times.
Approach is general and can be applied to a variety
of fluid problems including compressible, andincompressible ideal flows.
In porous media, the velocity potential is related to
the head and the stream function is related to the
path.
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Velocity Potential
The velocity potential is given by the head or fluid
pressure.
The gradient of the velocity potential function is used
to recover the velocity value at a point in the flow
field.
The velocity potential satisfies the governing mass
balance equation for steady-incompressible flow.
ThKh or
)(gradU
0or02
yx
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Streamline
A streamline is defined as a line that is tangent to the
velocity vector in a flow field.
Tangent means:0or vdxudy
u
v
dx
dy
y
dy
dx
v
u
streamline
x
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Stream Function
Conservation of mass requires that QABP=QACP.
OnceAis fixed, QRdepends solely on the location, P.
The volumetric flow through Ris called the stream
function,
x
y
R
B
C
P
A
),( yx
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Stream Functions and Streamlines
x
y
B
C
D
A
1
2
constant.aisBC
alongofvaluethee,Furthermor
.streamlineaisBCsegmentthe
andthen0If
But
;:Then
A.atfunctionstreamthe
ofvaluethebeLet;;
12
12
21
21
D
D
D
D
BC
BC
ACAB
A
BCACAB QQQ
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Potential and Stream Function Relationships
(1) The velocity is given by the gradient of the
velocity potential.
(2) Streamlines are tangent to velocity.
(3) Lines of constant are streamlines.
Law)s(Darcy';y
vv
xu
0
dx
ydy
xvdxudy
0
dy
ydx
xd
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Flow Net Mathematics
The last two relations supply the rules to construct a
flow net.
Since both equations equal the same constant, then
the partial derivatives in each term must be equal.
0
dy
xdx
y
0
dy
ydx
x
xyand
yx
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Cauchy-Riemann Conditions
These equalities are called the Cauchy-Riemann
Conditions for Ideal Flow. They are further expandedusing Darcys Law as:
Or:
x
hK
xyy
hK
yx xy
and
yKxh
xKyh
xy
1and1
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Streamtubes
Flow bounded by two streamlines is called a streamtube.
Discharge in a streamtube is the dif ferencein the in the values
of the bounding stream functions.
x
y
P2
P1
DQ
A
Y2
Y1
DD
121212 AAPPQ
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Irrotational Flow
Irrotational flow means that:
Substitute Cauchy-Reimann conditions to obtain
Or, in compact notation:
0)()(
x
h
yy
h
x
0)1
())(1
( KK
graddiv
0)1
()1
(
yKyxKx xy
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Results
Compare to the steady groundwater flow equation.
These two PDEs are the basis of numerical
generation of flow nets.
0)())(( ijijgraddiv KK
0)1
())(1
(
jiji
graddivKK
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Application
Numerical generation of flow nets is accomplished by
Generating discrete distributions of potential and stream
functions over the entire problem domain
Contouring the results to create a picture of the flow net.
Practical aspects:
Both governing PDEs are LaPlace equations. Thus a tool
that solves LaPlace problems will suffice for both equations
(although boundary conditions will be different)
l S i h
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Two Layer Flow System with
Sand Below
Ku/ Kl= 1 / 50
T L Fl S i h
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Two Layer Flow System with
Tight Silt Below
Flow nets for seepage from one side of a channel
through two different anisotropic two-layer systems. (a)
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SZ2005 Fig. 5.11
Flow nets in anisotropic media
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Flownets in Anisotropic Media
So far we have only talked about flownets inisotropic material. Can we draw flownets for
anisotropic circumstances?
Kx2h
x2Ky
2h
y2 0
For steady-state anisotropic media, with x and yaligned with Kx and Ky, we can write the flow equation:
dividing both sides by Ky:KxKy
2h
x22h
y2 0
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Flownets in Anisotropic Media
Next, we perform an extremely cool transformation
of the coordinates:Ky
Kx
12
x X 1
X2Kx
Ky
1
x 2
This transforms our governing equation to:
2
hX2
2
hy 2
0 Laplaces Eqn!
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Flownets in Anisotropic Media
Steps in drawing an anisotropic flownet:
1. Determine directions of max/min K. Rotate axes
so that x aligns with Kmax and y with Kmin
2. Multiply the dimension in the x direction by
(Ky/Kx)1/2and draw flownet.
3. Project flownet back to the original dimension by
dividing the x axis by (Ky/Kx)1/2
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Flownets in Anisotropic Media
Example:
KxKy
Kx= 15Ky
Ky
Kx
12
1
15
12
0.26
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Flownets in Anisotropic Media
KxKy
Kx= 15Ky
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Flownets in Anisotropic Media
Kx= 15Ky
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Flownets in Anisotropic Media
Kx= 15K
y
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Flownets in Anisotropic Media
Kx= 15K
y
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Flownets in Anisotropic Media
Kx= 15K
y
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Flownets in Anisotropic Media
Kx= 15K
y
25%
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Flownets in Anisotropic Media
Kx= 15K
y
25%
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Flownets in Anisotropic Media
Kx= 15K
y
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Flownets in Anisotropic Media
Kx= 15K
y
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Flownets in Anisotropic Media
Kx= 15K
y
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Flownets in Anisotropic Media
Kx= 15K
y
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Flownets in Anisotropic Media
Kx= 15K
y
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Flow Nets: an example
A dam is constructed on a permeable stratum
underlain by an impermeable rock. A row of
sheet pile is installed at the upstream face. If
the permeable soil has a hydraulicconductivity of 150 ft/day, determine the rate
of flow or seepage under the dam.
After Philip BedientRice University
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Flow Nets: an examplePosit ion: A B C D E F G H I JDistance
from
front t oe
(ft)
0 3 22 37.5 50 62.5 75 86 94 100
n 16.5 9 8 7 6 5 4 3 2 1.2
The flow net is drawn with: m = 5 head drops = 17
After Philip BedientRice University
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Flow Nets: the solution
Solve for the flow per unit width:
q= m K
= (5)(150)(35/17)
= 1544 ft3
/day per ft
total change in head,
Hnumber of head drops
After Philip BedientRice University
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Flow Nets: An Example
There is an earthen dam 13 meters acrossand 7.5 meters high.The Impounded water is6.2 meters deep, while the tailwater is 2.2
meters deep. The dam is 72 meters long. Ifthe hydraulic conductivity is 6.1 x 10-4centimeter per second, what is the seepagethrough the dam if the number of head drops
is = 21
K = 6.1 x 10-4cm/sec
= 0.527 m/day
After Philip BedientRice University
Flow Nets: the solution
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Flow Nets: the solution
From the flow net, the total head loss, H, is
6.2 -2.2 = 4.0 meters.There are (m=) 6 flow channels and
21 head drops along each flow path:
Q = (mKH/number of head drops) x dam length
= (6 x 0.527 m/day x 4m / 21) x(dam length)
= 0.60 m3/day per m of dam
= 43.4 m3/day for the entire 72-meterlength of the dam
After Philip BedientRice University
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Aquifer Pumping Tests
Why do we need to know T and S (or K and Ss)?-To determine well placement and yield
-To predict future drawdowns
-To understand regional flow
-Numerical model input
-Contaminant transport
How can we find this information?
-Flow net or other Darcys Law calculation
-Permeameter tests on core samples
-Tracer tests-Inverse solutions of numerical models
-Aquifer pumping tests