flownet_1.ppt

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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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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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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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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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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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11

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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12

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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13

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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14

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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15

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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16

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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17

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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18

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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19

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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20

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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21

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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22

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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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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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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Example:

KxKy

Kx= 15Ky

Ky

Kx

12

1

15

12

0.26


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KxKy

Kx= 15Ky


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Kx= 15Ky


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Kx= 15K

y


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Kx= 15K

y


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Kx= 15K

y


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Kx= 15K

y

25%


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Kx= 15K

y

25%


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Kx= 15K

y


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Kx= 15K

y


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Kx= 15K

y


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Kx= 15K

y


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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



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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



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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




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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



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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

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