Effects of Boundary Condition on Shape of Flow Nets ? ? ? ? ?

Effects of Boundary Condition on Shape of Flow

Nets?

?

?

?

?

Seepage and Dams

Flow nets for seepage through earthen dams

Seepage under concrete dams

Uses boundary conditions (L & R)

Requires curvilinear square grids for solution

After Philip BedientRice University

Flow Net in a Corner:

Streamlines are at right angles to equipotential lines

Flow to Pumping Center:

Contour map of the piezometric surface near Savannah, Georgia, 1957, showing closed contours resulting from heavy local groundwater pumping (after USGS Water-Supply Paper 1611).

h = 92 m

Luthy Lakeh = 100 m

Peacock PondOrphanage

The deadly radionuclide jaronium-121 was releasedin Luthy Lake one year ago. Half-life = 365 daysConcentration in Luthy Lake was 100 g/L.If orphans receive jaronium over 50 g/L, theywill poop their diapers constantly. How longwill it take for the contaminated water to getto Peacock Pond and into the orphanage watersupply? Will jaronium be over the limit?

Example: Travel Time

Travel time

∑∑∑∑ Δ=

Δ=

Δ=Δ=Δ

4

1

4

1

4

1

4

1

i

i

i

ii

i

iitotal q

ln

q

ln

v

ltt

∑∑ ΔΔ

=Δ=Δ4

1

24

1

iitotal lhK

ntt

ii

i

i

ii l

hK

l

hK

bw

Qq

ΔΔ

=ΔΔ

==

h = 92 m

Luthy Lakeh = 100 m

Peacock Pond

Contamination in Luthy Lake

Travel time is only 131 days!

i

iii l

wbhKQΔΔ

Δ=

€

n = 0.25, Δh = 2m, K = 3.6x10−3 m/s,

m900

m50

m100

m250

m500

4

3

2

1

==Δ=Δ=Δ=Δ

totallllll

1

2

3

4

2lΔ

1lΔ

3lΔ

4lΔNote: because Δl terms are squared,

accuracy in estimating Δl is important.

Travel time

h = 92 m

Luthy Lakeh = 100 m

Peacock Pond

Contamination in Luthy Lake

Travel time is only 131 days!

i

iii l

wbhKQΔΔ

Δ=

€

n = 0.25, Δh = 2m, K = 3.6x10−3 m/s,

m900

m50

m100

m250

m500

4

3

2

1

==Δ=Δ=Δ=Δ

totallllll

1

2

3

4

2lΔ

1lΔ

3lΔ

4lΔ

€

Δttotal simple =ltotal

vaverage

=nltotal

KJaverage

=nltotal

K(4 × Δh / ltotal )=

nltotal2

K(4 × Δh)= 80days <<131days Supposed we used averages?

€

Co exp(−λ t) = Co exp −ln2 ⋅ tt1

2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟=100exp(−ln2 ⋅131

365) = 78μgL−1

Two Layer Flow System with Sand Below

Ku / Kl = 1 / 50

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) Ku / Kl = 1/50. (b) Ku / Kl = 50. Source: Todd & Bear, 1961.

SZ2005 Fig. 5.11

Flow nets in anisotropic media

Flownets in Anisotropic MediaSo far we have only talked about flownets in isotropic material. Can we draw flownets foranisotropic circumstances?

€

Kx

∂ 2h

∂x 2+ Ky

∂ 2h

∂y 2= 0

For steady-state anisotropic media, with x and y aligned with Kx and Ky, we can write the flow equation:

dividing both sides by Ky:

€

Kx

Ky

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟∂ 2h

∂x 2+

∂ 2h

∂y 2= 0

Flownets in Anisotropic Media

Next, we perform an extremely cool transformationof the coordinates:

€

Ky

Kx

⎛

⎝ ⎜

⎞

⎠ ⎟

12

x = X ⇒1

∂X 2=

Kx

Ky

1

∂x 2

This transforms our governing equation to:

€

∂2h

∂X 2+

∂ 2h

∂y 2= 0 Laplace’s Eqn!


Steps in drawing an anisotropic flownet:

1.Determine directions of max/min K. Rotate axesso that x aligns with Kmax and y with Kmin

2. Multiply the dimension in the x direction by (Ky/Kx)1/2 and draw flownet.

3. Project flownet back to the original dimension by dividing the x axis by (Ky/Kx)1/2

Flownets in Anisotropic MediaExample:

Kx

Ky

Kx = 15Ky

€

Ky

Kx

⎛

⎝ ⎜

⎞

⎠ ⎟

12

=1

15

⎛

⎝ ⎜

⎞

⎠ ⎟1

2

= 0.26


Kx

Ky

Kx = 15Ky


Kx

Ky

Kx = 15Ky


Kx = 15Ky


Kx = 15Ky


Kx = 15Ky


Kx = 15Ky

25%


Kx = 15Ky

25%


Kx = 15Ky


Kx = 15Ky


Kx = 15Ky


Kx = 15Ky


Kx = 15Ky

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 hydraulic conductivity of 150 ft/day, determine the rate of flow or seepage under the dam.


Flow Nets: an examplePosition: A B C D E F G H I JDistancefromfront toe(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


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


Flow Nets: An Example

There is an earthen dam 13 meters across and 7.5 meters high.The Impounded water is 6.2 meters deep, while the tailwater is 2.2 meters deep. The dam is 72 meters long. If the hydraulic conductivity is 6.1 x 10-4 centimeter per second, what is the seepage through the dam if the number of head drops is = 21

K = 6.1 x 10-4cm/sec= 0.527 m/day After Philip Bedient

Rice University

Flow Nets: the solutionFrom 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-meter length of the dam


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 Darcy’s Law calculation-Permeameter tests on core samples-Tracer tests-Inverse solutions of numerical models-Aquifer pumping tests

Aquifer Equation, based on assumptions becomes an ODE for h(r) :

-steady flow in a homogeneous, isotropic aquifer

-fully penetrating pumping well & horizontal, confined aquifer of uniform thickness, thus essentially horizontal groundwater flow

-flow symmetry: radially symmetric flow

02 =∇ h (Laplace’s

equation)

01

=⎟⎠⎞

⎜⎝⎛

drdh

rdrd

r

Steady Radial Flow Toward a Well

02

2

2

2

=∂∂

+∂∂

yh

xh (Laplace’s

equation in x,y 2D)

(Laplace’s equation in

radial coordinates)

Boundary conditions-2nd order ODE for h(r) , need two BC’s at two r’s, say r1 and r2

-Could be -one Dirichlet BC and one Neumann, say at well radius, rw ,

if we know the pumping rate, Qw

-or two Dirichlet BCs, e.g., two observation wells.


We need to have a confining layer for our 1-D (radial) flow equation to apply—if the aquifer were unconfined, the water table would slope down to

the well, and we would have flow in two dimensions.

b

h1h2

Q

Island with a confined aquifer in a lake

Multiply both sides by r dr

0=⎟⎠

⎞⎜⎝

⎛drdh

rd

Integrate

€

rdh

dr= C1

01

=⎟⎠

⎞⎜⎝

⎛drdh

rdrd

r

Evaluate constant C1 using continuity. The Qw pumped by the well must equal

the total Q along the circumference for every radius r. We will solve Darcys Law for r(dh/dr).


dhr

dr

T

Qw =π2

ODEConstant C1:

Separate variables:

€

Qw = (2πr)Tdh

dr

rdh

dr=

Qw

2πT= C1

Integrate again:

€

Qw

2πT

dr

rr1

r2

∫ = dhh1

h2

∫

€

Qw = Q(r) = KAdh

dr= K(2πr)b

dh

dr

= (2πr)Kbdh

dr=(2πr)T

dh

dr

b

h1h2

Q

r2 r1

Steady-State Pumping Tests

€

Qw

2πT

dr

rr1

r2

∫ = dhh1

h2

∫

Integrate

€

Q

2πTln

r2

r1

⎛

⎝ ⎜

⎞

⎠ ⎟= h2 − h1( )

( )12

1

2

2

ln

hh

r

rQ

T−

⎟⎟⎠

⎞⎜⎜⎝

⎛

=π

Thiem Equation

h

r

What is the physical rationale for the shape of this curve (steep at small r, flat at high r)?

Why is the equation logarithmic?This came about during the switch to radial coordinates.


(Driscoll, 1986)

As we approach the well the rings get smaller. A is smaller but Q is the

same, so must increase. dl

dh

Think of water as flowingtoward the well through a series

of rings.

The Thiem equation tells us there is a logarithmic relationship between head and distance from the pumping well.

lch

rr

Q

TΔ

⎟⎟⎠

⎞⎜⎜⎝

⎛

= 2

10ln

1

1

π

hlc

one log cycle

(Schwartz and Zhang, 2003)

The Thiem equation tells us there is a logarithmic relationship between head and distance from the pumping well.

lch

rr

Q

TΔ

⎟⎟⎠

⎞⎜⎜⎝

⎛

= 2

10ln

1

1

π

( ) ( )xx log 3.2ln ≅

If T decreases, slope increases

lch

QT

Δ=

2 3.2

π

hlc

one log cycle

(Schwartz and Zhang, 2003)

( ) 303.210ln ≅


Can we determine S from a Thiem analysis?

No—head isn’t changing!

Documents

Effects of Boundary Condition on Shape of Flow Nets ? ? ? ? ?