effects of boundary condition on shape of flow nets lecture 6 hydrology program, new mexico tech,...

ERTH403/HYD503 Lecture 6

Hydrology Program, New Mexico Tech,

Prof. J. Wilson, Fall 2006 1

Effects of Boundary Condition

on Shape of Flow Nets

Seepage and Dams

Flow nets forseepage throughearthen dams

Seepage underconcrete dams

Uses boundaryconditions (L & R)

Requirescurvilinear squaregrids for solution

After Philip Bedient

Rice 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 localgroundwater pumping (after USGS !Water-Supply Paper 1611).




h = 92 m

Luthy Lake

h = 100 m

Peacock PondOrphanage

The deadly radionuclide jaronium was released

in Luthy Lake one year ago. Half-life = 365 daysConcentration in Luthy Lake was 100 µg/L.

If orphans receive jaronium over 50 µg/L, they

will poop their diapers constantly. How long

will it take for the contaminated water to get

to Peacock Pond and into the orphanage water

supply? Will jaronium be over the limit?

Example: Travel Time

Travel time

!

"ttotal = "ti1

4

# ="livi1

4

# =ni"liqi1

4

# = n"liqi1

4

#

!! ""

="="4

1

2

4

1

iitotall

hK

ntt

ii

i

i

ii

l

hK

l

hK

bw

Qq

!

!=

!

!==

h = 92 m

Luthy Lake

h = 100 m

Peacock Pond

Contamination in Kidnapper Bog

Travel time is only 129 years.

i

iii

l

wbhKQ!

!!=

,m/s10,m2,25.05!

=="= Khn

m900

m50

m100

m250

m500

4

3

2

1

=

=!

=!

=!

=!

totall

l

l

l

l

1

2

3

4

2l!

1l!

3l!

4l!

days! 131 time travel then,m/s106.3 ifBut 3=!=

"K

!

"ttotal simple =ltotal

vaverage=

nltotal

KJave arg e=

nltotal

K(4 #"h / ltotal )=

nltotal2

K(4 #"h)= 80days <<131days

Note: because of !l terms are squared,

accuracy in estimating !l is very important.




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 differentanisotropic 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 MediaExample:

Kx

Ky

Kx = 15Ky




Flownets in Anisotropic Media

Kx = 15Ky

Flownets in Anisotropic Media

Kx = 15Ky




Flow Nets: an example

• A dam is constructed on a permeablestratum underlain by an impermeablerock. A row of sheet pile is installed atthe upstream face. If the permeable soilhas a hydraulic conductivity of 150ft/day, determine the rate of flow orseepage under the dam.


Rice University

Flow Nets: an examplePosition: A B C D E F G H I JDistance

from

front 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


Rice University




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, H

number of head drops


Rice University

Flow Nets: An Example

There is an earthen dam 13 metersacross and 7.5 meters high.TheImpounded water is 6.2 meters deep,while the tailwater is 2.2 meters deep.The dam is 72 meters long. If thehydraulic conductivity is 6.1 x 10-4

centimeter per second, what is theseepage through the dam if the numberof head drops is = 21

K = 6.1 x 10-4cm/sec

= 0.527 m/day After Philip Bedient

Rice University




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 and21 head drops along each flow path:

Q = (mKH/number of head drops) x damlength = (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


Rice University

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

=!"

#$%

&

dr

dhr

dr

d

r

Steady Radial Flow Toward a Well

02

2

2

2

=!

!+

!

!

y

h

x

h (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 Neuman, 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=!"

#$%

&

dr

dhrd

Integrate1C

dr

dhr !=

01

=!"

#$%

&

dr

dhr

dr

d

r

Evaluate constant C1 using continuity.

The Qw pumped by the well must equal

the total Q along the circumference for every

radius r.


dhr

dr

T

Qw =!2

ODEConstant C

1:

Separate variables:

!

Qw = (2"r)Tdh

dr

rdh

dr=Qw

2"T= C

Integrate again:

!

Qw

2"T

dr

rr1

r2

# = dhh1

h2

#

!

Qw =Q(r) = KAdh

dr= K(2"r)b

dh

dr

= (2"r)Kbdh

dr=(2"r)T

dh

dr




b

h1h2

Q

r2 r1

Steady-State Pumping Tests

!

Qw

2"T

dr

rr1

r2

# = dhh1

h2

#

Integrate

( ) !!"

#$$%

&='

1

2

12ln

2

r

rhh

Q

T(

( )12

1

2

2

ln

hh

r

rQ

T!

""#

$%%&

'

=(

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 flowing

toward 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

r

rQ

T!

""#

$%%&

'

= 2

10ln

1

1

(

#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

r

rQ

T!

""#

$%%&

'

= 2

10ln

1

1

(

( ) ( )xx log 3.2ln !

If T decreases, slope increases

lch

QT

!=

2

3.2

"

#hlc

one log cycle

(Schwartz and Zhang, 2003)

( ) 303.210ln !





Can we determine S from a Thiem analysis?

No—head isn’t changing!

effects of boundary condition on shape of flow nets lecture 6 hydrology program, new mexico tech,...

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