Transcript Lecture_9_Flownets

Effects of Boundary Condition
on Shape of Flow Nets
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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 Bedient
Rice University
Flow Net in a Corner:
Streamlines Y
are at right
angles to
equipotential
F 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).
Example: Travel Time
Luthy Lake
h = 100 m
The deadly radionuclide jaronium-121 was released
in Luthy Lake one year ago. Half-life = 365 days
Concentration in Luthy Lake was 100 g/L.
If orphans receive jaronium over 50 g/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?
h = 92 m
Peacock Pond
Orphanage
Travel time
Contamination in Luthy Lake
Luthy Lake
h = 100 m
l1
Qi  K hi b
1
l 2
l3
l1  500m
l 2  250m
l3  100m
l 4  50m
l 4
wi
li
ttotal
4
4
li
ni li
li
  ti  

 n
vi
qi
qi
1
1
1
1
4
4
Qi
K hi
K h


wi b
li
li
4
n 4
2
ttotal   ti 

l

i
K

h
1
1
qi 
2
3
Note: because l terms are squared,
accuracy in estimating l is important.
4
n  0.25, h  2m,K  3.6 x103 m/s,
h = 92 m
Travel time is only 131 days!
l total  900m Peacock Pond

t total simple 
ltotal
v average

2
nltotal
nltotal
nltotal


 80days131days
KJaverage K(4  h /ltotal ) K(4  h)
Qi  K hi b
l1  500m
l 2  250m
l3  100m
l 4  50m
l 4
wi
li
1
l 2
l3
Travel time
Contamination in Luthy Lake
Luthy Lake
h = 100 m
l1
Supposed we used averages?
2


Co exp(t)  Co expln2  t t  100exp( ln2 131365)  78gL1

1 

2 
3
4

n  0.25, h  2m,K  3.6 x103 m/s,
h = 92 m
Travel time is only 131 days!
l total  900m Peacock Pond

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.
Flow nets in anisotropic media
SZ2005 Fig. 5.11
Flownets in Anisotropic Media
So far we have only talked about flownets in
isotropic material. Can we draw flownets for
anisotropic circumstances?
For steady-state anisotropic media, with x and y
aligned with Kx and Ky, we can write the flow equation:
 2h
 2h
Kx 2  Ky 2  0
x
y
K  2 h  2 h
x
 2 0

dividing both sides by Ky:  
2

y
K y x

Flownets in Anisotropic Media
Next, we perform an extremely cool transformation
of the coordinates: 1
K y  2
1
Kx 1
x

X


 
X 2 K y x 2
K x 
This transforms our governing equation to:

 2h  2h
 2 0
2
X
y
Laplace’s Eqn!
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)1/2 and draw flownet.
3. Project flownet back to the original dimension by
dividing the x axis by (Ky/Kx)1/2
Flownets in Anisotropic Media
Example:
Ky
Kx
1
Kx = 15Ky
K y  2  1 1 2
      0.26
K x  15
Flownets in Anisotropic Media
Ky
Kx = 15Ky
Kx
Flownets in Anisotropic Media
Kx = 15Ky
Flownets in Anisotropic Media
Kx = 15Ky
Flownets in Anisotropic Media
Kx = 15Ky
Flownets in Anisotropic Media
Kx = 15Ky
Flownets in Anisotropic Media
Kx = 15Ky
25%
Flownets in Anisotropic Media
Kx = 15Ky
25%
Flownets in Anisotropic Media
Kx = 15Ky
Flownets in Anisotropic Media
Kx = 15Ky
Flownets in Anisotropic Media
Kx = 15Ky
Flownets in Anisotropic Media
Kx = 15Ky
Flownets in Anisotropic Media
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.
After Philip Bedient
Rice University
Flow Nets: an example
Posit ion:
A B C D
E
F G H
I
J
Distance
f rom
f ront t oe
( f t)
0
3
22
37 .5
50
62 .5
75
86
94
10 0
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 Bedient
Rice University
Flow Nets: the solution
• Solve for the flow per unit width:
q=mK
total change in head, H
number of head drops
= (5)(150)(35/17)
= 1544 ft3/day per ft
After Philip Bedient
Rice University
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 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-meter
length of the dam
After Philip Bedient
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
Steady Radial Flow Toward a Well
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
2h  0
 2h  2h
 2 0
2
x
y
1 d  dh 
r   0
r dr  dr 
(Laplace’s
equation)
(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.
Steady Radial Flow Toward a Well
Island with a confined aquifer in a lake
Q
b
h2
h1
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.
Steady Radial Flow Toward a Well
ODE
1 d  dh 
r   0
r dr  dr 
Constant C1:
r
Multiply both sides by r dr
 dh 
d r   0
 dr 
Integrate
r
dh
 C1
dr
Qw  (2r)T
dh
dr
dh Qw

 C1
dr 2T
Separate variables:

Qw dr
 dh
2T r
Integrate again:
Qw
2T
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).
dh
dh
Qw  Q(r)  KA
 K(2r)b
dr
dr
dh
dh
 (2r)Kb
(2r)T
dr
dr

r2

r1
dr

r
h2
 dh
h1
Steady-State Pumping Tests
Q
r2
r
1
b
h2
h1
Qw
2T
r2

r1
dr

r
h2
 dh
h1

Integrate
Q r2 
ln  h2  h1
2T r1 
 r2 
Q ln 
r1 

T
2 h2  h1 
Thiem Equation
Steady-State Pumping Tests
Why is the equation logarithmic?
This came about during the switch to radial coordinates.
What is the physical rationale for
the shape of this curve (steep at
small r, flat at high r)?
h
r
Think of water as flowing
toward the well through a series
of rings.
As we approach the well the
rings get smaller. A is
smaller but Q is the
same, so dh must increase.
dl
(Driscoll, 1986)
The Thiem equation tells us there is a logarithmic relationship between
head and distance from the pumping well.
 10r1 

Q ln
r1 

T
2 hlc
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.
 10r1 

Q ln
r1 

T
2 hlc
lnx   2.3 logx 
hlc
ln10  2.303
2 .3 Q
T
2  hlc
one log cycle
If T decreases, slope increases
(Schwartz and Zhang, 2003)
Steady-State Pumping Tests
Can we determine S from a Thiem analysis?
No—head isn’t changing!