laplace's eqn & flow nets streamline
TRANSCRIPT
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 1
Laplace’s Eqn & Flow Nets
• Today
– Streamlines &Streamtubes
– Laplace’s Eqn
– Flow Nets
qx
qy
StreamlineStreamline:a line everywhere tangent
to the local veloctiy vector
In GW Hydrology, often
called a “flowline.”
There is no flow across a
streamline.
Consequently, no-flow
Neumann boundaries are
also streamlines.
ld
ldq
q
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 2
Streamtube:a region bounded by
streamlines
Because there is no flow across
the bounding surface, each cross-
section of the streamtube carries
the same mass flow.
So the streamtube is equivalent to
channel flow embedded in the
rest of the flowfield.
Streamtube
BBBAAA AqAq !! =
=
or ,streamtubea in
outflux massinflux mass
waterof nkssources/si internal no and
flow, ibleincompressor flow steady If
A
B
BA
BBAA
BA
AqAq
=
=
=
=
or ,
outflux volumetricinflux volumetric
)(via lecompressibslightly only fluid, isothermal (e.g.,
assumes &further goes modelour But
!
""
i.e., constant discharge
Continuity along a streamtube
Sim
ple
exam
ple,
a D
arcy
col
umn.
AA = area of streamtube at A
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 3
AB
QQQ
bwqbwqAqAq
AA
BBBAAAAAAA
BA
==
==
=
=
or ,or ,
outflux volumetricinflux volumetric
)(via lecompressibslightly only fluid, isothermal (e.g.,
Assuming
!
""
i.e., constant discharge
Continuity along a streamtube in 2D
AA = area of streamtube at A = bA wA
bA = depth or thickness of streamtube at A
wA = width of streamtube at A
flowline
Streamtube discharge
= Q
QA
QB
A and B are constant head
lines (equipotential lines).
Equipotentials and Gradient
Recall a line of equipotential
where head h = constant.
The gradient of h, or -!h, is the direction of
steepest decent down the potential surface, in
2D:h=100m
h=80mh=90mDirection of
hydraulic gradient.h=70m
Potentiometric map in 2D(Bradley and Smith, 1995)
-!h
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 4
Gradient and Darcy VelocityRecall Darcy’s Law
or .
If hydraulic conductivity is isotropic,
Then vectors and -!h are parallel.– i.e., they point in the same direction.
– They are simply scaled by the scalar K.
hKq !"=
hKq !"=
q
h=100m
h=80mh=90m
h=70m
Potentiometric map in 2D
-!h
Conductivity ellipse
Direction of
hydraulic gradient.
Specific discharge vector.
Under these conditions
equipotentials and streamlines
should be orthogonal.
Let’s take advantage of the concepts of
– Streamlines,
– streamtubes for steady flow without
sources/sinks, i.e. QA=QB, and
– specific discharge vectors, and therefore
streamlines, orthogonal to the equipotentials,
to develop a simple solution method for 2D
steady flow problems:
“The Flow Net.”
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 5
Conceptual Model:Steady flow in a homogeneous, isotropic aqufer
• General Aquifer Model
• Steady Flow
• Isotropic Aquifer
• Homogeneous Aquifer
– LaPlace’s Equation
• 2 Dimensional LaPlace’s
Equation, say in x,z, then:
• Add BC’s & Geometry
hKt
hSs
!"!=#
#
0=!"! hK
0=!"! hK
0or022=!=! hhK
0 2
2
2
2
=!
!+
!
!
z
h
x
h
Flow Net: A graphical method to solve these types of 2D flow problems.
Evolution of a governing equation:
Flow Net• A graphical solution method for 2D steady flow in a
homogeneous, isotropic aquifer– assuming no distributed internal sources/sinks.
• Can be extended tohomogeneous, anistropic aquifers,and even to somesimple heterogeneous situations.
• Also applies to vertically integrated, essentially horizontalflow aquifer models, if– flow is steady,
– transmissivity T is homogeneous & isotropic, &
– there are no internal sources/sinks, like recharge:
!
T"x,y
2h = 0 # " x,y
2h = 0 0
2
2
2
2
=!
!+
!
!
y
h
x
h
where h is the vertically averaged head. LaPlace’s Eqn.
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 6
"l1
"w1
"l2
"w2
Q
h1 h2 h3
h1-h2 ="h1 "h2
1
1
1
1
1
1 l
hbwK
dl
dhKAQ
!
!!==
Assume
thickness = b
1
1
11 l
wbhKQ!
!!=
Make head drops the same = "h
aspect ratio of net elements
Flow Net
flowline
equipotential
In element 1:
Element 1:
Discharge:
Flow in element i of a streamtube is:i
iii
l
wbhKQ!
!!=
2
2
2
1
1
121 l
wbhK
l
wbhKQQ
!
!!=
!
!!"=
2
2
2
1
1
1 l
wh
l
wh
!
!!=
!
!!
Flow Net
Q1
h1 h2 h3
h1-h2 ="h1 "h2
Q2
Element 1: Element 2:
Flow through each element is the same:
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 7
Gradient:
Let the head drop in each element of the tube
be the same: "h1 = "h2 = "hi = "h = constant.
Require that all flow nets be drawn this way.
2
2
2
1
1
1 l
wh
l
wh
!
!!=
!
!!
Flow Net
2
2
1
1
l
w
l
w
!
!=
!
!
So if , for all elements i, then Laplace’s equation is satisfied.i
i
l
w
l
w
l
w
!
!=
!
!=
!
!
2
2
1
1
A flow net must meet these requirements ("h and are constants).
If it does, it then provides a graphical solution to Laplace’s equation. lw !!
"h = constant
Leads to the Net Aspect Ratio constraint:
Usually pick the aspect ratio to be one, ;
makes it easier to draw a good flow net. 1)( =!! lw
Rules for Constructing Flow NetsIf only boundary conditions are known:
– Constant head boundaries (lakes, rivers) represent initial or finalequipotentials
– Impermeable (no-flow) boundaries are flowlines
– Draw an appropriate number of flow lines and flow tubes(with a pencil; usually not more than five or so flow tubes)
– Draw the equipotentials(keep the aspect ratio equal to one)
– Remember that equipotentials and flowlines must always intersect atright angles
– Adjust until the flow net is “square”• Keep =1, and this is why we use the term “square”
• Ok to end up with fractional “squares”
– A flowline should never intersect• another flow line
• an impermeable boundary
– Re K• Don’t need K to draw flow net as the net satisfies LaPlace’s Equation
– which is insensitive to conductivity
• Do need K to find fluxes and travel times.
lw !!
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 8
Drawing Method:
1. Draw to a convenient scale the geometry of the aquifer.
2. Establish constant head and no-flow boundary conditions
3. Draw one or two flow lines and equipotential lines near the boundaries.
4. Sketch intermediate flow lines and equipotential lines by smooth curves
adhering to right-angle intersections and square grids.
Where flow direction is a straight line, flow lines are an equal distance apart
and parallel.
5. Continue sketching until a problem develops. Each problem will indicate
changes to be made in the entire net. Successive trials will result in a
reasonably consistent flow net.
6. In most cases, 5 to 10 flow lines are usually sufficient. Depending on the
number of flow lines selected, the number of equipotential lines will
automatically be fixed by geometry and grid layout.
After Philip Bedient
Rice University
Example 1:
Hydraulic structure
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 9
Inflow to a lake used for water supply.
Aquifer parameters:
b = 20 m
K = 10-5 m/s
h = 92 m
Luthy Lake
h = 100 m
Peacock Pond
Antoniette Orphanage:
needs 50 m3/d
Example 2: Aquifer Flow
124m10x 2
!!= sT
h = 92 m
Luthy Lake
h = 100 m
Peacock Pond
i
i
i
iii
l
hbwK
l
hKAQ
!
!!=
!
!=
Since the flow net has aspect ratio of one:
hThbKQi !=!=
For the entire flow net,
hTmhTQQm
i
iT !=!== ""=
1
where “m” is the number
of flow tubes;
m = 2 in this example
For the entire flow net,
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 10
h = 92 m
Luthy Lake
h = 100 m
Peacock Pond
sT
2
4 m10 x 2
!=
!
"h =100 m - 92 m
4 = 2 m
m = 2
( )s
m 10x 8 m 2
s
m10x 2 2
3
4-
2
4- =!!"
#$$%
&=TQ
d
m69
s
m 10x 8
d
s64008
33
4-=!!
"
#$$%
&!"
#$%
&=TQ
hTmQT !=
Rules for Constructing Flow Nets
If only heads are known:– Interpolate equipotentials from head data
(use existing contours if available)
– Impermeable (no-flow) boundaries are flowlines
– Draw an appropriate number of flow lines and flow tubes(with a pencil; usually not more than five or so flow tubes)
– Draw the equipotentials(keep the aspect ratio equal to one)
– Remember that equipotentials and flowlines must always intersect atright angles
– Adjust until the flow net is “square”• Keep =1, and this is why we use the term “square”
• Ok to end up with fractional “squares”
– A flowline should never intersect• another flow line
• an impermeable boundary
– Re K• Don’t need K to draw flow net as the net satisfies LaPlace’s Equation
– which is insensitive to conductivity
• Do need K to find fluxes and travel times.
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 11
http://wapi.isu.edu/envgeo/aquifer_gw_review/flownets.htm
Example 3: Discharge
If T is known can calculate the aquifer discharge from the streamtube density.
Similar application: aquifer recharge.
T = 4 x 10-4 m2 s-1
1
2
3
9
8
7
6
54
QT=Qw
From flow net:
m = 9
"h=1m
From measurement:
Qw = 5000 m3/day
Example 4a: Inverse for T
hm
QT w
!=
hTmQQ Tw !==
Heat contours & flow lines around a pumping well.
Well pumping rate =
Invert for estimate of T:
/daym2500daym
m
12
5000
23
=!
="
=hm
QT w
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 12
SZ2005 Fig. 5.8
Example 4c: Pumping Centers
Flow net for unit with
- "h = constant
- constant b
Example 5: Variation in K
What is happening here?
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 13
2
2
1
1 !"
#$%
&
'
'=!
"
#$%
&
'
'
l
wT
l
wT
2
1
12
!"
#$%
&
'
'
!"
#$%
&
'
'
=
l
w
l
w
TT
In this case, "w1 = "w2; "l1 = 2"l2
112
2
1
5.01
1TTT ==
Example 5: Variation in K
K1, Q1, T1
K2, Q2, T2
hThbKQ iii !=!=
2
2
2
1
1
121 l
wbhK
l
wbhKQQ
!
!!=
!
!!"=
SZ2005 Fig. 5.9
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 14
SZ2005 Fig. 5.10
Example 7: Aquifers & Aquitards
SZ2005 Fig. 5.11
Next time: flow nets in anisotropic media
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 15
Next time:
Travel time
h = 92 m
Luthy Lake
h = 100 m
Peacock Lake
!! ""
="="4
1
2
4
1
iitotall
hK
ntt
Contamination in Luthy Lake!
Travel time is only 81 days!
Flow Nets
• Advantages– Simple
– Easy to do; gives quick understanding of flow regime
– Examines aquifer at much larger scale than core, slugor pumping tests
– Can give as accurate a result as the simpleconceptualization allows and data justifies
• Disadvantages– Assumptions are very constraining
– Especially• Steady State
• More or less homogeneous domain
• Two dimensional flow
ERTH403/HYD503 Lecture 6
Hydrology Program, New Mexico Tech,
Prof. J. Wilson, Fall 2006 16
• Next time
– Some Simple
Analytical Solutions
– 1D Steady Flow
• Confined & Water Table
Aquifers Bounded by
Streams, and
• Well Hydraulics
`
Proper Mathematical Statement
• Review
– Streamlines & Streamtubes
– Laplace’s Eqn
– Flow Nets