introduction to groundwater flow modelingwang/geo627/strandpp.pdf1) darcy’s law, continuity, and...
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Introduction to groundwater flow modeling: finite difference
methods
Tyson Strand
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1) Darcy’s law, continuity, and the groundwater flow equation
2) Fundamentals of finite difference methods
3) FD solution of Laplace’s equation
4) FD solution of Poisson’s equation
5) Transient flow
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1) Darcy’s law, continuity, and the groundwater flow equation
qx
qy
qy+dy
qx+dx
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0)()( =−+−++ yydyyyxxdxxx
qqdxqqdy
Everything that goes in, must come out
00==
−+
− ++
dxdydy
dxqq yydyyyxxdxxx
After dividing by the volume (area), let dx and dy go to zero
0=∂
∂+
∂∂
yq
xq yx Continuity
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Darcy’s law: isotropic medium
xhKqx ∂∂
−=
yhKqy ∂∂
−=
zhKqz ∂∂
−=
Scalar form
hKkzhj
yhi
xhK ∇−=
∂∂
+∂∂
+∂∂
−= )ˆˆˆ(
Vector form
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Darcy’s law + continuity (in 2D)
xhKqx ∂∂
−=yhKqy ∂∂
−= 0=∂
∂+
∂∂
yq
xq yx
02
2
2
2
=
∂∂
+∂∂
−=
∂∂
−∂∂
+
∂∂
−∂∂
yh
xhK
yhK
yxhK
x
02 =∇ h Laplace’s equation
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The groundwater flow equation
thSW
zhK
zyhK
yxhK
x szzyyxx ∂∂
=−
∂∂
∂∂
+
∂∂
∂∂
+
∂∂
∂∂
Sources and sinks
Transient flow term
Darcy’s law + continuity
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Simplifying assumptions
• Isotropic medium: K = Kxx = Kyy = Kzz W = f(x,y,t)
• Steady-state flow (Poisson’s equation W = f(x,y))
• Steady-state flow, no sources/sinks (Laplace’s equation)
thSWhK s ∂∂
=−∇2
KWh =∇2
02 =∇ h
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Fundamentals of finite difference methods
• Discretization of space
• Discretization of (continuous) quantities
• Discretization of time
• The first spatial derivative
• The second spatial derivative
• Boundary conditions and initial conditions
• Solving the problem
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Disctretization of space
(i,j)
(i,j+1)
(i,j-1)
(i-1,j) (i+1,j)
(i,j)
xdx
yLx = (Nx – 1)dx
dy
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Discretization of quantitiesEach spatial location (i,j) has associated with it a head value and a vector darcy flux
(i,j)
(i,j+1)
(i,j-1)
(i-1,j) (i+1,j)h(i,j)qx(i,j)qy(i,j)
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Discretization of time
t = n n+1∆t n+2∆t n+3∆t n+4∆t n+5∆t
tm+2 tm+3 tm+4 tm+5tm tm+1
Increasing time
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The first spatial derivative
∆−
+∆−
=
∆∆
≅∂∂ +−
xhh
xhh
xh
xh iiii
ji
11
),( 21
(i,j)
(i,j+1)
(i,j-1)
(i-1,j) (i+1,j)xhh
xh
xh ii
ji ∆−
=∆∆
≅∂∂ −1
),(
Upstream difference
Central difference
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The second spatial derivative: step 1
xhh
xh ii
i ∆−
≅∂∂ −
−
1
21
xhh
xh ii
i ∆−
≅∂∂ +
+
1
21
(i,j)
(i,j+1)
(i,j-1)
(i-1,j) (i+1,j)
NOTE: This is essentially a central differenceapproximation of the first derivative evaluatedat the mid-positions
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The second spatial derivative: step 2
∂∂
∂∂
=∂∂
xh
xxh2
2
( )211
11
21
21
2
2 2x
hhhx
xhh
xhh
x
xh
xh
xh iii
iiiiii
∆+−
=∆
∆−
−∆−
=∆
∂∂
−∂∂
≅∂∂ −+
−+−+
The second derivative is just the derivative of the derivative
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Boundary conditionsWhat are boundary conditions and why are they necessary?
1.0=∂∂xhConsider the simple problem:
We can solve this directly by separating variables and integrating:
Cxhxh +=→∂=∂∫ ∫ 1.01.0
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There are an infinitenumber of solutionsto this equation.
They are referred toas a family of curves
Cxh += 1.0h
x
We must specify avalue of h at a knownposition x in order tosolve for C.
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For a second (spatial) derivative, two boundary conditionsmust be specified. There are 3 types of bc’s that we can apply
1) Head is specified at a boundary
- Called Dirichlet conditions
2) Flow (first derivative of head) is specified at a boundary
- Called Neumann conditions
3) Some combination of 1) and 2)
- Called mixed conditions
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Initial conditions
Equivalent to boundary conditions, except that the boundaryis now temporal instead of spatial
Typically, the state of the system (i.e. head values) arespecified at time t = 0
thSW
zhK
zyhK
yxhK
x szzyyxx ∂∂
=−
∂∂
∂∂
+
∂∂
∂∂
+
∂∂
∂∂
Question: How many boundary and initial conditionsWould be needed to solve the above governing equation?
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Solving the problem
After Wang & Anderson, 1982
Set of differentialequations
(Mathematical model)
Set of algebraicequations
(discrete model)
Finite DifferenceFinite Element
Fieldobservations
Analytical solution (Not always possible)
Calculustechniques
Iterative ordirect methods
Approximatesolution
CompareIf possible
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Recall how we arrived at Laplace’s equation
xhKqx ∂∂
−=yhKqy ∂∂
−= 0=∂
∂+
∂∂
yq
xq yx
02
2
2
2
=
∂∂
+∂∂
−=
∂∂
−∂∂
+
∂∂
−∂∂
yh
xhK
yhK
yxhK
x
02 =∇ h Laplace’s equation
We are focusing on iterative methods
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Solving Laplace’s equation (2D)
02
2
2
2
=∂∂
+∂∂
yh
xh02 =∇ h
( )2),1(),(),1(
2
2 2x
hhhxh jijiji
∆
+−≅
∂∂ −+
( )2)1,(),()1,(
2
2 2y
hhhyh jijiji
∆
+−≅
∂∂ −+
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Let ∆x = ∆y
( )2),1(),(),1(
2
2 2x
hhhxh jijiji
∆
+−≅
∂∂ −+
( )2)1,(),()1,(
2
2 2y
hhhyh jijiji
∆
+−≅
∂∂ −+
+
( )[ ]),()1,()1,(),1(),1(2 41
jijijijiji hhhhhx
−+++∆ +−+−
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( )[ ]),()1,()1,(),1(),1(2 41
jijijijiji hhhhhx
−+++∆ +−+−
Set equal to zero
04 ),()1,()1,(),1(),1( =−+++ +−+− jijijijiji hhhhh
The above equation is the basic finite difference solution to Laplace’s equation
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Now we rearrange the previousequation so that we can implement
it into our regular grid
Solve for h(i,j)
4)1,()1,(),1(),1(
),(+−+− +++
= jijijijiji
hhhhh
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Now what do we do?
(i,j)
Specify boundary conditions
Guess at initial values for headat all (i,j) locations
Begin iterations
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What does it mean to “iterate”?
Given: We know the boundary conditions and haveinitial guess values for head(position)
Do: Update head values at every (i,j) locationbased on previous equation (including boundaries if applicable)
While: Convergence criteria are not met
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Convergence
1) Compare obtained head values at a giveniteration to those at the previous iteration
2) Continue to iterate as long as head valuescontinue to change within some preset limit
3) Test the solution (to be described)
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Testing a computational model
Comparison to other analytical results/field data (validation?)
Changing the grid spacing
Changing the convergence criterion
Mass balance check
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Validation/verification
1) Analytical solution is known (uncommon)
2) Comparison to field data
3) Test on simple situations for which 1 or 2are known
4) Check model predictions
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Possible model problems
Stability
Robustness
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Sample problem: Laplace’s equation
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Lecture 2• Poisson’s equation
– Digression: Inflow, outflow, and sign conventions• Finite difference form for Poisson’s equation• Example programs solving Poisson’s equation• Transient flow
– Digression: Storage parameters• Finite difference form for transient gw flow
equation (explicit methods & stability)• Example transient flow program • Implicit iterative methods • Example transient flow program, fully implicit
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Solving Poisson’s equation
−==∇
TyxR
KWh ),(2 What is W?
KW
yh
xh
=∂∂
+∂∂
2
2
2
2
byxRW ),(
−=
Where b is the thickness of the aquifer (z dimension)
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Inflow, outflow, and sign conventions
Think about the magnitudes of qout and qinWhat does it mean in terms of sources/sinks?
x
02
2
=∂∂xh
02
2
>∂∂xh
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Sign conventions
02
2
<∂∂xh
02
2
<∂∂xh
There must be recharge (or someother source) at the REV (steady-state assumption)
02
2
>∂∂xh There must be discharge (or some
other sink) at the REV(steady-state assumption)
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−==∇
TyxR
KWh ),(2
• You need to be careful as to how things are defined
• W MUST be defined discharge (sink) positive• It is more intuitive to define a recharge (source)
function R(x,y) that is positive for recharge (sources) and negative for discharge (sinks)
• This will also help give us an intuitive understanding of the transient gw flow equation
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Back to Poisson’s equation
TyxRh ),(2 −=∇ Proceeding as we did for Laplace’s equation
( )[ ]
TR
hhhhhx
jijijijijiji
),(),()1,()1,(),1(),1(2 41
−=−+++∆ +−+−
Solving for h(i,j)
( )
∆++++= +−+− T
Rxhhhhh ji
jijijijiji),(
2
)1,()1,(),1(),1(),( 41
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Solving Poisson’s equation numerically
• Basically, we can proceed exactly as we did for Laplace’s equation, using the previous finite difference approximation for h(i,j)
• Define boundary conditions• Set initial guess values• Iterate• Check results
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Sample problem: Poisson’s equation with uniform recharge/discharge
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Sample problem: Poisson’s equation with point source
recharge/discharge
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Solving transient flow problems
btyxR
thShK s
),,(2 −∂∂
=∇
We’ve learned how to take care of everything except the time derivative
thh
th
th m
jimji
∆
−=
∆∆
≈∂∂ +
),(1),( Forward difference approximation
thh
th
th m
jimji
∆
−=
∆∆
≈∂∂ −1
),(),( Backward difference approximation
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Central difference approximation
thh
th
th m
jimji
∆
−=
∆∆
≈∂∂ −+
2
1),(
1),(
= BAD NEWS
• The central difference approximation is unconditionally unstable (for time derivative)
• This method should NOT be used to estimate the first time derivative
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Storage parameters• S = storage coefficient, or storativity
– Dimensionless– Measures the volume of water expelled (absorbed) per
unit surface area per unit head change
• Ss = specific storage = S/b– Units [1/L] also called elastic storage coeff.– Measures the water volume per unit aquifer volume that
is expelled (stored) due to compressibility (matrix & water) per unit head change
– Used for confined units or the saturated parts of an unconfined unit
– b = aquifer thickness for confined units, use bs = thickness of the saturated region for unconfined systems
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Specific yield• Sy = specific yield
– Dimensionless– Used to describe unconfined systems– Ratio of water volume that drains from a saturated
region of aquifer due to gravity forces per unit aquifer volume
– Generally several orders of magnitude larger than bsSsexcept in very fine grained systems
• Relationship between Sy, Ss, and S
ssy SbSS +=Fetter, 1994
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Storage parameters (concl.)
• In unconfined systems, the specific storage can generally be neglected– Except in very fine grained media
• In confined systems, only the specific storage is pertinent (so long as head values remain above the upper confining unit)
• In general, for computational models, it is best to use the storage coefficient (storativity)
• Definitions of storage parameters are not always consistent in literature, be careful
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Transient flow• So, let’s consider the transient flow equation in
the form
TtyxR
th
TS
yh
xh ),,(
2
2
2
2
−∂∂
=∂∂
+∂∂
Assume the left hand size is negative (out > in):then, by the convention we’ve set , either rechargeis positive or the time rate of change of storage is negative (right?)
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TtyxR
th
TS
yh
xh ),,(
2
2
2
2
−∂∂
=∂∂
+∂∂
Let’s write the finite difference approximation for this equation
TR
thh
TS
yhhh
xhhh
mji
mji
mji
mji
mji
mji
mji
mji
mji
),(),(1),(
2)1,(),()1,(
2),1(),(),1(
)(2
)(2
−∆
−=
∆
+−+
∆
+−
+
−+−+
This is called the explicit representation, since all spatial termsare evaluated at the “old” time = m
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Let ∆x = ∆y, and solve for the head at (i,j) at time index m+1
[ ]mji
mji
mji
mji
mjim
jimji hhhh
xStT
SRth
xStTh )1,()1,(),1(),1(2
),(),(2
1),(
41 −+−++ +++
∆∆
+∆+
∆∆
−=
The explicit equation is easy to solve:
• Given a set of initial conditions (& bc’s)• The head at all (i,j) locations at time index
m+1 can be calculated directly from the head values at the previous time index m
• Notice, there is no iteration, once you have an initial condition it is a simple matter to step forward in time by an amount ∆t
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Stability of the explicit method• One must be careful to keep the time steps small
enough such that the explicit method remains stable
• If the time step is too large, fluctuations will develop in the head as a function of time, which will be amplified as the model continues to step through time
• The stability criteria are:
25.02 <∆∆xStT 5.02 <∆
∆xStT
2-D 1-D
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Example transient problem:Explicit solution, 1D
Initial
Final
We are interested in what happensin between the initial and final states
The head at the right boundary dropsfrom blue to red at time t = 0+
The red potentiometric surface is thefinal state
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Implicit methods
• Notice in the explicit formulation that we have evaluated all spatial derivatives at the old time (tm)
• Is that the best?• What if we evaluated spatial derivatives at
the new time (tm+1)?• What if we evaluated spatial derivatives
halfway between?
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2),1(),(),1(
2
1),1(
1),(
1),1(
2
2
)(2
)1()(
2x
hhhx
hhhxh m
jimji
mji
mji
mji
mji
∆
+−−+
∆
+−≈
∂∂ −+
+−
+++ αα
Where α can vary between 0 (fully explicit) and 1 (fully implicit)
A similar expression can be written for the second derivativewith respect to y
We simplify the algebra by defining the following and letting ∆x=∆y
4~ )1,()1,(),1(),1(
),(+−+− +++
= jijijijiji
hhhhh
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[ ] [ ]mji
mji
mji
mji hh
xhh
xyh
xh
),(),(21),(
1),(22
2
2
2 ~)1(4~4−
∆−
+−∆
≈∂∂
+∂∂ ++ αα
Put this approximation into the gw flow equation
[ ] [ ]TR
thh
TShh
xhh
x
mji
mji
mjim
jimji
mji
mji
),(),(1),(
),(),(21),(
1),(2
~)1(4~4−
∆
−=−
∆−
+−∆
+++ αα
Rearrange
( )[ ]TRx
hhhtTSxhh
tTSx m
jimji
mji
mji
mji
mji 4
~14
~4
),(2
),(),(),(
21),(
1),(
2 ∆+−−+
∆∆
=−
∆
∆+ ++ ααα
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Solving the previous equation for the new head value at (i,j)
( )[ ]
∆+−−+
∆∆
+
∆∆
+= ++
TRx
hhhtTSxh
tTSx
hmjim
jimji
mji
mji
mji 4
~14
~
4
1 ),(2
),(),(),(
21),(2
1),( αα
α
This is the fundamental implicit finite difference approximationα = 0 Fully explicitα = ½ Crank-Nicolsonα = 1 Fully implicit
The above expression can be solved by iterative or direct methods
We will continue to focus on iterative methods
The advantage of implicit methods is STABILITY
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To solve iteratively:- Sweep through the lattice, updating the (i,j) head
value at the new time by the above formula- Continue to sweep through the lattice, updating,
as long as the new head values keep changing- Stop when new head values do not change
within preset limits- (update as you go along:
called Gauss-Seidel iteration)- Repeat for the next time step
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Fully implicit, 1D sample problem
∆
∆+
∆∆
+= ++ m
imi
mi h
tTSxh
tTSx
h )(
21
)(21)( 2
~
21
1
1)(
~ +mihWhere is now defined by
+=
++
+−+
2~ 1
)1(1)1(1
)(
mi
mim
i
hhh
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References
• H. F. Wang and M. P. Anderson, 1982. Introduction to Groundwater Modeling, Academic Press, Inc.
• C. W. Fetter, 1994. Applied Hydrogeology, Macmillan College Publishing Company.
• F. W. Schwartz and H. Zhang, 2003. Fundamentals of Ground Water, Wiley publishing.