Example #2 MODFLOW
This example is the simple "OBLEO" aquifer problem. Figure 1 is a plan view map of the
hydrologic setting of the OBLEO aquifer. The aquifer is an unconfined aquifer with a single well
P-1. The aquifer is bounded to the North and South by mountains, and the East and West by two
large lakes. The water elevation in each lake is 0.0 meters above some datum, while the ground
elevation at the well is 100.0 meters above the same datum.
Zenith Foothills
Lake
Alpha
(z=0.0m)
P-1 (z=100.m)
Lake
Omega
(z=0.0m)
1100 km
Nadir Mountains
Figure 1. Obleo Aquifer Setting
Rainfall records indicate that recharge to the subsurface is approximately 250 mm/yr. The
aquifer is homogeneous and isotropic with a hydraulic conductivity of K=100m/day, and a
specific yield S = 0.25. The example constructs a computer model to help determine the safe
yield of the aquifer - assume minimum permissible saturated thickness is 500 m and that the
aquifer is 1000 m thick.
The OBLEO water board would like to exploit the aquifer, they would like to know (for a 30 year
planning period):
a) What is the maximum rate P-1 that can be used so that the saturated thickness is never less
than 500m?
b) How much of the pumped water is induced recharge from the two lakes? (Why would this
question be of interest?)
A model is one tool that can help answer these questions. Typically one uses some semblance of
the following steps:
Conceptual model - Translate the real system into a simplified system that can be represented in
a computer (or analog) model.
Mathematical model – Convert the conceptual model to a set of governing equations,
boundary and initial conditions.
Numerical model – Computer program implementing some numerical methods to obtain
solutions to the mathematical model, or good approximations.
The steps of mathematical and numerical modeling are “implicit” if the tool is
MODFLOW. One will be using a diffusion equation with spatially varying material
properties (transmissivity, storativity), with a prescribed set of initial conditions and
boundary conditions. MODFLOW “solves” the mathematical model using a blockcentered finite-difference approximation to the governing equation and various different
linear equation solvers. The time-difference model used is either fully-implicit or
alternating-direction implicit.
Data encoding - This step is where the analyst translates the conceptual model into data streams
(input files) for the computer program to process. Generally the analyst must supply geometry,
material properties, forcing terms (wells, recharge, etc.), and boundary conditions. The program
will generate approximations at different times of heads in the aquifer layers, and cell-to-cell
flow terms (using Darcy’s law).
Solution - Typically the input files need several iterations of manipulation until one is asking the
correct questions (of the model) and the output makes sense. Expect initial program runs to fail
from input that is incorrect, inconsistent, or just plain wrong. Generally the “solution” will be an
array of heads in each model layer, and possibly fluxes at specific locations. Usually a graphical
tool is used to render the arrays into something straightforward to understand.
Interpretation - This step is where the model results are interpreted as answers to the questions
that we built the model for in the first place. This is where we would include “What-if …?”
scenarios and their results. The engineer also uses this step to make recommendations or design
decisions.
Conceptual Model
The obleo aquifer is a single unconfined, homogeneous, isotropic aquifer with hydraulic
conductivity K= 36,500m/yr and a specific yield of S = 0.25. The aquifer is rectangular with
dimensions 1100km x 1100km in the horizontal direction and 0.1km in the vertical. Flow is
assumed to be essentially horizontal so that the Dupuit assumptions can be used.
The Zenith Foothills to the North are conceptualized as a no-flow boundary as are the Nadir
Mountains to the South. Lake Alpha to the east and Lake Omega to the West are conceptualized
as constant head boundaries. A single well, P-1 is planned for the Obleo Aquifer. P-1 is located
at the geographic center of the aquifer, equidistant from all four boundaries.
Mathematical Model
The solution domain can be expressed as:
D = { (x,y) | 0 <x<1300m,
The boundaries can be expressed as:
0<y<1100m
}
G1 = { (x,y) | x=0;
G2 = { (x,y) | 0<x,1300m;
G3 = { (x,y) | x=1300m;
G4 = { (x,y) | 0<x<1300m;
0<y<1100m
y=1100m
0<y<1100m
y=0
}
}
}
}
The head in the aquifer is governed by the following diffusion equation:
∂h
div (Kh grad(h) ) = S ∂t + q - r
q- pumping; r- recharge
subject to the following boundary conditions:
h(x,y,t) = H(left) on G1
∂h
-K∂y (x,y,t) = 0 on G2
h(x,y,t) = H(right) on G3
∂h
-K∂y (x,y,t) = 0 on G4
h(x,y,0) = f(x,y)
on D
(Fixed head)
(No-Flow)
(Fixed head)
(No-Flow)
(Initial condition)
Numerical Model
The solution to the mathematical model will be obtained using the Finite-Difference Method.
The numerical implementation will be using the USGS-MODFLOW code.
Row
1
-1
1
1
1
1
1
1
1
1
1
1
1
-1
2
-1
1
1
1
1
1
1
1
1
1
1
1
-1
3
-1
1
1
1
1
1
1
1
1
1
1
1
-1
4
-1
1
1
1
1
1
1
1
1
1
1
1
-1
-1
1
1
1
1
1
1
1
1
1
1
1
-1
-1
1
1
1
1
1
1
1
1
1
1
1
-1
-1
1
1
1
1
1
1
1
1
1
1
1
-1
-1
1
1
1
1
1
1
1
1
1
1
1
-1
-1
1
1
1
1
1
1
1
1
1
1
1
-1
10
-1
1
1
1
1
1
1
1
1
1
1
1
-1
11
-1
1
1
1
1
1
1
1
1
1
1
1
-1
1
2
Column
3
12
13
Figure 2. Boundary-array diagram
The Obleo aquifer is modeled using the Dupuit assumptions as a single aquifer layer. The
horizontal resolution is 1km x 1km; the vertical resolution is 0.1 km. Flow is assumed to be
essentially horizontal. Figure 2 is a plan-view diagram of the aquifer with the constant head (-1),
no-flow (0), and computational cells (1) indicated. This diagram is the boundary array that will
be required by the program. Each cell is 1000 meters x 1000 meters.
Figure 3 depicts the vertical resolution of the model. The “aquifer” is 1 km thick. Boundary
heads at the two lakes are 900 meters above the bottom of the aquifer, but the lake is the datum,
thus the bottom is located 900 meters below the lake.
z=+100 m
z=0.0 m
z=-900m
Figure 3. Vertical conceptual model and datum
The MODFLOW packages that are used include:
BASIC + BLOCK CENTERED FLOW (always)
RECHARGE + WELLS
SIP (or SSOR ...)
The data stream will be located in the following set of DATA FILES, that correspond to the
different packages.
FOR001 <--- BASIC
FOR011 <--- BCF
FOR012 <--- WELLS
FOR018 <--- RECHARGE
FOR019 <--- SIP
The version of MODFLOW used assumes FORMATTED data sets.
Initial conditions - Because we really don't know the initial conditions, we will run the model
once without any pumping to obtain an equilibrium solution. This solution will represent our
initial condition for the "what-if" modeling. An alternative is to run the model “steady state”
first and save the output for subsequent modeling. Either method should result in identical
output, but the transient to equilibrium is sometimes easier.
Figure 4 illustrates the concept. We simulate for 30 years “pre-development” then for 30 more
years to evaluate the pumping impact. We select different pumping rates until we can satisfy the
draw down requirement then report this simulation as the “design” case.
Figure 4. Concept of transient to equilibrium ,then apply stress.
Data files are built using a text editor - examples are shown below:
Data Encoding
FOR001 <--- BASIC
CE6361 OBLEO AQUIFER EXAMPLE PROBLEM
<= first two lines are notes to ourselves!
1 LAYER, 11 ROWS, 13 COLUMNS, UNCONFINED
1
11
13
2
5 <= layers, row, column, periods, time unit.
11 12 0 0 0 0 0 18 19 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 <= modules
0
0
<= memory handling
1
1(20i3)
3
IBOUND- 1
<=boundary array
-1 1 1 1 1 1 1 1 1 1 1 1 -1
-1 1 1 1 1 1 1 1 1 1 1 1 -1
-1 1 1 1 1 1 1 1 1 1 1 1 -1
-1 1 1 1 1 1 1 1 1 1 1 1 -1
-1 1 1 1 1 1 1 1 1 1 1 1 -1
-1 1 1 1 1 1 1 1 1 1 1 1 -1
-1 1 1 1 1 1 1 1 1 1 1 1 -1
-1 1 1 1 1 1 1 1 1 1 1 1 -1
-1 1 1 1 1 1 1 1 1 1 1 1 -1
-1 1 1 1 1 1 1 1 1 1 1 1 -1
-1 1 1 1 1 1 1 1 1 1 1 1 -1
100.
<= head at no flow cells, use unusual number.
1
1.(10f10.0)
3
SHEAD- 1
<=initial head array
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
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0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
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0.
0.
0.
0.
0.
0.
0.
0.
0.
30.
30.
0.
1
1
0.
1.
1.
<= st. period length, time step multiplier, time step length
FOR011 <--- BCF
0
1
11
0
<= some flags
<= simulation type (transient)
1.00
(10G10.3)
0
TRPY-
0
0
DELR1.00
0
1.00
1.00
1.00
1.00
DELC1.00
0
1.00
1.00
1.00
1.00
1.00
11
1.E3
1.00
1.00
1.E3
1.00
(10G10.3)
1.00
1.00
(10g10.3)
1.00
1.00
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
11 1.E3
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
11 1.00
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
(10G10.3)
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
(10G10.3)
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
(10G10.3)
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
1.00
1.00
11
1.00
1.00
11
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
1.00
1.00
1.00
1.00
0.250
0.250
SF1- 1
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
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0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
3.65
3.65
HYDCON3.65
1
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
3.65
-900.
-900.
BOT- 1
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
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-900.
-900.
-900.
-900.
-900.
-900.
-900.
0
3
3
3
-900.
-900.
-900.
-900.
-900.
-900.
-900.
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-900.
-900.
-900.
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-900.
-900.
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-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
-900.
FOR012 <--- WELLS
1
1
1
1
1
0
6
7 0.000
6
7 -1.E9
FOR018 <--- RECHARGE
1
0
18
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
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0.250
0.250
0.250
0.250
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0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
1.00
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
18 1.00
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0
0
(10G10.3)
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
(10G10.3)
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
3
RECH- 0
0.250
0.250
0.250
0.250
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0.250
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0.250
0.250
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0.250
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0.250
0.250
0.250
0.250
0.250
RECH- 0
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
0.250
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0.250
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0.250
3
FOR019 <--- SIP
50
1.
5
1.E-07
MXITER NPARM
0
.001
1
ACCL,ERR,IPCALC,WSEED
These “files” are not expected to work the first attempt. Essentially we build the model to be
sure the program reads the files correctly (even if the data are incorrect), then iteratively correct
the input data until we are modeling the situation of interest. The biggest issues will be
adjustments to the FOR019 file to get convergence with reasonable accuracy, changing the predevelopment period length, and move the output from pre-development into the starting head
array. The analyst keeps tracks of these changes in a simulation log and it is best to change only
one thing at a time.
A log of the changes to the model is illustrated below:
Simulation Number
run1
run2
run3
run 4
run 5
Remarks
files read OK, but supplied
template is missing two
columns.
geometry OK, model fails to
converge.
solution converges, but not
sure if year 0 is equilibrium.
Year 0 and Year 30 heads are
same; run3 changes verified.
Year 0 and Year 30 heads are
different (no pumping)
run 6
Year 0 and Year 30 heads are
same.
run 7
Set P1 = 3,650 cu.m./year
(about 2,000 gpm)
Set P1 = 36,500 cu.m./year
(about 20,000 gpm)
Set P1 = 365,000 cu.m./year
(about 200,000 gpm)
Set P1 = 3,650,000 cu.m./year
(about 2,000,000 gpm)
run 8
run 9
run 10
run 11
Set P1 = 36,500,000
cu.m./year (about 20,000,000
gpm)
run 12
Set P1 = 43,100,000
Suggested Changes
Add data in
FOR001,FOR011,FOR018 to
get model into 11X13
Adjust FOR019 – increase
MAXITER to 200.
Set P1=0 in second stress
period (no pumping). Set
FOR018 recharge to small
value. Set Starting heads to
900, year 0 and year 30
solution should match.
Set FOR018 recharge to
specified value.
Year 0 is not an equilibrium
solution. Use Year 30 heads
as start heads, and start
simulation at –240 years.
Model is ready for pump
evaluation. Year 0 solution is
an equlibrium state. Save this
output file as Case0 – it will
be part of appendix.
Drawdown is undetectable.
Drawdown is about 1 meter in
well cell.
Drawdown is about 5 meters
in well cell.
Drawdown is about 44 meters
in well cell. Head is 980
meters (> 500 meters).
Drawdown is about 412
meters in well cell. Head is
588 meters (> 500 meters);
we are close to our “safeyield” criterion.
Head in well cell is 509 meters
cu.m./year
(>500 meters). Save this
output file as Case 1.
COMPUTER OUTPUT: (after graphical post-processing)
Running the program produces pages of output. In the absence of a contouring tool one can take
the output, superimpose a grid and manually contour the output or use graphical post-processing.
Modern versions of MODFLOW handle a lot of this for the analyst.
Figure 5 is an example of hand-drawn contour maps using the output displayed in Figure 6.
Figure 7 is the same array in Figure 6 rendered using a graphics tool (in this case EXCEL).
Figure 5. Hand-drawn contour map using MODFLOW output (circa 1991).
Figure 6. Std. MODFLOW output file (used to draw Figure 5 and 7).
100
-100
-300
-500
-700
9
1
-900
7
5
3
5
13
11
9
1
Figure 7. Rendering of head array after 30 years of pumping. Note bottom of plot is bottom of aquifer.
The contour plots are useful for answering where (and when) the aquifer fails to meet the
minimum saturated thickness requirement (which in this example will always be at the well cell).
One could also use Darcy’s law and hand calculations to determine the amount of water flowing
into the aquifer from the lakes (how?). Instead of making this calculation by-hand, the model
also computer cell-by-cell flow terms and a water balance summary. Figure 8 is an excerpt from
the standard output file that shows this summary.
Figure 8. Water balance summary.
Typically we would present plots of Case00 (The pre-development condition) and Case01 (the
“management solution”). Figure 9 is a perspective (3D) rendering of the predevelopment
condition. The water table is mounded because of the uniform recharge (this result is easily
predicted with an analytical solution – review course notes for the solution). Figure 10 is the
same array rendered as a contour map (using EXCEL which is not the best contouring tool – the
contour intervals are indicated by different colors/shading).
10500
1060-1100
1060-1100
1020-1060
9500
980-1020
8500
980-1020
940-980
7500
940-980
900-940
6500
860-900
1020-1060
900-940
860-900
820-860
5500
780-820
4500
780-820
740-780
3500
740-780
2500
700-740
700-740
660-700
12500
11500
9500
500-540
10500
8500
7500
6500
5500
540-580
4500
500
3500
580-620
2500
1500
500
620-660
1500
10500
8500
6500
4500
2500
12500
500
8500
10500
4500
6500
500
2500
1100
1060
1020
980
940
900
860
820
780
740
700
660
620
580
540
500
820-860
660-700
620-660
580-620
540-580
500-540
Figure 9. Predevelopment Perspective Plot.
Figure 10. Pre-development contour map.
10500
1060-1100
1060-1100
9500
1020-1060
1020-1060
980-1020
8500
980-1020
940-980
7500
940-980
900-940
6500
860-900
5500
820-860
900-940
860-900
820-860
780-820
4500
780-820
740-780
3500
740-780
2500
700-740
700-740
660-700
1500
Figure 11. Management Solution Perspective Plot
12500
11500
9500
8500
7500
6500
5500
4500
10500
500-540
3500
540-580
2500
580-620
500
620-660
1500
10500
8500
6500
4500
2500
12500
500
8500
10500
4500
1100
1060
1020
980
940
900
860
820
780
740
700
660
620
580
540
500
6500
500
2500
Next we present the renderings for the “management solution” where the result of pumping
produces a water table that nearly hits our management objective (head > 500 meters). Figures
11 and 12 are the 3D plot and the contour map.
500
660-700
620-660
580-620
540-580
500-540
Figure 12. Management Solution contour map
Interpretation
Interpretation of the results is critical to the whole effort (there is really no point to modeling if
there is not going to be interpretation). Figures 11 and 12 actually help answer the second
question without any flux calculations. The contour map suggests that there is a hydraulic
“divide” between the well and the lakes and the lakes actually receive water from the aquifer.
This conclusion is supported by the flux calculations in MODFLOW that are repeated below.
The second column displays the rates and there is no flux into the model from the constant head
(lake) boundaries, but there is an outflow to the lakes amounting to about half of the volume rates
pumped. Thus in this aquifer scenario, the entire pump discharge over the planning period is
supplied entirely by recharge (precipitation).
Figure 13. Volumetric budget calculations from the program.
Thus to answer the two questions we arrive at:
a) What is the maximum rate P-1 that can be used so that the saturated thickness is never less
than 500m? P1 on the order of 43,100,000 cubic meters per year.
b) How much of the pumped water is induced recharge from the two lakes? (Why would this
question be of interest?) None of the pumped water is from the lakes. The aquifer is a net
supplier of water to the lakes.
Exercise:
The files on the server should produce the results used pt produce figures 9-13. So start from
these files and do the following:
1) Run the model and verify that the results are consistent with the figures above.
2) Determine the pumping rate that lowers the water table to 250 meters above bottom of the
aquifer.
3) Plot the water table 3D plot and contour map for this management solution.
4) Does this new management solution draw any water from the lakes?