EPA/500/R-93/01Q
Febnliiry 1993 '.
A MANUAL OF INSTRUCTIONAL PROBLEMS
i FOR THE U.S.G.S: MODFLOW MQPEL
•V)
by
Peter F. Andersen
GeoTrans, Inc.
4605,0 Manekin Plaza, Suite 100
Sterling, Virginia 22170
DYNAMAC CONTRACT
68-C8-0058
Project Officer
John E. Matthews
Extramural.Activities and Assistance Division -
Robert S. Kerr Environmental Research Laboratory
Ada, Oklahoma 74820
ROBERT S. KERR ENVIRONMENTAL RESEARCH
OEHCE OF RESEARCH AND
U. S/ENVWOhfMENTAL PROTECTION
S ' • P.O.BOX 1198
ADA, OKLAHOMA 74820
Printed on Recycled Paper
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DISCLAIMERS
'; ' , '% „ ' i
The information in this document has been funded in part by the Environmental Protection
Agency under DYNAMAC Contract No. 68-C8:OQ58. with .G^oTransJnc^ as # snb-icontractor.
Scott Huling and Randall Ross, Rotten S. Kerr Environmental ResparcrPLaboratory, will gswe
as Task Managers on thlis project. 'It has been subject,to, the Ag'^rKiy *s'peer andadmi,nistrar -
live tevi&#,'and it has beeh^p^rbved as an EPA document. Mentipn^f trade names or,,: ~-,^ ,,
commercM products does nd'tj&>nstitute endorsement or recomrrierjdatiQn for,:
This report utiliie's the U.S.G.S. Modular Three-Dimensional Ground-Water- Flow Model
(MODFLOW) for analyzing grpundwater flow under various^ hydrologic conditions. - ,r
MODFLOW is a public-domain code and may be used and copied freely. If errors-are-found
in the document or if you have suggestions for improvement, please contact the Center, for. ,
Subsurface: Modeling Support (CSMoS) at the Robert S. Kerr Environmental Research
Laboratory, -Ada, Oklahoma. ,
Center^or Subsurface Modeling Support :,,
RdbeftS^Kfirr Environmental Research Laboratory
P.O; Bbx4l98, Ada;'Oklahoma, 74820
(405)'436-8500 ;
11
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FOREWORD
i -. ,. ,v ,. - -
EPA is charged by Congress to protect the Nation's land, air and water systems. Under a
mandate of national environmental laws focused on air and water quality, solid waste, ;
management and, the control of toxic substances, pesticides, noise and radiation, the Agency ,,
strives to1 formulate and implement actions which lead to a compatible balance between
human activities and the ability of natural systems to support and nurture life.
, i*
The Robert S. Kerr Environmental Research Laboratory is the Agency's center of •-
expertise for the investigation of the soil and subsurface environment. Personnel at the ,
Laboratory are responsible for the management of research programs to: (a) determine, the
fate, transport, and transformation rates of pollutants in the soil, and the unsatarated and
saturated zones of the subsurface environment; (b) define the processes to be used in
characterizing the soil and subsurface environment as a receptor of pollutants; (c) develop
techniques for predicting the effects of pollutants on ground water, soil, and indigeneous
organisms; and (d) define and demonstrate the applicability and limitations of using natural ,
process, indigenous to the soil and subsurface environment, for the protection of this resource.
I*
EPA is involved in groundwater flow modeling to analyze and predict the movement of
water in the subsurface. Traditionally, groundwater flow models are rarely supported by
documents that assemble the practical application aspects of modeling. While it is important
to understand the theory behind the mathematical model, it is equally important to understand
the principles of modeling, model options, rules of thumb, and common mistakes from an
applications perspective. This manual was developed specifically for the U.S.G.S. modular
groundwater flow model (MODFLOW) and it illustrates by examples, the principles of
groundwater flow modeling and model options. The manual was developed to be used for
self-study or as a text for courses. Three diskettes are included which contain the input and
output data sets for each problem presented in the manual. A copy of the MODFLOW code is
not included. The information in this document should be of interest to both the beginner and
advanced modeler for hands-on experience with the practical application of MODFLOW.
Clinton W. Hall
Director
Robert S. Kerr Environmental
Research Laboratory
111
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CONTENTS
Foreword - iii
Figures iv
Tables vii
Acknowledgments ix
Introduction 1-1
The Theis solution 1-1
Anisotropy 2-1
Artesian-water table conversion 3-1
Steady-state 4-1
Mass balance 5-1
Similarity solutions in calibration 6-1
Superposition 7-1
Grid and time stepping considerations 8-1
Calibration and prediction , 9-1
Transient calibration 10-1
Representation of aquitards . 11-1
Leaky aquifers 12-1
Solution techniques and convergence 13-1
Head dependent boundary conditions 14-1
Drains 15-1
Evapotranspiration 16-1
Wells , 17-1
Cross-sectional simulations 18-1
Application to a water supply problem 19-1
Application to a hazardous waste problem . 20-1
References R-l
Appendix
A: Abbreviated Input Instructions for MODFLOW . A-l
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FIGURES
Number Page
1.1 Configuration of the model for simulating radial flow Parts a-d 1-4
1.2 Drawdown versus time for each model configuration 1-10
2.1 Drawdown versus time at the observation point located 55 m '
from the pumping well along the x-axis for the three
model configurations 2-10
2.2 Drawdown versus time at the observation point located 55 m
from the pumping well along the y-axis for the three model
configurations 2-11
2.3 Drawdown contours (ft) for the 10:1 anisotropic case modeled
in Part a £-13
2.4 Three-dimensional view of the drawdown for the 10:1 anisotropic
case modeled in Part a ]2-14
3.1 Drawdown versus time for the four MODFLOW configurations
and the analytical solution . .1 3-7
3.2 Drawdown versus distance at 2.19 days for the water table,
conversion, and artesian cases ,13-8
4.1 Configuration of the Problem 4 modeled domain . 4-2
5.1 Potentiometric surface map and hydraulic head array at time step 1 [5-3
5.2 , , Model wide mass balance at time step 1 5-4
5.3 . Printout of cell-by-cell flow terms for each component of the >. •
mass balance 5-5
3.4 Hand calculations for each component of the mass balance 5-10
6.1 Contour map of potentiometric surface, hydraulic head array, £
and mass balance output for Part a r 6,-3
-6(2 Contour map of potentiometric surface, hydraulic head array, „.,-•,'
.and mass balance output for Part b ? 6-4
6.3 Contour map of potentiometric surface, hydraulic head array, f,./-
and mass balance output for Part c 6-5
7.1 Contour map of potentiometric surface, hydraulic head array, j ?;
model wide mass balance, and individual specified head node
. . mass balance for Part a 7-4
7.2 Contour map of potentiometric surface, hydraulic head array, •_ r
model wide mass balance, and individual specified head node
. . mass balance for Part b 7-5
"7.3 Contour map of potentiometric surface, hydraulic head array, ; ^,1
• i . . model wide mass balance, and individual specified head node -_. /;
. mass balance for Part c 7-6
7.4 . Contour map of potentiometric surface, hydraulic head array, , <-••
model wide mass balance, and individual specified head node • . ; ^
mass balance for Part d , 7-7
8.1 Location of pumping wells, observation wells, and boundary,, i. i/j
conditions for Problem 8 -.;f. ........ 8-2
vi
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8.2 Drawdown (m) at 20 days for the 4 x 4 grid simulation of Part a 8-10
8.3 Drawdown (m) at 0.2 days for the 16 x 16 grid simulation of Part b . . . . 8-11
9.1 Geometry and potentiometric surface of the aquifer system 9-2
9.2 Hydraulic head arrays, potentiometric surface contour maps,
and mass balance summary for Part a 9-5
9.3 Hydraulic head arrays, potentiometric surface contour maps, and mass
balance summary for Part b using pumpage of -0.4 ftVs 9-6
9.4 Potentiometric surface contour map for Part b using pumpage
of -0.1 ftVs . 9-9
9.5 Potentiometric surface contour map for Part b using pumpage
of -0.5 ftVs 9-10
10.1 Grid and boundary conditions for coastal transient problem 10-3
11.4 Hydraulic head (ft) in the middle of the confining bed versus time
for 11-5 cases a, b, and d 11-5
11.2 Total flux (ftVs) and storage flux versus time from the confining
bed for the seven layer model 11-6
12.1 Drawdown versus time for the analytical, MODFLOW, and SEFTRAN
simulations 12-9
13.1 Model geometry, boundary conditions, and hydraulic conductivity
zonation for Problem 13 13-2
13.2 Iteration history for variations in the SIP seed parameter 13-9
13.3 Iteration history for variations in the SSOR acceleration parameter .... 13-10
14.1 Hydraulic head (ft) versus flow rate (ftVd) for each of the five
methods of representing the third type boundary conditions 14-9
15.1 Model configuration for Problem 15 15-3
16.1 Finite-difference grid, boundary conditions, and simplified
topography for Problem 16 16-2
16.2 Potentiometric surface (ft) for Problem 16, Part a 16-6
16.3 Potentiometric surface (ft) for Problem 16, Part b (net recharge) 16-7
16.4 Drawdown map (ft) for Problem 16, Part c 16*8
16.5 Drawdown (ft) map for Problem 16, Part d 16-9
16.6 Net recharge rates (in/yr) for the steady-state, non-pumping
-° scenario (Part a) 16-11
17.1 Drawdown versus distance for the fully penetrating well case
and the partially penetrating well case in the 20 m thick aquifer
at time = 50938 s 17-7
17.2 Drawdown versus distance for the fully penetrating well case
and the partially penetrating well case in the 40 m thick aquifer
at time = 50938 s 17-8
18.1 Layering and zonation used in the cross-sectional model 18-2
18.2 Hydraulic head arrays for the cross-sectional model 18-10
19.1 Geologic map of the Musquodoboit harbor region 19-2
19.2 Geologic map of the Musquodoboit harbor region 19-3
19.3 Model boundary and transmissivity zones used in the
^ " numerical model 19-13
19.4 Location of the river boundary condition, pumping well,
- x and observation wells used in the numerical model 19-14
vii
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19.5 Drawdown (ft) versus time (min) for the aquifer test conducted at
Musquodoboit Harbor 19-16
19.6 Comparison of modeled to observed drawdown (ft) data
for the base case 19-17
19.7 Comparison of modeled to observed drawdown in well 1
for the base case and for a 2-fold increase and reduction
in transmissivity 19-18
19.8 Comparison of modeled to observed drawdown in wells for
the base case and for storage coefficients of 0.1 and 0.005 19-19
19.9 Drawdown (ft) after 1000 days of pumping at 0.963 ftVs 19-21
19.10 Comparison of modeled drawdown for the drawdown-dependent
storage coefficient 19-23
19.11 Comparison of modeled to observed drawdown in well 1 for
the base case and for order of magnitude increase and decrease
in river conductance 19-24
20.1 Finite difference grid showing the location of specified head cells
for the steady-state model 20-3
20.2 Grid cells representing the impermeable clay cap, the slurry wall,
and the drain 20-9
20.3a Hydraulic head (ft) contours in the vicinity of the landfill for the
steady-state case (a) 20-19
20.3b Hydraulic head (ft) contours in the vicinity of the landfill for the
case involving a cap (b) 20-20
20.3c Hydraulic head (ft) contours in the vicinity of the landfill for the
case involving a cap and a slurry wall (c) 20-20
20.3d Hydraulic head (ft) contours in the vicinity of the landfill for the
case involving a cap and a drain (e) 20-21
20.4 Hydraulic head (ft) along column 19 of the model at 2.69 years for
each remedial alternative simulation 20-24
Vlll
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TABLES
Number Page
1 Verification of MODFLOW results 1-4
2 Packages used in the problem sets 1-5
1.1 Parameters used in Problem 1 1-2
1.2 Grid spacing (m) used for various model configurations 1-3
1.3 Calculations for determination of transmissivity and storage
coefficient for wedge-shaped domain 1-8
1.4 Drawdown versus time for each model configuration 1-9
2.1 Parameters used in problem 2 2-2
2.2 Grid spacing used in the various model configurations 2-3
2.3 Drawdown (m) at an observation point located 55 m from
the pumping well along the x axis 2-7
2,4 Drawdown (m) at an observation point located 55 m from
the pumping well along the y axis , 2-8
2.5 , Drawdown (m) at an observation point located 77.8 m from
the pumping well at a 45° angle between the x and y axis 2-9
3.1 Parameters used in problem 3 3-2
3.2 Grid spacing (ft) used in problem 3 3-3
3.3 Drawdown versus time for each model configuration 3-6
4.1 Initial head (SHEAD) at specified head cells 4-3
4,2 Hydraulic head (ft) at node (7,1), storage component of
mass balance, and iteration data for each time step and the
steady-state simulations 4-6
5.1 Comparison of model calculated and hand calculated rate
mass balance 5-12
8.1 Grid data 8-3
8.2 Comparison of results for various grid spacings in Part a 8-7
8.3 Comparison of results for various grid spacings in Part b 8-7
8.4 Comparison of results for variations in time stepping in Part c 8-8
8.5 Comparison of results for variations in closure criterion in Part d 8-8
8.6 Comparison of drawdowns (m) at well 2 for various time
derivatives and spatial approximations (analytical = 1.63) 8-12
9.1 River data 9-3
9.2 Calibration targets 9-3
10.1 Hydraulic head (ft) versus time (weeks after drought began)
at an observation well located at node (1,5) 10-2
10.2 Pre-drought groundwater levels (ft) within the model- domain 10-3
10.3 Groundwater levels resulting from a steady-state simulation
using a hydraulic conductivity of 850 ft/d 10-7
11.1 Hydraulic heads (ft) in the middle of the confining bed versus
time for all cases of Problem 11 11-7
12.1 Parameters and discretization data used in Problem 12 12-2
IX
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12.2 Time versus drawdown (analytical solution) at distances
of 117.4 m 12-3
12.3 Time versus drawdown at distances of 117,4 m for the
analytical solution, MODFLOW configuration, and SEFTRAN
radial solution 12-8
13.1 Sensitivity analysis on SIP seed and acceleration parameter 13-7
13.2 Sensitivity analysis on SSOR and acceleration parameter 13-8
14.1 Aquifer parameters and discretization data for Problem 14 14-2
14.2 Hydraulic head at node (1,4) for each of the five methods of
representing the third type boundary condition 14-7
14.3 Discharge for each of the five methods of representing the
third type boundary condition 14-8
14.4 Other uses for the head-dependent flux boundary conditions
in MODFLOW 14-10
15.1 Grid spacing used in the fine-gridded model (Part b) 15-4
15.2 Hydraulic head at the drain node (column 6) and drain flux for
variations in drain conductance (coarse model) 15-19
16.1 Comparison of hydraulic heads (ft) along row 10 for MODFLOW
and FTWORK 16-12
17.1 Parameters and discretization used in Problem 17 17-3
17.2 Drawdown versus distance at 50938 s for the fully penetrating,
partially penetrating, and stratified aquifer simulations 17-6
18.1 Bottom and top elevations (ft) in cross-sectional model 18-4
18.2 Initial heads in layer 6 18-5
18.3 Assumed saturated thickness (ft) of layer 1 in the cross-sectional
model 18-6
19.1 Input data for the water supply problem 19-15
19.2 Observed drawdown data from aquifer test 19-15
19.3 Modeled drawdown data for the base case 19-20
19.4 Drawdown (ft) versus time in observation well 1 for variations in
transmissivity, storage coefficient, and river leakance 19-25
20.1 Attributes of the drain used in Part d . 20-10
20.2 Comparison of MODFLOW results versus USGS2D results for the
steady-state case (Part a) and the wall and cap scenario (Part c)
at 6.08 yr 20-22
20.3 Hydraulic heads (ft) along column 19 of the model at 2.69 years for
each remedial alternative simulation 20-23
20.4 Drain discharge (ft3/s) versus time for the well, cap, and drain
scenario (Part d)and the cap and drain scenario (Part e) 20-25
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ACKNOWLEDGMENTS
The author wishes to thank the many individuals who have contributed to make this
document possible. In particular, I thank Joanne Elkins of GeoTrans, Inc. who was largely
responsible for typing this document. Also, I wish to thank Debbie Shackleford and Carol
House of Dynamac Corporation for the final preparations of the document. Finally, I thank
my students at the International Ground Water Modeling Center's (IGWMC) Applied
Groundwater Modeling short course for their many helpful suggestions.
XI
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INTRODUCTION
A recent report by the United States Environmental Protection Agency Groundwater
Modeling Policy Study Group (van der Heijde and Park, 1986) offered several approaches to
training Agency staff in the application of groundwater modeling. They identified the
problem that current training efforts tend to be of short duration (one week or less) with a
lack of in-house programs to reinforce training received in a formal setting. The study group
suggested, among other things, the alternative of self-study coupled with obtaining experience
under the guidance of a senior modeling specialist.
In order for groundwater modeling self study to be viable, a curriculum must exist that
allows the student to have hands-on experience with the practical application of models.
Available resources do not meet this need. Current groundwater modeling texts deal
primarily with the mathematics or theory of modeling. Code documentations usually discuss
the programming aspects and performance standards of particular models. They usually
include one or two test problems for verification purposes. Journal articles or U.S.
Geological Survey publications best fit the need for learning about the practical application of
models. However, these sources either do not give enough information to reproduce results
or involve data setup that is too complicated to allow a student to efficiently have hands-on
experience with the model.
This manual is intended to meet the need described above. Twenty documented problems,
complete with problem statements, input data sets, and discussion of results are presented.
The problems are designed to cover modeling principles, specifics of input/output, options
available to the modeler, rules of thumb, and common modeling mistakes.
Data set preparation time and execution time have been minimized by simplifying the
problems to small size and to focus only on the aspect that is under consideration. Model
grids are generally smaller and more homogeneous than would be used in practice, however,
the intent and result of each exercise are not compromised by the simplification.
This manual is developed for the U.S. Geological Survey modular groundwater model
(MODFLOW) by McDonald and Harbaugh (1988). MODFLOW is perhaps the most popular
groundwater flow model used by government agencies and consulting firms. MODFLOW
solves the partial differential equation describing the three-dimensional movement of
groundwater of constant density through porous material:
ic + - IK [K - w - s
r*dx ay yyay az zza* sat
-------
where:
, Kyy, and K^, are values of hydraulic conductivity along the x, y, and z coordinate
axes, which are assumed to be parallel to the major axes of
hydraulic conductivity (LT"1);
h is the potentiometric head (L);
W is a volumetric flux per unit volume and represents sources and/or
sinks of water (T"1);
Ss is the specific storage of the porous material (L "'); and
t is time (T).
Ss, K^, Kyy, and K^ may be functions of space and W may be a function of space and time.
This equation, combined with specification of boundary and initial conditions, is a
mathematical expression of a groundwater flow system. MODFLOW uses the finite
difference method to obtain an approximate solution to this equation. Hydrogeologic layers
can be simulated as confined, unconfined, or a combination of confined and unconfined.
External stresses such as wells, areal recharge, evapotranspiration, drains and streams can also
be simulated. Boundary conditions include specified head, specified flux, and head-dependent
flux. Two iterative solution techniques, the Strongly Implicit Procedure and Slice Successive
Over Relaxation, are contained within MODFLOW to solve the finite difference equations
(McDonald and Harbaugh, 1988).
The user of this manual should attempt to solve the problems as described in the problem
statement portion of each exercise. The model setup can be checked in the data set listing
given in the model input section of each problem. Results can be checked by the pertinent
portions given in the model output section. Some training on the structure and input of
MODFLOW as well as some training on the theory of groundwater modeling is assumed.
The user will need to refer to the MODFLOW manual on some occasions. The abbreviated
input instructions given in the MODFLOW manual are included as Appendix A to this
manual.
A secondary function of this manual is for verification purposes. Although the
MODFLOW code has been extensively applied, very little documentation of its testing and
verification is available in the literature. To address this situation, where possible, model
generated results are compared to analytical solutions, results of other models, or to
simulations with alternative boundary conditions or configurations. In addition to providing
an informal benchmarking of MODFLOW, these problems can be used to verify the correct
installation of the code on a particular computer system or to verify that certain user
modifications have not altered the integrity of the program. The results of the simulations
may vary slightly (approximately ±0.02 ft or m) from one computer to another. The results
obtained here were with a 386 microcomputer. Table 1 shows the problems that were run
and what types of verification were performed.
1-2
-------
All the packages of MODFLOW have been utilized at least twice in this series of
problems. Table 2 is a matrix showing which packages were utilized in individual problems.
Several parts exist to each problem. Input and output files are included on the attached
diskette for the data sets listed in the manual. Minor modifications, as described in the model
input section of each problem are not included as separate data sets. The diskettes included
with this document do not include a copy of MODFLOW. It is assumed the reader has
obtained a copy of MODFLOW and has the necessary computer hardware to execute the
program.
The problems given in the manual are intended to be useful without changes or additions.
However, the problems may also be useful as a stepping stone to more detailed analysis.
Rather than creating new data sets, the analyst can modify existing data sets to fill a
particular need.
1-3
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Table 1. Verification of MODFLOW results
Problem
No.
1
2
3
4
5
6
7
8
9
10
11.
12
13
14
15
16
17
18
19
20
Analytical or
Semianalytical
Tide Solution
Theis solution X
Anisotropy X
Artesian-water table X
conversion
Steady State
Mass balance
Similarity solutions in model
calibration
Superposition
Grid and time stepping X
considerations
Calibration and prediction
Transient calibration
Representation of aquitards
Leaky aquifers X
Solution techniques and
convergence
Head dependent boundary
conditions
Drains
Evapotranspiration
Wells
Cross-sectional simulations
Application to a water supply
problem
Application to a hazardous
waste site
Alternate
Boundary
Condition or
Numerical Model
Model Configuration
X
X
X
X
X
X
X
X
•
X
X
1-4
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Table 2. Packages used in the problem sets*
Problem
No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Well Drain River ET
X
X
X
X
X
-x
X
X X
X
X X X X
X
X X
X
X X
X
GHB Recharge
X
X
X
X
X
X
X
X
X
X
X
X
SIP
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Output
SSOR Control
X
X
X
X
X
X '
X X
X
X
X
X
X
X
X X
X
X
*The Basic and Block Centered Flow packages were used for all simulations. Packages
available in MODFLOW and their major function are:
Basic Overall model setup and execution
Block Centered Flow Calculates terms of finite difference equations for flow within
porous media
Well Specified flux condition (volumetric input)
Drain Head dependent flux condition limited to discharge
ET Evapotranspiration, head dependent flux condition limited to
discharge with a maximum specification of discharge
GHB General Head Boundary, head dependent flux condition
Recharge Specified flux condition (linear input)
SIP Strongly Implicit Procedure solution technique
SSOR Slice Successive Over Relaxation solution technique
Output Control Directs amount, type, and format of model output
1-5
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PROBLEM 1
The Theis Solution
INTRODUCTION
With the exception of Darcy's Law, perhaps the most widely used analytical technique by
hydrologists is the solution by Theis (1935). It is therefore fitting that the first problem
presented in this manual is a benchmark of MODFLOW with the Theis solution. Three
different model configurations for analyzing radial flow to a well are examined. The
techniques described in this problem can be generally applied to well test analysis and
representations of radial flow.
PROBLEM STATEMENT AND DATA
Theis' solution predicts drawdown in a confined aquifer at any distance from a well at any
time since the start of pumping given the aquifer properties, transmissivity and storage
coefficient.
The assumptions inherent in the Theis solution include:
1) The aquifer is homogeneous, isotropic, uniform thickness, and of infinite areal extent.
2) The initial potentioinetric surface is horizontal and uniform.
3) The well is pumped at a constant rate and it fully penetrates the aquifer.
4) Flow to the well is horizontal, the aquifer is fully confined from above and below.
5) The well diameter is small, storage in the wellbore can be neglected.
6) Water is removed from storage instantaneously with decline in head.
All of these assumptions, with the exception of infinite areal extent, can be easily
represented with the numerical model. Several options exist to represent the domain as
effectively infinite. The most frequently applied method is to extend the model domain
beyond the effects of the stress. The modeled domain is therefore usually fairly large and a
limited time frame is modeled. An increasing grid spacing expansion is used to extend the
model boundaries.
The model domain is assumed to be uniform, homogeneous, and isotropic. A single layer
is used to model the confined aquifer. A fully penetrating well located at the center of the
model domain pumps at a constant rate. The potentiometric surface of the aquifer is
monitored with time at an observation well 55 m from the pumping well. Specific details of
the problem are from Freeze and Cherry (1979) pp. 345, and are given in Table 1.1.
1-1
-------
Table 1.1. Parameters used in Problem 1
Initial head 0.0 m
Transmissivity 0.0023 m2/s
Storage coefficient 0.00075
Pumping rate 4 x 10"3 mVs
Final time 86400 s
Number of time steps 20
Time step expansion factor 1.3
SIP iteration parameters 5
Closure criterion 0.0001
Maximum number of iterations 50
Part a) Represent the entire aquifer domain by using the grid spacing shown in Table 1.2.
Place the well at the center of the domain, row 10, column 10. Run the model,
noting drawdown at each time step at an observation point 55 m from the
pumping well. The configuration of the model for part a and future parts b, e,
and d is shown in Figure 1.1.
1-2
-------
Table 1.2. Grid spacing (m) used for various model configurations
Row number, i
(=column number, j)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Part a
DELC (i)
(=DELR(j))
300
200
150
100
80
60
40
30
30
20
30
30
40
60
80
100
150
200
300
Part b
DLEC(i)
(=DELR(j))
20
30
30
40
60
80
100
150
200
300
Part c
DELC(i)
(=DELR(j))
1
1.413
2
2.83
4 -
5.65
8
11.3
12
14.62
20
28.3 •
40
56.5
80
110
150
200
252.89
1-3
-------
well
wedge
c quadrant
a (entire square)
Figure 1.1. Configuration of the model for simulating radial flow for parts a-d. Arrows
denote groundwater flow direction.
Part b) Because of symmetry, the aquifer domain can be represented as a quadrant. Set
up a second model covering only the lower right quadrant of the previous domain.
The grid spacing for this model is shown in Table 1.2. Position the well at the
upper left corner of the new model, row 1, column 1. Because only one-fourth of
the aquifer is simulated, the well discharge should also be reduced to one-fourth
the original discharge. Run the model and note drawdown at each time step at an
observation point 55 m from the pumping well.
Part c) Re-run part b with the grid spacing shown in Table 1.2. The overall model
domain is the same size as part b, but grid spacing is finer near the pumping well.
Run the model and note drawdown at each time step at an observation point 55 m
from the pumping well.
Part d) Another form of symmetry for this problem (radial flow) is a pie shaped wedge
with the well at the vertex of the wedge. Unfortunately this geometry is difficult
to represent because the finite difference method is based on orthogonality of rows
and columns. However, because the model is posed in terms of conductance (a
1-4
-------
function of grid spacing and transmissivity) and grid block storativity (a function
of storage coefficient and area) it is possible to adjust T and S in such a manner
to approximate the wedge. Using a 20 m wide row (DELC(l) = 20) and grid
spacing along a row (DELR) as in part b, calculate changes to transmissivity and
storage coefficient for a 10° pie wedge. Adjust the well discharge to account for
the reduced model domain and input these parameters into the model. Run this
one-dimensional model and note drawdown at each time step at an observation
point 55 m from the well.
1-5
-------
MODEL INPUT
The following is a listing of data sets used for part a.
*********************************
* Basic package *
*********************************
theis problem full grid
1/4/90 pfa
1 19 19 1 1
11 12 0 0 0 0 0 0 19 0 0 22
0 1
0 1
999.00
0 .OOOE+00
86400. 201.3000
headngd)
headng(2)
nIay, nrow,ncoI,nper,itmuni
iunit array
iapart,istrt
ibound(Iocat,iconst)
hnoflow
shead(locat.cnstnt)
perlen,nstp,tsmult
*********************************
* Block Centered Flow Package *
*********************************
0
11
300.0
30.00
80.00
11
300.0
30.00
80.00
0
0
.100E+01
.100E+OK7G11.4)
200.0
30.00
100.0
150.0
20.00
150.0
100.0
30.00
200.0
.100E+OK7G11.4)
200.0
30.00
100.0
.750E-03
.230E-02
150.0
20.00
150.0
100.0
30.00
200.0
12
80.00
30.00
300.0
12
80.00
30.00
300.0
60.00
40.00
60.00
40.00
40.00
60.00
40.00
60.00
iss,ibcfcb
Iayeon
trpy
delr(locat,cnstnt,fmtin,iprn)
delr array
delc
-------
The same data set was used in part b, except the model domain was reduced to a 10x10 grid
(NROW = 10, NCOL =10) in the BASIC package. Accordingly, only one-fourth of the grid,
(as shown in Table 1.2, part b) was used. In addition, the well discharge was moved to row
1, column 1 and reduced to Ixl0"3m3/s in the WELL package. The part c data set is identical
to part a, except grid spacing (DELC, DELR in the BCF package) is modified as shown in
Table 1.2 and the well location and discharge is as in part b. The data set for part d is shown
below, minus the SIP and output control files, which are identical to those of parts a-c. The
calculations for adjustments to transmissivity and storage coefficient are shown in Table 1.3.
***•**»**•*•*»•»»»*»*»**
*********
* Basic package *
******»*******»*»*»*»*»*»********
theis problem pie grid
1/4/91 pfa
1 1 10 1 1
11 12 0 0 0 0 0 0 19 0 0 22
0 1
0 1
999,00
0 .OOOE+00
86400. 201.3000
*********************************
* Block Centered Flow Package *
*********************************
0 -100E+01
11 .100E+OK7G11.4) 12
20.00 30.00 30.00 40.00 60.00 80.00 100.0
150.0 200.0 300.0
0 .200E+02
11 7.500E-04(7G11.4> 12
.22001-01 .2180 .4800 .7850 1.222 1.833 2,618
3.709 5.236 7.418
11 2.3001-03C7G11.4J 12
.4400E-01 .2180 .4800 .7850 1.222 1.833 2.618
3.709 5.236 7.418
*»*«*»*******»***»*****»*»*******
* Well package *
*********************************
1 0
1
1 1 1-.11111E-3
headng(l)
headng(2)
nlay,nrow,ncol,nper,itmuni
iunit array
iapart,istrt
ibound(Iocat,iconst)
hnoflo
istrt(locat.cnstnt)
perlen.nstp,tsmult
iss.ibcfcb
Iayeon
trpy
delr(locat,cnstnt,fmtin, iprrt}
delr array
delc(locat,cnstnt)
sf1(locat,cnstnt,fmtin,iprn)
sfl array
tran
-------
Table 1.3. Calculations for determination of transmissivity and storage coefficient for
wedge-shaped domain (part d)
Block
number j
1
2
3
4
5
6
7
8
9
10
Area
DELC
X
DELR
400
600
600
800
1200
1600
2000
3000
4000
6000
Radius to
block edge
10
40
70
10
170
250
350
500
700
1000
Individual
block area
of 10°
wedge
8.73
130.9
288.0
628.32
1466.1
2932.2
5236.0
11126.5
20944.0
44505.9
Wedge
area
-r
acutal area
0.022
0.218
0.480
0.785
1.222
1.833
2.618
3.709
5.236
7.418
Radius to
block
midpoint
5
25
55
90
140
210
300
425
600
850
10° arc
length
0.873
4.363
9.599
15.71
23.43
36.652
52.360
74.176
104.72
148.35
10° arc
length
.:-
actual
DELC
0.044
0.218
0.480
0.785
1.222
1.833
2.618
3.709
5.236
7.418
Adjusted transmissivity *
10° arc length
actual DELC
transmissivity
Adjusted storage coefficient =
wedge area
actual area
storage coefficient
1-8
-------
MODEL OUTPUT
Drawdowns versus time are tabulated in Table 1,4 for each of the four cases. Comparison
is also made to the analytical solution of Theis. A drawdown versus time plot is shown in
Figure 1.2 for the best comparison case (the refined quadrant) and the worst comparison case
(the coarse quadrant). .Other cases are not shown, but are generally very similar to the refined
quadrant case.
Table 1.4. Drawdown versus time for each model configuration
Time Step
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Time (sec)
137.1
315.3
547.1
848.6
1239.9
1748.9
2410.7
3271,1
4389.5
5843.4
7733.6
10190.7
13385.1
17537.7
22936.1
29954.0
39077.4
50937.7
66356.1
86400.
Analytic
0.009
0.044
0.086
0.129
0.170
0.210
0.249
0.288
0.326
0.364
0.401
0.438
0.475
0.512
0.549
0.586
0.622
0.659
0.695
0.731
Drawdown (m)
Full grid Quadrant
(case a) (case b)
0.017
0.048
0.085
. 0.126
0.167
0.208
0.248
0.288
0.327
0,365
0.403
0.441
0.479
0.516
0.553
0.591
0.628
0.665
0.704
0.744
0.010
0.030
0.059
0.092
0.128
0.165
0..203
0.240
0.278
0.315
0.353
0.390
0.427
0.464
0.501
0.538
0.575
0.613
0.651
0.691
Refined
Quadrant
(case c)
0.014
0.043
0.079
0.120
0.160
0.201
0.241
0.280
0.320
0.358
0.397
0.434
0.471
0.508
0.545
0.582
0.619
0.656
0.697
0.738
Pie Wedge
(case d)
0.013
0.039
0.074
0.114
0.155
0.197
0.237
0.277
0.316
0.354
0.392
0.429
0.467
0.504
0.540
0.577
0.614
0.651
0.691
0.733
1-9
-------
0.80
0.00
analytical
***** coarse quadrant
refined quadrant
20000 40000 60000 80000 100000
TIME (seconds)
Figure 1.2. Drawdown versus time for each model configuration.
DISCUSSION OF RESULTS
With the exception of the coarse-quadrant grid (case b), the MODFLOW results compare
well to the analytic solution. The numerical results are generally within 0.005 m of the
analytic. An exact comparison is not attained because of the approximations made in the
numerical model. These include: 1) use of a discrete rather than continuous spatial domain,
2) use of a discrete rather than continuous time domain, 3) use of an iterative solution with a
closure tolerance, and 4) artificial placement of boundaries.
The distant no-flow boundary is only a small factor in this analysis because it is placed far
enough from the stress so that drawdown at the boundary is very limited. There is a
significant departure from the Theis curve at the final time step, however, as the non-infinite
nature of the model domain becomes a factor. The comparison would continue to deteriorate
if the model were run for longer time.
1-10
-------
This problem illustrates three methods of modeling radial flow to a well. The first, placing
the well at the center of a rectangular grid, is the most intuitive approach to this problem, but
is not the most efficient The second method, the quadrant, recognizes symmetry of flow.
Some care must be taken in designing the grid. The third method, the pie wedge, also
recognizes symmetry but involves fairly labor intensive parameter adjustment to approximate
a wedge shaped grid.
The quadrant grid is a satisfactory approximation, provided it is sufficiently fine near the
pumping well. The predominant reason for the approximation error noted in the first
quadrant analyzed (case b) is because the block-centered grid "approach models a larger area
than a quadrant. There will always be an extra 1/2 grid block on the margins of the model
area and therefore extra storage in the model domain. The extra storage accounts for a
majority of the underprediction of drawdown in case b. When the size of the blocks on the
margins is reduced in case c, the error is also reduced.
The pie-wedge grid provides a reasonable approximation for this particular problem. The
user is cautioned that it is conceptually difficult and error-prone to develop the grid and
aquifer parameters for this type of configuration. Some approximation errors may become
more apparent if larger areas or greater wedge angles are used. Although this is an
appropriate methodology, its main reason for presentation in this manual is to reinforce the
user's understanding of the relationship between transmissivity, grid spacing, and
conductance.
1-11
-------
PROBLEM 2
Anisotropy
INTRODUCTION
Anisotropy is often encountered in aquifers, particularly in the vertical direction. Vertical
anisotropy is handled in MODFLOW through the VCONT term, which is used in the three-
dimensional simulations. Horizontal anisotropy can also occur and may result from fracture
networks or depositional environments. Although MODFLOW was designed as a porous
media model, the scale of many modeling efforts is such that fractured media or a karst
environment can be considered an equivalent porous media. This problem examines how
MODFLOW handles horizontal anisotropy, provides a check on model accuracy, and
illustrates some special considerations for modeling anisotropic aquifers.
PROBLEM STATEMENT AND DATA
This problem is very similar to the Theis problem (problem 1) with regard to assumptions,
model configuration, and hydraulic parameters. An effectively infinite confined aquifer is
assumed, with a fully penetrating well located at the center of the model domain pumping at
a constant rate. The aquifer is ten times as transmissive in the x-direction as in the y
direction. For parts a and b, the principal directions of the hydraulic conductivity tensor are
assumed to be aligned with the model grid. The potentiometric surface of the aquifer is
monitored at 3 points: 55 m from the pumping well in the x direction, 55 m from the
pumping well in the y direction, and 77.8 m from the pumping well along a diagonal at 45°
to the x and y axis. Specific details on the problem are nearly identical to the Theis problem
and are given in Table 2.1. The data sets from problem 1 can be easily modified rather than
creating new data sets. Note that area! anisotropy is handled with the TRPY term in the BCF
package.
2-1
-------
Table 2.1. Parameters used in Problem 2
Initial head
Transmissivity, T^
Transmissivity, Tyy
Storage coefficient
Pumping rate
Stress period length
Number of time steps
Time step expansion factor
SIP iteration parameters
Closure criterion
Maximum number of iterations
0.0 m
0.0023 m2/s
0.00023 m2/s
0.00075
4 x 10'3 m'/s
86400 s
20
1.3
5
0.0001
50
Part a) Represent the entire aquifer domain with the grid spacing shown in Table 2.2. Note
that this spacing is the same as problem 1, part a. Place the well at the center of the
domain, row 10, column 10. Run the model, noting drawdowns at each time step at
the 3 observation points described above.
2-2
-------
Table 2.2. Grid spacing used in the various model configurations
Row number, i
(=column number ,j)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
Part a
DELC(i)
(=DELR(j))
300
200
150
100
80
60
40
30
30
20
30
30
40
60
80
100
150
200
300
Part b
DELC(i)
(=DELR(j))
1
1.41
2
2.83
4
5.65
8
11.3
12
14.62
20
28.3
40
56.5
80
110
150
200
252.89
Part b) Represent a quadrant of the aquifer domain with the grid spacing shown in Table 2.2.
Note that this is the same spacing used in problem 1, part c. Place the well at the
upper left corner of the model, row 1, column 1 and reduce the pumping to one-
fourth the original value. Note drawdowns at each time step at the 3 observation
points.
Part c) In the previous parts to this problem, the principal directions of the hydraulic
conductivity tensor were aligned with the finite difference grid. That is, the
maximum T (0.0023 nf/s) was along the x axis while the minimum T (0.00023 m2/s)
was along the y axis. In this exercise, we will examine the error which occurs if the
grid is not aligned with the principal directions of the hydraulic conductivity tensor.
We will assume that the maximum T is still 0.0023 nr/s and the minimum T is still
0.00023 m2/s and at right angles to one another, however, the analyst has not aligned
the finite difference grid along these maximums and minimums. The grid is tilted
2-3
-------
20° off the principal directions of hydraulic conductivity. The transmissivity along
the x and y axis can be calculated from equations given by Bear (1972), page 140.
T 4. T T _ f
T » _£! *1 + _£ II cos 20 (2.1)
T + T T — T
*'* y
-------
MODEL INPUT
The
following is a listing of the input data sets for part b.
*********************************
*
*****
* Basic package
************************
ani sot ropy problem quadrant fine spacing
2/20/91 pfa
1 19 19 1 1
11 12 0 0 0 0 0 0 19 0 0 22
0 1
0 1
999.00
0 .OOOE+00
86400. 201.3000
*********************************
* Block Centered Flow Package *
*********************************
1.00
11.30
80.00
1.00
11.30
80.00
0
11
11
0
0
.100E+00
.100E+OU7G11.4)
1.41
12.00
110.0
2.00
14.62
150.0
2.83
20.00
200.0
.100E+OK7G11.4)
1.41
12.00
110.0
.750E-03
.230E-02
2.00
14.62
150.0
2.83
20.00
200.0
12
4.00
28.30
252.89
12
4.00
28.30
252.89
*********************************
1
1
1
*
***
0
it ^*^W^W li
Well package
'* ^ ifcWiMMfc 11 it iHt "toil
*
1 1 -.100E-02
wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww
* SIP package *
5.65
40.00
5.65
40.00
8.00
56.50
8.00
56.50
50 5
1.0000 .10000E-03 1.00000 1
*********************************
* Output Control package *
*********************************
10
0
0
-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
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
headngd)
headng(2)
nlay.ncol,nrow,nper,itmuni
iunit array
iapart,istrt
ibound(locat,iconst)
hnoflo
shead(Iocat,cnstnt)
perlen,nstp,tsmult
iss.ibcfcb
laycon
trpy
delr(locat,cnstnt,fmtin,iprn)
delr array
delc(locat,cnstnt,fratin,iprn)
dele array
sfK locat, cnstnt)
tran(locat,cnstnt)
mxuell,iwelcb
ittnp
layer,row,column,q
mxiter.nparm
accl,hclose,ipcalc,useed,iprsip
ihedfm.iddnfm.ihedun.
incode,ihddfl.ibudfl,
hdpr,ddpr,hdsv.ddsv
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode.ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode.ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
incode,ihddfl.ibudfl,
iddnun
icbcfKstep 1)
icbcfKstep 2)
icbcfKstep 3)
icbcfKstep 4)
icbcfKstep 5)
icbcfKstep 6)
icbcfKstep 7)
icbcfKstep 8)
icbcfKstep 9)
icbcfKstep 10)
icbcfKstep 11)
icbcfKstep 12)
icbcfKstep 13)
icbcfKstep 14)
icbcfKstep 15)
icbcfKstep 16)
icbcfKstep 17)
icbcfKstep 18)
icbcfKstep 19)
icbcfKstep 20)
2-5
-------
Part b is shown here because part a is nearly identical to that of problem 1, part a which was
shown previously in the problem 1 writeup. The only difference between the previous part a
data set and the current part a data set is that the layer wide anisotropy ratio (TRPY) is
changed from 1.0 to 0.1 to yield a transmissivity along a column of 1/10 that along a row.
The part b data set shown above is nearly identical to that of part c of Problem 1. Again the
layer wide anisotropy ratio is set at 0.1 for the current simulation. In part c, the same data
set as part a is used, however, the transmissivity along a row (TRAN) is changed to 0.00206
m2/s. Because we desire a transmissivity of 0.00047 mVs along the y axis (column), the layer
wide anisotropy ratio is set at 0.00047/0.00206 or 0.22816.
MODEL OUTPUT
Drawdown versus time is tabulated for the three observation points in Tables 2.3, 2.4, and
2.5 for the three cases. These results may be compared to the analytical solution of
Papadopulos (1965) for anisotropic aquifers. The results of these simulations are plotted in
Figures 2.1 and 2.2.
2-6
-------
Table 2.3, Drawdown (m) at an observation point located 55 m from the pumping well
along the x axis
Time
step
number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Time (sec)
137.1
315.3
547.1
848.3
1239.9
1748.9
2410.7
3271.1
4389.5
5843.4
7733.6
10190.7
13385.1
17537.7
22936.1
29954.0
39077.4
50937.7
66356.1
86400.0
Analytic
0.028
0.140
0.273
0.407
0.537
0.664
0.789
0.911
1.032
1.151
1.269
1.387
1.503
1.620
1.736
1.852
1.967
2.082
2.198
2.313
Drawdown (m)
Part a Part b Part c
0.050
0.154
0.293
0.447
0.600
0.744
0.880
1.009
1.133
1.255
1.375
1.495 -
1.614
1.732
1.851
1.969
2.087
2.205
2.324
2.446
0.044
0.135
0.252
0.379
0.509
0.636
0.762
0.886
1.008
1.129
1.249
1.369
1.487
1,605
1.722
1.839
1.957
2.074
2.193
2.315
0.036
0.109
0.203
0.303
0.401
0.497
0.590
0.681
0.772
0.861
0.950
1.038
1.126
1,214
1.301
1.388
1.474
1.561
1.649
1.738
2-7
-------
Table 2.4, Drawdown (m) at an observation point located 55 m from the pumping well
along the y axis
Time
step
number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Time (sec)
137.1
315.3
547.1
848.3
1239.9
1748.9
2410.7
3271.1
4389.5
5843.4
7733.6
10190.7
13385.1
17537.7
22936.1
29954.0
39077.4
50937.7
66356.1
86400.0
Analytic
0.000
0.000
0.001
0.006
0.022
0.050
0.092
0.148
0.215
0.292
0.377
0.468
0.565
0.665
0.769
0.876
0.984
1.094
1.204
1.316
Drawdown (m)
Part a Part b Part c
0.001
0.003
0.008
0.019
0.036
0.063
0.102
0.152
0.215
0.288
0.371
0.461
0.557
0.658
0.762
0.870
0.979
1.091
1.205
1.323
0.000
0.001
0.004
0.012
0.028
0.054
0.093
0.144
0.207
0.280
0.363
0.453
0.548
0.648
0.751
0.858
0.967
1.077
1.191
1.309
0.002
0.010
0.024
0.047
0.081
0.125
0.179
0.241
0.309
0.381
0.457
0.535
0.616
0.697
0.780
0.863
0.948
1.032
1.118
1.206
2-8
-------
Table 2.5. Drawdown (m) at an observation point located 77.8 m from the pumping well
at a 45° angle between the x and y axis
Time
step
number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Time (sec)
137.1
315.3
547.1
848.3
1239.9
1748.9
2410.7
3271.1
4389.5
5843.4
7733.6
10190.7
13385.1
17537.7
22936.1
29954.0
39077.4
50937.7
66356.1
86400.0
Analytic
0.000
0.000
0.001
0.004
0.017
0.041
0.078
0.129
0.192
0.265
0.347
0.436
0.530
0.629
0.732
0.837
0.945
1.054
1.164
1.276
Drawdown (m)
Part a Part b Part c
0.000
0.001
0.005
0.013
0.027
0.050
0.085
0.131
0.190
0.259
0.339
0.426
0.520
0.619
0,722
0.828
0.937
1.048
1.162
1.280
0.000
0.001
0.003
0.009
0.022
0.045
0.079
0.126
0.185
0.255
0.334
0,421
0.514
0.612
0.714
0.820
0.928
1.038
1.151
1.269
0.001
0.004
0.013
0.029
0.055
0.092
0.139
0.194
0257
0.325
0.398
0.473
0.552
0.632
0.713
0.796
0.879
0.963
1.049
1.137
2-9
-------
3.00 -i
•2.00 -
c
£
o
O
i^
Q
1.00 -
0.00
100
analytic
***** part a
x x x x x part b
+ + + + + part c
1000 10000
Time (seconds)
100000
Figure 2.1. Drawdown versus time at the observation point located 55 m from the
pumping well along the x-axis for the three model configurations.
2-10
-------
2.00 -i
O 1.00 -
TJ
Q
0.00
100
analytic
***** part a
x x x x x part b
+ + + + + part c
1000 10000
Time (seconds)
100000
Figure 2.2. Drawdown versus time at the observation point located 55 m from the
pumping well along the y-axis for the three model configurations.
2-11
-------
DISCUSSION OF RESULTS
The comparison of MODFLOW results with the analytical solution is again very good.
However, just as the overall grid design was important in the Theis problem, the directional
grid design becomes important for areally anisotropic problems. Note in Figures 2.3 and 2.4
that the drawdown contours form an ellipse with the major axis in the direction of highest
transmissivity.
The model results are in excellent agreement with analytical results along the y-axis, which
is in the direction of low transmissivity, for both the coarse and fine grids (see Table 2.4 and
Figure 2.2 for parts a and b). It appears from these results that the coarse and fine grids are
equally satisfactory. Along the x-axis, or direction of high transmissivity, there is a more
apparent difference between the results of the coarse and fine meshes. The results using the
fine mesh are very close to the analytical results, but the coarse mesh results consistently
show greater drawdown. This is not a boundary effect, the model boundary is located at
equivalent distances (1000 m) for both grids. Instead, the grid resolution influences the
results more in this direction because the drawdown and gradient to the pumping well are
greater than in the y-direction. This illustrates that for areally anisotropic problems, grid
design becomes even more important than for isotropic problems. As a general rule for aU
models, grids should be designed to match expected gradients. The grid should be able to
accommodate the vertical curvature of streamlines. Note that the results along the 45° angle
(Table 2.5) are similar to the results along the y-axis and are therefore not plotted. The
coarse and fine grids are also equally effective in providing satisfactory answers. Inspection
of Figure 2.3 shows the similarity between the results along the y-axis and along the 45°
angle.
The results of part c, where the grid was not aligned with the principal directions of the
hydraulic conductivity tensor, shows significant deviation from the analytical results. Note
that MODFLOW does not have the capability to accurately model a situation such as this.
The principal directions of the hydraulic conductivity tensor must be aligned with the x and y
directions of the model grid. Even a small misalignment, 20° in case c, can cause significant
errors. This becomes even more apparent for highly fractured systems where anisotropy
ratios may be greater than 10:1.
Area! anisotropy is handled in MODFLOW by the TRPY term, which establishes the ratio
of transmissivity along a column to transmissivity along a row. Note that this is a layer wide
term and a given anisotropy ratio is therefore assumed to exist layer wide.
2-12
-------
740.00 800.00 860.00 920.00 980.00 1040.00 1100.00 1160.00 1220.00
1080.00
1020,00
- 1080.00
- 1020.00
I ! I I I I I I I I I I I I I I I I I I ! l_ I
- 960.00
740.00 800.00 860.00 920,00 980.00 1040.00 1100.00 1160.00 1220.00
Figure 2.3. Drawdown contours (ft) for the 10:1 anisotropic case modeled in part a.
2-13
-------
Figure 2.4. Three-dimensional view of the drawdown for the 10:1 anisotropic case
modeled in part a.
2-14
-------
PROBLEM 3
Artesian-water table conversion
INTRODUCTION
When a confined aquifer is heavily stressed, its potentiometric surface may be drawn down
sufficiently such that the aquifer begins to dewater, or behave as a water-table aquifer. This
conversion takes place when the potentiometric surface falls below the top of the aquifer.
The primary change that takes place in a situation such as this is with the storage coefficient
(s); under confined conditions water is derived from pressure changes and S is fairly small,
while under water-table conditions water is derived from dewatering pore spaces and S is
usually fairly large. A secondary change is that if drawdown is sufficient to cause changes in
saturated thickness, the transmissivity of the aquifer will be reduced. MODFLOW has the
capability to model both these effects. This problem demonstrates the physical process of the
conversion, how it is implemented in MODFLOW simulations, and compares the numerical
results to an analytical solution.
PROBLEM STATEMENT AND DATA
The problem is essentially the same as the example presented by Moench and Prickett
(1972) who derived an analytical solution to the artesian-water-table conversion problem. The
assumptions inherent in the Theis solution are also a part of this solution. Of particular
interest to this problem, the thickness of the aquifer is assumed to be such that the dewatering
does not significantly reduce the aquifer transmissivity, all flow lines in the water table region
are assumed horizontal, and water is released instantaneously from storage. The model
domain is assumed to be effectively infinite; the grid is therefore extended to where the
effects of the stress are negligible.
A fully penetrating well located at the center of the aquifer pumps at a constant rate. The
potentiometric surface of the aquifer is monitored with time at an observation well 1000 ft
from the pumping well. Specific details on the problem are given in Table 3.1 and are from
Moench and Prickett (1972).
3-1
-------
Table 3.1. Parameters used in Problem 3
Initial head 0.0 ft
Transmissivity 2673.8 ft2/d
Storage coefficient (confined) 0.0001
Specific yield (unconfmed) 0.1
Pumping rate 33636 ft3/d
Stress period length 100 days
Number of time steps 25
Time step expansion factor 1.44
SIP iteration parameters 5
Closure criterion 0.001
Maximum iterations 50
Part a) Represent the entire aquifer domain by using the grid spacing shown in Table 3.2.
Place the well at node 1,1, and use one-fourth of the well discharge given in Table
3.1, because only 1/4 of the aquifer domain is modeled. Place the aquifer top at -1
ft. Use layer type 2 (LAYCON) so that the conversion only involves a change in
storage coefficient. Run the model and note drawdown with time at a point 1000 ft
from the pumping well.
Part b) Run the problem with the aquifer top set at -2 ft. Note drawdown versus time at a
point 1000 ft from the pumping well. Compare to part a.
Part c) Run the problem as confined (LAYCON = 0) with storage coefficient of 0.0001 and
note drawdown versus time at a point 1000 ft from the pumping well.
Part d) Rerun part c except use a storage coefficient of 0.1 and note drawdown versus time
at a point 1000 ft from the pumping well.
3-2
-------
Table 3.2. Grid spacing (ft) used in Problem 3
Row number i
(=column number, j) •
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
DELC(i)
(=DELR(j))
10
15
20
30
50
70
100
150
200
220
280
300
400
600
800
1000
1500
2000
3000
4000
6000
8000
10000
15000
20000
30000
3-3
-------
MODEL INPUT
The following is a listing of the input data sets for part a.
*********************************
* Basic package *
*********************************
ARTESIAN/WATER-TABLE CONVERSION PROBLEM HOENCH PRICKETT 1972
1/14/91 PFA
1 26 26 1 4
11 12 0 0 0 0 0 0 19 0 0 22
0 1
0 1
999.00
0 .OOOE+00
100.00 251.414
*********************************
* Block Centered Flow Package *
*********************************
10.00
150.0
800.0
8000.
10.00
150.0
800.0
8000.
0
11
11
0
0
0
0
.100E+01
.100E+OK7G11.4)
15.00
200.0
1000.
10000.
20.00
220.00
1500.
15000.
30.00
280.0
2000.
20000.
.100E+OK7G11.4)
15.00
200.0
1000.
10000.
.1006-03
.2674E+04
.100E+00
-.100E+01
20.00
220.0
1500.
15000.
30.00
280.0
2000.
20000.
12
50.00
300.0
3000.
30000.
12
50.00
300.0
3000.
30000.
70.00
400.0
4000.
70.00
400.0
4000.
100.0
600.0
6000.
100.0
600.0
6000.
*********************************
* Well package *
*********************************
0
1
1 -8409.09
1.0000
*********************************
* SIP package *
*********************************
50 5
.10000E-02 1.00000 1
headngd)
headng(2)
nlay,nrow,ncol,nper,itmuni
iunit array
iapart,istrt
ibound(locat,iconst)
hnoflo
shead(locat,cnstnt)
perlen,nstp,tsmult
iss.ibcfcb
Iayeon
trpy(locat,cnstnt)
delr( locat, cnstnt. ftntin.iprn)
delr array
delc(locat,cnstnt,fmtin,iprn)
dele array
sf1(locat,cnstnt)
t ran(Iocat,cnstnt)
sf2(locat,cnstnt)
top(Iocat.cnstnt)
mxwell,iuelcb
itmp
layer,row,column,q
mxiter,nparm
accl.hclose,ipcalc.wseed,iprsip
3-4
-------
*********************************
* Output Control package *
*********************************
-1
ihedfm.iddnfm,
incode.ihddfI.
hdpr,ddpr,hdsv
incode.ihddfI,
incode.ihddfI,
incode,ihddfI,
incode.ihddfI,
incode.ihddfI,
incode.ihddfI,
incode.ihddfI,
incode.ihddfI,
incode.ihddfI,
incode.ihddfI,
incode.ihddfI,
incode.ihddfI,
incode.ihddfI,
incode.ihddfI,
incode,ihddfI,
incode.ihddfI.
incode,ihddfI,
incode,ihddfI,
incode.ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI.
incode,ihddfI,
ihedun,iddnun
ibudfl.icbcfKstep 1)
,ddsv
ibudfl.icbcfKstep 2)
ibudfl.icbcfUstep 3)
ibudfl,icbcfl(step 4)
ibudfl.icbcfKstep 5)
ibudfl.icbcfKstep 6)
ibudfl.icbcfKstep 7)
ibudfl.icbcfKstep 8)
ibudfl.icbcfKstep 9)
ibudfl.icbcfKstep 10)
ibudfl.icbcfKstep 11)
ibudfl.icbcfKstep 12)
ibudfl.icbcfKstep 13)
ibudfl.icbcfKstep 14)
ibudfl.icbcfKstep 15)
ibudfl.icbcfKstep 16)
ibudfl.icbcfKstep 17)
ibudfl.icbcfKstep 18)
ibudfl.icbcfKstep 19)
ibudfl.icbcfKstep 20)
ibudfl.icbcfKstep 21)
ibudfl.icbcfKstep 22)
ibudfl.icbcfKstep 23)
ibudfl.icbcfKsteo 241
ibudfl.icbcfKstep 25)
3-5
-------
In part b, aquifer top (TOP) is set to -2 ft. In part c layer type (LAYCON) is changed to 0.
As a result, the secondary storage factor (SF2) and aquifer top (TOP) are no longer required.
In part d, the primary storage factor (SF1) is changed from 0.0001 to 0.1.
MODEL OUTPUT
Drawdown versus time is tabulated in Table 3.3 and plotted in Figure 3.1 for each of the
four cases. The results of parts a and b can also be compared to Moench and Prickett (1972)
which is reproduced on the table.
Table 3.3. Drawdown versus time for each model configuration
Time step
number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
• 23
24
25
Time
(days)
0.0072
0.0173
0.0317
0.0520
0.0806
0.1212
0.1785
0.2596
0.3743
0.5364
0.7657
1.090
1.548
2.196
3.113
4.409
6.241
8.832
12.50
17.68
25.00
35.36
50.01
70.72
100.0
Aquifer
Analytical
0.02
0.09
0.16
0.23
0.29
0.36
0.42
0.48
0.55
0.61
0.67
0.73
0.79
0.85
0.91
—
1.03
1.13
1.27
1.46
1.68
1.93
2.21
2.51
2.82
Drawdown (ft)
top at -1 ft Aquifer
MODFLOW Analytical
0.03
0.09
0.16
0.22
0.29
0.36
0.42
0.49
0.55
0.62
0.68
0.74
0.80
0.86
0.93
1.00
1.05
1.14.
1.27
1.44
1.65
1.90
2.17
2.46
2.77
0.04
0.17
0.30
0.42
0.55
0.66
0.78
0.90
1.01
1.13
1.24
1.35
1.46
1.58
1.69
1.80
—
2.02
2.17
. 2.35
2.57
2.83
3.10
3.40
3.71
top at -2 ft
MODFLOW
0.06
0.16
0.28
0.41
0.53
0.65
0.78
0.89
1.01
1.13
1.24
1.36
1.47
1.60
1.70
1.83
1.93
2.05
2.18
2.35
2.56
2.80
3.07
3.36
3.67
Confined
8=0.0001
0.16
0.47
0.84
1.22
1.61
1.98
2.36
2.72
3.08
3.43
3.78
4.14
4.50
4.85
5.20
5.55
5.90
6.26
6.61
6.96
7.37
7.72
8.07
8.42
8.79
Unconfined
S=0.1
0.00
0.00
0.00
0.00-
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.01
0.02
0.05
0.11
0.20
0.33
0.50
0.71
0.96
1.23
1.52
1.83
3-6
-------
o
o
analytic
***** MODFLOW
xxxxx MODFLOW
+++++ MODFLOW
***** MODFLOW
top at -1 ft)
top at -2 ft)
confined)
unconfined)
II III)
0.01
0.1 1 10
Time (days)
100
Figure 3.1. Drawdown versus time for the four MODFLOW configurations and the analytical
solution.
DISCUSSION OF RESULTS
This problem demonstrates the physical process of artesian/water table conversion as related to the
change in storage coefficient. MODFLOW results compare well to the analytical results for both
locations of aquifer top datum. It is apparent from Figure'3.1 that the time-drawdown plots for the
conversion cases are enveloped between the artesian and water-table time-drawdown plots. The
greater the distance from the initial potentiometric surface to the aquifer top, the closer the curve
becomes to the artesian case. The shape of the curve is generally similar prior to conversion to the
Theis curve for artesian conditions while after conversion the slope is similar to the unconfined curve.
Note that the storage coefficient is only related to the time-dependent nature of drawdown.
Figure 3.2 shows distance drawdown plots for the water-table, conversion, and artesian conditions
at 2.19 days. Note that the conversion curve is again enveloped between the artesian and water-table
curves. The water-table responds only near the well due to the large component of storage. The
3-7
-------
conversion case drawdown plot shows a fairly rapid response at distance, where the aquifer is under
artesian conditions. The well is, however, obtaining much of its withdrawal from the newly acquired
storage in the vicinity of the well.
Not shown in this exercise is the feature of MODFLOW which allows a confined aquifer
transmissivity to change to a saturated thickness based unconflned transmissivity. As can be seen
from Figure 3.2, most of a potential change in saturated thickness would be felt immediately near the
well for this problem. This is generally true for pumping well problems and it is often not necessary
to incorporate this added complexity. It may be necessary to account for both storage coefficient and
transmissivity conversion in relatively thin aquifers or in areas where the conversion is regional.
This problem deals with artesian to water-table conversion. It is also possible to convert from
water-table to artesian with MODFLOW. The conversion feature may also be used in a spatial sense:
parts of the model area may be under water-table conditions while others are under confined
conditions.
15-t
conversion
«-*-f artesian
* water-table
4000 6000
distance from well (ft)
8000
Figure 3.2. Drawdown versus distance at 2.19 days for the water table, conversion, and artesian
cases.
3-8
-------
PROBLEM 4
Steady-state
INTRODUCTION
Transient model simulations such as in the preceding problems involve flow into and out
of storage within the aquifer. The preceding problems considered only wells; in complex
aquifer systems other components, such as rivers, springs, evapotranspiration, and recharge,
may contribute or extract flow from the system.
When the aquifer is in equilibrium, flow is balanced between these various sources and
sinks and the system may be in a steady-state. In this exercise, the role of aquifer storage in
transient and steady state simulations is demonstrated. Several methods of simulating a
steady-state solution are attempted.
PROBLEM STATEMENT AND DATA
The modeled domain is discretized using a seven by seven uniformly spaced finite
difference grid of spacing 500 ft as shown in Figure 4.1. Specified head boundaries are
located along row 1 and along column 7. These boundaries may be conceptualized as two
rivers intersecting perpendicularly in the northeastern corner of the modeled groundwater
system. The hydraulic head values associated with these boundaries are given in Table 4.1.
Elsewhere, in the active part of the grid, use a starting head of 10.0 ft. Only a single aquifer
is modeled; therefore only 1 layer is used. The aquifer is treated as confined because it is
relatively thick and does not experience large changes in saturated thickness. The
transmissivity of the aquifer is 500 ftVd, while recharge occurs at a rate of 0.001 ft/d. A well
discharges at a rate of 8000 ft3/d at row 5, column 3,
The strongly implicit procedure (SIP) solution technique is used in this exercise. The
maximum number of iterations (MX1TER) used is 50, the number of iteration parameters
(NPARM) is 5, the acceleration parameter (ACCL) is 1.0, the head change criterion is 0,01,
IPCALC = 1, WSEED = 0,0, and IPRSIP = 1. A more detailed presentation of solution
techniques and convergence is presented in Problem 13.
4-1
-------
Row
2
3
5
6
7
Co umn'
/\ = constant head cell
= active cell
10 = assigned head (ft)
500 ft
Figure 4.1. Configuration of the Problem 4 modeled domain.
7
10
A
6
*
«
•
«
a
9
A
*
*
*
•
m
•
8
'A
o
*
«
«
»
«
6
A
a
«
*
a
a
*
4
A
a
a
#
«
*
a
2
A
a
«
*
*
•
a
0
A
3
A
6
A
8
A
12
A
15
A
20
A
4-2
-------
Table 4.1. Initial heads (SHEAD) at specified head cells
Row
1
1
1
1
1
1
1
2
3
4
5
6
7
Column
1
2
3
4
5
6
7
7
7
7
7
7
7
Head (ft)
10.
9.
8.
6.
4.
2.
0.
3.
6.
8.
12.
15.
20.
Part a) Run the model in a transient mode using a storage coefficient of 0.01. Five time
steps, a time step multiplier of 1.5, and stress period length of 365 days should be
specified in the BASIC package. Print the mass balance (budget) and head
distributions at all five time steps by using the OUTPUT CONTROL PACKAGE,
Part b) Run the model in a steady-state mode by invoking that option in the BCF package.
Run for 1 time step of 1 day in length. Use a time step multiplier of 1.0. Compare
the results to that of part a, time step 5.
Part c) Run the model in a steady-state mode as you did in part b, but ran for 1 time step of
365 days in length. Compare results to that of parts a and b.
Part d) Repeat part b, except use an initial head condition in the active part of the grid of
1000 ft. Compare results to that of part b.
4-3
-------
MODEL INPUT
The following is a listing of data sets used in problem 4 part a. In part b the time-stepping
parameters, PERLEN, NSTP, and TSMULT, are changed to 1.0, 1, and 1.0, respectively in
the BASIC package. The steady state flag (ISS) is changed to 1 and the storage coefficient is
eliminated in the BCF package. Part c uses the part b data sets, except PERLEN, the length
of the simulation, is set to 365 days in the BASIC package. Part d is identical to part b,
except the initial head (SHEAD) in the active area of the model is set to 1000 ft in the
BASIC package.
•**«»»**•*•*«•««*«**<
>****•*»•««
* Basic pack
steady state problem
5/28/91
11 12
-1-1-1-1
1111
1111
1111
1111
1111
1111
.00000
10.00
10.00
10.00
10.00
10.00
10.00
10.00
365.00
PFA
1 7 7
00000 18 19 00
0 0
1 1(4012)
-1-1-1
1 1-1
1 1-1
1 1-1
1 1-1
1 1-1
1 1-1
1 .1006*01(7011.4)
9.000 S.OOO
10.00 10.00
10.00 10.00
10.00 10.00
10.00 10.00
10.00 10.00
10.00 10.00
51 .5000
22
6.00
10.0
10.0
10.0
10.0
10.0
10.0
12
4.000 2.000 .0000
10.00 10.00 3.000
10.00 10.00 6.000
10.00 10.00 8.000
10.00 10.00 12.00
10.00 10.00 15.00
10.00 10.00 20.00
**»»***»»**********»***»*»*»»**»*
* Slock Centered Flow Package *
0 .1006+01
0 .500E+03
0 .5006+03
0 .100E-01
0 .500E+03
*********************************
* Met I package *
*********************************
0
5 3 -.SOOE+04
headngd)
headng<2)
nlay,nrow,ncol,nper, f tmuni
funit array
iapart,istrt
ibound(locat,iconst,frotin,iprn)
ibound array
hnoflo
sheed
-------
1.0000
A********************************
* SIP package *
••A******************************
50 5
.10000E-01 1.00000 1
mxiter.npartn
accl,hclose,ipcalc,wse«d,iprsip
»»*»***»»»»»***»****»»»»**»»*****
* Recharge package *
*********************************
0
0
.001
*********************************
* Output Control package *
*********************************
nrchop,i rchcb
inrech.inirch
rech(Iocat,cnstnt)
ihedfm,iddnfm,ihedun,iddnun
incode,ihddfl,ibudfl,icbcfl(step 1)
hdpr,ddpr,hdsv,ddsv
incode,ihddfl,ibudfl,icbcfl(step 2)
hdpr,ddpr,hdsv,ddsv
incode,ihddfl,ibodfl,icbcfl(step 3)
hdpr,ddpr,hdsv,ddsv
incode,ihddfl,ibudfl,icbcfl(step 4)
hdpr,ddpr,hdsv,ddsv
incode,ihddfl,ibodfl,icbcfl(step 5)
hdpr,ddpr,hdsv,ddsv
4-5
-------
MODEL OUTPUT
Hydraulic head, mass balance information, and iteration data are given in Table 4.2 for
each simulation in this problem set.
Table 4.2. Hydraulic head (ft) at node (7,1), storage component of mass balance, and
iteration data for each time step and the steady-state simulations
Hydraulic Into storage Out of storage No. of
Time step no. Time (days) head (7,1) (ft) (ft3/d) (ft3/d) iterations
1
2
3
4
5
Steady-state
Steady-state
Steady-state
(initial head =
1000 ft)
27.68
69.19
131.5
224.9
365.0
1.0
365.0
1.0
10.30
10.04
9.78
9.65
9.60
9.59
9.59
9.62
3227.9
713.37
217.01
65.43
14.47
0.00
0.00
0.00
1491.8
106.58
0.13
0.00
0.00
0.00
0.00
0.00
6
6
6
5
4
11
11
16
DISCUSSION OF RESULTS
In part a, the system was ran in a transient mode from an arbitrary initial condition in the
active part of the model area. After 1 year of flow (recharge, pumping well, flux to constant
heads, flux from constant heads, storage) the system reaches an equilibrium where heads no
longer change. Row into the system is perfectly balanced with flow out of the system. In
part b, the model was run in its steady-state model (ISS = 1) for a single 1 day time step.
Notice from Table 4.2 that the head at node (7,1) at 365 days for the transient simulation is
almost identical to the 1 day steady-state result Also note that the transient simulation shows
an asymptotic with time approach to the 1 day steady-state result. Further, notice that the
storage component decreases nearly to zero after 365 days for the transient simulation.
4-6
-------
In the 1 day steady state simulation, the problem is forced to steady-state in one time
eth
step by zeroing out the transient head change term ™ on the right-hand side of the
equation by setting the storage coefficient to zero:
= S _ (4.1)
9x2 9y2 3t
Set storage (S) to 0
d2" + d2h = o (4.2)
8x2 9y2
By eliminating time from the equation, the length of the simulation is immaterial.
Therefore, the hydraulic heads from a 1 day steady-state simulation and a 365 day steady-
state simulation (part c) are identical. Similarly, because the system is not responding to any
time related activity, the initial conditions are of no consequence. Therefore, the case (part d)
where initial conditions in the active part of the model were 1000 ft generates essentially the
same answers as when they were set to 10 ft. Part d required slightly more iterations to reach
the result, but within the accuracy of the iterative scheme, arrived at the same result. The
user is cautioned that although initial conditions are generally not important for steady-state
simulations, they could be important in certain non-linear situations where flux,
transmissivity, or saturation are a function of head. For example, for unconfmed simulations,
where the transmissivity is the product of hydraulic conductivity and saturated thickness, it is
important that the initial head be specified such that there is a finite saturated thickness.
4-7
-------
PROBLEM 5
Mass Balance
INTRODUCTION
Often modelers will use hydraulic heads or drawdowns derived from a model exclusively
without regard to other useful information that the model produces. The mass balance, which
is a volumetric accounting of all sources and sinks, is a very useful aspect of a model. The
mass balance can be used as a check on the conceptualization of an aquifer system, as a
check on the numerical accuracy of the solution, and to assess flow rates in discrete portions
of the aquifer. MODFLOW has a mass balance for model wide cumulative volumes,
volumetric rates for each time step for the entire model, and volumetric rates for individual
nodes. This problem demonstrates that the mass balance (or budget) is an algebraic
calculation based on simple hydraulic relationships.
PROBLEM STATEMENT AND DATA
The model domain is identical to that of problem 4 and uses the aquifer parameters and
general set-up of problem 4a (see Figure 4.1). The model input parameters for the SIP
package are also identical to that used in Problem 4.
Part a) Modify the data set from problem 4a to use the OUTPUT CONTROL PACKAGE to
print out the model wide mass balance and to save cell-by-cell budgets for the BCF,
WELL, and RECHARGE packages at timestep 1. Run the model. Using the
hydraulic heads generated for time step 1, manually compute the model wide rate
components into storage, out of storage, well discharge, out of constant heads, into
constant heads, and recharge. Hint: Use Darcy's law to compute constant head flux,
recall the definition of storage coefficient to determine rate change in storage.
Compare to the values computed by the model.
Part b) Run the POSTMOD program or equivalent to decipher the binary cell-by-cell
budgets. Compare the model computed values to your own calculations. How is the
cell-by-cell information useful?
5-1
-------
MODEL INPUT
The following is a listing of the input files for problem 5. Note that the cell-by-cell flags
are set in the individual packages as well as in the OUTPUT CONTROL PACKAGE.
*********************************
* Basic package *
mass balance probl<
5/28/91
11 12
1 1
1 1
1 1
1 1
1 1
1 1 1
.00000
10.00
10.00
10.00
10.00
10.00
10.00
10.00
365.00
Ml
PFA
1
0
0
1
1 1
1 1
1 1
1 1
1 1
1 1
1
0 0
1-1
1-1
1-1
1-1
1-1
1-1
0
7 7
0 18 19 0
0
1(4012)
0 22
1
.100E+OU7G11.4)
9.000
10
10
10
10
10
10
.00
.00
.00
.00
.00
.00
8.000
10.00
10.00
10.00
10.00
10.00
10.00
51.5000
A A A 4> J>A *^ * ^>A^
6.000
10
10
10
10
10
10
A*>*4>4>*>
.00
.00
.00
.00
.00
.00
4>4>4>4tA*
4
2
12
4.000
10
10
10
10
10
10
*4t*^A
.00
.00
.00
.00
.00
.00
4>AAA
2.000
10.
10.
10.
10.
10.
00
00
00
00
00
10.00
.0000
3.000
6.000
8.000
12.00
15.00
20.00
* Block Centered Flow Package *
*********************************
31
.100E+01
.500E+03
.500E+03
.100E-01
.500E+03
*********************************
* Well package *
*********************************
32
5 3 -.800E+04
*********************************
* Recharge package *
*********************************
33
0
.001
*********************************
* Output Control package *
*************************
headngd)
headng(2)
nlay,nrow,ncol,nper,itmuni
iunit array
iapart,istrt
ibound
-------
MODEL OUTPUT
The hydraulic head array and plot of the potentiometric surface at timestep 1 is given in
Figure 5.1. The model wide mass balance or budget is given in Figure 5.2. Printout of cell-
by-cell flow terms is given in Figure 5.3.
HEAD IN LAYER 1 AT END OF TIME STEP 1 IN STRESS PERIOD 1
1
2
3
4
5
6
7
10.00
9.95
9.89
9.72
9.59
9.98
10.30
9.00
9.43
9.49
9.14
8.50
9.54
10.18
a.oo
8.69
8.91
8.18
4.95
8.92
10.24
6.00
7.68
8.59
8.89
8.79
10.25
11.15
4.00
6.50
8.14
9.29
10.36
11.68
12.67
2.00
5.05
7.33
9.13
11.26
13.25
15.16
.00
3.00
6.00
8.00
12.00
15.00
20.00
Figure 5.1. Potentiometric surface map and hydraulic head array at time step 1.
5-3
-------
VOLUMETRIC BUDGET FOR ENTIRE MODEL AT END OF TIME STEP 1 IN STRESS PERIOD 1
CUMULATIVE VOLUMES L**3 RATES FOR THIS TIME STEP
L**3/T
IN:
STORAGE
CONSTANT HEAD
UELLS
RECHARGE
TOTAL IN
OUT:
STORAGE
CONSTANT HEAD
UELLS
RECHARGE
TOTAL OUT
IN - OUT
PERCENT DISCREPANCY
IN:
89340.
.10213E+06
.00000
.24910E+06
.44057E+06
41290.
.17802E+06
.22142E+06
.00000
.44073E+06
-164.50
-.04
STORAGE =
CONSTANT HEAD <
UELLS =
RECHARGE =
TOTAL IN =
OUT:
STORAGE
CONSTANT HEAD
UELLS
RECHARGE
TOTAL OUT
IN - OUT
PERCENT DISCREPANCY «
3227.9
3690.0
.00000
9000.0
15918.
1491.8
6432.0
8000.0
.00000
15924.
-5.9434
-.04
TIME SUMMARY AT END OF TIME STEP
SECONDS
TIME STEP LENGTH
STRESS PERIOD TIME
TOTAL SIMULATION TIME
1 IN STRESS PERIOD 1
MINUTES HOURS
.239136E+07
.239136E+07
.239136E+07
39855.9
39855.9
39855.9
DAYS
YEARS
664.265
664.265
664.265
27.6777
27.6777
27.6777
.757775E-01
.757775E-01
.757775E-01
Figure 5.2. Model wide mass balance at time step 1.
-------
Cell-by-eel I flow terms for CONSTANT HEAD.
Layer Row Co I urn Flow
1
1
1
1
1
1
2
3
4
5
6
7
1
2
3
4
5
6
7
7
7
7
7
7
27.371710
-216.143300
•344.624700
-837.609300
-1249.781000
-1527.405000
-1027.405000
-663.161900
-565.842700
372.070300
872.851000
2417.705000
Cell-by-ceU flou terms for FLOW FRONT FACE
Layer
Row
Coll.
Flow
i
1
1
1
•
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
3
3
3
3
3
3
4
4
I 4
4
4
4
5
5
5
5
5
5
6
6
6
6
6
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
&
1
2
3
4
5
6
1
2
3
4
5
6
27.371710
-216.143300
-344.624700
-837.609300
-1249.781000
-1527.405000
25.881840
-29.765600
-111.489900
-458.164400
-817.985900
-1135.757000
84.710740
177.319700
366.522000
•148.238300
-575.810500
-902.680800
66.183140
319.798200
1614.626000
47.723980
-538.281600
-1062.087000
-193.995200
-521.259900
-1983.955000
-728.688400
-660.206100
-999.219400
-161.909900
-321.031900
-658.988300
-450.047100
-493.980100
-955.145500
Figure 5.3. Printout of eell-by-cell flow terms for each component of the mass
balance.
5-5
-------
Figure 5.3. (Continued)
-by-cell flow term for
Layer ROM Colunrt
WELLS.
Flow
1
-8000.000000
Cell-by-cell flow terms for
RECHARGE.
Layer
Row
Column
Flow
4
I 2
1 2
1 2
1 2
2
1 2
3
1 3
3
3
3
3
4
4
4
4
4
4
5
5
1 5
I S
1 5
1 5
1 6
1 6
1 6
1 6
6
6
7
7
7
1 7
1 7
1 7
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
&
1
2
3
4
5
&
2SO
250
250
250
250
250
.000000
.000000
.000000
.000000
.000000
.000000
250.000000
250
250
250
250
250
,000000
.000000
.000000
.000000
.000000
250.000000
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
250
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
.000000
5-6
-------
Figure 5.3. (Continued)
Cell--by-cell flow terms for
STORAGE.
Layer
Row
Co I urn
Flow
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
2
2
2
2
2
2
3
3
3
3
3
3
4
4
4
4
4
4
5
5
5
5
5
5
6
6
6
6
6
6
7
7
7
7
7
7
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
4.944750
51.278940
118.394000
209.986700
316.178300
446.676100
9.620309
45.901760
98.253240
127.219000
168.408600
241.500700
24.923340
77.934650
164.465700
100.439600
64.387950
78.430820
36.879360
135.706500
456.149000
109.061000
-32.852960
-113.435900
1.834027
41.540440
97.746050
-22.577030
-152.119600
-293.945600
-27.415140
-16.454130
-21.300590
-103.878400
-241.357500
-466.493300
5-7
-------
Figure 5.3. (Continued)
Cell-by-cell flow terms for FLOW RIGHT FACE
Layer
Row
Column
Flow
2
2
2
2
2
•1 2
1 3
1 3
1 3
1 3
1 3
1 3
1 4
1 4
1 4
1 4
1 4
1 4
1 5
1 5
1 5
1 5
1 5
5
6
6
6
6
6
6
7
7
7
7
7
1 7
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
256.485000
371.518600
507.015400
587.828300
722.376200
1027.405000
200.837600
289.794300
160.340900
228.006800
404.605000
663.161900
293.446500
478.996600
-354.419400
-199.565400
77.734710
565.842700
547.061700
1773.824000
-1921.321000
-785.570900
-446.070800
-372.070300
219.797000
311.128500
-666.054100
-717.088600
•785.084000
-872.851000
60.675000
-26.827900
-457.112900
-761.021600
-1246.250000
-2417.705000
5-8
-------
DISCUSSION OF RESULTS
In addition to the hydraulic heads printed in Figure 5.1, MODFLOW provides the
comprehensive mass balance or volumetric budget shown in Figure 5.2. The budget has two
component, cumulative volume and rates for the time step. The cumulative mass balance
accumulates volumes (L3) over the entire length of the simulation. The rate mass balance
deals only with the current time step and divides the volume transferred to various sources
and sinks by the length of time step to yield a rate (L3/T). Because storage is considered in
the mass balance, inflow must always equal outflow. Storage can be viewed as an external
term: water comes in from storage when a well is pumped but goes out as storage when a
well injects.
There will nearly always be a slight difference between outflow and inflow which is
reflected in the in-out and percent discrepancy terms. Generally the percent discrepancy
should be less than 1 percent. Mass balance errors on the order of less than 10 percent are
usually the result of an unconverged solution, too high a closure criterion, too coarse grid
spacing, or too long a time step. Mass balance errors of greater than 10 percent may indicate
a conceptual problem.
The mass balance is actually a series of simple arithmetic calculations that are made using
the computed hydraulic heads. Figure 5.4 shows the hand calculations for each component of
the rate mass balance using the heads shown in Figure 5.1. The well rate is given in the
problem writeup. Recharge is the specified recharge rate integrated over the active area of
the grid. Note that constant head cells do not receive recharge. Constant head discharge is
simply Darcy's law from constant head cell to adjacent cell. Note that MODFLOW does not
consider flow from constant head to constant head in the mass balance. Because the storage
coefficient is the volume of water given up per unit surface area of aquifer per unit decline in
head, the volume from storage is the storage coefficient times the area of aquifer times the
decline in head. Table 5,1 compares the hand calculated mass balance sums with the model
results. The minor difference which occurs is due to truncation error. The hand calculated
values use heads accurate to the nearest hundredth of a foot, whereas the model's precision is
much greater.
Figure 5.3 shows the cell-by-cell printouts for each component of the mass balance. These
support the hand calculations. In addition to the terms shown in the model wide mass
balance, the cell-by-cell mass balance can calculate right face, front face, and bottom face
(multilayer simulations) fluxes. Because of shared faces, only three sides of a six-sided finite
difference cell are printed. This level of detail is useful for analyzing subregions of models,
for input to submodels, and to use in particle tracking programs such as MODPATH (Pollack,
1989).
5-9
-------
Mass Balance Computations for each component
Well Rate = -8000 ft'/d
(given)
Recharge = 0.001 ft/d x 500 ft x 6 x 500 ft x 6 = 9000 ft3/d
(constant head cells do not receive recharge)
Constant head discharge = q
= kia
(for all noted adjacent to constant head cells)
kAh (DELRXb)
DElT
TAh (along a column)
kAh DELC(b)
DELR
= TAh (along a row)
note that DELC = DELR and T = kb
row 1, column 1 = 500 (10-9.95)= 25
row 1, column 2 = 500 (9-9.43)= -215
row 1, column 3 = 500 (8-8.69)= -345
row 1, column 4 = 500 (6-7.68)= -840
row 1, column 5 = 500 (4-6.5)= -1250
row 1, column 6 = 500 (2-5.05)= -1525
row 2, column 7 = 500 (3-5.05)= -1025
row 3, column 7 = 500 (6-7.33)= -665
row 4, column 7 = 500 (8-9.13)= -565
row 5, column 7 = 500 (12-11.26)= 370
row 6, column 7 = 500 (15-13.25)= 875
row 7, column 7 = 500 (20-15.16)= 2420
(flow from constant head to constant head = 0, therefore flow at row 1,
column 7 = 0)
Sum of constant head discharge = -6430 ft3/d
Sum of constant head sources = 3690 ftVd
Figure 5.4. Hand calculations for each component of the mass balance.
5-10
-------
Figure 5.4. (Continued)
Storage = (S) (area) (drawdown)/At
= (0.01) (500 ft)2 (drawdown)/27.6778 d
= 90.325 ft2/d (drawdown)
row 2, column 1 = 90.325 (10-9.95) = 4.52
row 2, column 2 = 90.325 (10-9.43) = 51.49
row 2, column 3 = 90.325 (10-8.69) = 118.33
row 2, column 4 = 90.325 (10-7.68) = 209.55
row 2, column 5 = 90.325 (10-6.50) = 316.14
row 2, column 6 = 90.325 (10-5.05) = 447.11
row 3, column 1 = 90.325 (10-9.89) = 9.94
row 3, column 2 = 90.325 (10-9.49) = 46.07
row 3, column 3 = 90.325 (10-8.91) = 98.45
row 3, column 4 = 90.325 (10-8.59) = 127.36
row 3, column 5 = 90.325 (10-8.14) = 168.00
row 3, column 6 = 90.325 (10-7.33) = 241.17
row 4, column 1 = 90.325 (10-9.72) = 25.29
row 4, column 2 = 90.325 (10-9.14) = 77.68
row 4, column 3 = 90.325 (10-8.18) = 164.39
row 4, column 4 = 90.325 (10-8.89) = 100.26
row 4, column 5 = 90.325 (10-9.29) = 64.13
row 4, column 6 = 90.325 (10-9.13) = 78.58
row 5, column 1 = 90.325 (10-9.59) = 37.03
row 5, column 2 = 90.325 (10-8.50) = 135.49
row 5, column 3 = 90.325 (10-4.95) = 456.14
row 5, column 4 = 90.325 (10-8.79) = 109.29
row 5, column 5 = 90.325 (10-10.36) = -32.52
row 5, column 6 = 90.325 (10-11.26) = -113.81
row 6, column 1 = 90.325 (10-9.98) = 1.81
row 6, column 2 = 90.325 (10-9.54) = 41.55
row 6, column 3 = 90.325 (10-8.92) = 97.55
row 6, column 4 = 90.325 (10-10.25) = -22.58
row 6, column 5 = 90.325 (10-11.68) = -151.75
row 6, column 6 = 90.325 (10-13.25) = -293.56
row 7, column 1 = 90.325 (10-10.30) = -27.10
row 7, column 2 = 90.325 (10-10.18) = -16.26
row 7, column 3 = 90.325 (10-10.24) = -21.68
row 7, column 4 = 90.325 (10-11.15) = -103.87
row 7, column 5 = 90.325 (10-12.67) = -241.17
row 7, column 6 = 90.325 (10-15.16) = -466.08
Storage (source) = sum of positives = 3227.32 ft3/d
Storage (discharge) = sum of negatives = -1490.38 ft3/d
5-11
-------
Table 5.1. Comparison of model calculated and hand calculated rate mass balance
Model Hand Calculations
Inflows (ftVd)
Storage 3227.9 3227.3
Constant head 3690.0 3690.0
Recharge 9000.0 9000.0
Total inflow 15918.0 15917.0
Outflows (ft/d)
Storage
Constant head
Wells
Total outflow
1491.8
6432.0
8000.0
15924.0
1490.4
6430.0
8000.0
15920.0
The mass balance is a very useful aspect of the model. Because the program uses
computed heads to develop the mass balance, the mass balance provides a check on the
accuracy of the numerical solution. Although a good mass balance may not guarantee an
accurate solution, a poor mass balance generally indicates problems with the solution. In
addition, the information in the mass balance is useful to understand the relative importance
of flows into and out of the system.
5-12
-------
PROBLEM 6
Similarity Solutions in Model Calibration
INTRODUCTION
Model calibration involves matching modeled results to observed data. In the process of
obtaining a match, aquifer parameters are usually adjusted within reasonable ranges until a
satisfactory match is derived. Because subsurface properties are generally heterogeneous and
obtained from limited observations, they are somewhat inexact for modeling purposes.
Several "inexact" parameters usually are involved in the construction and calibration of a
model. This problem examines the interplay of two parameters, recharge and transmissivity,
and the ramifications of uncertainty in both parameters on model calibration.
PROBLEM STATEMENT AND DATA
The model domain is identical to that of problems 4 and 5 and uses the steady state
configuration of problem 4b, except the well is eliminated (see Figure 4.1).
Part a) Make a steady state simulation (1 stress period, 1 timestep of 1 day length) using the
following parameters:
Recharge = 0.001 ft/d
Transmissivity = 500 frVd
Part b) Make another steady-state simulation as you did in Part a, but lower the
transmissivity to 50 frVd, Compare these hydraulic heads to those of Part a.
Part c) Make another steady-state simulation with the following parameters:
Recharge = 0.0001 ft/d
Transmissivity = 50 ftVd
Compare these hydraulic heads to those of Part a.
6-1
-------
MODEL INPUT
The following is a listing of data sets used for Part a. It is identical to that used for
problem 4 part b, except the well package is eliminated.
*********************************
www www w w w w wwww w w w w w w w ww w
similarity solutions in calibration
5/28/91 PFA
1771
11 000000 18 19 000
0 0
1(4012)
1*111 _14
1111
1111
1111
1111
1111
1111
.00000
1-1
1-1
1-1
1-1
1-1
1-1
1 .100E+OK7G11.4)
10.00 9.000
10.00 10.00
10.00 10.00
10.00 10.00
10.00 10.00
10.00 10.00
10.00 10.00
1 .0000
8.000
10.00
10.00
10.00
10.00
10.00
10.00
11.0000
*^ A ***** *.Mfc * A A
6.000
10.00
10.00
10.00
10.00
10.00
10.00
****** *****
w w w w w w w w w
4
2
12
4.000
10.00
10.00
10.00
10.00
10.00
10.00
*********
2.000
10.00
10.00
10.00
10.00
10.00
10.00
.0000
3.000
6.000
8.000
12.00
15.00
20.00
1
* Block Centered Flow Package *
*********************************
0
0 .100E+01
0 .500E+Q3
0 .SOOE+03
0 .500E+Q3
a********************** »*>•*«*•**
Recharge package
0
0
.001
*********************************
* SIP package *
*********************************
50 5
1.0000 .10000E-01
headngd)
headng<2)
nlay,nrow,ncol,nper,itmuni
iunit array
japart,istrt
ibound(locat,iconst,fratin,iprn)
ibound array
hnoflo
shead(locat,cnstnt,fmtin,iprn)
shead array
1.00000E+00
perlen.nstp,tsmult
iss.ifacfeb
Iayeon
trpy
delr
-------
MODEL OUTPUT
Hydraulic head arrays, contour maps of potentiometric surface, and mass balance printout
for Paris a, b, and c are given in Figures 6.1, 6.2, and 6.3, respectively.
HEAD IN LAYER 1 AT END OF TINE STEP 1 ID STRESS PERICO 1
1
2
3
4
5
6
7
10.00
11.37
12.82
H.I?
15.30
16.12
16.54
9.000
10.79
12.42
13.89
IS. 11
16.00
16.47
8.000
9.860
11.70
13.36
14.76
1S.81
16.37
6.000
8.4S6
10.64
12.58
14.27
1S.5S
16.34
4.000
6.824
9.321
11.56
13.67
15.41
16.57
2.000
5.020
7.757
10.19
12.92
IS. 32
17.47
O.OOOOE+00
3.000
6.000
8.000
12.00
15.00
20.00
VOUWETIiC BUDGET FOR EHTItE MODEL AT BO OP TIME STEP 1 IN STRESS PERIOD 1
CUNULAT1VE VOLUMES L««3 MTES TOR THIS TIME STEP L**3/T
ID:
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL IN
OUT:
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL OUT
IN - OUT
PERCENT DISCREPANCY
Q.QOOOOE+00
1266.9
9000.0
10267.
O.OOOOOE+00
10263.
O.OOOOOE+00
10263.
4.2383
0.04
IN:
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL IN
OUT:
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL OUT
IN - OUT
PERCENT DISCREPANCY
O.OOOOOE+00
1266.9
9000.0
10267.
O.OOOOOE+00
10263.
O.OOOOOE+00
10263.
4.2383
Figure 6.1. Contour map of potentiometric surface, hydraulic head array, and mass
balance ouput for Part a.
0.04
6-3
-------
HEAD IN LAYER 1 AT END OF TIME STEP 1 IN STRESS PERIOD 1
1
I
3
4
5
6
7
10,00
28.82
43.81
55.31
63.65
69.08
71.76
9.000
27.66
42.30
53.47
61 .56
66.83
69.44
8.000
25.52
39.25
49.70
57.28
62.23
64.72
6.000
22.16
34.50
43.81
50.62
55.16
S7.48
4.000
17.62
27.77
35.42
41.23
45.29
47,56
2.000
11.5*
1S.55
23.89
21.58
32.20
34.92
O.QOOOE+00
3.000
6.000
8.000
12.00
15.00
20.00
VOLUMETRIC BUDGET FOR ENTIRE MODEL AT END OF TIME STEP 1 IN STRESS PERIOD 1
CUMULATIVE VOLUMES L«*3 RATES FOR THIS TIME STEP L**3/T
IN:
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL ill
OUT:
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL OUT
IN - OUT
PERCENT DISCREPANCY
O.OOOQQE-HJO
O.OOOOOE-KJO
9000.0
9000.0
0.000006*00
8999.7
O.QOQOOE+00
8999.7
0.31641
0.00
IN:
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL IN
OUT:
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL OUT
IN - OUT
PERCENT DISCREPANCY
O.OOOOOE+00
O.QOOOOE+QO
9000.0
9000.0
0.000006*00
8999,7
O.OOOOOE+00
8999.7
0.31641
0.00
Figure 6.2. Contour map of potentiometrie surface, hydraulic head array, and mass
balance output for Part b.
6-4
-------
HEAD IN LAYER 1 AT END OF TIME STEP 1 IN STRESS PERICO 1
1
2
3
4
5
6
7
10.00
11.37
12.82
14.17
15.30
16.12
16.54
9.000
10.79
12.42
13.89
15.11
16.00
16.47
8.000
9.360
11.70
13.36
14.76
15.81
16.37
6.000
8.4S6
10.64
12.S8
14.27
15.S8
16.34
4.000
6.824
9.321
11.56
13.67
15.41
16.57
2.000
5.020
7.757
10.19
12.92
15.32
17.47
O.OOOOE+00
3.000
6.000
S.OOO
12.00
15.00
20.00
VOLUMETRIC BUDGET FOR ENTIRE MODEL AT END OF TIME STEP 1 IN STRESS PERIOD 1
oiMuunve VOLUMES L*«3 RATES FOR THIS TIHE STEP L«*3/T
IN:
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL IN
OUT;
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL OUT
IK - OUT
PERCENT DISCREPANCY
O.QOOOQE+00
126.69
900.00
1026.7
O.OOQOQE+00
1026.3
O.OOOOOE-00
1026.3
0.42651
0.04
IK:
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL IN
OUT:
STORAGE
CONSTANT HEAD
RECHARGE
TOTAL OUT
IN - OUT
PERCENT DISCREPANCY
O.OOOOOE+00
126.69
900.00
1026.7
O.OOOOOE+00
1026.3
O.OOOOOE+00
1026.3
0.42651
0.04
Figure 6.3. Contour map of potentiometric surface, hydraulic head array, and mass
balance output for Part c.
6-5
-------
DISCUSSION OF RESULTS
The potentiomelric surface generated in Part a represents a balance between sources
(primarily recharge, some specified head) and sinks (specified head). Flow is generally
toward the specified head cells and gently slopes toward the potentiometric low at the
confluence of the two "rivers". The "rivers" are gaining, except for a small portion in the
southeastern corner which contributes flux to the groundwater system.
In Part b, the transmissivity is decreased, representing a much "tighter" aquifer. For the
given recharge rate, hydraulic heads and gradients increase. Note that again sources balance
the sinks, however all flow is now toward the rivers; recharge is the only source. If you
wished to calibrate this model by varying transmissivity you would therefore decrease
transmissivity if modeled heads were lower than observed and increase transmissivity if
modeled heads were too high.
In Part c, recharge is reduced by an order of magnitude in addition to the reduction in
transmissivity that was done in Part b. Identical heads and gradients are obtained for Part c
as in Part a. Although this result may be surprising, there is a simple mathematical
explanation of this phenomenon. If we look at the two-dimensional steady-state groundwater
flow equation (6.1), we can see that it relates hydraulic gradients to transmissivity (T) and a
source term, (R). Algebraic manipulation of (6.1) results in (6.2) which shows that the ratio
of source terms to transmissivity governs the computed hydraulic gradient.
___
ax2 ay2
.
a*2 ay2 T
Therefore, similar ratios of transmissivity and recharge will generate the same head
distribution. In Part a, the ratio of recharge to transmissivity was 0.001/500 or 2xlO"6; in Part
c the ratio was 0.0001/50 or 2xlO"6. Theoretically, infinite combinations of recharge and
transmissivity (as long as their ratio is the same) could cause identical head distributions.
This phenomenon is often referred to as "non-uniqueness" by hydrologists, but is referred to
more correctly as similarity solutions by mathematicians. The ramifications of this
phenomenon are quite important: a good match of modeled results to observed data does not
necessarily guarantee an accurate model. In order to narrow the range in hydraulic
parameters, supporting field data should be collected for the necessary parameters. Secondly,
the effect of parameter uncertainty should be evaluated by observing model response within
the range of parameter uncertainty.
Note that for this example another calibration target is potentially available; matching
observed stream baseflow to model results. This would provide additional assurance of model
accuracy.
6-6
-------
PROBLEM 7
Superposition
INTRODUCTION
A goal in groundwater modeling is often to examine the independent effect of a stress on
the system. Given the complexity of most hydrologic systems, including transients, parameter
uncertainty, and the interplay of these parameters, it is sometimes difficult to isolate the result
of one particular stress. This exercise illustrates a property of the groundwater flow equation
that allows the modeler to simplify problems and also use these simplifications to examine
problems involving multiple stresses and optimal pumping rates.
PROBLEM STATEMENT AND DATA
The model domain is identical to that of problems 4, 5 and 6 (see Figure 4.1). This
problem uses the aquifer parameters given in problem 6, part a.
Part a) Rerun Part a of Problem 6. Print out the individual specified head fluxes by
invoking that option in the BCF package.
Part b) Specify a well located at row 5, column 3 pumping of a rate of -8000 ft3/d and
run a steady-state simulation (1 stress period, 1 timestep of 1 day length). As in
Part a, printout the individual specified head fluxes. Observe the results and
compare to Part a.
Part c) Set up a "drawdown" model using the parameters and stresses of Part b. This
model will have an initial head of zero, recharge rate of zero, and specified heads
of 0 along row 1 and column 7. Run a steady-state simulation (1 stress period, 1
timestep of 1 day length). As in Parts a and b, printout the individual specified
head fluxes. On a node-by-node basis, add the heads of Part a and c and compare
the results to those of Part b. Perform a similar computation for the specified
head fluxes given in each output file.
Part d) Run the problem of Part c with twice the well rate. Compare the heads of Part c
to Part d.
7-1
-------
MODEL INPUT
The following is a listing of data sets for Part a.
*********************************
* Basic package *
*********************************
SUPERPOSITION PROBLEM PART A
12/8/89
1 7
11 0 0 0 0 0 0 18 19
0 0
0 22
1
1(4012)
-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
.OOOOOE+00
1.
10
10
10
10
10
10
10
.00
.00
.00
.00
.00
.00
.00
1
0.
0000
100E+OK7G11.4)
9.000
10.00
10.00
10.00
10.00
10.00
10.00
11
8.
10
10
10
10
10
10
.0000
000
.00
.00
.00
.00
.00
.00
6.000
10
10
10
10
10
10
.00
.00
.00
.00
.00
.00
4.
10
10
10
10
10
10
12
000
.00
.00
.00
.00
.00
.00
*********************************
*
Block Centered
Flow
Package *
*********************************
0
1
0
0
0
0
0.
0.
0.
0.
-1
100E+01
500E+03
500E+03
500E+03
2.000
10.00
10.00
10.00
10.00
10.00
10.00
O.OOOOE+00
3.000
6.000
8.000
12.00
15.00
20.00
1.0000
*********************************
* Recharge package *
*********************************
1 0
0 0
0 0.100E-02
*********************************
* SIP package *
*********************************
50 5
.10000E-01 1.OOOOOE+00 1
*********************************
* Output Control package *
*********************************
35
1
1
headngd)
headng<2)
nlay,nrow,ncol,nper,itmuni
iunit array
iapart,istrt
ibouncK locat,iconst,fmtin,iprn)
ibound array
hnoflo
shead(locat,cnstnt,fmtin,iprn)
shead array
perlen,nstp,tsmult
iss.ibcfcb
laycon
trpy(locat,cnstnt)
delr(locat,cnstnt)
delc(locat,cnstnt)
tran
-------
In Part b, the WELL package shown below is added. It is invoked by setting IUNTT(2) to
12 in the BASIC package.
************************
* Well package
************************
0
******
*
3-.1600E+05
mxweU.iwelcb
i trap
layer,row,col,q
The following is a listing of the data sets for Part c.
*********************************
SUPERPOSITION
3/15/90 PFA
1
11 12 0 0
0
1
-1-1-1-1-1-1-1
1111 1-1
1111 1-1
1111 1-1
1111 1-1
1111 1-1
111111-1
.00000
WYWWWWWWWWWWWWWWWWWWWWW
PROBLEM PART C
7 7 1
0 0 0 0 19 0 0 22
0
1(4012)
wwwwwwwww
4
2
o .oooe+oo
1.0000
11.0000
1
50
1.0
* Block Cantered Flow Package *
a**.******************************
-1
0 .1006+01
0 .5006+03
0 .500E+Q3
0 .500E+03
*********************************
* Well package *
*********************************
0
5
3 -.800E+Q4
*********************************
* SIP package *
*********************************
5
.01 1 0.0 1
headngd)
headng<23
nlay,nrow,ncol,nper,itmuni
iunit array
iapart,istrt
i bound(locat,iconst,fmtin,iprn)
fbound array
hnoflo
i st rt(Iocat,enstnt)
perlen,nstp,tsmult
iss.ibcfcb
Iayeon
trpy(locat,cnstnt)
delr(locat,cnstnt)
delcClocat,cnstnt)
tran(locat,cnstnt)
mxwell,iwelcb
itmp
layer,row,col,q
mxiter,nparm
accl,hclose,ipcalc,wseed,iprsip
In Part d, the data set of Part c is modified by changing the well rate (parameter Q) in the
WELL package from -8000 to -16000 ft3/d.
7-3
-------
MODEL OUTPUT
Hydraulic head arrays, contour maps of potentiometric surface, model wide mass balance,
and individual node mass balances are presented for Parts a, b, c, and d in Figures 7,1, 7.2,
7.3, and 7.4, respectively.
1
2
3
4
5
6
10.00
11. 37
12.82
H.17
15.30
16.12
9.00
10.79
12.42
13.89
15.11
16.00
8.00
9.86
11.70
13.34
14.76
15.81
6.00
8.46
10.64
12.58
14.27
1S.S8
4.00
6.82
9.32
11.54
13.67
15.41
2.00
5.02
7.76
10.19
12.92
15.32
.00
3.00
6.00
8.00
12.00
15.00
16.54 16.47 16.37 16.34 16.S7 17.47 20.00
IN:
STORAGE » .00000
CONSTANT HEAD « 1266.9
RECHARGE =• 9000.0
TOTAL IN » 10267.
OUT:
STORAGE = .00000
CONSTANT HEAD 10263.
RECHARGE .00000
TOTAL OUT 10263.
IN - OUT 4.2607
PERCENT DISCREPANCY
.04
KM
HOW
MM
MM
ROU
ROW
ROW
ROU
ROW
ROU
ROM
ROU
ROW
1
1
1
1
1
1
1
2
3
4
5
6
7
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
1
2
3
4
5
6
7
7
7
7
7
7
7
-685.0535
-894.2446
-929.9719
-1227.958
•1412.217
-1510.177
.0000000
-1010.177
-878.3768
-1092.589
-459.6479
•162.2080
1266.881
Figure 7.1. Contour map of potentiometric surface, hydraulic head array, model
wide mass balance, and individual specified head node mass balance for
Part a.
7-4
-------
1
2
3
4
5
6
7
10.00
9.58
9.24
8.87
8.67
9.16
9.59
9.00
9.01
8.77
8.21
7.49
8.71
9.51
8.00
8.18
8.11
7.19
3.87
8.18
9.75
6.00
7.10
7.81
8.09
8.11
9.90
11.05
4.00
5.90
7.U
8.74
10.08
11.76
12.99
2.00
4.56
6.83
8.83
11.24
13.54
15.68
.00
3.00
6.00
8.00
12.00
15.00
20.00
IN:
STORAGE » .00000
CONSTANT HEAD a 3478.1
WELLS = .00000
RECHARGE = 9000.0
TOTAL IN = 12478.
OUT:
STORAGE = .00000
CONSTANT HEAD = 4479.9
WELLS = 8000.0
RECHARGE = .00000
TOTAL OUT = 12480.
IN - OUT = -1.8662
PERCENT DISCREPANCY =
-.01
ROW
ROW
ROW
ROW
ROW
ROW
ROW
ROW
ROW
ROW
ROW
ROW
ROW
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
1
2
3
4
5
6
7
7
7
7
7
7
7
208.5836
-3.670284
-89.71581
-548.7798
-950.0516
-1279.053
.0000000
-779.0528
-416.2093
-413.4138
380.5986
728.3688
2160.528
Figure 7.2. Contour map of potentiometric surface, hydraulic head array, model
wide mass balance, and individual specified head node mass balance for
Part b.
7-5
-------
1
2
3
4
5
6
7
.00
-1.79
-3.58
-5.30
-6.63
-6.96
-6.96
.00
-1.78
-3.66
-5.69
-7.62
-7.29
-6.96
.00
-1.68
-3.58
-6.16
-10.89
-7.62
-6.63
.00
-1.36
-2.83
-4.50
-6.16
-5.69
-5.30
.00
-.92
-1.88
-2.83
-3.58
-3.66
-3.58
.00
-.46
-.92
-1.36
-1.68
-1.78
-1.79
.00
.00
.00
.00
.00
.00
.00
IN:
STORAGE = .00000
CONSTANT HEAD = 7993.9
WELLS = .00000
TOTAL IN = 7993.9
OUT:
STORAGE = .00000
CONSTANT HEAD = .00000
WELLS = 8000.0
TOTAL OUT = 8000.0
IN - OUT = -6.1274
PERCENT DISCREPANCY =
-.08
ROU
ROW
ROU
ROU
ROU
ROU
ROU
ROU
ROU I
ROU '
ROU
ROU
ROU
COL
COL
COL
COL
COL
COL
COL
2 COL
5 COL
I COL
5 COL
& COL
7 COL
1
2
3
4
5
6
7
7
7
7
7
7
7
893.6370
890.5743
340.2561
679.1781
462.1656
231.1239
.0000000
231.1239
462.1674
679.1755
340.2465
890.5767
893.6470
Figure 7.3. Contour map of potentiometric surface, hydraulic head array, model
wide mass balance, and individual specified head node mass balance for
Part c.
7-6
-------
1
2
3
4
5
6
7
.00
-3.57
-7,16
-10.60
-13.25
-13.92
-13.92
.00
-3.56
-7.31
-11.37
-1S.25
-14.58
-13.92
.00
-3.36
-7.17
-12.33
-21.79
-15.25
-13.25
.00
-2.72
-5.66
-8.99
-12.33
-11.37
-10.60
.00
-1.85
-3.75
-5.66
-7.17
•7.31
-7.16
.00
-.92
-1.85
-2.72
-3.36
-3.56
-3.57
.00
.00
.00
.00
.00
.00
.00
IN:
STORAGE = .00000
CONSTANT HEAD = 15988.
WELLS = .00000
TOTAL IN = 15988.
OUT:
STORAGE = .00000
CONSTANT HEAD .00000
WELLS 16000.
TOTAL OUT 16000.
IN - OUT -12.255
PERCENT DISCREPANCY
ROM
ROW
ROW
ROU
ROW
ROU
ROU
ROU
ROU
ROU
ROU
ROU
ROU
1
1
1
1
1
1
1
2
3
4
5
6
7
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
COL
1
2
3
4
5
6
7
7
7
7
7
7
7
RATE
RATE
RATE
RATE
RATE
RATE
RATE
RATE
RATE
RATE
RATE
RATE
RATE
1787.274
1781.149
1680.512
1358.356
924.3312
462.2479
.0000000
462.2479
924.3349
1358.351
1680.493
1781.153
1787.294
Figure 7.4. Contour map of potentiometric surface, hydraulic head array, model
wide mass balance, and individual specified head node mass balance for
Partd.
7-7
-------
DISCUSSION OF RESULTS
The potentiometric surface generated in Part a represents a balance between sources
(primarily recharge, some specified head) and sinks (specified head). Flow is generally
toward the specified heads and generally slopes toward the potentiometric low at the
confluence of the two "rivers." The "rivers" are gaining except for a small portion in the
southeastern corner, which contributes flux to the groundwater system. This flow reversal
may be verified in the cell-by-cell flux printout which indicates a positive specified head flux
for row 7, column 7.
In Part b, a well is added, resulting in lowered head and flow otherwise destined for the
specified head cells to be diverted to the well. This reduced specified head flux may be
observed in the model wide mass balance (10263 ft3/d OUT for Part a, 4479.9 ftVd OUT for
Part b; 1266.9 ft3/d IN for Part a, 3478.1 IN for Part b).
In Part c, the drawdown model shows only the effects of the pumping well. Pumpage
from the well is obtained by a diversion from the specified head cells. When matrix addition
is performed, the sum of heads at individual nodes in Part a and Part c equals the head at the
corresponding node in Part b. For example, at row 7, column 1:
16.54 + (-6.96) = 9.59
a + c = b
For all nodes the" sum of the background head (a) and head from the drawdown model (c)
is equivalent to the head in the composite model (b). Note that the mass balance components
are also additive in this sense. For example the flow from the specified head cell in row 4,
column 7 is:
-1092.59+ 679.18 = -413.41
a + c = b
Notice that although flow in Part c is positive or out of the specified head cell, the net
result of the well (Part b) is to reduce the amount of flow into the specified head cell. A
similar computation may be made for the components of the model wide mass balance:
-10263 + 1266.9 + 7993.9 = -4479 + 3478.1
a + a + c= b + b
This additive property of the groundwater flow equation for heads and fluxes is called the
principle of superposition.
7-8
-------
In Part d, the well rate is doubled, resulting in a doubling of the drawdown. For example
at row 7, column 1, head for part c was -6,96, for part d it was -13.92. This is consistent with
the principle of superposition in that the 16,000 ftVd discharge could be broken into two 8000
ft3/d discharges and the results summed. The results of this summation would be twice the
drawdown generated by the 8000 ft3/d discharge.
The principle of superposition implies that for any linear problem, the individual effect of a
stress can be modeled individually and then superimposed onto the natural flow system.
Several stresses can also be modeled individually and the results summed to develop a
composite result. Some advantages of using superposition in groundwater system are
discussed by Reilly et al. (1987). They summarize this discussion as follows:
Superposition enables us to simplify complex problems and to obtain useful
results despite a lack of certain information describing the groundwater system
and the stresses acting on it. Through the use of superposition, the problem can
be formulated in simpler terms, which saves effort and reduces data
requirements. Thus, if the technique is-applicable, it may be advantageous to
use superposition in solving many specific problems.
In order for superposition to be valid, the system (governing equation and boundary
conditions) must be linear. An unconfined system or head dependent boundary conditions
with abrupt flux change (drain, E-T, river) is non-linear and superposition will not strictly be
valid.
7-9
-------
PROBLEM 8
Grid and Time Stepping Considerations
INTRODUCTION
In finite difference models, the aquifer system which is described by a partial differential
equation representing a continuous domain is simplified to a series of algebraic equations
which represent discrete intervals of the system. Both space and time are broken into intervals
(discretized). Questions often arise regarding the proper level of discretization required for
accuracy. Another related question arises regarding the proper closure criterion to use for the
iterative solution of the system of equations. The objective of this exercise is to examine
various levels of grid spacing, time stepping, and closure criterion for a problem for which an
exact solution is known. Comparisons of relative accuracy and execution time as well as
general observations concerning selection of the parameters can be made.
PROBLEM STATEMENT
This problem has been modified from example 4 of Rushton and Tomlinson (1977), A
two-dimensional square aquifer with 15000 m sides has impermeable (no-flow) boundaries on
three sides and a fourth (the north side) held at a specified head of 0.0 rn. A well pumping at
15000 m3/d is located as shown in Figure 8.1. Three observation wells are used as illustrated
in Figure 8.1. The transmissivity of the aquifer is 2400 m2/d and the storage coefficient is
2.5x10"*. Five grid configurations will be examined in Parts a and b. The location of the
pumping and observation wells and additional data on each grid configuration are given in
Table 8.1. Notice that the wells are conveniently located at the center of finite difference
blocks.
In order to place the specified head boundary exactly on the edge of the model domain, the
general head boundary (GHB) package is used. The conductance parameter must be computed
to represent the conductance between the node center at row 1 to the northern edge of the
finite difference block of row 1, An example calculation for grid 1 is shown below:
(8.1)
L
where:
C = Conductance [L2/T]
T = Transmissivity in direction of flow [L2/T]
L = Length of flow path (node center to edge) [L]
W = Width of face perpendicular to flow [L]
8-1
-------
H = 0
=0
5000 m
X
f
5000 m
5000 m /
i /•
Well 1 Well 2
Well 3
Q - 15,000 m3/d
\
\
\
\
•v
\
$
^^^ ^^-^-
5000 m 5000 m 5000 m
Figure 8.1. Location of pumping wells, observation wells, and boundary conditions for
problem 8.
8-2
-------
Table 8.1. Grid data
Grid
Size Grid Spacing*
Pumping
Well
Well locations (row, column)
Well 1 Well 2 Well 3
4*4 2500 row, column 1
5000 row, column 2,3
2500 row, column 4
3,2
2,2
2,3
3,3
7*7 1750 row, column 1
2000 row, column 2
2500 row, column 3-5
2000 row, column 6
1750 row, column 7
5,3
3,3
3,5
5,5
10*10 1250 row, column 1
1666.7 row, columm 2-9
1250 row, column 10
7,4
4,4
4,7
7,7
16*16 500 row, column 1
1000 row, column 2-15
500 row, column 16
11,6
6,6
6,11
11,11
30*30 416.7 row, column 1-6 20,11
555.6 row, column 7-24
416.7 row, column 25-30
11,11
11,20
20,20
*spacing along a column is the same as along a row such that DELX(l) = DELY(l),
DELX(2) = DELY(2), etc.
8-3
-------
for row 1, column 1:
= TW m (2400 mVd)(2500 m) . ^
L (1250 m)
for row 1, column 2:
C = ™ = (2400mVd)(5000m) m
L (1250 m)
Note that L remains constant for a given grid because distance from center to edge is always
the same, but W changes due to varying column widths.
In each case, use the SIP solver, acceleration parameter = 1.0, closure criterion = 0.0001,
and maximum iterations = 50.
Part a) Set up the model for each of the grids (1-5) and run. Record drawdowns at
observation wells 1, 2, and 3 at the final time step. Record the total number of
iterations required for all time steps. Use the following time parameters:
time step multiplier = 1.414
number of time steps = 10
length of stress period = 20 days
Part b) Repeat part a, but use the following time parameters:
«
time step multiplier = 1.414
number of time steps = 10
length of stress period = 0.2 days
8-4
-------
Part c) Rerun one of the grids used in part a, changing only the number of time steps.
Record drawdowns at observation wells 1, 2, and 3 as well as the total number of
iterations for all time steps.
Run the following cases:
1. 1 time step
2. 2 time steps
3. 3 time steps
4. 5 time steps
5. 7 time steps
6. 10 time steps
7. 20 time steps
8. 30 time steps
Part d) Rerun one of the grids used in part a, changing only the closure criterion. Record
drawdowns or observations wells 1, 2, and 3 as well as the total number of
iterations for all time steps. Run the following cases:
1 HCLOSE = 0.0001
2 HCLOSE = 0.001
3 HCLOSE = 0.01
4 HCLOSE = 0.1
5 HCLOSE = 0.5
6 HCLOSE = 1.0
8-5
-------
MODEL INPUT
The following is a listing of data sets used in part a for grid 1.
*********************************
* Basic package *
*********************************
GRID AND TIME STEPPING CONSIDERATIONS
4/20/90 PFA
4 4
11 12
.00000
20.000
2500.
2500.
1
0 0
0
0
0
11
11
0
0
0 17
1
1
0 19 0 0 0
,0006+00
101.4140
A********************************
* Stock Centered Flow Package *
*********************************
0
.1001*01
.100E+01C7C11.4)
5000. 5000.
.1001+01(7011.4)
5000. 5000.
.250E-03
.240E+04
2500.
2500.
12
12
* Well package *
******»**»*******»*»»»******»***»
0
3 2-.150QE+05
*********************************
* General Head Boundary package *
****.*****.*/*********************»*
1 .OOOE+00 .480E+04
2 .OOOE+00 .960E+04
3 .OOOE+00 .96QE+04
4 .OOOE+00 .480E+04
1.0000
a********************************
* SIP package *
*********************************
50 5
.10000E-03 1.00000 1
headngd)
headng(Z)
nlay.rsrow, ncol, riper, i tmuni
iunit array
iapart.istrt
ibound(locat,iconst)
hnoflo
shead(tocat,cnstnt)
perten.nstp,tsmult
iss.ibcfcb
laycon
trpy(locat,cnstnt)
delr
-------
MODEL OUTPUT
Drawdown, iteration and CPU data are given in Tables 8.2, 8.3, 8.4, and 8.5 for parts a, b,
c, and d, respectively. A comparison is also made to analytical results obtained from the
image well technique.
Table 8.2 Comparison of results for various grid spacings in part a
Drawdown (m)
Observation Well
Grid
1
2
3
4
5
analytic2
# Nodes
16
49
100
256
900
Total
Iterations
47
74
96
124
143
CPU1
5.64
8.12
13.02
30.33
106.54
1
1.956
1.971
1.962
1.955
1.952
2.04
2
1.493
1.537
1.546
1.550
1.550
1.63
3
2.806
2.816
2.807
2.801
2.797
2.95
1 PRIME 550 computer
2 Image well solution given in Rushton and Tomlinson (1977)
Table 8.3. Comparison of results for various grid spacings in part b
Drawdown (m)
Observation Well
Grid
1
2
3
4
5
analytic
# Nodes
16
49
100
256
900
Total
Iterations
22
24
26
32
38
CPU
5.13
6.03
7.56
12.77
36.63
1
0.0162
0.0153
0.0126
0.0097
0.0085
0.0047
2
0.0010
0.0007
0.0007
0.0006
0.0006
0.0001
3
0.0162
0.0153
0.0126
0.0097
0.0085
0.0047
8-7
-------
Table 8.4. Comparison of results for variations in time stepping in part c
Drawdown (m)
Observation Well
#Time
Steps
* Of CPU 1
Iterations
2 3
1
2
3
5
7
10
20
30
11
18
24
31
37
47
72
92
3.92
4.09
4.23
4.46
4.70
5.05
6.05
7.03
1.586
1.784
1.860
1.921
1.943
1.956
1.962
1.962
1.169
1.335
1.403
1.459
1.481
1.493
1.499
1.499
2.233
2.530
2.650
2.748
2.785
2.806
2.816
2.816
Table 8.5. Comparison of results for variations in closure criterion in part d
Drawdown (m)
Observation Well
HCLOSE
0.0001
0.001
.001
.01
.05
1.0
# of
TT v*
Iterations
47
42
36
25
10
10
CPU
5.64
4.94
4.83
4.65
4.48
4.48
1
1.956
1.956
1.956
1.945
1.653
1.653
2
1.493
1.493
1.493
1.482
1.177
1.177
3
2.806
2.805
2.805
2.787
2.259
2.259
8-8
-------
DISCUSSION OF RESULTS
In part a, all the grid configurations provide reasonable approximations to the drawdown as
shown in Table 8.2. This is because the solution is close to steady-state and steep hydraulic
gradients near the pumping well do not exist. Successively finer spacings generally tend to
decrease drawdown directly along rows and columns and increase drawdown diagonal to rows
and columns. The answers generally converge toward a solution, but still differ from the
analytical solution. Note that CPU time is directly related to the number of nodes. The CPU
times stated herein are for comparative purposes and should not be used to estimate execution
times for other problems. The drawdowns at the final time step are shown in Figure 8.2 for
the coarse grid case.
Drawdowns for the final timestep of part b are shown in Figure 8.3 for the 16 x 16 grid.
The accuracy of the answer is highly dependent upon the grid configuration used (Table 8.3).
This is because in early time, the gradients are much steeper in the vicinity of the pumping
well. A fine grid can approximate this rapid spatial variation much better than a coarse grid.
Notice that the grid design can take on vastly different configurations depending on the intent
of the modeling. As a general rule, the grid should be designed to match the curvature of the
drawdown cone.
In part c, the number of time steps is shown in Table 8.4 to be important. The results after
1 time step are very inaccurate; approximately 4 time steps are required for acceptable results.
This is consistent with Priekett and Lonnquist (1971) who recommend performing 3-4 time
steps before relying on results. Just like with grid discretization, it is important to discretize
time increments to approximate steep gradients in early time. Comparison of CPU times
indicates that the time required for modeling four time steps is not great when compared to
the initial time required for one time step. This is because in all cases some time will be
spent reading the data and initiating execution of the program. Note that the results tabulated
in Table 8.4 are for the coarse 4x4 grid.
The results of part d, shown in Table 8.5, indicate that the optimal closure criterion for this
problem is 0.01. Little, if anything, is gained by a smaller closure criterion. A general rule of
thumb is that the closure criterion should be an order of magnitude smaller than the desired
accuracy. It is interesting to note that order of magnitude changes in closure criterion are not
excessively time consuming. However, some complex problems reach a threshold where
further convergence is no longer possible. Note that the results when using closure criterions
of 0.5 and 1.0 are identical because the closure criterion is satisfied after the first iteration of
each time step.
8-9
-------
15000
5000 1 0000 1 5000
15000
10000
5000
i i i i i i i i i i
- 10000
- 5000
5000
10000
15000
Figure 8.2. Drawdown (m) at 20 days for the 4x4 grid simulation of Part a.
8-10
-------
15000
0
5000
10000
10000
5000
0
I I I I I I I I I
15000
15000
0
V
I I I I
5000
10000
10000
5000
0
15000
Figure 8.3. Drawdown (m) at 0.2 days for the 16 x 16 grid simulation of Part b.
8-11
-------
To assess the reason for the seemingly large error between analytic and numerical results, the
finite element code SEFTRAN (GeoTrans, 1988) was run for comparative purposes.
SEFTRAN allows usage of a backward and central difference scheme for approximation of
the time derivative. MODFLOW uses only a backward difference scheme. Drawdowns at
well 2 for each grid are shown in Table 8.6. Notice that for this particular problem the
numerical method (finite element or finite difference) does not seem as important as the
approximation of the time derivative in matching the analytical result.
Table 8.6. Comparison of drawdowns (m) at well 2 for various time derivatives and
spatial approximations (analytical = 1.63)
Grid
Finite-
Difference
Backward
Difference
Finite-
Element
Central
Difference
Finite-
Element
1
2
3
4
5
1.493
1.537
1.546
1.550
1.550
1.620
1.565
1.556
1.553
1.729
1.671
1.662
1.659
8-12
-------
PROBLEM 9
Calibration and Prediction
INTRODUCTION
Groundwater models are usually applied either to conceptualize and understand a
hydrologic system or to predict the outcome of a future change to the system. In order to
provide some assurance that the model reflects the behavior or appearance of the flow system,
it must be calibrated prior to use as a predictive tool. Calibration involves matching modeled
results to observed data. This usually includes hydraulic heads, drawdowns, induced
discharge and/or induced recharge. In the process of obtaining a match, aquifer parameters,
such as transmissivity, leakance, storage coefficient, or the attributes of boundary conditions
are adjusted within reasonable ranges until a satisfactory match is obtained. Once the
modeler is convinced that the model replicates current system behavior, and that it is capable
of replicating future behavior, it may be used in a predictive mode. This problem provides an
exercise in system conceptualization, a simple model calibration, and use as a predictive tool.
PROBLEM STATEMENT
The idealized flow system shown in Figure 9.1 is a small, confined aquifer which is
strongly controlled by the river which runs across it. The aquifer is approximately 100 ft
thick and is composed primarily of silly sand. The river is not in direct hydraulic connection
with the aquifer, but acts as a leaky boundary condition which can gain or lose water to the
aquifer. Other boundary conditions are no flow, which surround the square and define the
areal extent of the aquifer. Evapotranspiration and small domestic users in the area may be
neglected, although precipitation recharge is significant Stage data for the river as well as
river bed elevation determined in an earlier study are shown in Table 9.1.
Part a) Given constraints of uniform transmissivity and recharge, and additional data
below, obtain a steady state calibration (history match) based on the
potentiometric surface map of Figure 9.1 and the calibration targets shown in
Table 9.2.
grid size: 15 x 15
Ax = Ay: 500 ft
river base flow at western model boundary: 10 cfs
river base flow at eastern model boundary: 111/8 cfs
River bed conductance: 0.01 ftVs
Part b) A source of contamination has been discovered in the northeastern corner of the
aquifer. At the same time an industry is trying to gain permission to pump
groundwater from a well located at row 13, column 4 of the modeled area. What
is the maximum pumping rate that should be allowed to prevent the industry from
contaminating its own water supply?
9-1
-------
X
PROPOSED WELL LOCATION
RIVER LOCATION
Figure 9.1. Geometry and potentiometric surface of the aquifer system.
9-2
-------
Table 9.1. River data
Bottom
Elevation
Row
4
4
4
4
4
5
6
7
8
9
9
9
9
9
9
Column
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Stage (ft)
100.0
100.0
100.0
99.0
99.0
98.0
97.0
96.0
95.0
94.0
94.0
94.0
94.0
93.0
93.0
(ft)
90.0
90.0
90.0
89.0
89.0
88.0
86.0
86.0
85.0
84.0
84.0
84.0
84.0
83.0
83.0
Table 9.2. Calibration targets
Row
14
11
13
8
4
9
2
11
7
3.
2
Column Head (ft)
1 124.0
4 119.9
13 113.9
1 116.1
12 113.0
6 114.0
3 108.5
10 111.7
14 107.6
8 111.3
15 115.6
9-3
-------
MODEL INPUT
The following is a listing of data sets used in Part a.
*****•««»*•««******•******««««««•
* Basic package *
a********************************
CALIBRATION AND PREDICTION
3/15/90 PFA (PART A)
1 15 15 1 1
11 0 0 14 0 0 0 18 19 0 0 0
0 0
0 1
.00000
0 .100E+03
86400. 11.0000
*********************************
* Block Centered Flow Package *
*********************************
0 .100E+01
0 .500E+03
0 .500E+03
0 .100E-01
w^ w A ft wft ww wft ft ft ft ft ft ftftwft ft ft ft ft ft ft ft ft ft ft ft ftw
* River package "*
15
15
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
4
4
4
4
4
5
6
7
8
9
9
9
9
9
9
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
100.0000
100.0000
100.0000
99.0000
99.0000
98.0000
97.0000
96.0000
95.0000
94.0000
94.0000
94.0000
. 94.0000
93.0000
93.0000
.0100
.0100
.0100
.0100
.0100
.0100
.0100
.0100
.0100
.0100
.0100
.0100
.0100
.0100
.0100
90.0000
90.0000
90.0000
89.0000
89.0000
88.0000
87.0000
86.0000
85.0000
84.0000
84.0000
84.0000
84.0000
83.0000
83.0000
*********************************
*
Recharge package
*
*********************************
1
0
0
0
0 .200E-07
************
***************
SIP package
50 5
1.0000 .10000E-01 1.00000
headngd)
headng(2)
nlay.nrow.ncol,nper,itmuni
iunit array
iapart,istrt
ibound(locat,iconst)
hnoflo
shead(Iocat,cnstnt)
perlen.nstp.tsmult
iss.ibcfcb
Iayeon
trpy(locat,cnstnt)
delrdocat,cnstnt)
delc(locat,cnstnt)
tran(locat,cnstnt)
nurivr.jrivcb
itnp
layer, row,colon,stage,cond.rbot
layer, row,colum,stage,cond.rbot
layer,row,coltwn,stage,cond.rbot
layer, row,colum,stage,cond,rbot
layer, row,colum, stage,cond, root
layer,row,coltwn,stage,cond,rbot
layer, row,colum,stage,cond, rbot
Iayer,row,col inn,stage,cond,rbot
layer,row,column,stage,cond,rbot
Iayer,row,column,stage,cond,rbot
Iayer,row,coluran,stage,cond,rbot
layer,row,column,stage,cond,rbot
Iayer,row,column,stage,cond,rbot
layer,row,column,stage,cond,rbot
layer,row,colurcn.stage,cond,rbot
nrchop,i rchcb
inrech.inirch
rech•••*•«««••«««««•««•
* Well package *
a********************************
0
13
4-.4000
raxwell.iwelcb
itrnp
layer,roM,col,q
9-4
-------
MODEL OUTPUT
Hydraulic head arrays, mass balance summaries, and potentiometrie surface contour maps
for Parts a and b are given in Figures 9.2 and 9.3, respectively.
10
11
14
15
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
108.8
108.2
107.1
105. A
108.6
111.4
113.9
116,1
11S.1
119.8
121.3
122.5
123.4
124.0
124,3
108.9
108.3
107.1
105.3
108.4
111.2
113.6
115.9
ii?.«
119.6
121.0
122.3
123.2
123.8
124.1
109.1
108.S
10T.2
105,2
108,2
110.8
113.2
115.4
117,3
119.1
120.6
121.8
122.8
123.4
123.8
109.5
108.8
107.3
105.0
107,7
110.1
112.4
114.6
116.1
118.3
119.9
121.2
122.2
122.9
123.2
110.1
109.4
107.9
105.4
107.1
109.1
111.3
113.4
115.4
117.3
119,0
120.4
121.4
122.2
122.5
110.8
110,2
109.0
1Q7.3
105.S
107.5
109.7
111.9
114.0
116,0
117.8
119.3
120.5
121.3
121.7
111.7
111.2
110.2
108.8
107.2
105.3
107.6
109.9
112.2
114.4
116.4
118.1
119.5
120.3
120.8
112.6
112.1
111.3
110.1
108.5
106.8
104.9
107.4
109.9
112.5
114.9
116,9
118,3
119.3
W.8
113.4
113.0
112.2
111.1
109.7
108.0
106,1
104.2
107.1
110.4
113.3
115.5
117.2
118.3
118.9
114.1
113.8
113.0
111.9
110.5
108.8
107.0
105,1
103.5
108.3
111.7
114.3
116.1
117.3
117.9
114.7
114.4
113.6
112.5
111.1
109.4
107.4
105.1
102.4
107.0
110.5
113.2
115.2
116.5
117,1
111.2
114.9
114.1
113.0
111.5
109.7
107.6
105.0
102.0
106.2
109.7
112.5
114.5
115.8
116.4
115.6
115.2
114.5
113.3
111.8
109.9
107.6
104.9
101.7
105.7
109.2
111.9
113.9
11S.3
115.9
115.8
115.4
114.?
113,5
112.0
110.0
107.6
104,8
101.3
105.4
108.8
111.5
113.5
114.?
115.6
115.9
115.6
114.8
113.6
112.1
110.1
107.6
104.7
101.2
105.2
108.6
111.3
113.4
114.7
115.4
CUMULATIVE VOLUMES L**3
RATES FOR THIS TIME STEP
L**3/T
II:
STORAGE .00000
CONSTANT HEAD .00000
RECHARGE 97200.
RIVER LEAKAGE .00000
TOTAL ill 97200.
OUT:
STORAGE .00000
CONSTAHT HEAD .00000
RECHARGE .00000
RIVER LEAKAGE 96977.
TOTAL OUT 96977.
IK - OUT 222.98
PERCENT DISCREPANCY
IN:
STORAGE
CONSTANT HEAD
RECHARGE
RIVER LEAKAGE
TOTAL IN
OUT:
STORAGE
CONSTANT HEAD
RECHARGE
RIVER LEAKAGE
TOTAL OUT
IN • OUT
PERCENT DISCREPANCY
.00000
.00000
1.1250
.00000
1.1250
.00000
.00000
.00000
1.1224
1.1224
.2SB08E-02
Figure 9.2. Hydraulic head arrays, potentiometrie surface contour maps,
balance summary for Part a.
.23
and mass
9-5
-------
10
11
12
13
K
15
1
2
3
4
5
6
7
3
9
10
11
12
13
14
15
106.4
105.8
104.7
102.9
103.7
103.9
103.7
103.0
101.8
100.3
98.5
96.6
95.1
94.5
94.4
106.5
105.9
104.7
102.9
103.6
103.9
103.6
102.9
101.7
100.1
98.0
95.7
93.8
93.6
93.8
106.;
106.1
104.2
102.;
103.
103.
103.
102.
101.
99.
97.
93.
90.
91.
92.
' 107.1
106.4
1 105.0
» 102.8
103.5
103.6
103.4
102.8
101.6
99.8
? 96.8
P 91.8
2 81.0
r 89.8
7 92.5
107.6
106.9
105.5
103.2
103.3
103.4
103.2
102.7
101.8
100.3
98.1
95.0
91.6
93.3
94.5
108.3
107.7
106.5
104.8
102.7
102.9
102.9
102.6
102.0
101.0
99.7
98.1
96.7
96.9
97.2
109.1
108.6
107.6
106.1
104.2
102.1
102.3
102.3
102.1
101.7
101.0
100.3
99.8
99.7
99.8
109.9
109.5
108.5
107.2
105.6
103.6
101.3
101.6
101.8
102.0
102.0
101.9
101.8
101.9
101.9
110.7
110.3
109.5
108.2
106.7
104.8
102.8
100.5
101.1
102.0
102.6
103.0
103.3
103.5
103.6
111.4
111.0
110.2
109.1
107.6
105.8
103.8
101.7
99.7
101.7
103.0
103.8
104.4
104.7
104.9
112.0
111.6
110.9
109.8
108.3
106.5
104.5
102.1
99.5
101.7
103.2
104.4
105.1
105.6
105.8
112.5
112.1
111.4
110.3
108.8
107.0
104.9
102.4
99.5
101.7
103.5
104.8
105.7
106.2
106.5
112.9
112.5
111.8
110.7
109.2
107.3
105.1
102.5
99,5
101.8
103.6
105.0
106.0
106.7
107.0
113.1
112.8
112.0
110.9
109.4
107.5
105.2
102.5
99.3
101.8
103.7
105.2
106.3
107.0
107.3
113.3
112.9
112.1
111.0
109.5
107.6
105.3
102.5
99.3
101.8
103.8
105.3
106.4
107.1
107.5
CUMULATIVE VOLUMES LM3
IK:
STORAGE * .00000
CONSTANT HEAD • .00000
WELLS : .00000
RECHARGE * 97200.
RIVER LEAKAGE * .00000
TOTAL IN > 97200.
OUT:
STORAGE ' .00000
CONSTANT HEAD • .00000
WELLS ' 34S60.
RECHARGE = .00000
RIVER LEAKAGE * 62597.
TOTAL OUT = 97157.
IN - OUT - 43.211
PERCENT DISCREPANCY »
.04
RATES FOR THIS TIME STEP L*«3/T
IN:
STORAGE » .00000
CONSTANT HEAD ' .00000
WELLS » .00000
RECHARGE = 1.1250
RIVER LEAKAGE = .00000
TOTAL IN > 1.1250
OUT:
STORAGE ' .00000
CONSTANT HEAD - .00000
WELLS > .40000
RECHARGE * .00000
RIVER LEAKAGE » .72450
TOTAL OUT » 1.1245
IN - OUT * .50008E-03
PERCENT DISCREPANCY »
.04
Figure 9.3. Hydraulic head arrays, potentiometric surface contour maps, and mass
balance summary for Part b using pumpage of -0.4 ftYs.
9-6
-------
DISCUSSION OF RESULTS
The first step in this problem is to perform the steady state history match or calibration.
One could attempt to calibrate the model by trying various combinations of T and R until a
match was achieved. This would be costly, time consuming, and would not ensure that the
right combination of T and R had been used (see Problem 6),
The modeler should realize that the only discharge is to the river and the only source is
recharge. Therefore, to be in steady state, these two must balance. Recharge must therefore
equal 1.125 cfs (the river gain equals 11.125 cfs - 10 cfs). Spreading over the modeled area:
1.125 il/(15xl5)(500ft x 500ft) = 2xlQ-8 —
Since recharge is now known, we must calibrate by varying transmissivity. A first cut
estimate of transmissivity can be obtained by recognizing that flow to the river is known, as
is the gradient. Assuming that flow is slightly less from the northeastern corner, we can write
Darcy's law as:
q=kia
0.5ft Vs = k 8 (bXlength of river)
2500
kb = T - °-5(2500) . 0.021
(8X7500)
This first cut estimate will not match the steady state distribution. Further adjustment
yields T = 0.01 tf/s.
A trial and error procedure is used to compute the allowable discharge from the well. It
should be obvious that the answer must be somewhere between 0.0 cfs and 1.125 cfs. Figure
9.4 shows the results of an 0.1 cfs simulation, which hardly is noticeable. Figure 9.5 shows a
0.5 cfs simulation, where all flow is toward the well. Finally, using a discharge of 0.4 cfs, a
slight ridge forms near the river. These results are presented as the maximum allowable
discharge shown in Figure 9.3. Using an optimization package a maximum rate of 0.42 cfs
was obtained for this problem.
9-7
-------
This is a highly idealized problem where many assumptions have been made. Some of the
assumptions particular to this problem include:
1) The river discharge measurements are precise and do not change with time,
2) The system is in a steady state condition where heads and thus the magnitude and
location of the "ridge" do not change with time.
3) The river characteristics, conductance and stage, are precisely known.
4) The no-flow boundaries surrounding the model are true hydrologic features
(aquifer extent or pinchout) and do not change upon imposition of the stress.
Note that these assumptions would be violated in most practical situations. A "factor of
safety" has not been built in to the calculation of permissible withdrawal. A sensitivity
analysis would be required to assess parameter uncertainty and ramifications of modeling
assumptions. A more rigorous analysis than the one performed for this demonstration
problem would probably need to be conducted for a real world problem with similar
contamination potential.
9-8
-------
Figure 9.4. Potentiometric surface contour map for Part b using pumpage of -0.1 ft?/s.
9-9
-------
Figure 9.5. Potentiometric surface contour map for Part b using pumping of -0.5 frVs.
9-10
-------
PROBLEM 10
Transient Calibration
INTRODUCTION
Most modeling studies deal with a steady-state calibration such as the one performed in
the previous problem. It is often desirable and sometimes necessary to perform a transient
calibration. This problem gives an example of a transient calibration and cites a common
misapplication of the transient calibration process.
PROBLEM STATEMENT
A regional coastal area has been experiencing a drought for the past six months.
Hydrograph data (shown in Table 10.1) indicates that water levels have dropped as much as 5
ft in the unconfined aquifer since the drought began. Water resource officials are interested
in the amount of net recharge reduction that has occurred. A numerical model is being used
to assess the situation.
Because all flow is toward the coast, a simple one-dimensional model is being used. A
great deal of confidence exists in the specific yield value of 0.1 and the pre-drought recharge
rate of 20 in/yr. Hydraulic conductivity is assumed to be uniform within the aquifer and has
been estimated to be 850 ft/d. The aquifer base is uniformly at -120 ft. The model is a
single row of 15 nodes, each of which is 1 mile in length. The coastal boundary is simply a
constant head of 0.0 ft on the right side of the model (column 15) as shown in Figure 10.1.
Elsewhere, the nodes in the model are active. Pre-drought water levels which remained fairly
steady for a number of years are shown in Table 10.2.
Set up the model and determine the recharge rate reduction that has caused the observed
groundwater level decline at node (1,5) shown in Table 10.1.
10-1
-------
Table 10.1. Hydraulic head (ft) versus time (weeks after drought began) at an
observation well located at node (1, 5)
Week
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
Head (ft)
61.7
61.5
61.3
61.1
60.9
60.7
60.5
60.2
60.0
59.8
59.6
59.4
59.2
59.0
58.8
58.6
58.4
58.2
58.0
57.8
57.6
57.4
57.2
57.0
56.8
56.7
10-2
-------
123456789
10 ii 12 13 14. 15
( scale in mile )
0123
Constant Head Node
Figure 10.1. Grid and boundary conditions for coastal transient problem.
Table 10.2. Pre-drought ground water levels (ft) within the model domain
Node (column)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Head (ft)
67.24
66.71
65.75
64.06
61.91
59.19
55.86
51.91
47.27
41.89
35.70
28.59
20.42
10.99
0
10-3
-------
MODEL INPUT
The input files that correctly model the transient behavior at the observation well are
shown below.
*********************
* Basic package *
*********************************
INITIAL CONDITIONS PROBLEM (TRANSIENT)
6/28/91 PFA
1 1 15 1
11 000000 18 00 21 22
0 0
1 1(4012)
11111111111111-1
999.00
1 .100E+OK7G11.4)
67.24 66.71 65.65 64.06
51.91 47.27 41.89 35.70
.0000
183.00 261.0000
12
61.91
28.59
*********************************
***
0 0
0 .100i+01
0 .528E+04
0 .528E+04
0 -100E+00
0 .13QE+04
0 -.1206+03
Block Centered Flow Package
*********************************
* Recharge package *
*********************************
59.19
20.42
55.86
10.99
1 0
0 0
0 .1601-02
************ «*«**»*****»*«* ******
* SSOR package *
50
1.0000 .10000E-01
headngd)
headng(2)
nlay,nrow,ncol,nper,itmuni
ii*iit array
iapart,$strt
ibound
-------
**»***»»«»*
*
0
1
0
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
Output
0
1
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Control package *
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
ihedfm,iddnfm,
incode.ihddfl,
hdpr,ddpr,hdsv
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode,ihddfl,
incode.ihddfl.
incode.ihddfl.
incode.ihddfl,
Incode.ihddfl,
incode.ihddfl,
incode,ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl.
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
fncode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
ihedun,
ibudfl,
r.ddsv
ibudfl
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl
ibudfl,
ibudfl,
ibudfl
ibudfl,
ibudfl,
ibudfl
ibudfl
ibudfl
ibudfl
ibudfl
ibudfl
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl
ibudfl
iddnun
icbcfKstep 1)
icbcf Kstep 2)
icbcfKstep 3)
icbcf Kstep 4)
icbcfKstep 5}
icbcfKstep 6)
icbcfKstep 7)
icbcfKstep 8)
icbcfKstep 9)
icbcfKstep 10)
icbcfKstep 11)
icbcfKstep 12)
icbcfKstep 13)
icbcfKstep U)
icbcfKstep 15)
icbcfKstep 16)
icbcfKstep 17)
icbcfKstep 18)
icbcfKstep 19)
icbcfKstep 20)
icbcfKstep 21)
icbcfKstep 22)
icbcfKstep 23)
icbcf Kstep 24)
icbcfKstep 25)
icbcfKstep 26)
10-5
-------
DISCUSSION OF RESULTS
The analyst should have obtained a value of 7 in/yr (0.0016 ft/d) for recharge, a 13 in/yr
reduction from the pre-drought condition. A common error in model calibration was made if
a value of 0.44 in/yr (0.0001 ft/d) was obtained. Prior to running the transient simulation, the
modeler should have checked the reasonableness of the given parameters in obtaining the
initial conditions. This could be done by running a steady-state simulation with the pre-
drought recharge rate and checking the result with the water levels that "had remained fairly
steady for a number of years". If the 850 ft/d value for hydraulic conductivity were used, the
modeler would have noted the heads shown in Table 10.3, which are about 20 ft too high.
Because hydraulic conductivity was only an estimate, while other parameters had a fair
amount of confidence associated with them, hydraulic conductivity should have been adjusted.
A value of 1300 ft/d would have given the desired head distribution. The analyst would have
then derived 7 in/yr for recharge by simple trial and error after the hydraulic conductivity
adjustment was made.
Performing a transient calibration without a prior steady-state calibration or isolation of
the stress and response (superposition model) is a common mistake. In this example this
resulted in the transient response being a combination of seeking equilibrium with the given
hydraulic conductivity (rising water levels) and the response to the recharge reduction (falling
water levels). For this reason this recharge reduction had to be much higher than if the
proper hydraulic conductivity had been used. Franke et al. (1987) discuss this aspect of the
importance of initial conditions.
10-6
-------
Table 10.3. Groundwater levels resulting from a steady-state simulation using a
hydraulic conductivity of 850 ft/d
Node (column) Head (ft)
1 943
2 , 93.6
3 92.2
4 90.0
5 87.2
6 83.5
7 79.0
8 . 73.7
9 67.4
10 60.1
11 51.5
12 41.6
13 30.1
14 16.4
15 0
10-7
-------
PROBLEM 11
Representation of Aquitards
INTRODUCTION
In multiaquifer simulations, the modeler has to choose the most appropriate way of
representing confining beds that separate aquifers. Aquitards can be modeled implicitly as a
leakage term or explicitly as a separate model layer. This problem provides insight into
choosing the method of representing aquitards and how to choose the proper level of
discretization if the confining bed is modeled explicitly.
PROBLEM STATEMENT
A one-dimensional vertical leakage conceptual model will be evaluated; these principles
can be extended areally to three-dimensional applications. Two 50 ft thick aquifers are
separated by a 100 ft thick confining bed. At time t=0, the head in the lower aquifer is
instantaneously lowered to -10 ft. This is simulated using a constant head boundary condition
in the bottom aquifer. The head in the overlying aquifer is also held constant at a head of 0.0
ft. Different methods of representing the confining bed are evaluated, aquitard properties are
varied, and the magnitude of the lower boundary condition is changed. The properties of the
hydrologic system are shown below:
Aquifers
hydraulic conductivity ±= 2x1 G"5 ft/s
thickness = 50 ft
specific storage = lxlO"7/ft
Aquitard
hydraulic conductivity = IxlO"8 ft/s
thickness = 100 ft
specific storage = 5xlO*6/ft
Areal dimensions = Ax = Ay = 100 ft
Part a) Set up and run the model"such that the confining bed is represented as a
separate layer. A 3 layer, 1 row, 1 column model will therefore be set up.
You will need to compute VCONT between the aquifers and the aquitard using
equation 51 (page 5-13) from the MODFLOW documentation:
VCONT= *
AZ,/2 AZ/2 (11,1)
~K~ +~KT -
11-1
-------
Run the model for a simulation time of 1 year, broken into 25 time steps with
time step multiplier of 1.3. Plot hydraulic head in the confining bed and flux
into storage as a function of time.
Part b) Represent the confining bed as 3 separate layers of 25, 50, and 25 ft
thicknesses. Represent each of the aquifers by 2 layers of 25 ft thickness. A 7
layer, 1 row, 1 column model will therefore be used. VCONT and storage
coefficients will need to be recomputed. Run the model and plot results as you
did in part a.
Part c) Rerun part b, lowering the head in the lower aquifer to -100 ft. Compare the
response to that of part b.
Part d) Rerun part b, raising the hydraulic conductivity of the aquitard by a factor of 2,
and compare the response to part b.
Part e) Rerun part b, dividing the specific storage of the aquitard by a factor of 2.
11-2
-------
MODEL INPUT
The following is a listing of model input for part b.
*********************************
* Basic package *
*********************************
ONE DIMENSIONAL RESPONSE IN AN AOUIFER/AQUITARD SYSTEM
1/16/91 PFA
1 1 1
19 0 0 22
7
11 0 0
0
0
0
0
0
0
0
0
999.00
0
0
0
0
0
0
0
.31536e*08
1
0 0 0 0 I
0
-1
-1
1
1
1
-1
-1
.OOOE+00
.OOOE+00
.OQOE+00
.OOOE+00
.OOOE+00
-.100E+03
-.100E+03
251.3
000
*********************************
* Block Centered Flow Package *
1.0000
0
1 0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
50
«
0
0 0
.100E+01
.100E+03
.1001*03
.250E-05
.500E-03
.flOOE-06
.250E-05
.500E-03
.800E-09
.125E-03
.250E-06
.267E-09
.250E-03
-500E-06
.267E-09
.125E-03
.250E-06
.800E-09
.250E-05
.500E-Q3
.800E-06
.250E-05
.500E-03
*1
*
Al
5
10000E-01
*********************************
SIP package
1.00000
headngd)
headna(2)
nIay,nron,ncoI,nper,itmuni
iunit array
iapart,istrt
ibound layer 1( locat,iconst)
ibound layer 2(locat,iconst)
ibound layer 3Clocat,iconst)
ibound layer 4(locat,iconst}
ibound layer 5(locat,iconst)
ibound layer 6(locat,iconst)
ibound layer 7(locat,iconst)
hnoflo
shead layer 1(locat,cnstnt)
shead layer 2(locat,cnstnt)
shead layer 3
-------
*»***»»»***•**»***»***«*»**•*••*«
* Output Control package *
ft********************************
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
ihedfm,iddnfm,ihedun,iddnun
tncode,ihddfl,ibudfl,icbcfl(step 1)
hdfl.ddfl.hdsv.ddsv
incode,ihddfl.ihudft.fcbcfUstep 2)
ineode, ihddfl.ibudfl.icbcfUstep 3)
incode,ihddfl,ibudfl,icbcfl
-------
5.00 -i
0.00
* * * * * 3 layer model
x >< x x x 7 layer model
doubled k
T i i i i i 111 i r i i nil] T i i i i i 111
1 10 .100 1000
time (days)
Figure 11.1. Hydraulic head (ft) in the middle of the confining bed versus time for
cases a, b, and d.
11-5
-------
10000^1
1000-
H—
o
00
I
UJ
X
100"
10-
0.1
* >< x x x total leakage
i i i i i confining bed storage
1 10
time (days)
100
Figure 11.2. Total flux (ftVs) and storage flux versus time from the confining bed for
the seven layer model.
11-6
-------
Table 11,1. Hydraulic heads (ft) in the middle of the confining bed versus time for all
cases of Problem 11
Time
(days)
0.155
0.357
0.620
0.961
1.41
1.98
2.73
3.70
4.97
6.62
8.77
11.6
15.2
19.9
26.0
34.0
44.3
57.7
75.2.
97.9
127.5
165.9
215.8
280.7
365.
3 layer
model
(case a)
0.05
0.12
0.21
0.32
0.46
0.63
0.85
1.11
1.42
1.79
2.20
2.65
3.12
3.58
4.01
4.36
4.63
4.81
4.91
4.97
4.99
5.00
5.00
5.00
5.00
10 layer
model
(case b)
0.01
0.04
0.10
0.20
0.36
0.59
0.91
1.32
1.80
2.34
2.91
3.46
3.95
4.34
4.63
4.82
4.92
4.97
4.99
5.00
5.00
5.00
5.00
5.00
5.00
100ft
decline
(case c)
0.11
0.40
0.98
1.99
3.59
5.92
9.10
13.16
18.01
23.43
29.10
34.58
39.47
43.44
46.33
48,18
49.22
49.71
49,91
49,98
50.00
50.00
50.00
50.00
50.00
Kx2
(case d)
0.04
0.13
0.30
0.57
0.94
1.41
1.96
2.55
3.15
3.70
4.16
4.51
4.75
4.88
4.96
4.99
5.00
5.00
5.00
5.00
5.00
5.00
5.00
5.00
5.00
Ss/2
(case e)
0.04
0.13
0.31
0.57
0.94
1.41
1.96
2.55
3.15
3.70
4.16
4.51
4.75
4.89
4.96
4.99
5.00
5.00
5.00
5.00
5.00
5.00
5.00
5.00
5.00
11-7
-------
DISCUSSION OF RESULTS
There are three ways of representing an aquitard in multiaquifer simulations. The first
and simplest is the quasi-three-dimensional approach. In this situation, the aquitard is not
explicitly represented. It is simply incorporated as a leakage term (VCONT) between
adjacent aquifers. This effectively ignores storage within the confining bed and assumes an
instantaneous response in the unstressed aquifer. This case was not ran, but it should be clear
that the flux between the stressed and unstressed aquifers would be constant throughout time.
This is because the only source of water is the unstressed aquifer and the head difference
between the two aquifers does not change with time. Applying Darcy's law, the leakage
would be IxlO"5 cfs. This analysis may be appropriate for steady-state simulations or systems
with very thin confining beds with limited storage properties.
A second approach is to discretize the confining bed as a separate layer. This considers
the storage within the aquitard, but generally does not provide a good approximation of
gradient within the confining bed. As Figure 11.1 indicates, the response time is not as
accurate as with a finer gridded confining bed. Because the single layer does not approximate
the gradient well, leakage is also in error. This method is appropriate for approximations of
response time, if accuracy is not a prime consideration. Note from Figure 11.1 that the
storage factor only dictates the response time. Equilibrium heads are the same for all cases.
The third method involves several layers within the confining bed to approximate the
gradient. This provides a better estimate of the response time, although more gridding than
the 3 layers used in this example would probably provide a better approximation. The
modeler must weigh the benefits of including gridding in an area where there is probably
limited data and interest in hydraulic heads.
Bredehoeft and Finder (1970) presented an approximation of response time in multilayer
systems. Using equation 11.2, for dimensionless time of less than 0.1, response is entirely
from the aquitard, while at dimensionless time greater than 0,5, the aquitard is in equilibrium
and flux is from the unstressed aquifer.
K't
(11.2)
where tD is dimensionless time
K' is aquitard hydraulic conductivity
t is time
S/ is aquitard specific storage
b is aquitard thickness
For this problem, equation 11.2 indicates that all response should be from the aquitard
until an elapsed time of 5.8 days and the aquitard should be in equilibrium at 29 days. Figure
11.1 supports the 29 day time with the aquitard head at about 4.7 ft or 94% of its final value.
Figure 11.2 shows that the upper aquifer begins to contribute at approximately 3 days. This
slight discrepancy between 3 days and 5.8 days is probably due to grid discretization.
11-8
-------
Equation 11.2 shows that magnitude of response is not time related. This was
demonstrated in part c where the 100 ft decline generated a response 10 times greater than the
10 ft decline, but at the same time. The equation also indicates that response time is directly
proportional to hydraulic conductivity and inversely proportional to specific storage. For this
reason, doubling of hydraulic conductivity in part d generated identical answers as a halving
of specific storage in part e. Response time is cut in half for these two cases as illustrated in
Figure 11.1.
Equation 11.2 and the relationships presented in the problem should be useful for
designing model grids and determining necessity of vertical discretization. Note that the
intent of the model may influence greatly its final configuration: a steady- state multiaquifer
water supply model may not require discretization of aquitards; a transient model to assess
contaminant advection through several layers may require significant discretization.
11-9
-------
PROBLEM 12
Leaky Aquifers
INTRODUCTION
Large grids are often required to accurately model transient behavior of aquifers that are
adjacent to aquitards with significant storage properties. In this case a majority of discharge
from the aquifer may actually be obtained from confining bed storage with the aquifer serving
as merely a conduit to flow. This problem demonstrates the applicability of MODFLOW to
simulate leaky aquifer problems, presents an application of a large transient problem, and
provides a benchmark of MODFLOW with an analytic solution and another numerical model.
PROBLEM STATEMENT AND DATA
The modeled domain consists of a 50 ft thick aquifer that is overlain by a 50 ft thick
aquitard. A well fully penetrating the aquifer is pumped at a constant rate and drawdown is
noted at an observation well. The assumptions inherent in the Theis solution are all
applicable, except the assumption of total confinement. In order to minimize the total number
of nodes in the problem, only a quadrant of the entire domain is modeled. Aquifer
parameters and discretization data are given in Table 12.1.
12-1
-------
Table 12.1. Parameters and discretization data used in Problem 12
aquifer hydraulic conductivity, K
aquifer specific storage, Ss
aquifer thickness, b
aquitard hydraulic conductivity, K'
aquitard specific storage, Ss
well discharge, Q
length of stress period
number of time steps
time step multiplier
closure criterion
initial head
NROW = NCOL
DELX (n) = DELY (n)
top layer constant head, all others active
Part a vertical discretization (m)
Part b vertical discretization (m)
0.001 m/s
0.0001 m'1
50m
0.00001 m/s
0.0016 nv1
6.283 mVs (1.571 m3/s for quadrant)
787900 s
30
1.414
0.001 m
0
25
1, 1.5, 2, 3, 5, 8, 12, 18,25, 25, 34.8, 46.2,
69, 100, 150, 200, 250, 250, 250, 250,
250, 250, 250, 250, 300
48, 2, 2, 3, 4, 6, 9, 12, 14, 2
2 layers aquifer, 7 layers aquitard, 1 layer
upper aquifer
50,50,2
Part a) Using a fully three-dimensional grid with fine grid spacing in the aquitard (Table
12.1), set up and ran the model for the stress period length of 787900 s, 30
timesteps, and timestep multiplier of 1.414. Note drawdown versus time in the
pumped aquifer at an observation point 117.4 m from the pumping well. Compare
the results to the analytical solution shown in Table 12.2. Note that transmissivities,
storage coefficient, VCONTs, and -well discharges will have to be apportioned to
accommodate the grid spacing.
Part b) Using the coarse three-dimensional grid (3 layer model shown in Table 12.1), set up
and run the model for a stress period length of 787900 s, 30 time steps, and time
step multiplier of 1.414. Note drawdown versus time at an observation point 117.4
m from the pumping well. Compare the results to the analytical solution and part a.
12-2
-------
Table 12.2. Time versus drawdown (analytical solution) at distances of 117.4 m
Time (103 sec)
0.1689
0,2488
0.3619
0.5217
0.7476
1.067
1.519
2.158
3,061
4339
6.144
8.698
12.31
17.42
24.64
34.84
49.28
69.69
98.56
139.4
197.1
278.7
394.1
557.2
787.9
Drawdown (m) at r = 1 17.4
Analytical Solution
Short time solution
0.2838
0.7524
1.442
2.427
3.645
5.047
6.585
8.217
9.908
11.64
13.38
15.14
16.89
Long time solution
15,49
18.23
20.89
23.42
25.73
27.72
29.34
30.54
31.33
31.76
31.94
32.00
12-3
-------
MODEL INPUT
The following is a listing of the input file for part a.
*********************************
* Basic package *
*********************************
hantush verification problem
6/25/91 pfa
10
11 12 0
0
0
0
0
0
0
0
0
0
0
0
999.00
0
0
0
0
0
0
0
0
0
0
.78790E+06
25 25 1
00000 19 00 22
0
-1
1
1
1
1
1
1
1
1
1
.OOOE+00
.OOOE+00
.OOOE+00
.OOOE+00
.OOOE+00
.OOOE+00
.OOOE+00
.OOOE+00
.OOOE+00
.OOOE+00
301.4140
* Block Centered Flow
1
1 1i It it A A A irtt 1i ii
Package *
*********************************
0
00000
0
11
1.000
18.00
150.0
250.0
11
1.000
18.00
150.0
250.0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
00000
.100E+01
.100E+01(7G11.4)
1.500 2.000 3.000
25.00 25.00 34.80
200.0 250.0 250.0
250.0 250.0 300.0
.100E+OK7G11.4)
1.500 2.000 3.000
25.00 25.00 34.80
200.0 250.0 250.0
250.0 250.0 300.0
.200E-03
.200E-02
.1427E-5
.224E-01
.140E-03
.7692E-6
.192E-01
.120E-03
.7407E-6
.144E-01
.900E-04
.1333E-5
.960E-02
.600E-04
.200E-05
.640E-02
.400E-04
.2857E-5
12
5.000
46.20
250.0
12
5.000
46.20
250.0
8.000
69.00
250.0
8.000
69.00
250.0
12.00
100.0
250.0
12.00
100.0
250.0
headngd)
headng(2)
nlay,nrou,ncol,nper,itmuni
iunit array
iapart,istrt
ibound layer 1(locat,iconst)
ibound layer 2
-------
0
0
0
0
0
0
0
0
0
0
0
.480E-02
.300E-04
.4006-05
.320E-02
.20QE-Q4
.990E-05
.2001-03
.200E-02
.400E-04
.480E-02
.480E-01
sf1 layer 7{locat,cnstnt)
Iran layer 7(local,enstnt)
vcont layer 7-8(locat,cnstnl)
sfl layer SClocat.cnstnt)
tran layer 8{locat,enstnt5
vcont layer 8-9{local,cnstnl)
sf1 layer 9{local,cnstnt)
tran layer 9(local,cnstnt)
vcont layer 9-10(locat,cnslnl)
*f1 layer 10( local, cmlnt)
Iran layer 10
-------
For part b, the following data set is used.
*********************************
* Basic package *
*********************************
hantush verification problem part b
6/25/91 pfa
3
11 12 0
0
0
0
0
999.00
0
0
0
.78790E+06
25 25 1
00000 19 00 22
0
-1
1
1
.OOOE+00
.OOOE+00
.OOOE+00
1
301.4140
wwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww
* Block Centered Flow Package *
0
000
0
11
1.000
18.00
150.0
250.0
11
1.000
18.00
150.0
250.0
0
0
0
0
0
0
0
0
0
.100E+01
.100E+OK7G11.4)
1.500 2.000 3.000
25.00 25.00 34.80
200.0 250.0 250.0
250.0 250.0 300.0
.100E+OK7G11.4)
1.500 2.000 3.000
25.00 25.00 34.80
200.0 250.0 250.0
250.0 250.0 300.0
.200E-03
.200E-02
.400e-06
.800e-01
.500e-03
.396e-06
,500e-02
12
5.000
46.20
250.0
12
5.000
46.20
250.0
.500e-01
w w w ww w w w w w ww w w w ww w ww w w w w w w w w w w w w w
* Well package *
1
1
3
W W w w ww w w w w w w w w w w w w w w w w w w ww w w w w w w w
0
1 1-1.57075
8.000
69.00
250.0
8.000
69.00
250.0
12.00
100.0
250.0
12.00
100.0
250.0
*********************************
SIP package
50 5
1.0000 .10000E-02 1.00000
headngd)
headng(2)
nlay,nrow,ncol,nper,itmuni
iunit array
iapart,istrt
ibound layer 1(locat,iconst)
(bound layer 2(locat,iconst)
ibound layer 3(locat,iconst)
hnoflo
shead layer 1(locat,cnstnt)
shead layer 2(locat,cnstnt)
shead layer 3(locat,cnstnt)
perlen,nstp,tsmult
iss.ibcfcb
laycon array
trpy(locat,cnstnt)
delrdocat,cnstnt,fmtin, iprn)
delr array
delc(locat,cnstnt,fmtin,iprn)
dele array
sf1 layer 1(locat,cnstnt)
tran layer 1(locat,cnstnt)
vcont layer 1-2(locat,cnstnt
sf1 layer 2(locat,cnstnt)
tran layer 2(locat,cnstnt)
vcont layer 2-3(locat,cnstnt)
sf1 layer 3(locat,cnstnt)
tran layer 3(locat,cnstnt)
mxuell.iuelcb
itmp
layer,row,column,q
mxiter,nparm
accl,hclose,ipcalc,wseed,iprsip
12-6
-------
*********************************
* Output Control package *
0 0
1 1
9 0
9 0
9 0
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
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 -
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
ihedfm,iddnfm,
incode.ihddfl,
hdpr,ddpr,hdsv
hdpr,ddpr,hdsv
hdpr,ddpr,hdsv
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode,ihddfL,
incode.ihddfl,
incode.ihddfl,
incode,ihddfI,
incode,ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode,ihddfI,
incode.ihddfl,
incode,ihddfI,
incode,ihddfI,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl.
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode,ihddfI,
incode.ihddfl,
incode.ihddfl,
ihedun,iddnun
ibudfl.icbcfUstep 1)
,ddsv(layer 1)
,ddsv(layer 2)
,ddsv(layer 3)
ibudfl.icbcfUstep 2)
ibudfl.icbcfUstep 3)
ibudfl.icbcfUstep 4)
ibudfl.icbcfUstep 5)
ibudfl.icbcfUstep 6)
ibudfl.icbcfUstep 7)
ibudfl.icbcfUstep 8)
ibudfl.icbcfUstep 9)
ibudfl.icbcfUstep 10)
ibudfl.iebcfUstep 11}
ibudfl.icbcfUstep 12)
ibudfl.icbcfUstep 13)
ibudfl.icbcfUstep 145
ibudfl.icbcfUstep 15)
ibudfl.icbcfUstep 16)
ibudfl.icbcfUstep 17)
ibudfl.icbcfUstep 18}
ibudfl.icbcfUstep 19)
ibudfl.icbcfUstep 20)
ibudfl.icbcfUstep 21}
ibudfl.icbcfUstep 22}
ibudfl.icbcfUstep 23)
ibudfl.icbcfUstep 24)
ibudfl.icbcfUstep 25}
ibudfl.icbcfUstep 26}
ibudfl.icbcfUstep 27)
ibudfl.icbcfUstep 28}
ibudfl.icbcfUstep 29}
ibudfl.icbcfUstep 30}
MODEL OUTPUT
Table 12.3 compares results of the Han tush analytical solution to the 10 and 3 layer
MODFLOW simulation. Comparison is also made to SEFTRAN (GeoTrans, 1988) a finite
element model. This data is plotted in Figure 12.1.
12-7
-------
Table 12.3. Time versus drawdown at distances of 117.4 m for the analytical solution,
MODFLOW configuration, and SEFTRAN radial solution.
Time
(103 sec)
0.1689
0.2488
0.3619
0.5217
0.7476
1.067
1.519
2.158
3.061
4.339
6.144
8.698
12.31
17.42
24.64
34.84
49.28
69.69
98.56
139.4
197.1
278.7
394.1
557.2
787.9
Drawdown (m) at r = 1 17.4 m
Analytical 10 layer 3 layer
Solution MODFLOW MODFLOW
0.2838
0.7524
1.442
2.427
3.645
5.047
6.585
8.217
9.908
11.64
13.38
15.14
16.89
15.49
18.23
20.89
23.42
25.73
27.72
29.34
30.54
31.33
31.76
31.94
32.00
Short time solution
0.48
0.98
1.72
2.70
3.88
5.24
6.74
8.33
9.98
11.69
13.42
15.16
16.92
Long time solution
18.68
20.44
22.20
23.97
25.72
27.42
28.98
30.17
31.38
32.11
32.55
32.79
0.59
1.27
2.32
2.79
5.64
7.79
10.15
12.62
15.10
17.50
19.71
21.67
23.32
24.65
25.68
26.49
27.20
27.90
28.65
29.45
30.25
30.98
31.57
31.98
32.21
SEFTRAN
0.1266
0.4710
1.119
2.066
3.251
4.619
6.122
7.713
9.359
11.04
12.75
14.47
16.18
17.89
19.60
21.30
23.00
24.69
26.35
27.88
29.15
30.06
30.60
30.85
30.93
12-8
-------
10-
c
2
O
~D
Z
O
L
Q
1 ~
0. 1
x x
analytic
* * * * * SEFTRAN
x x x x x MODFLOW 3 I
+++++MODFLOW 10
a y er
layer
1 00
1 000 1 0000 1 00000 1 000000
T I me ( seconds)
Figure 12.1. Drawdown versus time for the analytical, MODFLOW, and SEFTRAN
simulations.
DISCUSSION OF RESULTS
The Hantush (1960) analytical solution to this problem is stated as two solutions, one for
early time and one for late time. The results of the 10-layer MODFLOW simulation
compares well with both short time and long time solutions. There is some apparent over-
prediction in early times when compared to the analytical solutions. The three-layer
MODFLOW simulation does not compare well in early-time, although there is excellent
agreement in late time. The poor early time comparison is because the level of vertical
discretization is not fine enough to approximate the steep gradient within the confining bed
during early time. As the gradient dissipates, the model is able to approximate leakage much
better. The configuration of a model will therefore depend upon when in time following a
12-9
-------
stress that accurate answers are desired. For a general purpose model, the fine discretization
is most accurate, but a price is paid in terms of number of nodes and execution time. A
10-layer finite element model called SEFTRAN (GeoTrans, 1988) was run with a similar
level of areal discretization. This provides comparison with another numerical model. Note
from Figure 12.1 that SEFTRAN under-predicts drawdown in early time. The answers
compare well with both the analytic and MODFLOW 10-layer results.
Output from MODFLOW is extremely voluminous for large problems such as this. In
order to keep the output file within reasonable size, the option to print only certain layers was
invoked in the OUTPUT CONTROL PACKAGE. As such, only layers 9 and 10 were
printed. Layer 10 was used for plotting results because it represents the bulk of the aquifer
and is away from the steep gradient near the aquifer-aquitard interface. An option to print
only observation nodes would be useful for applications such as these. This is not available
in MODFLOW, but is an easy modification to make.
12-10
-------
PROBLEM 13
Solution Techniques and Convergence
INTRODUCTION
Aquifer systems that are either very heterogeneous or that have complex boundary
conditions are generally difficult to model numerically. The choice of solution technique
parameters, or even the solution technique itself may govern whether the model will converge
and give reasonable results. The purpose of this exercise is to give the user some insight into
methods and parameter adjustments for making difficult solutions converge. MODFLOW
includes two iterative solution techniques: Strongly Implicit Procedure (SIP) and Slice
Successive Over Relaxation (SSOR). Both techniques will be utilized and adjustments to
iteration parameters will be made to achieve a solution.
PROBLEM STATEMENT AND DATA
The three-layer system shown in Figure 13.1 is bounded on its east side in layer 1 by a
specified head boundary set at 160 m. All other external boundaries are implicitly no-flow.
The aquifer receives recharge of 2.5xlO"10m/s on a portion of layer 1 (Zone 3), elsewhere
layer 1 is considered to be overlain by impermeable material. Grid spacing is uniform in the
horizontal, 15 columns by 10 rows, each block being 1000 m on a side. In the vertical, layer
1 is considered to be unconfmed, with a bottom elevation of 150 m. Layers 2 and 3 have
uniform thicknesses of 100 m and 50 m, respectively. For computing VCONT between layer
1 and layer 2, assume layer 1 is 20 m thick. Hydraulic conductivity zonation is shown in
Figure 13.1.
Set up the model and obtain a solution using the SIP solution technique for Part a and
using the SSOR technique for part b. Use a closure criterion of 0,01 and do not allow more
than 50 iterations for this steady-state problem.
13-1
-------
7 , 8 , 9 , 10 , 11 , 12 ,13 , 14 , 15
3|4!5|6t7[8[9, 10 | 11 ,12 113 r14 115
LAYER 2
a
3
4
LAYER 3 1
6
?
8
9
10
1|2|3|4i5;6i7 i 8 i 9 | 1Q..1.11 | 12 | 13 I 14 ) 1S
CONDUCTIVITY ZONES
ZONE 1 K = 10-s
ZONE 2 K s 2x10'"
ZONE 3 K = 1Q'»
ZONE 4 K = 10-11
Figure 13.1. Model geometry, boundary conditions, and hydraulic conductivity zonation
for Problem 13.
13-2
-------
MODEL INPUT
The following is a listing of data sets for Problem 13, part a.
*********************************
* Basic package *
*********************************
SOLUTION TECHNIQUES AND CONVERGENCE
0 0
PFA
11
1
1
1
1
1
1
1
1
1
1
999
3/7/90
3
00000
0
1
11111111
11111111
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
0
0
.00
10
0
0
15
18 19 0
1(4012)
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
0 .160E+03
0 .160E+03
0 .160E+03
86400.
11
.
0000
headngd)
headng(2)
nlay,nrou,ncol,nper,itiTuni
iunit array
iapart,istrt
ibouncK locat,iconst,fmtin.iprn)
ibound array (layer 1)
ibound layer 2(locat,iconst)
ibound layer 3(locat,iconst)
hnoflo
shead layer Klocat.cnstnt)
shead layer 2(locat,cnstnt)
shead layer 3(locat,enstnt)
perlen,nstp,tsmult
*********************************
* Block Centered Flow Package *
*********************************
1
0 0
0
0
0
11
.1000E-04
.1000E-04
.1000E-04
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-OS
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.1000E-05
.1000E-05
.2000E-OS
.1000E-05
.1000E-05
.2000E-05
0
.100E+01
.100E+04
.100E+04
.100E+OK7G11
.1000E-04 .
.1000E-04 .
.2000E-05 .
.2000E-05 .
.2000E-05 .
.2000E-05 .
.2000E-05 .
.2000E-05 .
.4)
1000E-04
1000E-04
2000E-05
2000E-05
2000E-05
2000E-05
2000E-05
2000E-05
.2000E-OS .2000E-05
.2000E-05 .
.2000E-05 .
.2000E-05 .
.2000E-05 .
.2000E-05 .
.2000E-05 .
.2000E-05 .
.1000E-05 .
.1000E-05 .
.1000E-05 .
.1000E-05 .
2000E-05
2000E-05
2000E-OS
2000E-05
2000E-05
2000E-05
2000E-05
1000E-05
1000E-05
1000E-05
1000E-05
.1000E-04
.1000E-04
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.1000E-05
.2000E-05
.1000E-05
.2000E-05
12
.1000E-04
.1000E-04
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.1000E-05
.2000E-05
.1000E-05
.2000E-05
.1000E-04
.1000E-04
.2000E-05
.2000E-05
.2000E-05
-2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.1000E-05
.2000E-05
.1000E-05
.2000E-05
iss.ibcfcb
1 aye on
trpy
delr( locat, cnstnt)
delc( locat , ens tnt)
k( locat, cnstnt , f mt in, iprn)
.1000E-04 k array (layer 1)
.1000E-04
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.2000E-05
.1000E-05
.2000E-05
.1000E-05
.2000E-05
13-3
-------
0
11
.1670E-06
.1670E-06
.1670E-06
.2000i-12
.2000E-12
-2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.20006-12
.2000E-12
.2000E-12
.2000E-12
-2000E-12
.2000E-12
.20006-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.20006-12
.16706-07
,16706-07
-2000E-12
.16706-07
.16706-07
.2000E-12
11
.10005-02
. 1000E-02
.1000E-02
.1000E-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.1000E-08
.1000E-08
.1000E-08
.1000E-08
.1000E-08
.10006-08
.1000E-08
. 1000E-08
.1000E-08
.1000E-08
.1000E-08
.1000E-08
.1000E-08
.1000E-08
.1000E-08
.1000E-03
.1000E-03
.1000E-08
.10001-03
. 1000E-03
.1000E-08
11
. 1330E-06
. 1330E-06
.13306-06
.20006-12
.2000E-12
.20006-12
.20006-12
.20006-12
.20006-12
.1506*03
.1006*01(7(511.4)
.16706-06
.16706-06
.20006-12
.20006-12
.2000E-12
.2000E-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.16706-07
.16706-07
.16706-07
.16706-07
.16706-06
.1670E-06
.20006-12
.20006-12
.20006-12
.20006-12
.2000E-12
.20006-12
.20006-12
.20006-12
.2000E-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.16706-07
.16706-07
.16706-07
.1670E-07
.16706-06
.16706-06
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.16706-07
.20006-12
.16706-07
.20006-12
.1006*01(7011.4)
.10006-02
.10006-02
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
-1000E-08
.1000E-08
.10006-08
.10006-08
.10006-08
.1000E-08
.1000E-08
.10006-08
.10006-03
.10006-03
.10006-03
.10006-03
,10006-02
.10006-02
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.1000E-08
.10006-08
.10006-08
.10006-08
.10006-03
.1000E-03
.10006-03
.10006-03
.10006-02
.10006-02
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-03
.10006-08
.10006-03
.10006-08
.1006*01(7011.4)
.13306-06
.13306-06
.20006-12
.20006-12
.20006-12
.20006-12
.13306-06
.13306-06
.20006-12
.20006-12
.20006-12
.20006-12
.13306-06
.13306-06
.20006-12
.20006-12
.20006-12
.20006-12
12
.16706-06
.16706-06
.20006-12
.20006-12
.20006*12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.16706-07
.20006-12
.16706-07
,20006-12
12
.10006-02
.10006-02
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-03
.10006-08
.10001-03
.10006-08
12
.13306-06
.13306-06
.20006-12
.20006-12
.20006-12
.20006-12
.16706-06
.16706-06
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.16706-07
.20006-12
.16706-07
.20006-12
.10006-02
.10006-02
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-03
.10006-08
.10006-03
.10006-08
.13306-06
.13306-06
.20006-12
.20006-12
.20006-12
.20006-12
.16706-06
.16706-06
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.20006-12
.16706-07
.20006-12
.16706-07
.20006-12
.10006-02
.10006-02
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-08
.10006-03
.10006-08
.10006-03
.10006-08
.13306-06
.13306-06
.20006-12
.20006-12
.20006-12
.20006-12
bot layer 1(loc«t,cnatnt)
vcont 1-2 (locat.cnatnt.firtin.iprn)
vcont array (layer 1-2)
trans (locat, crwtnt, fmtin, iprn)
trans array (l»y«r 2)
vcont 2-3
-------
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.1330E-07
.1330E-07
.2000E-12
.1330E-07
.1330E-07
.2000E-12
11
.5000E-03
.5000E-03
.5000E-03
.5000E-04
.5000E-04
.5000E-04
.5QOOE-04
.5000E-04
.SOOOE-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.500QE-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.2000E-12 .
.2000E-12 .
.2000E-12 .
.2000E-12 .
.2000E-12 .
.2000E-12 .
.2000E-12 .
.2000E-12 .
.2000E-12 .
.2000E-12 .
.1330E-07 .
.1330E-07 .
.1330E-07 .
.1330E-07 .
.100E+OK7G11
.5000E-03 .
.5000E-03 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
.5000E-04 .
2000E-12
2000E-12
2000E-12
2000E-12
2000E-12
2000E-12
2000E-12
2000E-12
2000E-12
2000E-12
1330E-07
1330E-07
1330E-07
1330E-07
.4)
5000E-03
5000E-03
5000E-04
5000E-04
SOOOE-04
5000E-04
5000E-04
5000E-04
5000E-04
5000E-04
5000E-04
5000E-04
5000E-04
5000E-04
5000E-04
5000E-04
5000E-04
5000E-04
5000E-04
5000E-04
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.1330E-07
.2000E-12
.1330E-07
.2000E-12
.5000E-03
.5000E-03
.5000E-04
.5000E-04
.SOOOE-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.SOOOE-04
.5000E-04
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.1330E-07
.2000E-12
.1330E-07
.2000E-12
12
.5000E-03
.5000E-03
.5000E-04
.5000E-04
.SOOOE-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.SOOOE-04
.5000E-04
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.1330E-07
.2000E-12
.1330E-07
.2000E-12
.5000E-03
.5000E-03
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.SOOOE-04
.5000E-04
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.2000E-12
'
.2000E-12
.2000E-12
.2000E-12
.2000E-12
.1330E-07
.2000E-12
.1330E-07
.2000E-12
transdocat, cnstnt, fmtin, iprn)
.5000E-03 trans array (layer 3)
.5000E-03
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.SOOOE-04
.5000E-04
.SOOOE-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.5000E-04
.SOOOE-04
.5000E-04
13-5
-------
1
0
18
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.2500E-09
.2500E-09
.0000
.2500E-09
.2500E-09
.0000
*********************************
* Recharge package *
*********************************
0
.100E+OK7G11.4)
.0000 .0000
.0000 .0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
12
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.2500E-09 .2500E-09 .2500E-09 .2500E-09
.2500E-09 .2500E-09 .0000 .0000
.2500E-09 .2500E-09 .2500E-09 .2500E-09
.2500E-09 .2500E-09 .0000 .0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.2500E-09
.0000
.2500E-09
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.2500E-09
.0000
.2500E-09
.0000
nrchop.irchcb
inrech.inirch
rechdocat.cnstntjfmtin.iprn)
rech array
1.0000
«««««**««»***«««*««>«»«>««*****
* SIP package *
*********************************
50 5
.10000E-01 1.00000E-00 1
mxiter.nparm
accl.hclose,ipcalc,wseed,iprsip
13-6
-------
For part b, the SSOR package shown below is invoked by setting IUN1T (21) to some finite
value and shutting off the SIP package by setting IUN1T (19) to 0.
* SSOR package
********************
50
1.7200 .10000E-01
mxiter
accl,hclose,iprsor
MODEL OUTPUT
For part a, a sensitivity analysis on the SIP seed and acceleration parameter was
conducted. The results are shown in Table 13.1. In part b, a sensitivity analysis on the
SSOR acceleration parameter was conducted. The results are shown in Table 13.2.
Table 13.1. Sensitivity analysis on SIP seed and acceleration parameter
Maximum head change
after x iterations
SEED
0.0009*
0.0001
0.005
0.00001
0.00005
0.0009*
0.0009*
0.0009*
0.0009*
0.0009*
ACCL
1.0
1.0
1.0
1.0
1.0
0.8
1.2
1.4
1.6
1.7
5
8.637
18.23
4.251
52.64
22.14
7.031 .
10.25
11.96
13.97
15.17
20
.1.607
0.975
1.527
-3.048
-0.730
1.804
1.401
1.195
-1.582
4.507
50
0.3456
0.0134
0.4036
-0.1268
—
0.3970
0.2821
0.2184
0.4613
8.643
Mass
balance
% error
22
0.23
55.93
-0.31
-0.02
34.22
14.13
8.92
5.18
0.28
Total
iterations
50
50
50
50
42
50
50
50
50
50
*model calculated seed
13-7
-------
Table 13.2. Sensitivity analysis on SSOR acceleration parameter
Maximum head change
after x iterations
ACCL
1.0
2.0
1.4
1.85
1.8
1.7
1.72
1
2.697
7.634
4.403
6.751
6.468
5.918
6.026
11
0.9858
51.97
1.505
18.70
13.26
6.761
7.720
50
0.2212
234.
0.1087
0.0748
~
—
—
Mass
balance %
error
62.35
-194.14
11.45
0.79
-0.11
0.28
0.04
Total
iterations
50
50
50
50
47
35
31
DISCUSSION OF RESULTS
This problem exhibits a large variation in hydraulic conductivities both horizontally and
vertically. Consequently, it is difficult to solve numerically. In part a, the initial run with
model calculated SIP seed and acceleration parameter of 1.0 does not converge in 50
iterations. A trial-and-error process of optimizing the seed is attempted. By trying order of
magnitude variations in the seed, a convergent solution is discovered fairly quickly. The
iteration history for seeds of 0.005 and 0.00005 is shown in Figure 13.2. Notice that
decreasing the seed induces an oscillation in early time, but makes the solution convergent.
A second alternative of keeping the model calculated seed, while varying the acceleration
parameter was also attempted. The results of this scheme were less promising; convergence
was not achieved although an acceptable mass balance was attained.
In part b the SSOR solution technique was used. A trial-and-error process was again used
to optimize the acceleration parameter (Table 13.2). A few trials were required to discover
that a convergent solution could be attained between 1.65 and 1.8. The iteration history for
acceleration parameters of 1.0, 1.85 and 1.72 is shown in Figure 13.3. Notice that the high
acceleration factor induces oscillation whereas the low acceleration factor causes an
asymptotic approach. The SSOR solution technique works fairly well for this problem
because most of the heterogeneity is in the vertical. Because SSOR makes a direct solution
to slices in the vertical, the heterogeneity is primarily solved for by direct methods.
13-8
-------
30 n
20-
O)
c
D 10
_C
u
"D
O
0
f— 0
-10-
seed=0.005
seed=0.00005
i T i i i i i I i I i II I r I iri|iiriririi[iiriTiii!]irTii FI i r |
0 10 20 30 40 50
iteration number
Figure 13.2. Iteration history for variations in the SIP seed parameter.
13-9
-------
20-i
15-
10-
Q)
C
O
"U
0
CD
_c
5-
0-
-5-
** * * accl=1.0
i ' ' ' accl=1.85
x x x x accl=1.72
1 U n i i i i i i i
0
10
20
30
40
50
iteration number
Figure 13.3. Iteration history for variations in the SSOR acceleration parameter.
13-10
-------
PROBLEM 14
Head-Dependent Boundary Conditions
INTRODUCTION
The RIVER, DRAIN, GENERAL HEAD, and EVAPOTRANSPIRATION packages of
MODFLOW are all head-dependent flux or third type boundary conditions. Although their
names imply specific types of sources or sinks, these packages are mathematically very
similar and can be used for a variety of hydrologic conditions other than those their names
suggest. This exercise illustrates the similarity of the packages, compares results of each to
one another as a verification, and gives insight to the utility of parameters used in the
packages.
PROBLEM STATEMENT AND DATA
In order to evaluate these boundary conditions, a single layer, 7 node by 7 node
unconfined aquifer is modeled in parts a-d. All cells in the domain are active and a well
pumps in the upper left-hand corner (node 1,1). A head-dependent flux boundary condition
runs along column 4 for the entire length of the system. The boundary will be treated in five
different ways in this exercise. Details on the model specific to all configurations are given
in Table 14.1.
Part a) Model the third type boundary condition as a river running down the center of
column 4. The river has the following characteristics:
Elevation = 0.0 ft
Width = 20 ft
Riverbed hydraulic conductivity = 0.1 ft/d
Riverbed thickness = 1 ft
River bottom elevation = -2.0 ft
Run the model for the 1 year simulation period described in Table 14.1. Note
hydraulic head and boundary discharge at row 1, column 4 for each time step.
You will need to invoke the cell-by-cell print flag in both the river package and
the output control package.
Part b) Model the third type boundary condition as a general head boundary running
down the center of column 4. The boundary has the following characteristics:
Elevation = 0.0 ft
Conductance = 200 ftVd
Run the model and note the results as you did in part a.
14-1
-------
Table 14.1. Aquifer parameters and discretization data for Problem 14
Initial head 10.0 ft
Hydraulic conductivity 10 ft/d
Aquifer base -50 ft
Storage coefficient 0.1
Grid spacing (uniform) 100 ft
Pumping rate 2500 ft3/d
Stress period length 365 days
Time steps 20
Time step multiplier 1.2
SIP iteration parameters 5
Maximum number of iterations 50
Acceleration parameter 1.0
Closure criterion 0.001
Part c) Model the third type boundary condition as a drain running down the center of
column 4. The drain has the following characteristics:
Elevation = 0.0 ft
Conductance = 200 ft2/d
Run the model and note the results as you did in parts a and b.
Part d) Model the third type boundary condition as a line of ET nodes running down
column 4. These nodes will have the following characteristics:
Maximum ET rate = 0.2 ft/d
Extinction depth = 10 ft
ET surface elevation = 10 ft
Run the model and note the results as you did in parts a-c. You"will need to store
the cell-by-cell ET rates for each time step and then run POSTMOD to put into an
ASCH form.
Part e) Model the system described above using a two-layer model. The top layer will be
the same as in parts a-d, except a third type boundary will not be explicitly
included. Instead, the bottom layer will represent the third type boundary
condition. The bottom layer will be inactive except along column 4, which will
be constant head of 0.0 ft. The bottom layer will be confined and have a
transmissivity of 100 ft2/d. Calculate a VCONT between layers 1 and 2 to give a
conductance of 200 ftVd. Run the model and note results as you did in parts a-d.
14-2
-------
MODEL INPUT
The following is a listing of the input data sets for part a.
*********************************
* Basic package *
*********************************
Third type boundary condition verification problem
2/6/91 PFA
17714
11 12 0 14 0 0 0 0 19 0 0 22
0 0
0 1
999.00
0 .100E+02
365.00 201.2000
*********************************
* Block Centered Flow Package *
*********************************
0 0
1
0 .100E+01
0 .100E+03
0 .100E+03
0 .100E+00
0 .100E+02
0 -.500E+02
*********************************
* Well package *
*********************************
0
1 1 -.250E+04
*********************************
* River package *
-1
1
2
3
4
5
6
7
4
4
4
4
4
4
4
.0000
.0000
.0000
.0000
.0000
.0000
.0000
200.0000
200.0000
200.0000
200.0000
200.0000
200.0000
200.0000
-2.0000
-2.0000
-2.0000
-2.0000
-2.0000
-2.0000
-2.0000
1.0000
*********************************
* SIP package *
*********************************
50 5
.10000E-02 1.00000 1
headngd)
headng<2)
nlay,nrou,ncol,nper,itinuni
iunit array
iapart,istrt
ibound
-------
*********************************
* Output Control package *
*********************************
ihedfm,iddnfm,
incode,ihddfI,
hdpr,ddpr,hdsv
incode.ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
incode,ihddfI,
ihedun,
ibudfl,
,ddsv
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
iddnun
icbcfKstep 1)
icbcfKstep 2)
icbcfKstep 3)
icbcfKstep 4)
icbcfKstep 5)
icbcfKstep 6)
icbcfKstep 7)
icbcfKstep 8)
icbcfKstep 9)
icbcfKstep 10)
icbcfKstep 11)
icbcfKstep 12)
icbcfKstep 13)
icbcfKstep 14)
icbcfKstep 15)
icbcfKstep 16)
icbcfKstep 17)
icbcfKstep 18)
icbcfKstep 19)
icbcfKstep 20)
14-4
-------
In part b, the RIVER package is substituted with the following GENERAL HEAD
package. The flag in the IUNIT array is changed from using the RIVER package to using the
GENERAL HEAD package.
7
7
1
*********************************
* General Head Boundary package *
*********************************
1
2
3
4
5
6
7
4 .000
4 .000
4 .000
4 .000
4 .000
4 .000
4 .000
.200E+03
.200E+03
.200E+03
.200E+03
.200E+03
.200E+03
.200E+03
mxbnd.ighbcb
itnp
layer,row,column,boundary.cond
In part c, the RIVER package is substituted with the following DRAIN package. The flag
in the lUNTT array is changed from using the RIVER package to using the DRAIN package.
• **»«O*******'O***O**«1>«*«I>******«
* Drain package *
*********************************
-1
1
2
3
4
5
6
7
.000
.000
.000
.000
.000
.000
.000
.200E+03
.200E+03
.200E+03
.200E+03
.200E+03
.200E+03
.200E+03
mxdrn.idrncb
itmp
layer,row,col.elevation,cond
In part d, the RIVER package is substituted with the following ET package. The flag in
the IUNIT array is changed from using the RIVER package to using the ET package.
*****•***•**••****•«**•**********
1
1
15
20.00
20.00
20.00
20.00
20.00
20.00
20.00
0
0
WWW
30
1
wwwwwwwww
1
.100E+OK7G11.4)
20.00
20.00
20.00
20.00
20.00
20.00
20.00
.200E+00
.100E+02
20.00
20.00
20.00
20.00
20.00
20.00
20.00
1
10.00
10.00
10.00
10.00
10.00
10.00
10.00
12
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
20.00
nevtop
insurf,inevtr,inexdp,inievt
surf(locat,cnstnt,fmtin,iprn)
surf array
evtr(locat,cnstnt)
exdp(locat.cnstnt)
14-5
-------
In part e, a two-layer model is constructed. The following BASIC and BCF files are used
in conjunction with the previously shown WELL, SIP and OUTPUT CONTROL packages.
*********************************
* Basic package *
third type boundary condition verification
2/6/91 PFA
27714
11 12 0 0 0 0 0 0 19 0 0 22
0 0
0 1
1 1(4012) 2
1000
000
000
000
000
000
000
000-
000-
000-
000-
000-
000-
999.00
000
365.00
1 0
.100E+02
.OOOE+00
201.2000
*********************************
* Block Centered Flow Package *
*********************************
•1
0 .100E+01
0 .100E+03
0 .100E+03
0 .1006+00
0 .100E+02
0 -.500E+02
0 .200E-01
0 .100E+00
0 .500E+03
headngd)
headng(2)
nlay,nrou,ncol,nper,itmuni
iunit array
iapart,istrt
ibound(locat,iconst)layer 1
ibound(locat,iconst,fmtin,iprn)layer
ibound array (layer 2)
hnoflo
shead(locat,cnstnt)layer 1
shead(locat,cnstnt)layer 2
perlen.nstp,tsmult
iss.ibcfcb
laycon(1,2)
trpydocat.cnstnt)
delr(locat.cnstnt)
delc(locat,cnstnt)
sfKlocat.cnstnt) layer 1
hy
-------
MODEL RESULTS
Table 14.2 shows hydraulic head versus time at node (1,4) for each of the five parts to
this problem. Table 14.3 shows discharge versus time at node (1,4). Hydraulic head versus
flow is plotted in Figure 14.1.
Table 14.2. Hydraulic head at node (1, 4) for each of the five methods of representing
the third type boundary condition
Time Step
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Ellapsed
Time
(days)
1.955
4.301
7.117
10.495
14.549
19.414
25.252
32.258
40.655
50.753
62.858
77.385
94.817
115.74
140.84
170.96
207.11
250.49
302.54
365.
Hydraulic Head
River GHB
(part a) (part b)
8.510
7.540
6,716
5.926
5.129
4.313
3.481
2.643
1.816
1.023
0.286
-0.371
-0.930
-1.382
- -1.727
-1.972
-2.175
-2.337
-2.448
-2.520
8.510
7.540
6.716
5.926
5.129
4.313
3.481
2.643
1.816
1.023
0.286
-0.371
-0.930
-1.382
-1.727
-1.972
-2.133
-2.231
-2.285
-2.311
(ft)
Drain
(partc)
8.510
7.540
6.716
5.926
5.129
4.313
3.481
2.643
1.816
1.023
0.286
-0.495
-1.440
-2.540
-3.843
-5.399
-7.264
-9.505
-12.20
-15.43
E-T
(partd)
8.510
7.540
6.716
5.926
5.129
4.313
3.481
2.643
1.816
1.023
0.286
-0.495
-1.440
-2.540
-3.843
-5.399
-7.264
-9.505
-12.20
-15.43
Constant
head
(pane)
8.510
7.540
6.716
5.926
5.129
4.313
3.481
2.643
1.816
1.023
0.286
-0.370
-0.930
-1.382
-1.727
-1.972
-2.133
-2.231
-2.285
-2.310
14-7
-------
Table 14.3. Discharge for each of the five methods of representing the third type
boundary condition
Time Step
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Elapsed
Time
(days)
1.955
4.301
7.117
10.495
14.549
19.414
25.252
32.258
40.655
50.753
62.858
77.385
94.817
115.74
140.84
170.96
207.11
250.49
302.54
365
Discharge (ft3/d)
River GHB Drain
(part a) (part b) (part c)
-1702.071
-1507.910
-1343.224
-1185.109
-1025.712
-862.600
-696.128
-528.506
-363.187
-204.534
-57.278
74.103
186.031
276.494
345.353
394.344
400.000
400.000
400.000
400.000
-1702.071
-1507.910
-1343.224
-1185.109
-1025.712
-862.600
-696.128
-528.506
-363.187
-204.534
-57.278
74.103
186.031
276.494
345.353
394.344
426.648
446.221
457.037
462.149
-1702.071
-1507.910
-1343.224
-1185.109
-1025.712
-862.600
-696.128
-528.506
-363.187
-204.534
-57.278
0
0
0
0
0
0
0
0
0
E-T
(part d)
-1702.071
-1507.910
-1343.224
-1185.109
-1025.712
-862.600
-696.128
-528.506
-363.187
-204.534
-57.278
0
0
0
0
0
0
0
0
0
Constant
head
(part e)
-1702.071
-1507.910
-1343.224
-1185.109
-1025.712
-862.600
-696.128
-528.506
-363.187
-204.534
-57.296
74.088
186.017 .
276.478
345.343
394.331
426.631
446.210
457.025
462.099
14-8
-------
~o
o
0)
X
5-
0-
-5-
-15-
-20-
AH
boundary
conditions
GENERAL HEAD
and constant
head boundary
RIVER
•DRAIN and
ET
TT—I 1 1 1 1 \ 1 1 1 I 1 1 1 1 1 1 1—T
•2000 -1500 -1000 -500 0
Discharge (cfd)
500
Figure 14.1. Hydraulic head (ft) versus flow rate (ft3/d) for each of the five methods of
representing the third type boundary condition.
DISCUSSION OF RESULTS
The head-dependent boundary conditions in MODFLOW are all very similar. They allow
leakage into or out of the system depending upon the difference in head between the aquifer
and some constant head external to the system. The amount of leakage is controlled by a
conductance term, which establishes the degree of hydraulic connection between the aquifer
and the external source/sink. This exercise demonstrates the equivalency of the boundaries
and highlights some of the differences.
14-9
-------
The general head boundary is, as the name implies, a general leakage boundary. Flux is
directly proportional to head difference for the entire range of saturated conditions. The drain
is essentially a one-way general head boundary. Flow can only be out of the aquifer.
Evapotranspiration is posed in terms of rates, but it is equivalent to a drain with an upper
limit on flux out of the aquifer. Flow cuts off at a certain depth and can only reach a certain
upper threshold. Finally, the river is a general-head boundary with an upper threshold of flux
into the aquifer. Flow can be into or out of the aquifer, but it can only inflow to an upper
limit.
The head-dependent flux boundaries can be used for other hydrologic conditions than their
names suggest. Table 14.4 shows some other uses for these boundaries.
The head-dependent flux boundaries are similar in behavior to a constant-head boundary
in an adjacent aquifer. In part d, a VCONT parameter of 200 ft2/d/(100 ft)2 = 0.02/d was
equivalent to a conductance and therefore gave similar answers as the general head boundary.
The excellent comparison of the head-dependent flux boundaries to one another provides
assurance that they are all implemented in the same fashion. The additional comparison to
the two-layer model with adjacent constant head nodes provides a further check that they are
implemented correctly in the model.
Table 14,4. Other uses for the head-dependent flux boundary conditions in MODFLOW
General Head
Boundary
Rivers
Drain
Intermittent streams
River
Adjacent aquifers
Evapotranspiration
Drains with
maximum flow
limitation
Exterior model Springs Welands
boundaries
Adjacent aquifers Ditches
14-10
-------
PROBLEM 15
Drains
INTRODUCTION
The drain package of MODFLOW is a third-type or head-dependent flux boundary
condition. This exercise demonstrates the utility of the package, provides guidance on
computing the conductance term, and compares this boundary to a more detailed
characterization of the drain.
PROBLEM STATEMENT AND DATA
This problem is a simple, one-dimensional flow system which is intersected by a drain.
As shown in Figure 15.1, the system is a 120 ft wide strip of a confined aquifer, 1200 ft long
with a potentiometric surface which slopes linearly from 10 ft at one end to 0 ft at the other.
The potentiometric surface is established by constant head cells at each end of the model
domain. A drain with an effective width of 4.44 ft is placed midway between the two
constant head nodes and covers the entire 120 ft strip. The head in the drain is 2.0 ft. A
range of conductance values for the drain will be tested and compared to a detailed
characterization of the drain as a specified-head condition. The aquifer is isotropic with
transmissivity of 100 ft2/d.
Part a) Set up a coarse-gridded model consisting of 1 layer, 1 row, and 11 columns of
120 ft length. Constant heads of 10 and 0 ft are placed at columns 1 and 11,
respectively. Compute the drain conductance as:
(15,1)
where:
C is conductance, L2/T
L is length of drain, L (120 ft)
W is width of drain, L (4.44 ft)
K. is hydraulic conductivity of material surrounding drain, LTT (varies)
M is thickness of material surrounding drain, L (1 ft).
Make several steady-state runs of the model, varying K from 0.0001 ft/d to 100
ft/d. Note hydraulic head in the block containing the drain and the drain flux rate
for each K.
Part b) Set up a fine-gridded model consisting of 6 layers, 9 rows, and 60 columns. Use
the column spacing shown in Table 15.1. Row spacing is uniform at 13.33 ft
15-1
-------
Layer spacing is 1 ft, 1 ft, 2 ft, 3 ft, 5 ft, and 8 ft, from top to bottom. Constant
heads are placed in all layers in columns 1 and 60 at values of 10 ft and 0 ft,
respectively. Two specified-head cells per row (in columns 30 and 31) are used
to model the drain. These are set at a head of 2.0 and a hydraulic conductivity of
1000 ft/d to approximate a gravel. A 1 ft-thick-filter layer surrounds the drain on
its side and base as shown in Figure 15.1. Note that the specified-head cells are
only in layer 1 while the filter is in layer 1 and 2. Run the model and obtain a
hydraulic conductivity for the filter layer which gives an equivalent flux as the 0.1
ft/d hydraulic conductivity used in part a.
15-2
-------
120'
CONSTANT HEAD =10 ft.
DRAIN ELEVATION = 2 ft.
CONSTANT HEAD = Oft.
(M
COARSE DRAIN MODEL
(CONSTANT HEAD)
FILTER MATERIAL
DRAIN DETAIL FOR FINE
GRIDDED MODEL
Figure 15.1. Model configuration for Problem 15.
-------
Table 15.1. Grid spacing used in the fine-gridded model (Part b)
Column
No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Width (ft)
40
40
40
40
40
30
30
30
30
24
24
24
24
24
20
20
20
20
20
20
13.33
13.33
6.67
6.67
6.67
5.11
4
2
1
1.22
Column
No.
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Width (ft)
1.22
1
2
4
5.11
6.67
6.67
6.67
13.33
13.33
20
20
20
20
20
20
24
24
24
24
24
30
30
30
30
40
40
40
40
40
15-4
-------
MODEL INPUT
The data set for part a with a K of 0.1 ft/d is shown below.
*********************************
* Basic package *
*********************************
DRAIN PROBLEM COARSE GRID
1/21/91 PFA
1 1 11 14
11 0 13 00000 19 000
0 0
1 1(4012) 2
-1111111111-1
999.00
1 .1QQE+OK7G11.4) 12
10.00 .0000 .0000 .0000 .0000
.0000 .0000 .0000 .0000
1.0000 11.0000
*********************************
* Block Centered Flow Package *
••» **««««««**«««•***«««*«««««»<«<
1 0
0
0 .100E+01
0 .12QE+03
0 .120E+Q3
0 .100E+Q3
*********************************
* Drain package *
*********************************
1
6 .2QQE+Q1 ,53281+02
*********************************
* SIP package *
*********************************
50 5
1.0000 .10000E-02 1.00000 1
headngd)
headng(2)
nlay,nrowfncol,nper,Itrauni
iunit array
iapart,fstrt
iboundClocat,iconst,fmtfn,iprn)
ibound array
hnoflo
sheadClocat,cnstnt, f nitIn,iprn)
.0000 .0000 shead array
perIen,nstp,tsmuIt
iss.ibefcb
Iayeon
trpy(locat,cnstnt)
delr(locat,cnstnt)
delc(locat,cnstnt)
transClocat,cnstnt)
mxdrn,idrncb
itmp
layer,row,col,elevation,cond
mxiter,nparra
accl,hclose,ipcalc,useed,iprsip
15-5
-------
The data set for part b with a K of 0.1 ft/d is shown below.
drain problem
6/27/91
11 0
-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
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
pfa
6
0 0
0
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 1
1 1
1 1
1 1
1 1
1 1
1 1
1 1
*********
*
********4
fine grid
9
0 0 0 0 19
0
1(4012)
111 11
111 11
111 11
111 11
111 11
111 11
111 11
1111 11
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
1111
1111
1111
1111
1111
1111
1111
1111
1111
(4012)
1111
1111
1111
1111
1111
1111
1111
1111
1111
11 1 1
1111
1111
1111
1 1 1
1 1 1
1 1 1
1 1 1
1111
1(4012)
1111
1111
1111
1111
1111
1111
1 1 1
1 1 1
1 1 1
1111 11
1111 11
1111 11
111 11
111 11
111 11
11 111
11 111
11 111
(4012)
11 1111
11 1111
11 111
11 111
11 111
11 111
11 111
11 111
1111111
1111111
Basic package *
** 1t1t1t^t1t1t1t W** 1t1rtt 1t1t1t4t1t1l1t1t 4t
60
0 0
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
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22
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4
1
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1-1
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1-1
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1111111111 1-1-1 111111111
I 1-1
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1-1
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1 -1
2
111111111111111111111
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111111111111111111111
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111111111111111111111
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111111111111111111111
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1111111111111111111111
1-1
15-6
headngd)
headng(2)
nlay,nroH,ncol,nper,itmuni
iunit array
iapart.istrt
i bound layer Klocat.iconst.fmtin, ipr
ibound layer 2(locat,iconst(fmtin,ipr
ibound layer 3(locat,iconst,fmtin,ipr
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1 .100E+OK7G11.4)
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1 .100E+OK7G11.4)
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*********************************
* Block Centered Flow Package *
12
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5.000
5.000
5.000
5.000
12
,000
.000
,000
.000
.000
,000
,000
5.000
5.000
5.000
000
000
000
000
000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
.000
.000
.000
.000
.000
.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
30.00
24.00
20.00
4.000
4.000
20.00
24.00
30.00
5.000
5.000
000
000
000
000
000
5.000
5.000
5.000
000
000
000
000
000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
•5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
30.00
24.00
13.33
2.000
5.110
20.00
24.00
40.00
.000
.000
.000
.000
.000
.000
5.000
5.000
5.000
5.000
5.000
000
000
000
000
5.000
.000
.000
.000
.000
.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
000
000
000
000
000
000
000
5.000
5.000
5.000
5.000
5.000
perlen.nstp.tsmult
iss.ibcfcb
laycon array
trpy
-------
0.100
5.000
5,000
5.000
5.000
S.OOO
S.OOO
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
S.OOO
5.000
5.000
11
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
S.OOO
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
S.OOO
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
5.000
0.100
5.000
S.OOO
.10001+04
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
.10001+04
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
.10006+04
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
.1QOOE+04
5.000
5.000
5.000
5.000
.10001+04
5.000
5.000
5.000
5.000
5.000
5.000 -
5.000
5.000
.1000E+04
5.000
5.000
5.000
5.000
S.OOO
5.000
5.000
5.000
.1000E+04
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
.1000E+04
5.000
S.OOO
5.000
5.000
0.100
5.000
S.OOO
5.000
S.OOO
S.OOO
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
0.100
5.000
5.000
S.OOO
5.000
S.OOO
S.OOO
5.000
5.000
0.100
5.000
5.000
5.000
5.000
.100E+OU7G11.4)
5.000
S.OOO
S.OOO
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
S.OOO
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
S.OOO
5.000
S.OOO
5.000
5.000
0.200
5.000
5.000
5.000
S.OOO
5.000
S.OOO
S.OOO
5.000
0.200
5.000
S.OOO
5.000
S.OOO
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
0.200
5.000
5.000
S.OOO
5.000
S.OOO
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
12
5.000
S.OOO
5.000
S.OOO
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
5.000
5.000
S.OOO
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
S.OOO
S.OOO
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5. 000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
,
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
S.OOO
5,000
5.000
5.000
' 5.000
5.000
5.000
S.OOO
5.000
5.000
S.OOO
5.000
S.OOO
5.000
5.000
5.000
S.OOO
S.OOO
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
vcont layer 1-2Clocat,cnstnt,fmtin,ip
vcont array
15-15
-------
5.000
5.000
5.000
5.000
5.000
5.000
0,100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
11
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.200
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
Q. 100
5.000
5.000
5.000
5.000
.100E+OK7G11.4)
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0,100
5.000
5.000
5.000
5.000
5.000
5.000
5,000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5,000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5,000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5,000
5.000
5.000
12
5.000
5.000
5,000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5,000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5,000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
tran layer 2(locat,cnstnt,fmtin,iprn)
tran array
15-16
-------
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
11
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
0.100
5.000
5.000
5.000
5.000
.100E+OK7G11.4)
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
, 3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
12
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
s: ooo
5.000
5.000
5.000
5.000
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
5.000
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
vcont layer 2-3(locat,cnstnt,fmtin,ip
vcont array
15-17
-------
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
0
0
0
0
0
0
0
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
.100E+02
.200E+01
.150E+02
.125E+01
.250E+02
.769E+00
.400E+02
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
0.192
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
3.330
tran layer 3(locat,cnstnt)
vcont layer 3-4(locat,cnstnt)
tran layer 4(locat,cnstnt)
vcont layer 4-5(locat,cnstnt)
tran layer S(locat,cnstnt)
vcont layer 5-6(locat,cnstnt)
tran layer 6(locat,cnstnt)
50 5
1.0000 .10000E-02
**************
SIP package *
****************************
0.00002
mxiter,nparm
accl.hclose,ipcalc.wseed,iprsip
*********************************
* Output Control package *
*********************************
ihedfm,iddnfm,ihedun,iddnun
incode,ihddfI,ibudfI,icbcfI
hdpr,ddpr,hdsv,ddsv
15-18
-------
MODEL OUTPUT
Hydraulic head at the drain node and drain flux is shown for part a in Table 15.2.
Table 15.2. Hydraulic head at the drain node (column 6) and drain flux for variations
in drain conductance (coarse model)
K
K
K
K
K
K
K
= 0.0001
= 0.001
= 0.01
= 0.1
= 1.0
= 10.0
= 100
Constant head
Hydraulic head
(ft)
5.00
4.96
4.65
3.29
2.21
2.02
2.00
2.00
K
K
K
K
K
K
K
= 0.0001
= 0.001
= 0.01
= 0.1
= 1.0
= 10.0
= 100
Constant head
Drain Flux
(ft3/d)
0.160
1.577
14.105
68.542
111.62
119.11
119.92
120.0
DISCUSSION OF RESULTS
This problem compares two methods of representing a drain. The first method, using the
MODFLOW DRAIN package, represents the drain as an external constant head which is
separated from the aquifer by a conductance term. Leakage from the aquifer is due to the
head difference between the aquifer and the external drain elevation and is controlled by the
user-specified conductance terra. Conductance is composed of a number of resistance or head
loss producing factors including: hydraulic conductivities in the vicinity of the drain,
thickness of material surrounding the drain, size of drain, and drain penetration into the
aquifer. Conductance is usually somewhat difficult to quantify because it is a combination of
so many factors. Equation 15.1 simplifies the conductance term by establishing a drain size
(width x length) and assuming that the resistance to flow into the drain is controlled by a
filter around the drain. It is further assumed that the properties of the filter can be quantified.
Regardless of quantification problems, the hydraulic conductivity becomes the variable
factor in equation 15.1. The sensitivity analysis for this problem (Table 15.2) shows that a
hydraulic conductivity value of 0.001 ft/d effectively shuts off discharge whereas a value of
10 ft/d causes a direct connection between the drain and aquifer.
The configuration of the fine-gridded model attempts to represent the drain as a set of
distinct, quantifiable hydraulic conductivity zones in the aquifer. The drain itself is
conceptualized as a gravel zone 2.44 ft wide and 1 ft deep. A wetted perimeter of 4.44 ft is
therefore modeled. The gravel is surrounded by a 1 ft-thick layer of porous material. The
15-19
-------
hydraulic conductivity of this material was varied in part b of this exercise to match a drain
discharge equivalent to the case where K was equal to 0,1 ft/d in the coarse-gridded drain
model, A six-layer aquifer is modeled in part b to further characterize the vertical gradients
near the drain.
The hydraulic conductivity which best matches the 68.55 ftVd discharge from the coarse
model is also 0.1 ft/d. The drain flux for this case is 70.76 ftVd. Note that the six-layer
model has difficulty in converging due to the large variations in hydraulic conductivity from
block to block. It is interesting how similar the fluxes generated by these two methods of
representing the drain are. Although some of this can be attributed to the linearity and
simplicity of the system as well as the thinness of the aquifer, it is apparent that the drain
package is a viable means of characterizing drains. The DRAIN package can be used
provided that hydraulic conductivity in the drain vicinity can be calculated. Note that the
drain behaves as a constant head if the material in the vicinity of the drain is of greater
conductivity than the aquifer. In many instances, the conductance is a calibrated parameter
that is determined as a part of the modeling exercise.
15-20
-------
PROBLEM 16
Evapotranspiration
INTRODUCTION
Evapotranspiration is a component of the water budget which is often subtracted from an
overall precipitation recharge rate prior to inclusion in the groundwater model This may be
physically appropriate, such as when the water table is sufficiently beneath the subsurface to
minimize the effect of evapotranspiration. In other instances, it is not implicitly included due
to data limitations. This problem illustrates the utility of the evapotranspiration module and
shows how excluding it from an analysis can affect the calibration and predictive capability of
the model. The problem also gives an example of how a well may "capture" water otherwise
lost to evapotranspiration.
PROBLEM STATEMENT AND DATA
The problem domain is a coastal environment covering a regional area of 90 square miles.
For the purposes of this analysis, the limestone aquifer extends approximately 8 miles inland
and ends abruptly. Groundwater flow lines define the northern and southern extent of the
model domain and form no-flow boundaries in those areas. There is some topographic relief
in the area, with land surface elevation changing from 0 ft at the coast to 18 ft in the
southwest corner of the domain. A uniformly spaced 20 row by 18 column, finite-difference
grid with 2640 ft spacing is used. Boundaries and topographic elevations are simplified as
shown in Figure 16,1. The aquifer is unconfined with base of -200 ft and hydraulic
conductivity of 1340 ft/d.
Part a) Run the model in a steady-state mode with the EVAPOTRANSPIRATION option
and parameters given below:
Maximum ET rate = 50 in/yr
ET extinction depth = 8 ft
ET surface = land surface array from Figure 16.1
Recharge = 25 in/yr
Closure criterion =0.01 ft
Save the output hydraulic heads for later use as an initial condition. Do this using
the Hdsv parameter in the Output Control package. Plot the potentiometric
surface.
Part b) Subtract the total evapotranspiration rate component in the mass balance of part a
from the recharge rate used in part a. Use this as a net uniformly distributed
recharge rate, eliminating the EVAPOTRANSPIRATION package. Run the
model, plot the potentiometric surface and compare to part a.
Part c) Using the results of part a as an initial condition, run the model with a well
pumping at row 4, column 5, at a rate of 535,000 ft3/d. Plot the steady-state
drawdown,
16-1
-------
Part d) Using the results of part a as an initial condition, run the model with a well
pumping at row 17, column 5, and a rate of 535,000 ft3/d. Plot the steady-state
drawdown.
WELL, PART C
CONSTANT
HEAD
WELL, PART D
Figure 16.1. Finite-difference grid, boundary conditions, and simplified topography for
Problem 16.
16-2
-------
MODEL INPUT
The following is a listing of the data set used for part a.
* Basic package *
*********************************
EVAPOTRANSPIRATION PROBLEM
1/18/91 PFA
11 0
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
99.00
1
0 0 15
0
1
1111
1111
1111
1111
1111
1111
1111
1111
1111
1111
1111
1111
1111
1111
1111
1111
1111
1111
1111
1111
20
0
0 18
18
19 0
0 22
0
1(4012)
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-
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
0 .OOOE+00
,0000
11.0000
*****•« *********** ****************
* Block Centered Flow Package
*********
0
*************************
.100E+Q1
,2646+04
.2641+04
.134E+04
.200E+03
*****************
* Evapotranspiration package *
*********************************
12
10.00
4.000
10.00
4.000
10.00
4.000
10.00
4.000
10,00
4.000
10.00
4.000
1
1
• 15
14.00
8.000
2.000
14.00
8.000
2.000
14.00
8.000
2.000
14.00
8.000
2.000
14.00
8.000
2.000
14.00
8.000
2.000
0
. 1
1
1
.10QE+01C7G11.4)
14.00
6.000
2.000
14.00
6.000
2.000
14.00
6.000
2.000
14.00
6.000
2.000
14.00
6.000
2.000
14.00
6.000
.0000
12.00
6.000
2.000
12.00
6,000
2.000
12.00
6.000
.0000
12.00
6.000
.0000
12.00
6.000
.0000
12.00
6.000
.0000
12.00
4.000
.0000
12.00
4.000
.0000
12.00
4.000
.0000
12.00
4.000
.0000
12.00
4.000
.0000
12.00
4.000
.0000
10.00
2.000
10.00
2.000
10.00
2.000
10.00
2.000
10.00
2.000
10.00
2.000
8.000
2.000
8.000
2.000
8.000
2.000
8.000
2.000
8.000
2.000
8.000
2.000
headngC1)
headng(2)
nlay.nrow.rvcol ,nper, i tmuni
iunit array
iapart,istrt
ibound(loeat,iconst,fmtin,iprn)
ibound array
hnoflo
shead(locat,cnstnt)
perLen,nstp,tsnult
iss.ibcfcb
laycon
trpy(locat.enstnt)
delr(local.cnstnt)
dele
-------
14.00
8.000
2.000
14.00
8.000
.0000
16.00
8.000
.0000
16.00
8.000
.0000
16.00
8.000
.0000
16.00
10.00
.0000
16.00
10.00
.0000
16.00
10.00
.0000
18.00
12.00
.0000
18.00
12.00
.0000
18.00
12.00
.0000
18.00
12.00
.0000
18.00
12.00
.0000
18.00
12.00
.0000
0
0
14.00
6.000
.0000
14.00
6.000
.0000
14.00
6.000
.0000
16.00
6.000
.0000
16.00
8.000
.0000
16.00
8.000
.0000
16.00
8.000
.0000
16.00
10.00
.0000
18.00
10.00
.0000
18.00
10.00
.0000
18.00
10.00
.0000
18.00
10.00
.0000
18.00
10.00
.0000
18.00
10.00
.0000
.114E-01
.8001+01
12.00
6.000
.0000
14.00
6.000
.0000
14.00
6.000
.0000
14.00
6.000
.0000
16.00
6.000
.0000
16.00
6.000
.0000
16.00
8.000
.0000
16.00
8.000
.0000
18.00
8.000
.0000
18.00
8.000
.0000
18.00
8.000
.0000
18.00
8.000
.0000
18.00
8.000
.0000
18.00
8.000
.0000
12.00
4.000
.0000
12.00
4.000
.0000
12.00
4.000
.0000
14.00
4.000
.0000
14.00
4.000
.0000
14.00
6.000
.0000
16.00
6.000
.0000
16.00
6.000
.0000
16.00
6.000
.0000
16.00
6.000
.0000
16.00
6.000
.0000
16.00
6.000
.0000
16.00
6.000
.0000
16.00
6.000
.0000
10.00
4.000
10.00
4.000
12.00
4.000
12.00
4.000
12.00
4.000
14.00
4.000
14.00
4.000
14.00
4.000
16.00
4.000
16.00
4.000
16.00
4.000
16.00
4.000
16.00
4.000
16.00
4.000
10.00
2.000
10.00
2.000
10.00
2.000
10.00
2.000
12.00
2.000
12.00
2.000
12.00
2.000
14.00
2.000
14.00
2.000
14.00
2.000
16.00
2.000
16.00
2.000
16.00
2.000
16.00
2.000
8.000
2.000
8.000
2.000
8.000
2.000
10.00
2.000
10.00
2.000
10.00
.0000
12.00
.0000
12.00
.0000
12.00
.0000
14.00
.0000
14.00
,.0000
14.00
.0000
14.00
.0000
14.00
.0000
*********************************
* Recharge package *
*********************************
0
0
.571E-02
1.0000
*•••«*««***«««*•••*•***•*•**•****
* SIP package - *
««••••*•*•*••**••*•****«***«•••*«
50 5
.10000E-01 1.00000 1
*********************************
* Output Control package *
*********************************
9
0
1
30
1
1
evtr(locat,cnstnt)
exdp(Ioca t,cnstnt 5
nrchop,irchcb
inrech.infrch
rech(locat,cnstnt)
mxiter,nparm
accl,hclose,ipcal.c,wseeci, iprsip
ihedfm,iddnfm,ihedun,iddnun
incode,ihddfI,ibudfI,icbcfI
hdpr,ddpr,hdsv,ddsv
16-4
-------
In part b, the evapotranspiration package is turned off by setting IUNIT(5) to 0 in the
BASIC package. The recharge rate (RECH) is changed to 0.001214 ft/d in the RECHARGE
package.
The following is a partial listing of the part c data set. Note that the initial heads
(SHEAD) are the unstressed steady-state results from part a and are read from an external
binary file (unit 32) in the BASIC package. A negative unit number directs the model to read
an unformatted file. The other packages, BCF, ET, RECHARGE, SIP, and OUTPUT
CONTROL are identical to part a, and are not shown here.
*********************************
* Basic package *
*********************************
EVAPOTRANSPIRATION PROBLEM
1/18/91 PFA
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
999.00
-32 .100E+01
1.0000 11.0000
1 20 18
2 0 0 15 0 0 18 19 0 0 22
0 1
1 1(4012)
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
1111
1111
1111
1111
1111
1111
1111
1111
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
11111-1
1111-1
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
1111 1-1-1-1- -1
1111111111 1-1-1-1- -1
*********************************
* Well package *
*********************************
0
4 5 -.535E+06
headngd)
headng(2)
nlay.nrow.ncol.nper, it muni
iunit array
iapart,istrt
ibound(local,(const,fmtin,iprn)
ibound array
hnoflo
shead(Iocat,cnstnt)
perlen,nstp,tsmult
mxwell,iuelcb
itirp
layer,row,col,q
The data set for part d is identical to part c, except the row location for the well (WELL
Package) is changed to 17.
16-5
-------
MODEL OUTPUT
Figures 16.2 and 16.3 are the potentiometric surface for part a and b, respectively. Figures
16.4 and 16.5 are drawdown plots for parts c and d, respectively.
52800
10560
21120
31680
42240 -
31680 -
21120 -
10560 -
42240
i i i i i i
i i i i i t i i
52800
- 42240
- 31680
- 21120
- 10560
10560
21120
31680
42240
Figure 16.2. Potentiometric surface (ft) for Problem 16, Part a.
16-6
-------
52800
0
10560 21120 31680 42240
42240
31680
21120
10560
0
1 I I
I i I
I i I i
I I I t I I l
52800
42240
31680
21120
10560
0 10S60 21120 31680 42240
Figure 16.3. Potentiometric surface (ft) for Problem 16, Part b (net recharge).
0
16-7
-------
52800
10560
21120 31680
42240
42240
31680
21120
10560
0
i i i i [ I
1 I I I 1 I T
52800
42240
31680
21120
10560
i ii LI ill i i i i i i i ii
0
0 10560 21120 31680 42240
Figure 16.4, Drawdown(ft) map for Problem 16, Part c.
16-8
-------
52800
10560 21120 31680 42240
42240 -
31680 -
21120 -
10560
i i i i i i i i i i i i i i i i i
i i i i i
52800
- 42240
- 31680
- 21120
- 10560
0 10560 21120 31680 42240
Figure 16.5. Drawdown (ft) map for Problem 16, Part d.
16-9
-------
DISCUSSION OF RESULTS
The steady-state results for this problem indicate that the system is dominated by
evapotranspiration. The only source is precipitation recharge and 79 percent of this is
discharged by evapotranspiration. The remainder (21 percent) discharges to the sea. There is
quite a variation in net recharge areally across the system, as shown in Figure 16.6.
Basically, the lower left corner of the model area is a recharge area because land surface
elevation is much greater than the water-table elevation. Toward the northeast, recharge
becomes progressively less until it reaches 0 where the water table is 4 ft below land surface.
This occurs when
4 ft
ET = _ x 50 in/yr = 25 in/yr and recharge = 25 in/yr.
8 ft J
North and east of this line is all net discharge where ET is greater than 25 in/yr. Notice that
the water-table elevation is a subdued representation of the topography.
In part b, only the net recharge rate (0.001214 ft/d) is applied evenly across the region. As
expected from Figure 16.6, this approach is entirely inappropriate because variations in
recharge and discharge areas are ignored. Overall, heads are much lower than in part a and
water levels tend to follow the coastal boundary.
In part c, a pumping well is placed in the northwest corner of the model, and the model is
run to steady state. A cone of depression develops around the well, with the 0.2 ft contour
extending less than half the north-south distance of the model. The maximum drawdown at
the well node is 1.19 ft. In part d, a well is placed in the southwest corner of the model.
This well pumps at the same rate as the part c well, is in the same column as in part c, and is
an equivalent distance from the southern boundary as the part c well was from the northern
boundary. Intuitively, the drawdowns in parts c and d should be very similar. Comparison of
the results of these two simulations show significant differences, however. The 0.2 ft contour
of part d extends greater than half the north-south distance, and the maximum drawdown at
the well is 1.57 ft. The reason for the discrepancy is due to the recharge-discharge
relationship in the aquifer.
In part c (Figure 16.4), the well is located in an area where the water table is close to land
surface and evapotranspiration is occurring. When the well is turned on, less discharge
occurs as evapotranspiration because the water table is now drawn down. Although the well
is a new discharge from the system, the previous discharge from evapotranspiration is
reduced. Therefore, the system does not see the full impact of the discharging well. In part
d, however, the well is placed in an area where evapotranspiration is not significant. The
well responds with greater drawdown (Figure 16.5) because discharge from evapotranspiration
is not significantly reduced. The system in this case sees the full impact of the well.
16-10
-------
52SOO
42240
31680
21120
10560
10560
1 I
RECHARGE
AREA
257yr.
RECHARGE
21120
31630
42240
DISCHARGE
AREA
CONSTANT
HEAD
I I I L
52800
42240
31680
21120
10560
10560
21120
31680
42240
Figure 16.6. Net recharge rates (in/yr) for the steady-state, non-pumping scenario
(Part a)
16-11
-------
An error exists with using the non-stressed conceptual model as a base for the stressed
simulations. Recall that the northern and southern boundaries were no-flow because they
were flow lines. When the aquifer is stressed, this approximation is invalidated. For the
purposes of this analysis, this error does not change the conclusions previously stated,
however it does highlight the fact that the modelers should always be aware of the
assumptions inherent to the original model before proceeding to predictive simulations.
This exercise highlights the importance of including evapotranspiration in simulations
where it is an important component of the water budget and causes natural variation in
recharge and discharge areas. Attempting to calibrate a groundwater model with a uniform
net areal recharge rate, as is often done, would be inappropriate in this situation. This
exercise also illustrates an interesting consideration for well siting.
As a check of the MODFLOW results, the hydraulic heads for part a were compared to a
similar simulation using the FTWORK (Faust et al, 1989) code. As shown in Table 16.1, the
results of the codes are nearly identical.
Table 16.1. Comparison of hydraulic heads (ft) along row 10 for MODFLOW and
FTWORK
Column
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
MODFLOW
7.41
7,27
6.99
6.59
6.08
5.49
4.84
4.15
3.43
2.73
2.03
1.39
0.80
0.33
0.00
FTWORK
7.44
7.30
7.02
6.62
6.10
5,50
4.85
4.15
3.43
2.73
2.03
1.39
0.80
0.33
0.00
16-12
-------
PROBLEM 17
Wells
INTRODUCTION
The WELL package of MODFLOW allows the user to specify withdrawal or injection
from the modeled area. Wells are assumed to be placed at the center of grid blocks and to
fully penetrate the layer for which they are specified. This problem examines situations of
partially penetrating wells and multiple aquifer wells. A commonly used rule of thumb is
assessed numerically.
PROBLEM STATEMENT AND DATA
The model domain is essentially the same as that used for the fmely-gridded quadrant of
the Theis problem (problem Ic). Instead of the fully penetrating well assumed for problem 1,
a partially penetrating well will be analyzed as a part of this problem. In addition, a stratified
aquifer with a well fully penetrating 2 layers of varying transmissivity will be assessed. A
second layer will therefore be required to model these features. Table 17.1 is a listing the
physical parameters and discretization data used in the model.
Part a) Re-run the single layer model used in problem Ic for comparison purposes.
Part b) A well which penetrates only the upper 50% of the model domain is required.
Because all wells in MODFLOW are assumed to fully penetrate a model layer,
the system will be split into two layers of equal thickness and the well specified
for the upper layer. Set up the two layer model Apportion transmissivity and
storage coefficient evenly between the two layers. Assume the entire aquifer
thickness is 20 m (10 m per layer) and calculate a VCONT based on an isotropic
hydraulic conductivity. Run the model and compare the distance-drawdown
relationship at time = 50938 s to the distance drawdown relationship at the same
time for the fully penetrating case.
Part c) Assume the thickness of the aquifer is 40 m (20 m per layer) with the same
transmissivities and storage coefficients as part b. VCONT is therefore the only
parameter which must be recalculated and input to the model. Run and compare
distance-drawdown at time - 50938 s to parts a and b.
Part d) Assume that the aquifer system is stratified as 2 layers. The top layer is 10 m in
thickness with a transmissivity of 0.002 nr/s and the bottom layer is 30 m thick
with a transmissivity of 0.0003 nr/s. Storage coefficient is the same as in part a
and is distributed based on thickness (equivalent specific storages are used). Note
that the net storage coefficient and transmissivity are consistent for parts a-d.
Recompute VCONT and input to the model.
A fully penetrating well will be used in this application. Because of differences in
thickness and hydraulic conductivity of the units, discharges from each layer must be
scaled in some fashion. A common method is to use a weighted average:
17-1
-------
. QT
T
where: QN is the well discharge from layer N
QT is the total well discharge
TN is the transmissivity of layer N
TT is the total transmissivity.
Using the same discharge as in parts a-c, apportion flux to the wells. Run the model and
compare the distance drawdown relationships at time = 50938 s for the two aquifers and
to the one-layer simulation.
17-2
-------
Table 17.1. Parameters and discretization used in Problem 17
Initial head
Transmissivity
Storage coefficient
Pumping rate
Final time
Number of time steps
Time step expansion factor
SIP iteration parameters
Clousre criterion
Maximum number of iterations
Number of rows, columns
Number of layers
Grid spacing (m):
Row number , i
(=column number, j)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
0.0 m
0.0023 m2/s
0.00075
4 x 10"3 m3/s (1 x 10"3 m3/s for quandrant)
86400s
20
1.3
5
0.0001
50
19
2
DELC (i)
(=DELR(j))
1
1.143
2
2.83
4
5.65
8
11.3
12
14.62
20
28.3
40
56.5
80
110
150
200
252.89
17-3
-------
MODEL INPUT
The data set for part b is shown below.
*********************************
* Basic package *
*********************************
partially penetrating well problem quadrant fine spacing
6/25/91 pfa
1 1
11 12
999.00
86400.
0 0
2
0
0
0
0
0
0
19
0 0 0 0 0 19
1
1
1
.OOOE+00
.OOOE+00
201 .3000
19
0
0 22
*********************************
* Block Centered Flow Package *
*********************************
1.00
11.30
80.00
1.00
11.30
80.00
0
11
11
0
0
0
c
0
.100E+01
.100E+OU7G11.4)
1.41
12.00
110.0
2.00
14.62
150.0
2.83
20.00
200.0
.100E+OU7G11.4)
1.41
12.00
110.0
.375E-03
.115E-02
.115E-04
.375E-03
.115E-02
2.00
14.62
150.0
2.83
20.00
200.0
12
4.00
28.30
252.89
12
4.00
28.30
252.89
5.65
40.00
5.65
40.00
8.00
56.50
8.00
56.50
**************************1
* Well package *
*********************************
0
1
1 -.100E-02
1.0000
****************
* SIP package *
*********************************
50 5
.10000E-03 1.00000 1
headngd)
headng(2)
nlay,nrow,ncol,nper,itmuni
iunit array
iapart,istrt
ibound layer Klocat, iconst)
ibound layer 2(locat,iconst)
hnoflo
shead layer 1(locat.cnstnt)
shead layer 2(locat,cnstnt)
perlen.nstp.tsmult
iss.ibcfcb
laycon
trpy
-------
*********************************
* Output Control package *
*********************************
10
0
0
-1
-1
-1
-1
-1
-1
-1
-1
-1
10
ihedfm,iddnfm,
incode.ihddfl,
hdfl.ddfl.hdsv
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl.
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl,
incode.ihddfl.
ihedun,
ibudfl,
,ddsv
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
ibudfl,
iddnun
icbcfKstep 1)
icbcfKstep 2)
icbcfKstep 3)
icbcf Ustep 4)
icbcf Ustep 5)
icbcfKstep 6)
icbcfKstep 7)
icbcfKstep 8)
icbcfKstep 9)
icbcfKstep 10)
icbcfKstep 11)
icbcfKstep 12)
icbcfKstep 13)
icbcfKstep 14)
icbcfKstep 15)
icbcfKstep 16)
icbcfKstep 17)
icbcfKstep 18)
icbcfKstep 19)
icbcfKstep 20)
17-5
-------
The part a data set was described in problem 1. In part c, the part b data set is modified
by changing VCONT to 0.2875 E-5/s. In part d the following parameters are used:
T, = 0.002 m2/s S, = 1.875 E-4
T2 = 0.0003 m2/s S2 = 5.625 E-4
VCONT = 6.557 E-7/s
Q! = 8.696 E-4 m3/s
Q2 = 1.304 E-4 m3/s
MODEL OUTPUT
Drawdown versus distance at time = 50938 s is tabulated for parts a-d in Table 17.2.
These results are plotted in Figures 17.1 and 17.2.
Table 17.2. Drawdown versus distance at 50938 s for the fully penetrating, partially
penetrating, and stratified aquifer simulations
distance
(m)
0
1.207
2.913
5.328
8.743
13.57
20.30
30.04
41.69
55.00
72.31
96.46
130.61
178.86
247.11
342.11
522.11
747.11
873.56
fully
penetrating
1.890
1.628
1.438
1.288
1.158
1.040
0.928
0.821
0.732
0.656
0.581
0.502
0.419
0.335
0.251
0.174
0.107
0.057
0.030
Drawdown (m)
20 m aquifer 40 m aquifer
partially penetrating partially penetrating
pumped unpumped pumped unpumped
2.583
2.067
1.708
1.442
1.236
1.072
0.939
0.824
0.733
0.656
0.581
0.502
0.419
0.335
0.251
0.173
0.106
0.056
0.029
1.197
1.188
1.169
1.134
1.080
1.007
0.918
0.819
0.732
0.656
0.581
0.502
0.419
0.335
0.251
0.173
0.106
0.056
0.029
2.778
2.257
1.884
1.597
1.360
1.161
0.992
0.849
0.745
0.662
0.584
0.503
0.421
0.336
0.252
0.174
0.107
0.057
0.029
1.006
1.003
0.996
0.982
0.959
0.922
0.868
0.797
0.724
0.655
0.582
0.503
0.420
0.336
0.052
0.174-
0.107
0.056
0.029
Stratified
top bottom
1.890
' 1.628
1.438
1.288
1.158
1.040
0.928
0.822
0.733
0.656
0.581
0.502
0.419
0.335
0.252
0.174
0.107
0.057
0.030
1.887
1.625
1.436
1.285
1.155
1.037
0.926
0.819
0.730
0.654
0.579
0.500
0.417
0.333
0.250
0.172
0.106
0.056
0.030
17-6
-------
2.50 n
2.00 H
'—S
E
-—^
c 1.50 H
O
"D
o1-OOH
0.50 -
0.00
fully penetrating well
*-+-+-*HC partially penetrating (pumped)
(i
X X X X X
< partially penetrating (unpumped)
r i i r i i i i
i i i i r i
10
distance
100
(m)
1000
Figure 17.1. Drawdown versus distance for the fully penetrating well case and the
partially penetrating well case in the 20 m thick aquifer at time = 50938 s.
17-7
-------
2.50 1
2.00 -
1.50 -
O
"O
1.00 -
0.50 -
0.00
fully penetrating well
* * * * partially penetrating (pumped)
*-* partially penetrating (unpumped)
X X X X X
I tI I 111
10
distance
100
(m)
1000
Figure 17.2. Drawdown versus distance for the fully penetrating well case and the
partially penetrating well case in the 40 m thick aquifer at time = 50938 s.
17-8
-------
DISCUSSION OF RESULTS
This problem illustrates a method of modeling partially penetrating wells. A separate layer
is used over the uncased part of the well while the rest of the aquifer is modeled with another
layer. In many situations it may not be necessary to incorporate this level of complexity.
This was shown in parts b and c, which support the rule of thumb that partial penetration
effects vanish at a distance of 0.5 to 2 times the aquifer thickness. For part b, the twenty m
thick aquifer, partial penetration effects are minimal at 20m while for part c, the forty m thick
aquifer, the effects are minimal at 40m from the well.
A multiaquifer well is modeled in part d. Two wells are actually required because
MODFLOW assumes one well per layer. Well discharge was apportioned based on a
weighted average of the transmissivities. This method results in the same drawdown in the
well nodes for the two aquifers, as well as in the rest of the aquifers. The weighted average
methodology is an intuitive approach which is commonly used. It does not account for some
of the complexity inherent in natural systems. Bennett et ah, (1982) and McDonald (1984)
describe the dynamics of multiaquifer wells and how they may be incorporated in numerical
models.
17-9
-------
PROBLEM 18
Cross-Sectional Simulations
INTRODUCTION
When conceptualizing flow in a three-dimensional system, it is often useful to simplify the
system to a two-dimensional cross-section. In other instances, such as in modeling flow
beneath a dam, the entire analysis may lend itself to a cross-sectional representation. This
exercise shows how to set up a cross-section, illustrates a method of modeling layers of non-
uniform thickness and extent, and discusses advantages of certain solution techniques.
PROBLEM STATEMENT AND DATA
The area to be modeled is near a major river system. A two-dimensional vertical cross-
section is useful to conceptualize the flow system, determine reasonable ranges of aquifer
parameters, assess model boundaries, and to determine the most influential parameters in the
system. Specifically for this problem, the model was used to assess whether aquifer thinning
and facies changes could account for a steep hydraulic gradient in that area.
The two-dimensional model domain is shown in Figure 18.1. Most apparent from this
illustration is the highly variable layer thicknesses and pinchouts of certain layers. Partly due
to the pinchouts and variable thicknesses, some of the layers have a pronounced dip
associated with them.
The model domain is six layers and 27 columns. Because it is a vertical section, a single
row is used. The top layer is unconfined, all others are convertible. To avoid calculating
unique transmissivities manually for each block, a fully convertible option (LAYCON=3) is
used such that both aquifer tops and bottoms are read in. A groundwater divide is located on
the left side of the model domain. It is implicitly modeled as a no-flow boundary. A
specified head boundary condition is used in layer 6 that allows leakage into and out of the
overlying system. The river is assumed to penetrate layers 1, 2 and 3 and is modeled as
specified head. A divide is assumed beneath the river such that all flow discharges up into
the river. The remainder of the right boundary is therefore also assumed implicitly to be no-
flow. The upper boundary is the water table and receives recharge of 1.315 in/yr. Because
some layers may be desaturated, recharge is assumed to be to the highest active layer.
The layer pinchouts are handled by specifying a minimal thickness of 0.5 ft in the area
where the bed is absent and assigning a hydraulic conductivity typical of an areally adjacent
layer. Therefore, layer 5 has a hydraulic conductivity of 2.8 x 10'5 ft/d for columns 1 through
9 and 28 ft/d for columns 10 through 27. Vertical leakance terms are computed from these
hydraulic conductivities and layer thicknesses. Order of magnitude values of hydraulic
conductivity are used, with a horizontal to vertical anisotropy of 10 to 1. Hydraulic
conductivities are shown below.
18-1
-------
Surficial Deposits, "A"
Clay aquitard, "B"
Gravel aquifer, "C"
Sand aquifer, "D"
Dense clay aquitard, "E"
Leaky Clay aquitard, "F"
= 0.28 ft/d
= 0.028 ft/d
= 28 ft/d
= 0.28 ft/d
= 0.000028 ft/d
= 0.0028 ft/d
Bottom elevations of each layer are given in Table 18.1. Note that the top elevations for
the underlying layer are identical to the bottom elevation for the overlying layer. Initial
conditions for the model are 290 ft in the river nodes and 500 ft elsewhere in layers 1
through 5. Hydraulic heads in layer 6 are given in Table 18.2. Horizontal grid spacing is
uniform at 1050 ft. For purposes of computing VCONT's for layer 1, the assumed saturated
thickness of layer 1 is given in Table 18.3.
450
ELEVATION (MSL)
200
12345678 9101112131415161718192021222324252627
MODEL BLOCK NUMBER
Figure 18.1. Layering and zonation used in the cross-sectional model.
18-2
-------
Part a) Set up and run the model in a steady-state mode. Use the SIP solution
technique with an acceleration parameter of 1.0, 5 iteration parameters, closure
criterion of 0.01, a maximum of 50 iterations and model calculated seed. Note
the number of iterations required for convergence and the iteration history. In
case of non-convergence, adjust the SIP seed to obtain a solution.
Part b) Run the model using the SSOR solution technique with acceleration parameter
1.0. Note the number of iterations required for convergence and the iteration
history.
18-3
-------
Table 18.1. Bottom and top elevations (ft) in cross-sectional model
Column
No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
Layer 1
Bottom
Layer 2
Top
425
420
425
406
397
390
375
360
335
325
320
314
310
310
314
319
318
318
317
317
316
313
310
310
310
300
288
Layer 2
Bottom
Layer 3
Top
415
410
415
396
387
380
365
350
325
315
310
304
300
300
304
309
308
308
307
307
306
303
300
300
300
290
287
Layer 3
Bottom
Layer 4
Top
395
390
385
380
367
360
355
330
290
277
277
277
211
111
277
277
277
277
277
277
277
111
111
277
111
277
277
Layer 4
Bottom
Layer 5
Top
327
329
330
332
335
334
330
322
289
276.5
276,5
276.5
276.5
276.5
276.5
276.5
276.5
276.5
276.5
276.5
276.5
276.5
276.5
276.5
276.5
276.5
276.5
Layer 5
Bottom
Layer 6
Top
265
266
267
268
270
272
274
276
276
276
276
276
276
276
276
276
276
276
276
276
276
276
276
276
276
276
276
Layer 6
Bottom
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
200
18-4
-------
Table 1S.2. Initial heads in layer 6
Column
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
Head (ft)
320
319.6
319.2
318.8
318.5
318.1
317.7
317.3
316.9
316.5
316.2
315.8
315.4
351.0
314.6
314.2
313.9
313.5
313.1
312.7
312.3
311.9
311.5
311.2
310.8
310.4
310.0
18-5
-------
Table 18.3. Assumed saturated thickness (ft) of layer 1 in the cross-sectional model
Column
1
2
• 3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
Thickness (ft)
25
20
15
19
3
10
25
20
30
35
40
51
55
55
51
46
52
52
48
48
44
37
40
10
10
15
2
18-6
-------
MODEL INPUT
The data sets for part a are shown below.
*********************************
* Basic package *
***•••*****•**•*«««*******•******
CROSS SECTIONAL MODEL SIP SOLUTION
MAY 22, 1990
6 1 27 1 4
11 0 0 0 0 0 0 18 19 0 0 22
0 0
1
11111
1
22222
1
33333
1
1
55555
0
.00000
1
500.0
500.0
500.0
500.0
1
500.0
500.0
500.0
500.0
1
500. 0
500.0
SOO.O
500.0
0
0
1
320.0
317.3
314.6
311.9
100.00
1(4012)
1111111111
1(4012)
2222222222
1(4012)
3333333333
1(4012)
1(4012)
5555555555
•6
.100E+OK7G11.4)
500.0 500.0
500.0 500.0
500.0 500.0
500.0 500.0
.100E+QK7G11.4)
500.0 500.0
500.0 500.0
500.0 500.0
500.0 500.0
.1001+01(7611.4)
500.0 500.0
500.0 500.0
500.0 500.0
500.0 500.0
.50QE+03
.500E+Q3
.1QOE+QK7C11.4)
319.6 319.2
316.9 316.5
314. 2 313.9
311.5 311.2
11.0000
1111111
2222222
3333333
5555555
500.0
500.0
500.0
500.0
500.0
500.0
500.0
500.0
500.0
500.0
500.0
500.0
318.8
316.2
313.5
310.8
2
1111-1
2
2222-2
2
3333-3
2
2
55555
12
500.0
500.0
500.0
500.0
12
500.0
500.0
500.0
500.0
12
500.0
SOO.O
500.0
500.0
12
318.5
315.8
313.1
310.4
500.0
500.0
500.0
290.0
500.0
500.0
500.0
290.0
500.0
500.0
500.0
290.0
318.1
315.4
312.7
310.0
1
13333
0
0
0
11
.2800
.2800
.2800
.2800
11
425.0
360.0
314.0
313.0
*********************************
* Block Centered Flow Package *
*«****«**i» *•*•*•*««•*••*********
.100E+Q1
.105E+04
.105E+04
.1001+01(7611.4) 12
.2800 .2800 .2800 .2800 .2800
.2800 .2800 .2800 .2800 .2800
.2800 .2800 .2800 .2800 .2800
.2800 .2800 .2800 .2800 .2800
.100E+OK7G11.4) 12
420.0 417.0 406.0 397.0 390.0
335.0 325.0 320.0 314.0 310.0
319.0 318.0 318.0 317.0 317.0
310.0 310.0 310.0 300.0 288.0
500.0
500.0
500.0
500.0
500.0
500.0
500.0
500.0
500.0
317.7
315.0
312.3
.2800
.2800
.2800
375.0
310.0
316.0
headngd)
headng{2>
nlay.nron,ncol,nper,itmuni
iunit array
iapart,istrt
ibound layer 1{locat,iconst,fmtin,ipr
ibound array
ibound layer 2(locat,iconst.fmtin,ipr
ibound array
ibound layer Sdocat.ieonst.frotin.ipr
ibound array
ibound layer 4(locat,iconst,fmtin,ipr
ibound array
ibound layer 5(locat,iconst,fmtin,ipr
ibound array
ibound layer 6(locat,iconst)
hnoflo
shead layer 1(locat,cnstnt,fmtin,iprn
shead array
shead layer 2(locat,cnstnt,fmtin,iprn
shead array
shead layer 3{locat,cnstnt,frntin,fprn
shead array
shead layer 4(locat,cnstnt)
shead layer 5(locat,cnstnt)
shead layer 1
-------
11
.44806-03
.4667E-03
.3709E-03
.4088E-03
11
.28006-01
.28006-01
.28006-01
.28006-01
11
415.0
350.0
304.0
303.0
11
.5589E-03
.55896-03
.55856-03
.55856-03
11
425.0
360.0
314.0
313.0
11
28.00
28.00
28.00
28.00
11
395.0
330.0
277.0
277.0
11
.82116-03
.68296-02
.2036
.2113
11
415.0
350.0
304.0
303.0
11
.2800
.2800
28.00
28.00
11
327.0
322.0
276.5
276.5
11
.90316-07
.12176-06
5.600
5.600
11
395.0
330.0
277.0
277.0
11
.28006-04
.28006-04
28.00
28.00
.1006*01(7611.4)
.46676-03 .48706-03
.43086-03 .41486-03
.38366-03 .36846-03
.40006-03 .50916-03
.1006*01(7011.4)
.28006-01 .28006-01
.28006-01 .28006-01
.28006-01 .28006-01
.28006-01 .28006-01
.1006+01(761 1.4)
410.0 407.0
325.0 315.0
309.0 308.0
300.0 300.0
.1006+OU7G11.4)
.5589E-03 .55886-03
.55806-03 .55796-03
.55826-03 .55836-03
.55876-03 .55876-03
.1006+01(761 1.4)
420.0 417.0
335.0 325.0
319.0 318.0
310.0 310.0
. 1006+01 (7G1 1.4)
28.00 28.00
28.00 28.00
28.00 28.00
28.00 28.00
.1006+01(701 1.4)
390.0 385.0
290.0 277.0
277.0 277.0
277.0 277.0
.1006+01(7011.4)
.91506-03 .10146-02
.1556 .1455
.1723 .1778
.2383 .2383
. 1006+01 (7G1 1.4)
410.0 407.0
325.0 315.0
309.0 308.0
300.0 300.0
.1006+01(701 1.4)
.2800 .2800
28.00 28.00
28.00 28.00
28.00 28.00
.100E+OK7G11.4)
329.0 330.0
289.0 276.5
276.5 276.5
276.5 276.5
. 1006+01 (7G1 1.4)
.88886-07 .88886-07
.43086-06 5.600
5.600 5,600
5.600 5.600
. 1006+01 (7G1 1.4)
390.0 385.0
290.0 277.0
277.0 277.0
277.0 277.0
.1006+01(701 1.4)
.28006-04 .28006-04
.28006-04 28.00
28.00 28.00
28.00 28.00
.47066-03
.40006-03
.36846-03
.50916-03
.28006-01
.28006-01
.28006-01
.28006-01
396.0
310.0
308.0
300.0
.55916-03
.55826-03
.55836-03
.55876-03
406.0
320.0
318.0
310.0
28.00
28.00
28.00
28.00
380.0
277.0
277.0
277.0
.11636-02
.1672
.1778
.2383
396.0
310.0
308.0
300.0
.2800
28.00
28.00
28.00
332.0
276.5
276.5
276.5
.87496-07
5.600
5.600
5.600
380.0
277.0
277.0
277.0
.28006-04
28.00
28.00
28.00
12
.54376-03
.37096-03
.37846-03
.48706-03
12
.28006-01
.28006-01
.28006-01
.28006-01
12
387.0
304.0
307.0
290.0
12
.55896-03
.55856-03
.55836-03
.55936-03
12
397.0
314.0
317.0
300.0
12
28.00
28.00
28.00
28.00
12
367.0
277.0
277.0
277.0
12
.17396-02
.2036
.1836
.4148
12
387.0
304.0
307.0
290.0
12
.2800
28.00
28.00
28.00
12
335.0
276.5
276.5
276.5
12
.86156-07
5.600
5.600
5.600
12
367.0
277.0
277.0
277.0
12
.28006-04
28.00
28.00
28.00
.50916-03
.36136-03
.37846-03
.46676-02
.28006-01
.28006-01
.28006-01
.28006-01
380.0
300.0
307.0
287.0
.55896-03
.55876-03
.55836-03
.55456-02
390.0
310.0
317.0
290.0
28.00
28.00
28.00
28.00
360.0
277.0
277.0
277.0
.21376-02
.2383
.1836
.5333
380.0
300.0
307.0
287.0
.2800
28.00
28.00
28.00
334.0
276.5
276.5
276.5
.90326-07
5.600
5.600
5.600
360.0
277.0
277.0
277.0
.28006-04
28.00
28.00
28.00
.44806-03
.36136-03
.38896-03
.28006-01
.28006-01
.28006-01
365.0
300.0
306.0
.55946-03
.55876-03
.55846-03
375.0
310.0
316.0
28.00
28.00
28.00
355.0
277.0
277.0
.22316-02
.2383
.1898
365.0
300.0
306.0
.2800
28.00
28.00
330.0
276.5
276.5
.10006-06
5.600
5.600
355.0
277.0
277.0
.28006-04
28.00
28.00
vcont layer 1-2(locat,cmtnt,fat1n,1p
vcont array
hy layer 2iprn)
bot array
vcont layer 2-3(locat,cnstnt,fMtin,ip
vcont array
top layer 2(locat>cnstnt,fmtin>iprn)
top array
hy layer 3
-------
11
265.0
276.0
276.0
276.0
11
.8939E-07
.1198E-06
.7368E-05
.7368E-05
11
327.0
322.0
276.5
276.5
11
.2800E-02
.2800E-02
.2800E-02
.2800E-02
11
200.0
200.0
200.0
200.0
11
265.0
276.0
276.0
276.0
.100E+OK7G11.4)
266.0 267.0
276.0 276.0
276.0 276.0
276.0 276.0
.100E+OK7G11.4)
.8797E-07 .8795E-07
.4070E-06 .7368E-05
.7368E-05 .7368E-05
.7368E-05 .7368E-05
.100E+OK7G11.4)
329.0 330.0
289.0 276.5
276.5 276.5
276.5 276.5
.100E+OK7G11.4)
.2800E-02 .2800E-02
.2800E-02 .2800E-02
.2800E-02 .2800E-02
.2800E-02 .2800E-02
.100E+OK7G11.4)
200.0 200.0
200.0 200.0
200.0 200.0
200.0 200.0
.100E+OK7G11.4)
266.0 267.0
276.0 276.0
276.0 276.0
268.0
276.0
276.0
276.0
.8658E-07
.7368E-05
.7368E-05
.7368E-05
332.0
276.5
276.5
276.5
.2800E-02
.2800E-02
.2800E-02
.2800E-02
200.0
200.0
200.0
200.0
268.0
276.0
276.0
12
270.0
276.0
276.0
276.0
12
.8524E-07
.7368E-05'
.7368E-05
.7368E-05
12
335.0
276.5
276.5
276.5
12
.2800E-02
.2800E-02
.2800E-02
.2800E-02
12
200.0
200.0
200.0
200.0
12
270.0
276.0
276.0
276.0 276.0 276.0 276.0
272.0
276.0
276.0
276.0
.8929E-07
.7368E-05
.7368E-05
.7368E-05
334.0
276.5
276.5
276.5
.2800E-02
.2800E-02
.2800E-02
.2800E-02
200.0
200.0
200.0
200.0
272.0
276.0
276.0
276.0
274.0
276.0
-276.0
.9870E-07
.7368E-05
.7368E-05
330.0
276.5
276.5
.2800E-02
.2800E-02
.2800E-02
200.0
200.0
200.0
274.0
276.0
276.0
W W W W W WWW WWWWWWWWWWWWWW WW WWWWWWWWW
* Recharge package *
3
0
0
0
0
.003E-01
****************************
SIP package *
****************************
50 5
1.0000 .10000E-01
0.00001
1
*********************************
* Output Control package *
*********************************
000
1 1 0
000
bot layer 5(locat.cnstnt,fmtin,iprn)
bot array
vcont layer 5-6(locat,cnstnt,fmtin,ip
vcont array
top layer 5
-------
In part b, the IUNTT array is modified in the BASIC package to use the SSOR solution
technique and the following SSOR package is used.
*********************************
* SSOR package *
*********************************
50 mxiter
1.0000 .10000E-01 1 accl.hclose.iprsor
MODEL OUTPUT
Hydraulic head arrays for the model are shown in Figure 18.2.
HEAD IN LAYER 1 AT END OF TINE STEP 1 IN STRESS PERIOD 1
1
16
2
17
3
18
4
19
5
20
6
21
7
22
8
23
9
24
10
25
11
26
12 13
27
14 15
350.6 347.6 344.5 341.3 337.8 334.2 330.0 325.1 319.5 313.5 304.8 290.0
HEAD IN LAYER 2 AT END OF TIME STEP 1 IN STRESS PERIOD 1
1
16
2
17
3
18
4
19
5
20
6
21
•»07 1.
7
22
*«•» /
8
23
T9*. /_
9
24
V7*» a
10
25
11
26
*i*T •*
12 13
27
•**/, 4 TZrt /
14 15
ic*. * ICT n
349.8 346.8 343.7 340.5 337.0 333.4 329.3 324.4 318.9 312.9 304.2 290.0
HEAD IN LAYER 3 AT END OF TIME STEP 1 IN STRESS PERIOD 1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
16 17 18 19 20 21 22 23 24 25 26 27
"iwia" 398* 1""395'7*"392*9""389"9""387*i""381 "9""375*9""372*3""369*5 "366.'7""363*6""359*9*"356*6""352*4"
349.3 346.3 343.2 339.9 336.5 332.8 328.7 323.8 318.3 312.4 303.6 290.0
HEAD IN LAYER 4 AT END OF TIME STEP 1 IN STRESS PERIOD 1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
16 17 18 19 20 21 22 23 24 25 26 27
'sw'a' "39s! i" *395". f" '392'.9' '389"9" "isK i" 'is*'."*' "375 ".9" "372'.3' '369*5' '366'.7""363!6"359'.9"356'.o"352".4"
349.3 346.3 343.2 339.9 336.5 332.8 328.7 323.8 318.3 312.4 303.6 290.0
HEAD IN LAYER 5 AT END OF TIME STEP 1 IN STRESS PERIOD 1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
16 17 18 19 20 21 22 23 24 25 26 27
'366!^'359!6''357!7''356!6""354!4'"352!8''356!6""346!8''345!4'"369!5' 366.7 363.6 359.9 356.6 352.4
349.3 346.3 343.2 339.9 336.5 332.8 328.7 323.8 318.3 312.4 303.6 290.0
HEAD IN LAYER 6 AT END OF TIME STEP 1 IN STRESS PERIOD 1
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
16 17 18 19 20 21 22 23 24 25 26 27
'32o!6"'3i9!6-'3i9!2*'iisis'"iisis""iis!i""iify'iifii'"^^"'iidis"^^'"sis.a 31*5.4 sis.6 3*4.6
314.2 313.9 313.5 313.1 312.7 312.3 311.9 311.5 311.2 310.8 310.4 310.0
Figure 18.2. Hydraulic head arrays for the cross-sectional model.
18-10
-------
DISCUSSION OF RESULTS
This problem resembles a true field application more than the previous problems. It is, in
fact, based on an actual field study. Because this is an actual hydrostratigraphic system,
heterogeneity and variable thicknesses are a factor in the analysis. This complicates the
model set-up considerably. Preprocessing capabilities become more necessary.
The vertical leakance parameter (VCONT) incorporates both hydraulic conductivity and
vertical grid spacing. Because VCONT is read as a two- dimensional array, each grid cell
can conceivably have a unique thickness. This is somewhat counter to the standard
orthogonality of the finite difference method, but can be used provided the grid distortion is
not too great. Notice that vertical grid spacing is never used explicitly in MODFLOW; it is
always posed in terms of VCONT.
A layer must always exist, therefore, a layer cannot simply vanish when a pinchout occurs.
Instead, this example models the layer as thinning to a minimal thickness and then taking on
the properties of an adjacent layer. In a more general application, properties of layers can be
zoned; thinning of layers is not always required.
Several statements regarding desaturation appears in the model output: NODE (1,1,1)
GOES DRY AT ITERATION 9. This indicates that the head in that particular layer has
fallen below the specified aquifer base. This may be a physical reality (the case here) or a
result of an oscillatory iteration history. Dry nodes in the latter case are a problem because
nodes are not allowed to resaturate in MODFLOW. A "domino effect" may ensue once
nodes begin to dry up as a result of oscillatory iteration; the flow system is altered as a result
of a dry node, followed by more dry nodes, etc. This type of behavior can be minimized by
specifying acceleration parameters and seeds such that an asymptotic solution is approached
from a condition of higher head.
The SSOR solution technique is superior to SIP in this particular application. Because the
cross-section is taken along a row, the model solves the entire vertical slice by direct means.
Some iteration is performed (7) because of non-linearities due to the upper water table and
the dry nodes. Considerably more iterations would have resulted if the cross-section had been
oriented along a column. In that case a "slice" would consist of six nodes (1 row x 6 layers)
and 27 slices would have been solved. The SIP solution technique requires some adjustment
to the seed before it will converge. A convergent solution using 27 iterations was achieved
using a seed of 0.00001.
Cross-sectional models can often be useful in conceptualization exercises such as this. The
user should be careful to align the cross-section along a relatively straight streamtube that has
minimal change in width.
18-11
-------
PROBLEM 19
Application of a Ground water Flow Model to a
Water Supply Problem
INTRODUCTION
Groundwater flow models arc often used in water resource evaluations to assess the long-
term productivity of local or regional aquifers. This exercise presents an example of an
application to a local system and involves calibration to an aquifer test and prediction using
best estimates of aquifer properties. Of historical interest, this problem is adapted from one
of the first applications of a digital model to a water resource problem (Finder and
Bredehoeft, 1968). The specific objective of their study was to assess whether a glaciofluvial
aquifer could provide an adequate water supply for a village in Nova Scotia.
PROBLEM STATEMENT AND DATA
The aquifer is located adjacent to the Musquodoboit River, 1/4-mile northwest of the
village of Musquodoboit Harbour, as shown in Figure 19.1. The aquifer is a glaciofluvial
deposit consisting of coarse sand, gravel, cobbles, and boulders deposited in a typical U-
shaped glacial valley cut into the slates and quartzites of the Meguma group and the granite
intrusives of Devonian age. The contrast in permeability between the granitic and
metamorphic rocks and the glaciofluvial valley fill is so great (approximately 106) that the
bedrock is considered as impermeable in the aquifer analysis. The aquifer, which is up to 62
feet thick, is extensively overlain by recent alluvial deposits of sand, silt, and clay. The
alluvial deposits are less permeable and act as confining beds. A cross-section through the
valley is given in Figure 19.2.
A pumping test was conducted to evaluate the aquifer transmissivity and storage
coefficient, and to estimate recharge from the river. The test was ran for 36 hours using a
well discharging at 0.963 cubic feet per second (432 gallons per minute) and three
observation wells (see inset of Figure 19.2 for locations). The test was discontinued when the
water level in the pumping well became stable. Initial estimates of aquifer parameters were
calculated using the Theis curve and the early segment of the drawdown curves for the
observation wells. The results were somewhat variable, ranging from 1.15 ftVs to 1.45 ft2/s.
A quasi-steady state formula for estimating transmissivity yielded results on the order of 0.3
ftVs. Because of the close proximity of boundaries, the pumping test results are difficult to
analyze using usual analytical methods.
A listing of the data set for the MODFLOW model is provided on page 19-4. The aquifer
is treated as confined, with transmissivity zones to account for thickness and facies changes.
The ratio between zones of transmissivity (1,2, and 4) are given in the data set; absolute
values of transmissivity are not given. A map of the transmissivity zones and model
boundaries is given in Figure 19.3.
19-1
-------
A uniform value of storage coefficient is used in the analysis. The model is used to
simulate drawdown, hence an initial head condition of 0.0 ft is used. Recharge is not
specified because only drawdown is simulated. A river is simulated using the RIVER
package. Its location is shown in Figure 19.4. Other pertinent data is given in Table 19.1.
CENOZOIC
PLEISTOCENE AND RECEHT
RECENT ALLUVIUM
GLACIO-FLUVIAL DEPOSITS
PALEOZOIC
DEVONIAN
GRANITE
SCALE 1:12,600
Figure 19.1. Geologic map of the Musquodoboit Harbor region. Inset is the well
configuration for the pump test conducted on this aquifer (from Finder
and Bredehoeft, 1968).
19-2
-------
LEGEND
CENOZOIC
PLEISTOCENE AND RECENT
RECENT ALLUVIUM
GLACIO-FLUVIAL DEPOSITS
PALEOZOIC AND PRECAMBRIAN
GRANITE AND SLATE
N
B
Scale
Horizontal:
Vertical:
1 in
1 in
948 ft
130 ft
Figure 19.2. Geologic cross-section through the Musquodoboit Harbor region (from
Finder and Bredehoeft, 1968).
19-3
-------
*********************************
* Basic package *
*********************************
Musquodoboit Harbor problem
1/6/92 pit
1 44 55 1 1
11 12 0 14 0 0 0 0 19 0 0 22
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1 1
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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
0
0
0
1
1
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1
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1
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1
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0
0
0
0
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
0
0
0
0
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
0
0
0
0
0
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
0
0
0
0
0
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
0
0
0
0
0
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
0 0
0 0
0 0
0 0
0 1
1 1
1 1
1 1
1 1
1 1
1 1
1 1
1 1
1 1
1 1
1 1
1 0
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
0
0
0
0
1
1
1
1
1
1
1
1
1
1
1
0
0
0
1
1
1
1
0
0
1
1
1
1
1
1
1
1
1
0
0
0
0
0
1
1
1
1
1
1
1
1
1
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
1
0 0
0 0
0 0
0 0
0 0
1 1
1 1
1 1
1 1
1 1
1 1
0 0
0 0
0 0
0 0
0 0
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0 0
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1 1
1 1
1 1
1 1
1 1
1 1
1 1
1 1
1 1
19-4
headngd)
headngC25
nlay,nron,neol,nperfitmuni
iunit array
iapart,istrt
ibotindC locat, iconst, fmtin, I prn)
-------
111111
000000
111111
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1 1 1
000
1111
1 1
0 0
1 1 1
000
1 1
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000000
100000
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000.
0
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.QOOE+00
151.4140
****************»******»»***»•«••
* Block Centered Flow Package *
IT********************************
0
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t.ooo
1.000
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hnoflo
shead(locat,cnstnt)
perlen,nstp,tsmuU
iss,ibcfcb
tayeon
t rpy(Iocat,cnstnt)
delr(locat,cnstnt)
delc(Iocat,cnstnt)
sfKlocat, cnstnt)
trans(locat,cnstnt,fmtin,iprn)
trans array
19-5
-------
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1.000
19-6
-------
1.000
1.000
1.000
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1.000
1.000
1.000
1.000
1.000
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2.000
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1.000
1.000
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2.000
1.000
1.000
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2.000
2.000
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1.000
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1.000
1.000
1.000
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19-7
-------
1.000
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4.000
4.000
4.000
1.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
2.000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
2.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
2.000
1.000
1.000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
2.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
2.000
1.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
2.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
2.000
1.000
1.000
.0000
.0000
4.000
4.000
4.000
2.000
1.000
1.000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
2.000
.0000
4.000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
4.000
4.000
4.000
1.000
1.000
.0000
.0000
4.000
4.000
4.000
2.000
1.000
.0000
.0000
4.000
4.000
4.000
2.000
1.000
.0000
.0000
4.000
4.000
4.000
2.000
1.000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
.0000
.0000
4.000
19-8
-------
4.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
2.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
4.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
.0000
2.000
.0000
2.000
.0000
4.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
2.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
4.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
2.000
2.000
.0000
2.000
2.000
.0000
.0000
.0000
2.000
2.000
.0000
2.000
.0000
4.000
4.000
4.000
2.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
4.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
2.000
2.000
.0000
2.000
.0000
.0000
.0000
.0000
2.000
2.000
.0000
2.000
.0000
4.000
4.000
4.000
2.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
2.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
.0000
2.000
2.000
.0000
2.000
.0000
.0000
.0000
.0000
2.000
2.000
.0000
2.000
.0000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
2.000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
2.000
.0000
.0000
.0000
.0000
4.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
.0000
2.000
2.000
.0000
2.000
.0000
.0000
.0000
.0000
2.000
2.000
.0000
2.000
.0000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
1.000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
2.000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
2.000
.0000
.0000
.0000
2.000
2.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
2.000
4.000
4.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
.0000
2.000
.0000
.0000
2.000
.0000
4.000
4.000
4.000
1.000
.0000
.0000
2.000
4.000
4.000
4.000
1.000
.0000
.0000
2.000
4.000
4.000
4.000
1.000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
2.000
4.000
4.000
4.000
.0000
.0000
.0000
2.000
2.000
4.000
4.000
.0000
.0000
.0000
2.000
2.000
2.000
4.000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
2.000
2.000
2.000
2.000
.0000
.0000
.0000
2.000
.0000
2.000
.0000
.0000
19-9
-------
.0000
.0000
.0000
.0000
.0000
.0000
2.000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
2.000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
2.000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
.0000
2.000
.0000
.0000
* Well package *
*********************************
29
32 -.963
*********************************
* River package *
*********************************
49
49
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
0
18
18
19
19
20
20
20
21
21
22
22
22
23
23
24
24
24
24
25
25
25
25
26
26
27
27
28
28
28
28
28
27
27
27
27
28
29
29
30
30
31
31
32
32
33
33
33
33
33
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
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30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0'
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
.02
-10.
-10.
•10.
-10.
-10.
-10.
-10.
-10.
•10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
-10.
•10.
-10.
-10.
-10.
-10.
nwwell, iwelcb
itmp
layer,row,col,q
nwrivr,irivcb
itmp
layer,row,col,stage,cond,rbot
19-10
-------
*********
SIP package
50 5
1.0000 .10000E-03
1.00000
mxiter,nperm
accl.hclose,ipcalc.wseed,iprsip
*********************************
* Output Control package *
9 33
1 1
1 0
1 1
1 0
1 1
1 0
1 1
1 0
1 1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
ihedfm,iddnfm,ihedun,
incode.ihddfl.ibudfl,
hdpr,ddpr,hdsv,ddsv
incode.ihddfl.ibudfl,
hdpr,ddpr,hdsv,ddsv
ineode.ihddfl.ibudfl,
hdpr,ddpr,hdsv,ddsv
ineode.ihddfl.ibudfl,
hdpr, ddpr,hdsv,ddsv
incode,ihddfl,ibudfl,
hdpr,ddpr,hdsv »ddsv
incode.ihddfl.ibudfl,
hdpr,ddpr,hdsv,ddsv
incode.ihddfl.ibudfl,
hdpr,ddpr,hdsv, ddsv
incode.ihddfl.ibudfl,
hdpr,ddpr,hdsv,ddsv
incode.ihddfl.ibudfl,
hdpr,ddpr,hdsv,ddsv
incod*,ihddfl,1budfi,
hdpr,ddpr,hdsv,ddsv
incodt.fhddfl.fbudfl,
hdpr,ddpr,hdsv,ddsv
incode.ihddfl.ibudfl,
hdpr,ddpr,hdsv,ddsv
incode.ihddfl.ibudfl,
hdpr,ddpr,hdsv,ddsv
incode.ihddfl.ibudfl,
hdpr,ddpr,hdsv,ddsv
incode.ihddfl.ibudfl,
hdpr,ddpr,hdsv,ddsv
iddnun
icbcfKstep 1)
icbcfKstep 2}
icbcfKstep 3}
icbcfKstep 4}
icbcflCstep 5)
icbcfKstep 6)
icbcfKstep 7}
icbcfKstep 8)
icbcfKstep 9)
icbcfKstep 10)
icbcfKstep 11)
icbcfKstep 12)
icbcfKstep 13}
icbcfKstep 14)
icbcfKstep 15)
19-11
-------
Part a) Run the model with the data set provided. Plot the drawdowns at the observation
wells and compare to the field data shown in Table 19.2 and Figure 19.5,
Estimate better values of transmissivity and storage coefficient. Do not change
location of transmissivity zones. Compare results and continue to adjust T and S
until you are satisfied with the results.
Part b) Make a predictive run for 1000 days at the same pumping rate with the values of
T and S that were obtained in Part a.
Part c) Make some conclusions:
How good is your history match?
What additional changes might improve it?
How important is river leakage?
How appropriate is the confined model approximation?
How much confidence do you have in your prediction?
Is the system at steady-state at 1000 days?
What are some weaknesses in this calibration/prediction procedure?
What does the modeling indicate regarding the feasibility of using this aquifer as a
water supply?
19-12
-------
10
15
30
35
55
5-
10 -
15
20-
35-
30-
40-
A
+
A
+
A
+
+
O
0
A
+
O
O
A
+
+
O
o
A
A
+
O
O
o
o
A
+
+
O
o
o
o
o
A
+
0
o
o
A
+
O
o
o
o
o
A
+
O
o
o
o
o
A
-t-
+
O
o
o
0
o
o
o
+
+
+
0
o
o
o
o
0
o
A
+
+
+
o
0
o
o
0
o
A
+
+
•
O
O
o
o
o
o
A
A
+
+
0
o
0
o
0
o
o
o
A
+
+
A
A
O
O
o
o
o
o
o
o
A
+
+
A
A
A
0
O
O
0
O
O
0
O
o
A
+
A
A
A
A
A
O
O
O
O
O
O
O
0
o
o
A
+
A
A
A
A
A
A
O
O
O
O
O
O
O
0
O
O
A
+
+
+
A
A
A
A
A
A
A
A
A
A
O
O
0
o
0
o
0
o
o
o
0
o
A
+
+
+
+
A
A
A
A
A
A
A
A
A
O
O
O
O
O
O
0
o
o
o
o
o
o
o
A
+
+
+
+
A
A
A
A
A
A
A
A
O
0
O
o
o
o
o
o
o
o
0
o
o
o
A
+
+
+
+
A
A
A
A
A
A
A
A
0
O
0
O
O
0
o
0
o
o
o
0
o
o
0
o
o
A
+
+
+
A
A
A
A
A
A
A
O
0
o
0
0
o
o
o
o
o
o
0
o
o
o
o
o
o
o
A
+
+
+
A
A
A
A
A
A
A
O
O
O
O
O
O
0
o
o
o
o
0
o
o
o
0
o
0
o
A
+
+
+
+
A
A
A
A
A
A
O
O
O
O
O
0
0
0
o
o
o
o
o
o
o
o
o
o
0
o
A
+
+
+
+
A
A
A
A
A
O
O
O
0
O
0
O
0
O
O
O
o
o
0
o
o
o
o
0
o
A
A
+
+
+
A
A
A
A
O
O
0
O
0
O
0
O
0
O
o
o
0
o
o
o
o
0
o
A
A
+
+
+
A
A
A
A
O
O
O
O
O
O
O
O
0
o
0
o
0
o
o
o
o
o
A
A
+
+
+
A
A
A
O
O
0
o
o
o
o
o
o
o
0
o
0
o
o
o
o
o
A
A
+
+
+
+
A
A
O
O
O
O
o
o
o
o
o
o
o
o
o
o
o
o
o
A
A
+
+
+
+
A
A
O
O
O
O
O
O
O
O
0
O
O
O
O
O
o
o
o
A
A
+
+
+
+
A
A
O
O
O
O
0
o
0
o
o
0
o
o
o
o
o
o
o
A
A
+
+
+
+
A
A
A
O
O
0
O
0
O
0
O
O
0
O
0
o
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o
o
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o r-i
A T-2
+ r-4
Figure 19.3 Model boundary and transmissivity zones used in the numerical model. Inactive cells contain no symbol; a
circle represents a relative transmissivity of 1, a triangle represents a relative transmissivity of 2, and a plus
represents a relative transmissivity of 4.
-------
10
15
I
20
I
25
1 -
5-
10 -
15-
30
I
35
i
40
45
i
50
I
A A
ao
as-
30-
35-
40-
55
I
A RIVER
* PUMPING WELL
® OBSERVATION WELL
Figure 19.4 Location of the river boundary condition, pumping well, and observation wells use in the numerical model.
-------
Table 19.1. Input data for the water supply problem
Grid: 44 rows, 55 columns, 1 layer
Grid Spacing: Uniform 100 ft
Initial Head: 0.0 ft
Transmissivity: Non-uniform spatially, 3 zones
Storage Coefficient: Uniform spatially
Closure Criterion: 0.001
Numbr of time steps 10
Time Step Multiplier: 1.414
Length of Simulation: 36 hours
Production Well Location: row 29, column 32
Pumping Rate: 0.963 ft3/s (432 gpm)
River Stage: 0.0
River Conductance: 0.02 ftVs
River Bottom Elevation: -10 ft
Table 19.2. Observed drawdown data from aquifer test
Time (min)
1
4
10
40
100
400
1000
2000
3000
Drawdown (ft)
Well 1 Well 2
0.17
0.26
0.33
0.48
0.57
0.79
0.99
' 1.19
1.33
0.04
0.12
0.16
0.22
0.29
0.51
0.70
0.86
0.98
Well 3
0.00
0.01
0.02
0.08
0.14
0.30
0.50
0.68
0.78
19-15
-------
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10 100 1000
TIME (minutes)
10000
Figure 19.5. Drawdown (ft) versus time (min) for the aquifer test conducted at
Musquodoboit Harbor. The top line represents observation well 1, the
bottom line represents observation well 3.
MODEL INPUT
Input data sets for the model were given on page 19-3. For part a, the transmissivity
multiplier that gave the best match for Finder and Bredehoeft (1968) was 0.0685 while the
storage coefficient was 0.06. In part b, the simulation time (PERLEN) was changed to 1000
days, with 30 time steps (NSTEP), and time step multiplier (TSMULT) of 1.414 in the
BASIC Package.
MODEL OUTPUT
A comparison of modeled to observed drawdown data for part a is given in Figure 19.6.
Various combinations of transmissivity and storage coefficients yield the drawdowns shown in
19-16
-------
Figures 19.7 and 19.8, respectively. The drawdown data for the base case is given in Table
19.3. A plot of drawdown for part b is given in Figure 19.9.
10:3
1-
O
Q
0.1 -
Q
0.01
0.1
x x x x x well 1
+ + +• + + well 2
* * * * * well 3
— observed data
I I T I I III * I* 1 I I I I II
I I I Til I ll| I 1 T I II I Ij
10 100 1000
TIME (minutes)
10000
Figure 19.6. Comparison of modeled to observed drawdown (ft) data for the base case.
19-17
-------
1-
o
Q
Of 0.1
Q
0.01
0.1
X
+
+ + + + + base case
x x,x xx 2 times T
***** 0.5 times T
observed data
1111
1
I I i I I
10 100 1000
TIME (minutes)
10000
Figure 19.7. Comparison of modeled to observed drawdown in well 1 for the base case
and for a 2-fold increase and reduction in transmissivity.
19-18
-------
1-
o
Q
0.1 -
0.01
+ + + + + base case
X X X X X S = 0.1
5=0.005
observed data
0.1
10 100 1000 10000
TIME (minutes)
Figure 19.8. Comparison of modeled to observed drawdown in wells for the base case
and for storage coefficients of 0.1 and 0.005.
19-19
-------
Table 19.3. Modeled drawdown data for the base case. Transmissivities were 0.0685
, 0.137 frVs, and 0,2740 ft2/s and storage coefficient was 0.06.
Time (min)
4.98
12.0
22.0
36.1
56.0
84.1
124
180
260
372
531
456
1074
1524
2160
Drawdown (ft)
Well 1 Well 2
0.03
0.09
0.17
0.27
0.37
0.47
0.57
0.67
0.77
0.86
0.95
1.04
1.14
1.23
1.33
0.00
0.01
0.03
0.07
0.12
0.18
0.25
0.33
0.41
0.50
0.59
0.68
0.77
0.86
0.96
Well 3
0.00
0.00
0.01
0.02
0.04
0.07
0.11
0.17
0.23
0.30
0.38
0.46
0.55
0.64
0.74
19-20
-------
Figure 19.9. Drawdown (ft) after 1000 days of pumping at 0.963 ftVs.
19-21
-------
DISCUSSION OF RESULTS
This problem illustrates one type of calibration or history match. In this case aquifer
parameters were adjusted in the model to match observed drawdown from a short-term
transient event. Within the constraints of the problem statement, adjustments to transmissivity
and storage coefficient resulted in the match shown in Figure 19.6. The match is good ta late
time, but not in early time. Adjustments to transmissivity change the magnitude of drawdown
at a given time (Figure 19.7) while adjustments to storage coefficient changes the shape of
the curve before equilibrium is attained (Figure 19.8). The poor match in early time appears
to he the result of the storage properties. A decrease in storage coefficient would have the
desired effect of increasing drawdown in early time, but would also increase it beyond
observed values in later time. To circumvent this dilemma, Finder and Bredehoeft (1968)
introduced a time-dependent storage coefficient to approximate drainage of the aquifer
system. The initial value of storage coefficient of 0.003 was allowed to increase linearly with
time to a maximum of 0.06 after 10 minutes of pumping. This is not a standard application
and requires either numerous restart simulations or a code modification. Another
approximation is to specify a partially convertible aquifer (LAYCON=2) in the BCF package.
A closer match (see Figure 19.10) to early time behavior is obtained with a primary storage
factor (SF1) of 0.003, a secondary storage factor (SF2) of 0.06, and an aquifer top elevation
(TOP) of -0.1 ft. Both the time-dependent storage adjustment by Finder and Bredehoeft and
the current magnitude-dependent adjustment are fairly crude approximations to what appears
to be a delayed yield effect.
The slightly imperfect match to late-time data for the MODFLOW model base case is the
result of using Finder and Bredehoeft's (1968) late-time storage coefficient without regard to
the early-time factor that they used. As was illustrated in Figure 19.8, a higher constant value
of storage coefficient (0.1) results in a better late-time match.
River leakage is important because steady-state flow conditions depend on the quantity of
water entering the system through the river bed. When the system is at steady state, the
pumpage will be balanced by river recharge. The system is close to steady state after 125
days (timestep 24) of pumping as may be seen from the storage contribution (0.0055 ftVs)
relative to the river leakage (0.9574 ft3/s) in the mass balance. The model is more sensitive
to river conductance in late time than in early time. This is shown in Figure 19.11. The
results of all sensitivity simulations for well 1 are given in Table 19.4.
Several potential weaknesses exist in this calibration procedure. The aquifer test provides
confidence in parameters close to the pumping well, but less confidence in the
characterization distant from the well. The variability in thickness and facies is apparent in
the cross-section of Figure 19.2, and yet the representation is fairly simple. Because no wells
exist to monitor the effect on the other side of the river, it is difficult to have complete
confidence in the characterization of the aquifer/river interaction. This is important because
the degree of connection will ultimately govern the productivity of the aquifer. Finally, the
need to introduce the delayed-yield effect is not satisfying. Although it is likely that delayed
yield is occurring, the representation in the model is very crude. The transient calibration
procedure performed here is well suited for a localized aquifer system where the ultimate
19-22
-------
source of water is close to the pumping well. Additional confidence in the calibration could
be obtained through a steady-state history match to water levels through the aquifer.
The prediction indicates that the aquifer can supply the village with the desired quantity of
water with minimal drawdown in the aquifer. The long term drawdown was shown in Figure
19.9. The results obtained by Pinder and Bredehoeft (1968) and Finder and Frind (1972) are
similar to the current results. This good comparison provides confidence in the applicability
of MODFLOW to a field problem.
10-a
1-
o
Q
0.1 -
o
0.01
x x x x x well 1
+ + + + + well 2
* * * * * well 3
observed data
0.1
1 10 100 1000
TIME (minutes)
10000
Figure 19.10. Comparison of modeled drawdown for the drawdown-dependent storage
coefficient.
19-23
-------
10q
1 -
o
Q
0.1 i
Q
0.01
x *
+ + + + + baSQ case
X X X X X C = 0.2
* * *** C=0,002
observed data
I I I I I I ll| I I I I 1 I'll | I I I ! (Ill] I I I I I I I I ] I I I I I I I 11
0.1 1 10 100 1000 10000
TIME (minutes)
Figure 19.11. Comparison of modeled to observed drawdown in well 1 for the base
case and for order of magnitude increase and decrease in river
conductance.
19-24
-------
Table 19.4, Drawdown (ft) versus time in observation well 1 for variations in
transmissivity, storage coefficient, and river leakance
Time (min) Base Case
4.98
2.0
22.0
36.1
56.0
84.1
124
180
260
372
531
756
1074
1524
2160
0.03
0.09
0.17
0.27
0.37
0.47
0.57
0.67
0.77
0.86
0.95
1.04
1.14
1.23
1.33
Transmissivity
2X 1/2X
0.04
0.09
0.15
0.21
0.27
0.33
0.38
0.43
0.48
0.53
0.59'
0.64
0,70
0,76
0.83
0.02
0.07
0.16
0.28
0.44
0.61
0.79
0.98
1.16
1.34
1.51
1.67
1.83
1.99
2.14
Drawdown (ft)
Storage Coefficient
0.1 0.005
0.01
0.05
0.10
0.17
0.26
0.35
0.44
0.54
0.63
0.73
0.82
0.91
1.00
1.09
1.19
0.33
0.57
0.75
0.88
1.00
1.11
1.22
1.33
1.44
1.55
1.67
1.80
1.93
2.06
2.18
River Leakance
0.2 0.002
0.03
0.09
0.17
0.26
' 0.36
0.45
0.53
0.60
0.66
0.70
0.74
0:78
0.80
0.82
0.84
0.03
0.09
0.17
0.27
0.37
0.48
0.58
0.68
0.78
0.88
0.99
1.10
1.21
1.33
1.47
19-25
-------
PROBLEM 20
Application of a Groundwater Flow Model to a
Hazardous Waste Site
INTRODUCTION
Despite its inability to assess the mechanisms of contaminant transport, a groundwater flow
model can provide useful information on various remedial alternatives for hazardous waste
sites. Specifically, the flow model can give information on the hydraulic effects, such as flow
rates, drawdowns, and flow directions resulting from the remedial alternatives. A
contaminant transport model is necessary if data on concentration reduction, mass fluxes, and
travel times are desired. This exercise shows how a flow model may be applied to a site,
how model input may be adjusted to represent various remedial alternatives, and how model
output is interpreted to assess the remediation.
This problem is derived from an analysis of conceptual remedial designs at a hazardous
waste site (Andersen et al., 1984). The study used an early version of the U.S. Geological
Survey groundwater flow model, referred to as USGS2D (Trescott et al., 1976). The model
was used to assess the relative effectiveness of a low permeability cap, an upgradient drain,
and an upgradient slurry wall. Various combinations of these features were analyzed for their
hydraulic merits. Combined with engineering and economic considerations, the results of the
groundwater modeling formed the basis of the design which was eventually proposed.
PROBLEM STATEMENT AND DATA
The waste site is underlain by shallow unconsolidated materials of the Cohansey
Formation. The Cohansey consists of an upper sandy zone that varies from 0 to 30 ft in
thickness and a lower silty zone that is 10 to 20 ft thick. A strong contrast in permeability
between the two units is apparent from the location of groundwater seeps at the contact
between the two units. Underlying the Lower Cohansey is a clay unit of very low
permeability. Of most importance at the waste site is the potential for contaminated
groundwater migration in the Upper Cohansey and subsequent discharge into surface waters.
Further details on the hydrogeology of the site is given in Andersen et al. (1984).
A groundwater flow model was calibrated based on observed groundwater levels and
discharge measurements from an adjacent stream. The model considers two-dimensional
unconfined flow in the Upper Cohansey and uses natural hydrologic features as boundary
conditions. A finite-difference grid (Figure 20.1) was designed using smaller spacing (30 ft)
near the landfill site where detail was required and larger spacing away from the site near the
boundary. Uniform values of hydraulic conductivity (42.5 ft/d), recharge (24 in/yr), and
specific yield (0.28) were used, but a non-uniform aquifer bottom elevation was used.
Part a) Use the data set given on page 20-4 to run the steady-state model. Save the
hydraulic heads on disk for later use in the transient remedial simulations.
20-1
-------
Part b) Simulate the effect of an impermeable clay clap by making adjustments to the
recharge array. The extent of the cap is shown in Figure 20.2. Run for 30.8 yrs
with 20 time steps and a time step multiplier of 1.5. Print out mass balances of
all time steps and hydraulic heads at time steps 8, 11, 14, 16 and 20. Use a
specific yield of 0.28.
Part c) Simulate the effect of an impermeable clay cap and a low permeability slurry
wall. Make adjustments to the hydraulic conductivity array to simulate the wall.
The grid cells that contain the wall will have a conductivity of IxlO"6 times that of
the aquifer. The extent of the wall is shown in Figure 20.2. Use the same time
stepping and printout specifications as were given in part b.
Part d) Simulate the effect of an impermeable clay cap, a low permeability slurry wall,
and a drain. Use the parameters shown in Table 20.1 in the DRAIN package to
simulate the drain. The extent of the drain is shown in Figure 20.2. Use the
same time stepping and printout specifications as were given in part b.
Part e) Simulate the effect of an impermeable cap and a drain. Use the same time
stepping and printout specifications as were given in part b.
20-2
-------
9
1 -
10 -
15-
20-
25-
30
35-
5 10 15
i 1 I
20
I
25 30
1 1
35
i
•CONSTANT HEAD
Figure 20.1 Finite-difference grid showing the location of specified head cells for the steady-state model.
-------
>•••••«••*«*«*««»»«»»*••»••****
* Basic package *
*********************************
LIPARI LANDFILL SIMULATIONS
1/10/92 PFA steady state run
1 39 37
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0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00 0.00
0.00 0.00
0.00 0.00
0.00 0.00
0.00 0.00
0.00 0.00
132.00 132.00 132.00 132.00 132.00
132.00 132.00 125.00 125.00 125.00
120.00 120.00 120.00 120.00 120.00
120.00 120.00 120.00 120.00 120.00
120.00 120.00
131.00 131.00 131.00 131.00 131.00
131.00 131.00 123.00 123.00 123.00
118.00 118.00 118.00 118.00 118.00
118.00 118.00 118.00 118.00 118.00
118.00 118.00
130.00 130.00 130.00 130.00 130.00
130.00 130.00 120.00 120.00 120.00
110.00 110.00 110.00 110.00 110.00
110.00 110.00 110.00 110.00 110.00
110.00 110.00
130.00 130.00 130.00 130.00 130.00
130.00 130.00 120.00 120.00 120.00
120.00 120.00 110.00 110.00 105.00
105.00 105.00 105.00 105.00 105.00
98.00 98.00
126.00 126.00 126.00 126.00 126.00
headngd)
headng(2)
nlay,nrow,ncol,nper,itmuni
iunit array
iapart,istrt
ibound(locat,iconst,fmtin,iprn)
ibound array
hnoflo
shead(Iocat,const,fmt i n,iprn)
20-4
-------
126.00
115.00
110.00
105.00
123.00
123.00
115.00
110.00
105.00
120.00
120.00
110.00
110.00
100.00
118.00
118.00
105.00
105.00
105.00
116.00
116.00
105.00
105.00
95.00
115.50
115.50
105.00
105.00
95.00
115.00
115.00
105.00
105.00
95.00
114.50
114.50
105.00
105.00
95.00
114.00
114.00
105.00
105.00
95.00
113,50
113.50
110.00
105.00
95.00
113.00
113.00
105.00
105.00
95.00
112.50
112.50
105.00
105.00
95.00
112.00
112.00
105.00
105.00
95.00
111.50
111.50
101.20
105.00
95.00
111.30
111.30
110.00
105.00
95.00
111.00
111.00
126.00
115.00
110.00
105.00
123.00
123.00
115.00
110.00
100.00
120.00
120.00
110.00
110.00
95.00
118.00
118.00
105.00
105.00
95.00
116.00
116.00
105.00
105.00
90.00
115.50
115.50
105.00
105.00
90.00
115.00
115.00
105.00
105.00
90.00
114.50
114.50
105.00
105.00
90.00
114.00
114.00
105.00
105.00
90.00
113.50
113.50
105.00
105.00
90.00
113.00
113.00
105.00
105.00
90.00
112.50
112.50
105.00
105.00
90.00
112.00
112.00
101.20
105.00
90.00
111.50
111.50
105.00
105.00
90.00
111.30
111.30
105.00
105.00
90.00
111.00
111.00
126.00
115.00
110.00
100.00
123.00
123.00
115.00
110.00
95.00
120.00
120.00
110.00
110.00
86.00
118.00
118.00
105.00
105.00
85.50
116.00
116.00
105.00
105.00
85.00
115.50
115.50
105.00
105.00
84.50
115.00
115.00
105.00
105.00
84.00
114.50
114.50
105.00
105.00
84.00
114.00
114.00
105.00
105.00
84.00
113.50
113.50
105.00
105.00
84.00
113.00
113.00
105.00
105.00
84.00
112.50
112.50
101.20
105.00
84.00
112.00
112.00
105.00
105.00
84.00
111.50
111.50
105.00
105.00
84.00
111.30
111.30
105.00
105.00
84.00
111.00
111.00
126.00
115.00
110.00
93.00
123.00
123.00
115.00
110.00
88.00
120.00
120.00
110.00
110.00
86.00
118.00
118.00
105.00
105.00
85.50
116.00
116.00
105.00
105.00
85.00
115.50
115.50
105.00
105.00
84.50
115.00
115.00
105.00
105.00
84.00
114.50
114.50
105.00
105.00
84.00
114.00
114.00
105.00
105.00
84.00
113.50
113.50
105.00
105.00
84.00
113.00
113.00
105.00
105.00
84.00
112.50
112.50
101.20
105.00
84.00
112.00
112.00
105.00
105.00
84.00
111.50
111.50
105.00
105.00
84.00
111.30
111.30
105.00
105.00
84.00
111.00
111.00
126.00
110.00
110.00
93.00
123.00
123.00
115.00
110.00
88.00
120.00
120.00
110.00
110.00
86.00
118.00
118.00
105.00
105.00
85.50
116.00
116.00
105.00
105.00
85.00
115.50
115.50
105.00
105.00
84.50
115.00
115.00
105.00'
105.00
84.00
114.50
114.50
105.00
105.00
84.00
114.00
114.00
105.00
105.00
84.00
113.50
113.50
105.00
105.00
84.00
113.00
113.00
105.00
105.00
84.00
112.50
112.50
105.00
105.00
84.00
112.00
112.00
100.20
105.00
84.00
111.50
111.50
100.20
105.00
84.00
111.30
111.30
105.00
105.00
84.00
111.00 •
111.00
115.00
110.00
110.00
123.00
115.00
115.00
110.00
120.00
110.00
110.00
110.00
118.00
110.00
105.00
105.00
116.00
110.00
105.00
105.00
115.50
110.00
105.00
105.00
115.00
110.00
105.00
105.00
114.50
110.00
105.00
105.00
114.00
110.00
105.00
105.00
113.50
110.00
105.00
105.00
113.00
110.00
105.00
105.00
112.50
110.00
105.00
105.00
112.00
110.00
105.00
105.00
111.50
110.00
105.00
105.00
111.30
110.00
100.20
105.00
111.00
110.00
115.00
110.00
110.00
123.00
115.00
110.00
105.00
120.00
110.00
110.00
110.00
118.00
105.00
105.00
105.00
116.00
110.00
105.00
105.00
115.50
110.00
105.00
105.00
115.00
110.00
105.00
105.00
114.50
110.00
105.00
105.00
114.00
110.00
105.00
105.00
113.50
110.00
105.00
105.00
113.00
110.00
105.00
105.00
112.50
110.00
105.00
105.00
112.00
110.00
105.00
105.00
111.50
110.00
105.00
105.00
111.30
110.00
105.00
105.00
111.00
110.00
115.00
110.00
110.00
123.00
115.00
110.00
105.00
120.00
110.00
110.00
105.00
118.00
105.00
105.00
105.00
116.00
105.00
105.00
105.00
115.50
105.00
105.00
105.00
115.00
110.00
105.00
105.00
114.50
110.00
105.00
105.00
114.00
110.00
105.00
105.00
113.50
110.00
105.00
105.00
113.00
110.00
105.00
105.00
112.50
110.00
105.00
105.00
112.00
110.00
105.00
100.00
111.50
110.00
105.00
100.00
111.30
102.20
105.00
100.00
111.00
102.20
20-5
-------
110.00
105,00
95.00
110.80
110.80
110.00
105.00
95.00
110.60
110.60
110.00
105.00
95.00
110.40
110.40
108.00
105.00
95.00
109.00
109.00
107.00
105.00
95.00
108.00
108.00
107.00
105.00
95.00
107.00
107.00
106.00
100.20
101.00
106.00
106.00
108.00
108.00
96.00
105.00
108.00
108.00
108.00
92.00
103.00
108.00
108.00
102.00
90.00
102.00
108.00
108.00
102.00
100.00
101.00
108.00
108.00
102.00
100.00
100.00
108.00
108.00
100.00
100.00
100.00
106.00
106.00
101.00
101.00
100.00
106.00
106.00
102.00
102.00
100.00
104.00
104.00
105.00
105.00
90.00
110.80
110.80
105.00
105.00
90.00
110.60
110.60
105.00
105.00
90.00
110.40
110.40
105.00
105.00
90.00
109.00
109.00
105.00
105.00
90.00
108.00
108.00
105.00
105.00
90.00
107.00
107.00
105.00
102.00
90.00
106.00
106.00
108.00
100.20
86.00
105.00
108.00
108.00
102.00
87.00
103.00
108.00
108.00
100.00
90.00
102.00
108.00
108.00
100.00
100.00
101.00
108.00
108.00
100.00
100.00
100.00
108.00
108.00
100.00
100.00
100.00
106.00
106.00
101.00
101.00
100.00
106.00
106.00
102.00
102.00
100.00
104.00
104.00
105.00
105.00
84.00
110.80
110.80
105.00
105.00
84.00
110.60
110.60
105.00
105.00
84.00
110.40
110.40
105.00
105.00
84.00
109.00
109.00
105.00
105.00
84.00
108.00
104.20
105.00
105.00
84.00
107.00
104.20
105.00
102.00
85.00
106.00
104.20
108.00
108.00
86.00
105.00
104.20
108.00
102.00
87.00
108.00
104.20
108.00
99.00
90.00
108.00
103.20
108.00
100.00
100.00
108.00
103.20
108.00
100.00
100.00
108.00
104.20
108.00
100.00
100.00
106.00
102.20
106.00
101 .00
101.00
100.00
106.00
106.00
102.00
102.00
100.00
104.00
103.00
105.00
105.00
84.00
110.80
110.80
105.00
105.00
84.00
110.60
110.60
105.00
105.00
84.00
110.40
110.40
105.00
105.00
84.00
109.00
104.20
105.00
105.00
84.00
108.00
108.00
105.00
105.00
84.00
107.00
107.00
105.00
102.00
84.00
106.00
106.00
108.00
108.00
86.00
105.00
108.00
108.00
102.00
87,00
108.00
108.00
108.00
97.00
90.00
108.00
108.00
108.00
100.00
100.00
108.00
108.00
108.00
100.00
100.00
108.00
108.00
ios:oo
100.00
100.00
106.00
106.00
106.00
101 .00
101.00
106.00
106.00
103.00
102.00
102.00
104.00
104.00
103.00
105.00
101.00
84.00
110.80
110.80
105.00
105.00
84.00
110.60
110.60
105.00
105.00
84.00
110.40
104.20
105.00
105.00
84.00
109.00
109.00
105.00
105.00
84.00
108.00
108.00
105.00
105.00
84.00
107.00
107.00
105.00
102.00
84.00
106.00
106.00
108.00
100.00
86.00
105.00
108.00
108.00
98.00
87.00
108.00
108.00
102.00
96.00
90.00
108.00
108.00
102.00
100.00
100.00
108.00
108.00
102.00
100.00
100.00
108.00
108.00
102.00
100.00
100.00
• 106.00
106.00
103.00
101.00
101.00
106.00
106.00
102.00
102.00
102.00
104.00
104.00
103.00
100.20
105.00
110.80
110.00
100.20
105.00
110.60
103.20
105.00
105.00
110.40
108.00
105.00
105.00
109.00
107.00
105.00
105.00
108.00
107.00
105.00
105.00
107.00
106.00
105.00
102.00
106.00
108.00
108.00
100.00
108.00
108.00
108.00
95.00
108.00
108.00
102.00
94.00
108.00
108.00
102.00
100.00
108.00
108.00
102.00
100.00
108.00
108.00
102.00
100.00
106.00
106.00
103.00
101.00
106.00
106.00
102.00
102.00
104.00
104.00
103.00
105.00
105.00
110.80
102.20
105.00
105.00
110.60
110.00
100.20
105.00
110.40
108.00
100.20
105.00
109.00
107.00
105.00
105.00
108.00
107.00
105.00
105.00
107.00
106.00
105.00
102.00
106.00
108.00
108.00
96.00
108.00
108.00
108.00
93.00
108.00
108.00
102.00
93.00
108.00
108.00
102.00
100.00
108.00
108.00
102.00
100.00
108.00
108.00
102.00
100.00
106.00
106.00
103.00
101.00
106.00
106.00
102.00
102.00
104.00
105.00
103.00
105.00
100.00
110.80
110.00
105.00
100.00
110.60
110.00
105.00
100.00
110.40
108.00
105.00
100.00
109.00
107.00
100.20
100.00
108.00
107.00
100.20
100.00
107.00
106.00
105.00
102.00
106.00
108.00
108.00
96.00
108.00
108.00
108.00
96.00
108.00
108.00
102.00
90.00
108.00
108.00
102.00
100.00
108.00
108.00
102.00
100.00
108.00
108.00
102.00
100.00
106.00
106.00
101.00
101.00
106.00
106.00
102.00
102.00
104.00
105.00
103.00
20-6
-------
103.00
103.00
100.00
103.00
103.00
103.00
103.00
100.00
102.00
102.00
102.00
102.00
100.00
100.00
100.00
100.00
100.00
86400.
103.00
103.00
100.00
103.00
103.00
103.00
103.00
100.00
102.00
102.00
102.00
102.00
100.00
100.00
100.00
100.00
100.00
11.
103.00
103.00
100.00
103.00
103.00
103.00
103.00
100.00
102.00
102.00
102.00
102.00
100.00
100.00
100.00
100.00
100.00
103.00
103.00
100.00
103.00
103.00
103.00
103.00
101.00
102.00
102.00
102.00
102.00
100.00
100.00
100.00
100.00
100.00
103.00
103.00
100.00
103.00
103.00
103.00
103.00
102.00
102.00
102.00
102.00
102.00
100.00
100.00
100.00
100.00
100.00
103.00
103.00
103.00
103.00
103.00
102.00
102.00
102.00
102.00
100.00
100.00
100.00
100.00
103.00
103.00
103.00
103.00
103.00
102.00
102.00
102.00
102.00
100.00
100.00
100.00
100.00
103.00
103.00
103.00
103.00
103.00
102.00
102.00
102.00
102.00
100.00
100.00
100.00
100.00
*********************************
* Block Centered Flow Package *
*********************************
1
1
,OOOE+01 3.OOOE+01 3.OOOE+01
.OOOE+01 3.OOOE+01 3.OOOE+01
0.
0.
0.
0.
0. 0. 0.
0 0.100E+01
11 0.100E+OU8E10.0)
2.OOOE+01 6.000E+02 3.000E+02 2.000E+02 1.400E+02 1.000E+02 8.OOOE+01 6.OOOE+01
3.OOOE+01 3.OOOE+01 3.OOOE+01 3.OOOE+01 6.OOOE+01 8.OOOE+01 8.OOOE+01 8.OOOE+01
8.OOOE+01 8.OOOE+01 8.OOOE+01 3.OOOE+01 8.OOOE+01 8.OOOE+01 8.OOOE+01 8.OOOE+01
3.OOOE+01 3.OOOE+01 3.OOOE+01 3.OOOE+01 8.OOOE+01 1.000E+02 1.400E+02 2.000E+02
3.000E+02 4.000E+02 4.000E+02 5.200E+02 2.OOOE+01
11 0.100E+OK8E10.0)
2.OOOE+01 4.000E+02 4.000E+02 4.000E+02 4.000E+02 3.000E+02 2.000E+02 1.400E+02
1.000E+02 8.OOOE+01 6.OOOE+01 3.OOOE+01 3.OOOE+01 3
3.OOOE+01 3.OOOE+01 3.000E+01 3.OOOE+01 3.000E+01 3
3.OOOE+01 6.000E+01 6.000E+01 8.000E+01 1.000E+02 1.000E+02 1.000E+02 1.000E+02
1.000E+02 1.000E+02 1.000E+02 1.000E+02 1.000E+02 1.000E+02 2.OOOE+01
0 4.92E-04
11 1.0(20F4.0)
0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100. 0. 0. 0. 0. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100. 0. 0. 0. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100.100. 0. 0. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100. 95. 95. 0. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100.100. 95. 90. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100. 95. 90. 85. 0.
0.100,100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100. 95. 90. 85. 80. 0.
0. 95.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.100.100.100.100.100.101.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.100.100.100.100.101.102.101.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.100.100.100.101.102.103.102.101.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.100.100.101.102.103.104.103.102.101.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.100.101.102.103.104.105.104.103.102.101.101.101.101.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.101.102.103.104.105.105.104.103.102.101.101.101.101.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.101.103.104.105.105.105.104.103.102.101.101.101.101.
20-7
perlen.nstp,tsmult
iss.ibcfcb
Iayeon
trpy(locat,cnstnt)
delr(locat,cnstnt)
delr array
delcdocat,cnstnt)
dele array
hy(locat,cnstnt)
botdocat, cnstnt, fmt in, iprn)
bot array
-------
100.
0.
100.
0.
100.
0.
100.
0.
100.
0.
100.
0.
100.
0.
100.
0.
100.
0.
100.
0.
100.
0.
101.
0.
101.
0.
101.
0.
101.
0.
100.
0.
100.
0.
0.
0.
0.
0.
0.
0.
0.
100.
90.
100.
90.
100.
90.
100.
90.
100.
90.
100.
90.
100.
90,
100.
90,
100,
91,
100,
92,
100.
96.
100,
97.
101,
97,
101,
98,
100.
98,
100,
0,
100.
0,
0.
0,
0,
0,
0,
0,
0.
100.100.
95.100.
100.100.
95.100.
100.100.
95.100.
100.100.
95.ICC.
100.1CU.
95.100.
100.100.
95.100.
100.100.
95.100.
100.100.
96.100.
100.100.
97.100.
100.100.
98.100.
100.100.
99.101.
100.100.
100.102.
101.100.
100.102.
101.100.
99.101.
99. 99.
99.100.
99. 99.
99.100.
100, 0.
99.100.
0.
99.
0.
99.
0.
0.
0.
100.100
100.100
100.100
ICO.100
100.100
1C3.100
100.100,
1CQ.1CQ
100.100
100.100
100.100
100.100
100.100
100.100
100.100
100.101
100.100
100,101
100.100
100.101
98. 98
102.103
98. 98
103.104
98. 98
106.105
98. 98
103.104
98. 98
101.102
0. 0
101.102
0. 0
101.102
0. 0
99. 99
0. 0
99
0.
0.
0.
,100.100,
,100.101,
.10C.SOC,
,:oc.ioi,
,100.10T,
,100.10,.',
.'oc.ioo,
,101.102,
,100.100,
,101.102,
,100.100,
,101.102,
,100.100,
,101.103,
,100.100,
,102.104,
,100.100,
.103.105,
, 99. 98
,103.105,
, 98. 96,
,105.105.
, 97. 95
,105.105
, 0. 0
,105.104,
. 0. 0
,104.104,
. 0. 0
,102.102.
, 0. 0
,102.102,
. 0. 0
,102.102
. 0. 0,
. 0.
. 0.
. 0.
. 0.
. 0.
. 0.
100.100.
103.104.
100,100.
1C3.104.
100.100.
103.104.
100.100,
103.104.
100.100.
103.105.
100.100.
103.105,
100.100.
105.105,
100.100.
105.104,
100. 98.
105.104.
98. 96,
104.104,
96. 94.
104.104.
94. 92,
104.104,
0. 0,
104.103,
0. 0.
104.104,
0. 0.
102.102,
0. 0,
102.102,
0. 0,
102.102.
0. 0.
, 0.
0.
, 0.
, 0.
, 0.
, 0.
100. 98.
105.105.
100. 98.
105.105.
100. 98.
105.105,
100. 98,
105.104.
100. 98.
105.104,
100. 98,
104.104,
100. 98.
104.104,
100. 98.
104.104.
98. 96.
104.104,
93. 92.
104.104,
91. 90,
104.104,
91. 0,
103.103,
0. 0,
103.103,
0. 0,
104.104,
0. 0,
102.102.
0. 0,
102.102,
0. 0,
102.102.
0. 0.
0.
0.
0.
0.
0.
0.
93
104
93
104
95
104
93
104
93
104
93
104
93
104
93
104
93
103
93
103
90
103
88
103
0
103
0
104
0
102
0
102
0
102
0
0
0
0
0
0
0
85. 83
104.103
85. 83.
'C4.103.
85. 83.
103.102.
85. 83.
103.102.
. 85, 83.
,103.102.
, 85. 83.
,103.102.
85. 83.
,103.102.
, 85. 83.
,103.102.
85. 83.
,103.102.
85. 83.
,102.102.
, 35. 83.
,103.102.
, 85. 83.
,103.102.
, 0. 0.
,103.102.
, 0. 0.
,104.103.
, 0. 0.
,102.102.
, 0. 0.
,102.102,
, 0. 0,
,102.102.
, 0. 0.
, 0.
, 0.
, 0.
. 0.
, 0.
, 0.
ao. o.
.102.101.101.
. 80. 0.
.102.101.101.
80. 0.
102.101.101.
80. 0.
102.101.101.
80. 0.
102.101
80. 0
102.101
80. 0
102.101
0
.101
.101
.101
80.
101.101.
80. 0.
101.
80.
101
.101
.101
101.101
0.
101.101.101
80. 0.
102.101
80. 0,
102.101
0. 0.
102.102.101.
0. 0.
103.103.103,
0. 0.
102.102.102,
0. 0.
102.102.102,
0. 0.
102.102.102
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
*********************************
* Recharge package *
*********************************
1 0
0 0
0 6.34E-8
*********************************
* SIP package *
*««»**•»••>**»*«•«»•«•••«*•**»**•
50 5
1.0 .01 0 .002 1
0.
0.
0.
0.
0.
0.
101.100.
101,100.
101.100.
101.100.
101.100.
101,100.
101.101.
101.101.
101.101.
101.101.
101.101.
101.101.
101.101.
102.102.
101.101.
101.101.
101. 0.
0. 0.
0, 0.
0. 0.
nrchop,i rchcb
inrech.inirch
rech*«**************«
* Output Control package *
*********************************
30
1
1
ihedfm,iddnfm,ihedun,iddnun
incode,ihddfI,ibudfI,icbcfI
hdpr,ddpr,hdsv,ddsv
20-8
-------
IS
JO-
a-
33-
* *
*
&
a
*
4
•
*
X
A
a
+
4
M
&
«
•V
4-
+
*
&
9
+
+•
4-
4-
¥
A
&
*
4-
4-
+
+
+
*
A
«
*
+
+
•»•
+
+
4-
M
d
•
+
•)•
M
a
9
4-
+
4-
+
4-
+
+
4-
+
*
a
a
a
4-
4-
4-
+
+
4-
4-
4-
*
A
A
«
0
*
+
+
+
4-
4
X
&
A
£
•
a
4-
4-
+•
X
A
A
ffl
«
4-
X
A
i
+
* CONSTANT HEAD
A DRAIN
O WALL
+ CAP
o ioa rr
Figure 20.2. Grid cells representing the impermeable clay cap, the slurry wall, and the
drain.
20-9
-------
Table 20.1. Attributes of the drain used in Part d
Row
17
17
18
18
19
19
20
20
21
21
22
22
23
23
24
24
25
25
26
26
27
27
28
28
29
30
31
32
33
34
Column
19
20
18
21
17
21
16
22
16
22
15
22
14
23
13
23
12
24
11
24
11
25
11
26
11
11
11
11
11
11
Elevation (ft)
101.2
101.2
101.2
100.2
101.2
100.2
102.2
100.2
102.2
100.2
102.2
100.2
103.2
100.2
104.2
100.2
104.2
100.2
104.2
100.2
104.2
100.2
104.2
100.2
104.2
104.2
103.3
103.2
104.2
102.2
Conductance
(ft?/s)
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
20-10
-------
MODEL INPUT
The input data for the steady-state simulation of part a was given in the problem statement.
For part b, the input is shown below. Note that the initial conditions are read from the binary
file created from the steady-state run.
*********************************
* Basic package *
*********************************
LIPARI LANDFILL SIMULATIONS
1/16/92 PFA part b cap only
1 39 37
11 0 0 0 0 0 0 18 19 0
0 22
0 0
1 K40I23
0000000000000000000000000000000000000
011111111
011111111
0111 11
0-111 11
0-111 11
0-111 11
0-111 11
0-1111 11
0-1111 11
0-1111 11
0-
o-
0-
0-
0-
0-
0-
0-
0-
0*
0*
o-
0-
Q-
0-
0-
0-
0-
0-
111 11
111111
111111
111111
111111
111111
111111
111111
111111
111111
11111
111111
111111
111111
111111
111111
111 11
111 11
1111111111
1111
1111
1111
1111
1111
1111
1111
1111
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
1111
11111
1111
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
111111 11
111111 1
111111 1
111 1 1
111 111
111 111
111 111
111 111
111 111
111111111
111111111
11111 111
111 111 11111 111
0-1 111 111 11111 111
0-1 111 111 11111 111
11111
11111
1 1
4
4 *
It
-
-
-
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
100000
-1-1 0000
1-1000
11-100
111-10
111-10
111-10
111-10
1111-10
1111-10
1111-10
1111-10
1111-10
1111-10
1111-10
11111-10
11111-10
11111-10
1-1 0
1-1 0
1-1 0
1-1 0
1-1 0
1-1 0
1-1 0
-1-1 0
-000
1 1 1 1-1- -000
-1-1-1-1-1 0- 0000
00000000000
00000000000
0-1 111 111111111 111111 1-1-1 00000000000
0-1 1 1 111111111 111111-10000000000000
00-1 1 111111111 1 1-1-1-1-1 00000000000000
0 (
3*1 1 1. . 1.1-1. 1.1. 1»1.1.1« *1 ^ f
JOOOOOOOOOOOOOOOOO
00-1 1 -10000000000000000000000000000000
0 0-1-1- 00000000000000000000000000000000
0000000000000000000000000000000000000
99.
•31 1.
9.72e08 201.5000
headng(l)
headng(2)
rUay,nron,ncol ,nper, i tmuni
iunit array
iapart.istrt
ibound(locat,i const,fmt1n,iprn)
ibound array
hnoflo
sheadClocat,cnstnt)
perlen,nstp,tsmult
20-11
-------
****************
*********
* Block Centered Flow Package *
*********************************
0 0
1
0 0.100E+01
11 0.100E+OK8E10.0)
2.OOOE+01 6.OOOE+02 3.OOOE+02 2.OOOE+02 1.400E+02 1.000E+02 8.OOOE+01 6.OOOE+01
3.OOOE+01 3.OOOE+01 3.OOOE+01 3.OOOE+01 6.OOOE+01 8.OOOE+01 8.OOOE+01 8.OOOE+01
8.OOOE+01 8.OOOE+01 8.OOOE+01 8.OOOE+01 8.OOOE+01 8.OOOE+01 8.OOOE+01 8.OOOE+01
iss.ibcfcb
Iayeon
trpy(locat,cnstnt)
delr(locat.cnstnt)
delr array
3.OOOE+01 3.OOOE+01 3.OOOE+01 3.
3.OOOE+02 4.OOOE+02 4.OOOE+02 5.
11 0.100E+OK8E10.0)
2.000E+01 4.000E+02 4.000E+02 4,
1. OOOE+02 8.OOOE+01 6.000E+01 3.
OOOE+01 8.000E+01
200E+02 2.OOOE+01
1.000E+02 1.400E+02 2.OOOE+02
OOOE+02 4.OOOE+02 3.OOOE+02 2.OOOE+02 1.400E+02
3.OOOE+01 3.OOOE+01 3.OOOE+01
OOOE+01 3.OOOE+01 3.OOOE+01 3.OOOE+01
OOOE+01 3.OOOE+01 3.OOOE+01 3.OOOE+01
1.
OOOE+02
OOOE+01
OOOE+01
OOOE+01
OOOE+02
delc(local,cnstnt)
dele array
0.
0.
0.
0.
0.
0.
0.
0.
0. 0. 0.
3.OOOE+01 6.000E+01 6.OOOE+01 8.OOOE+01 1.OOOE+02 1.OOOE+02
1.OOOE+02 1.OOOE+02 1.OOOE+02 1.OOOE+02 1.OOOE+02 1.OOOE+02 2
0 .28
0 4.92E-04
11 1.0C20F4.0)
0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0
0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100. 0. 0. 0. 0. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100. 0. 0. 0. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100.100. 0. 0. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
.100.100.100.100.100.100.100. 95. 95. 0. 0.
.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
.100.100.100.100.100.100.100.100. 95. 90. 0.
.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100. 95. 90. 85. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100. 95. 90. 85. 80. 0.
0. 95.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80
.100.100.100.100.100.100.100.100
.100.100.100.100.100.100.100.100. 98. 93
.100.100.100.100.100.100.100.100.100.100
.100.100.100.100.100.100.100.100. 98. 93
.100.100.100.100.100.100.100.100.100.101
.100.100.100.100.100.100.100.100. 98
100.100.100.100.100.100.
0.100.100.400.100.100.
100.100.100.100.100.100.
0.100.100.100.100..100.
100.100.
0. 90.
100.100.
95.100.100.100.
.100.
0. 90. 95.
100.100.100.
0. 90. 95.
100.100.100.
. 85
.100
. 85
100.100.
83. 80.
100.100.
. 83. 80.
,100.100.100.
, 85. 83. 80.
100.100.100.
0.
100.100.100.100.100.
0. 90. 95.100.100.
100.100.100.100.100.
0. 90. 95.100.100.
85.
.100.100.100.101.102.103.104.105.104.103
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83
0. 90. 95.100.100.100.100.101.102.103.104.105.105.104.103
100.100.100.100.100.100.100.100.100.100.100
0. 90. 95.100.100.100.100.101
100.100.100.100.100.100.100.100
0. 90. 95.100.100.100.100.101
.100.100.100.100.100.100
.101
.100
0.
102.101.10V
. 95.
.100.
. 95,
100.100.100.100.
100.100.100.100.
.105
.100
.105
.100.100.100. 98.
.103.104.105.105.
.100.100.100. 98.
98. 93. 85
105.105.104
98. 93. 85
105.104.104
93
104
93
83. 80.
85
104,
85,
0.
103.102.101.101,
83. 80. 0.
103.102.101.101,
80. 0.
102.101.101
80. 0.
83
103
83
.103.104
.100.100
.103.104
100.100
0. 90
100.100
0. 90
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.101.102.103.104.105.104.104.103.102.102.101.101
100.100.100.100.100.100.100.100.100.100.100.
0. 90. 95.100
100.100.100.100
90. 95.100
sfUlocat.cnstnt)
hy(locat.cnstnt)
bot (I ocat, cnstnt, f tnt i n, i prn)
bot array
0.
100.100.100.100.
0.
100.100.100.100.
.100.
.100.
0. 90. 95.100.100.100.100.100.100.100.100.101.102.101.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.100.100.100.100.101.102.103.102.101.100.100.100.100.100.
100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
100.100.100.100.101.102.103.104.103.102.101.100.100.100.100.
100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
102.101.101.101.101.
80.
101.101.
101.101.
101.100.
101.100.
100.100.100.100.102.103.104.105.105.104.103.102.102.101.101.101.100.
,101.100.
98. 93. 85. 83. 80. 0.
100.100.101.102.103.105.105.104.104.103.102.102.101.101.101.100.
100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
0. 90. 95.100.100.100.101.102.103.105.104.104.104.103.102.102.101.101.101.100.
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*********************************
SIP package
*********************************
50
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1
0
0
0
0
0
0
0
1
0
1
0
1
0
1
0
1
1
1
0
1
0
1
0
1
0
1
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
incode,ihddfl,ibudfl,icbcfl
-------
0.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.00
1.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.00
1.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.00
1.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001,000.00
0.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.00
1.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.00
1.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001E-61E-6
1.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001E-61.001E-6
1.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.0011-61.001.001.00
11-61.001.001.001.001.001.001.001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.00
1E-61.001.001.001.001.001.001.001.001.001.001.001.001.001.001.000.00
0,001.001.001.001.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.001.00
1E-61.001.001.001.001.OOJ.001.001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.001.001.00
1.001E-61.001.001.001.001.001,001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.001.001.001.00
1.001E-61.001.001.001.001.001.001.001.001,001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.001.001.001.001.00
1.001.001E-61.001.001.001.001.001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.001.001.001.001.001.00
1.001.0011-61.001.001.001.001.001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.001.001.001.001.001.00
1.001.001.0011-61.001.001.001.001.001.001.001.001.001.001.001.000.00
0.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.001.001.001.001.001.00
1.001.001.001.001E-61.001.001.001.001.001.001.001.001.000,000.000.00
0.001.001.001.001.001.001.001.001.001.001.0011-61.001.001.001.001.001.001.001.00
1.001.001.001.001.001.001.001.001.001.001.001.001.001.000.000.000.00
0.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.001.001,001.001.001.00
1.001.001.001.001.001.001.001.001.001.001.000.001.000.000.000.000.00
0.001.001.001.001.001.001.001.001,001.001.001E-61.001.001.001.001.001.001.001.00
1.001.001.001.001.001.000.000.000.000.000.000.000.000.000.000.000.00
0.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.001.001.001.001.001,00
1.001.001.001.001.001.000.000.000.000.000.000.000.000.000.000.000.00
0.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.001.001.001.001.001.00
1.001.001.001.001.001.000.000.000.000.000.000.000.000.000.000.000.00
0.001.001.001.001.001.001.001.001.001.001.001E-61.001.001.001.001.001.001.001.00
1.001.001.001.000.000.000.000.000.000.000.000.000.000.000.000.000.00
0.000.001.001.001.001.001,001.001.001.001.001.001.001.001.001.001.001.001.001.00
1.001.001.000.000.000.000.000.000.000.000.000.000.000.000.000.000.00
0.000.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.001.000.00
0.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.00
0.000.001.001.001.001.000.000.000.000.000.000.000.000.000.000.000.000.000.000.00
0.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.00
0.000.001.001.001.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.00
0.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.00
0.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.00
0.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000.000,00
11 1.0{20F4,0) bot(locat,cnstnt,fmtin,iprn)
0. 0, 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. bot array
0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
0.100.100.100.100.100.100.100.-100.100.100.100.100.100.100.100.100,100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100. 0. 0. 0. 0. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100,100.100,100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100. 0. 0. 0. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100.100. 0. 0. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100. 95. 95. 0. 0.
0.100.100.100.100.100.1X10.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100.100. 95. 90. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100.100.100.100.100.100.100. 95. 90. 85. 0.
0.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.100.100.100.100,100.100.100.100.100. 95. 90. 85. 80. 0.
0, 95.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.100.
100.100.100.IOff.100.100.100.100.100.100.100. 98. 93. 85. 83. 80. 0.
20-16
-------
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
100
0
101
0
101
0
101
0
101
0
100
0
100
0
0
0
0
0
0
0
0
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95,100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
,100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 95.100
.100.100.100
. 90. 96.100
.100.100.100
. 91. 97.100
.100.100.100
. 92. 98.100
.100.100.100
. 96. 99.101
.100.100.100
. 97.100,102
.101.101.100
. 97.100.102
.101.101.100
. 98. 99.101
.100. 99. 99
, 98. 99.100
.100. 99. 99
. 0. 99.100
,100.100. 0
. 0. 99.100
0
0.
0. 99.
0. 0.
0
99
0
0. 99. 99
0. 0. 0
0. 0. 0
0. 0. 0
,100.100.
,100.100.
100.100.
.100.100,
,100.100,
,100.100,
,100.100,
,100.100,
,100.100,
,100.100,
,100.100,
,100.100,
.100.100,
.100.100,
.100.100,
,100.100.
.100.100.
.100.100,
,100.100,
,100.100,
.100.100.
,100.100,
.100.100,
,100.100,
,100.100,
,100.100,
,100.100,
,100.100.
,100.100.
,100.100,
.100.100,
,100.100.
.100.100,
,100.100.
,100.101,
.100.100,
.100.101,
,100.100.
,100.101,
, 98. 98.
,102.103.
, 98. 98,
,103.104,
, 98. 98,
,104.105,
, 98. 98,
,103.104,
, 98. 98,
,101.102,
. 0. 0.
,101.102.
, 0. 0.
,101.102
, 0.
, 99.
0.
99.
. 0.
, 0.
, 0.
0,
99
0.
0
0
0
0,
100.100.
100.100.
100.100.
100.100,
100.100,
100.100.
100.100,
100.100,
100.100,
100.100,
100.100,
100.100,
100.100,
100.100.
100.100,
100.100,
100.101,
100.100,
100.101,
100.100,
100.101.
100.100,
100.101.
100.100,
100.102,
100.100,
101.102,
100.100.
101.102,
100.100,
101.102
100.100
101.103
100.100
102.104
100.100
103.105
99. 98
103.105
98. 96
105.105
97. 95
105.105,
0. 0,
105.104
0. 0,
104.104,
0. 0,
102.102,
0. 0,
102.102,
0. 0.
102.102,
0. 0,
0. 0,
100.100
100.100,
100.100,
100.100
100.100
100.100
100.100
100.100
100.100
100.100
100.100
100.100
100.101
100.100
101.102
100.100
102.103
100.100
103.104
100.100
103.104
100.100
103.104
100.100
103.104
100.100
103.104
100.100
103.105
100.100
103.105
100.100
105.105
100.100
105.104
100. 98
105.104
98. 96
104.104
96. 94
104.104
94. 92
104.104
0. 0
104.103
0. 0
104.104
0. 0
102.102
0. 0
102.102
0. 0
102.102
0. 0
0. 0
.100.100.100.
.100. 98. 93.
.100.100.100.
.100. 98. 93,
.100.100.100.
.100. 98. 93.
.100.100.101.
.100. 98. 93.
.100.101.102.
.100. 98. 93,
.101.102.103.
.100. 98. 93.
.102.103.104.
.100. 98. 93.
.103.104.105.
.100. 98. 93.
.104.105.105.
.100. 98. 93.
.105.105.105.
.100. 98. 93,
.105.105.104,
.100. 98. 93.
.105.105.104.
.100. 98. 93.
.105.105.104.
.100. 98. 93.
.105.104.104.
.100. 98. 93.
.105.104.104.
.100. 98. 93,
.104.104.104,
.100. 98. 93.
.104.104.104,
.100. 98. 93,
.104.104.104,
. 98, 96. 93,
,104,104.103,
. 93. 92. 93,
.104.104.103.
, 91. 90. 90,
,104.104.103,
.91. 0. 88.
.103.103.103,
, 0. 0. 0.
.103.103.103,
. 0. 0. 0,
.104.104.104.
. 0. 0. 0,
.102.102.102
. 0. 0
.102.102.102
. 0. 0. 0
.102.102.102
0. 0
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
100.100,
85. 83,
100.100,
, 85. 83
,100.100
, 85. 83
,100.100
85. 83
,101.100
, 85. 83
,102.101
, 85. 83
.103.102
, 85. 83
,104.103
, 85. 83
,104.103
, 85. 83
,104.103
, 85. 83
,104.103
, 85. 83
,104.103
. 85. 83
,103.102
, 85. 83
,103.102
. 85. 83
,103.102
. 85. 83
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. 85. 83
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. 85. 83
.103.102
. 85. 83
.103.102
. 85. 83
.102.102
, 85. 83
,103.102
. 85. 83
,103.102
, 0. 0
,103.102
, 0. 0
,104.103
, 0. 0
102.102
0. 0
102.102
0. 0
102.102
0. 0
100.100.100.
80. 0.
100.100.100,
, 80. 0.
,100.100.100,
, 80. 0.
,100.100.100,
, 80. 0.
.100.100.100,
, 80. 0.
,100.100.100,
, 80. 0.
,101.100.100,
. 80. 0.
,102.101
, 80. 0.
.102.101
, 80. 0.
,102.101.101
, 80. 0.
,102.101.101
0.
.101
.101
80.
102.101,
0,
101
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
80,
.102.101
80. 0.
102.101.101
80. 0.
102.101.101
, 80. 0.
102.101
, 80. 0
,102.101
80. 0
,101.101
, 80. 0.
,101.101.101
, 80. 0
,101.101
, 80. 0.
.102.101.101
, 80. 0.
.102.101.101
, 0. 0.
,102.102.101
, 0. 0.
,103.103.103
, 0. 0.
,102.102,102
, 0. 0.
,102.102.102
, 0. 0.
,102.102.102
0. 0.
0. 0. 0,
.101
.101
.101
.101
.101
0.
0.
0.
0.
0.
0.
0.
0.
0.
0.
100.100.
100.100.
100.100.
100.100.
100.100.
100.100.
100.100.
101.101.
101.101.
101.101.
101.100.
101.100.
101.100.
101.100.
101.100.
101.100.
101.101.
101.101.
101.101.
101.101,
101.101.
101.101.
101.101.
102.102.
.101.101.
101.101.
101. 0.
0. 0.
0. 0.
0. 0.
20-17
-------
In part d, the DRAIN package shown below is added to represent the drain. It is invoked
in the BASIC package in the IUNTT array. All other parameters are the same as in part c.
*********************************
* Drain package *
*********************************
30
30
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
17
17
18
18
19
19
20
20
21
21
22
22
23
23
24
24
25
25
26
26
27
27
28
28
29
30
31
32
33
34
19
20
18
21
17
21
16
22
16
22
15
22
14
23
13
23
12
24
11
24
11
25
11
26
11
11
11
11
11
11
101,
101.
101.
100.
101.
100.2
102.2
100.2
102.2
100.2
102.2
100.2
103.2
100,2
104.2
100.2
104.2
100.2
104.2
100.2
104.2
100.2
104.2
100.2
104
104
103,
103,
104,
102.2
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
10.0
nwdrn.tdrncb
itmp
layer,row,cot.elevation,cond
In part e, the BCF package of part b and the DRAIN package of part d are used. The other
packages are the same as those used in part d.
20-18
-------
MODEL OUTPUT
Hydraulic head contours at 2.69 years in the vicinity of the landfill are shown for four of
the five cases in Figures 20.3 a-d. Hydraulic heads along column 19 of the model are given
in Table 20.2 for the steady state (part a) and the wall and cap scenario (part c). These are
compared to the results of the original study, which used the USGS2D code (Trescott et al.,
1976). Shown in Table 20.3 are hydraulic heads along column 19 for each remedial
alternative. These are plotted in Figure 20.4. Finally, the drain discharge versus time is
shown for each scenario involving a drain in Table 20.4.
1200 1400 1600 1800 2000 2200 2400 2600 3000
1810 -
1610 r
1410 -
1210 -
1010 -
810 -
610 -
410 -
210 -
1810
1610
1410
1210
- 1010
1200 1400 1600 1800 2000 2200 2400 2600 2800
- 810
- 610
- 410
- 210
3000
Figure 20.3a, Hydraulic head (ft) contours In the vicinity of the landfill for the steady-
state case (a).
20-19
-------
1200 1400 1600 1800 2000 2200 2400 2600 2800 3000
1810 -
1610 -
1410 -
1210 -
1010 -
1200 1400 1600 1900 2000 2200 2400 2600 2800 3000
1810
1610
1410
1210
- 1010
810
610
410
210
- 810
- 610
- 410
- 210
Figure 20.3b. Hydraulic head (ft) contours in the vicinity of the landfill for the case
involving a cap (b)«
1200 1400 1600 1800 2000 .2200 2400 2600 2800 3000
1810 -
1610 -
1410 -
1210 r
1010 -
1810
- 1610
7 1410
1210
- 1010
810 -
610
410 -
210
- 810
- 610
- 410
- 210
1200 1400 1600 1808 2000 2200 2400 2600 2800 3000
Figure 20.3c. Hydraulic head (ft) contours in the vicinity of the landfill for the case
involving a cap and a slurry wall (c).
20-20
-------
1200 1400 1600 1800 2000 2200 2400 2600 2900 3000
1810 -
1610 -
1410 -
1210 -
1010 -
810 -
610
410 -
210
1810
- 1610
-: 1410
-1210
-1 1010
l I i l i l i l i i l ( i l i i i
1200 1400 1600 1800 2000 2200 2400 2600 2B00 3000
- 810
610
- 410
- 210
Figure 20.3d, Hydraulic head (ft) contours in the vicinity of the landfill for the case
involving a cap and a drain (e).
20-21
-------
Table 20.2 Comparison of MODFLOW results versus USGS2D results for the steady-
state case (Part a) and the wall and cap scenario (Part c) at 6.08 yr.
Hydraulic heads (ft) along column 19 of each model are shown
Row
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Steady
MODFLOW
127.19
126.71
125.90
124.72
123.30
122.07
121.13
120.41
119.84
119.38
119.08
118.87
118.66
118.44
118.23
118.00
117.76
117.52
117.28
117.03
116.79
116.54
116.29
116.03
115.64
115.11
1 14.47
113.63
112.67
111.68
110.60
109.30
107.75
105.77
103.00
State
USGS2D
127.20
126.72
125.91
124.72
123.30
122.07
121.13
120.41
119.84
119.33
119.07
118.85
118.65
118.44
118.22
117.99
117.75
117.52
117.27
117.03
116.73
116.53
116.23
116.03
115.64
115.13
114.47
113.63
112.67
111.67
110.60
109.30
107.75
105.77
103.00
Well and
MODFLOW
127.43
126.98
126.26
125.22
124.02
123.05
122.37
121.90
121.56
121.33
121.21
121.13
121.06
121.01
120.98
120.96
117.65
103.52
103.52
103.51
103.51
103.50
103.49
103.49
103.47
103.46
103.43
103.41
103.40
103.40
103.41
103.46
103.57
103.58
103.00
Cap
USGS2D
127.43
126.98
126.26
125.22
124.02
123.05
122.37
121.90
121.57
121.33
121.21
121.13
121.06
121.01
120.98
120.95
117.65
103.50
103.49
103.49
103.49
103.48
103.48
103.47
103.45
103.44
103.42
103.40
103.39
103.39
103.41
103.45
103.57
103.58
103.00
20-22
-------
Table 20.3. Hydraulic heads (ft) along column 19 of the model at 2.69 years for each
remedial alternative simulation
Row
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Cap
Part b
127.16
126.67
125.84
124.61
123.12
• 121.81
120.80
120.01
119.38
118.85
118.50
118.26
118.01
117.76
117.50
117.22
116.93
116.64
116.34
116.04
115.75
115.45
115.15
114.85
1 14.39
113.78
113.06
112.13
111.11
110.09
109.05
107.85
106.53
105.03
103.00
Wall, Cap
Partc
127.40
126.95
126.22
125.18
' 123.98
123.01
122.33
121.86
121.53
121.30
121.17
121.09
121.03
120.98
120.94
120.92
117.72
104.87
104.87
104.98
104.85
104.84
104.82
104.80
104.76
104.71
104.63
104.53
104.41
104.29
104.16
103.99
103.85
103.69
103.00
Wall, Cap, Drain
Part d
127.00
126.47
125.58
124.22
122.55
121.04
119.32
118.85
118.04
117.36
116.90
116.58
116.24
118.90
115.54
101.20
117.60
104.64
104.63
104.62
104.61
104.59
104.57
104.55
104.52
104.46
104.38
104.28
104.18
104.08
103.98
103.84
103.75
103.64
103.00
Cap, Drain
Part e
127.00
126.47
125.58
124.22
122.55
121.04
119.82
118.85
118.05
117.37
116.91
116.58
116.25
115.91
115.55
101.20
101.37
101.49
101.59
101.68
101.78
101.87
101.96
102.05
102.19
102.34
102.49
102.64
102.80
102.93
103.05
103.22
103.44
103.52
103.00
20-23
-------
125.00 -i
120.00 -
115.00 -
0)
>
0)
L.
-------
Table 20.4. Drain discharge (ftVs) versus time for the well, cap, and drain scenario
(Part d) and the cap and drain scenario (Part e).
Time
(yrs)
0.012
0.022
0.038
0.061
0.096
0.149
0.228
0.347
0.525
0.792
1.29
1.79 '
2.69
Wall, Cap, Drain
Part d
0.18341
0.17723
0.17258
0.06823
0.16312
0.15915
0.15411
0.15068
0.14740
0.14305
0.14137
0.14122
0.13893
Cap, Drain
Part e
0.33867
0.32242
0.30579
0.28748
0.26566
0.24384
0.22018
0.19714
0.17784
0.16098
0.15083
0.14626
0.14244
DISCUSSION OF RESULTS
This problem is an example of how a flow model may be applied to assess the
effectiveness of various remedial alternatives for sites where contaminant migration is the
major concern. The purpose of the original study was to provide input to engineering
decisions. Due to data and time constraints many simplifying assumptions were made. The
original model configuration has generally been preserved; little, if anything, has been
changed to accommodate new technology or new knowledge of the behavior of various
remedial alternatives.
The cap is simulated simply by limiting recharge. In this case, the cap was assumed to be
completely impermeable. A more reasonable approximation is to allow some recharge based
on calculations of cap effectiveness. It was further assumed that any precipitation on the cap
would be collected before running off and having the opportunity to recharge the aquifer.
The effect on the water table of the cap (part b) is to decrease the hydraulic head beneath the
cap by one to two feet (see Figures 20.3b and 20.4). Of more importance is the limitation of
percolation through the unsaturated zone and subsequent transport of contaminants to the
water table. The cap was simulated in all cases because there was no doubt whether it should
be used.
20-25
-------
The slurry well is simulated by assigning a hydraulic conductivity of IxlO'6 times that of
the aquifer within the grid block representing the well. Assuming a 30 ft wide grid block,
this is equivalent to a 2 ft wide wall of hydraulic conductivity 2.83 x W6 ft/d or IxlO"9 cm/s.
This is a very low hydraulic conductivity and essentially represents the wall as impermeable.
The purpose of the wall is to reduce flow through the landfill area and to reduce head beneath
the cap. As shown in Figure 20.3c, the wall is effective in deflecting water around the
landfill. Combined with the drain, the wall allows the drain to collect primarily clean water,
thereby limiting treatment costs and reducing the amount of water the drain must transmit.
The wall also allows the water table in the landfill area to drop more slowly and causes the
amount of solute discharging downgradient to be spread over time. When the wall is used
without a drain, some upgradient buildup occurs as the water is diverted around the landfill
area. A minor conceptual problem is apparent from Figure 20.4. A buildup of head occurs
on the node representing the wall as a result of the recharge on the low permeability wall.
Because of the relatively coarse discretization, the entire 30 ft width of the grid block is of
low permeability and cannot absorb the recharge. It would probably be most appropriate to
assume that water would run off the wall into the more permeable aquifer material.
Consequently, recharge could have been redistributed to other nodes in this model or, more
accurately distributed in a finer gridded model.
The drain is simulated using the DRAIN package of MODFLOW. In the original study,
constant head cells were used to represent the drain. This was done primarily because a drain
package was not a part of the USGS2D code, but also to assess the maximum amount of flow
which could be diverted to the drain. In this application, the DRAIN package was used, but a
relatively high conductance based on cell area and aquifer hydraulic conductivity was input.
The high conductance value had the net effect of making the drain very similar to a constant-
head node. An added benefit of using the DRAIN package was the differentiation in the
mass balance between true drain discharge and other constant-head discharge.
The drain causes an immediate lowering of the head in the vicinity of the landfill.
However, without the slurry wall, a gradient is established where water flows from the
landfill area into the drain. Because water comes from both inside and outside the landfill
area, flow rates are initially almost twice as high as when a wall is in place (see Table 20.4).
The drain without the wall is the most effective of all the alternatives in lowering head in the
landfill area. Notice in Figure 20.3d that parts of the Upper Cohansey are desaturated after
only 2.69 years.
A possible weakness of the model configuration of the drain was the need to specify the
drain elevation close to the aquifer base. Two-dimensional flow simulated by the model
begins to lose some accuracy near the drains where, due to drain placement, flow becomes
vertical. Andersen et al., (1984) discuss some of the problems associated with drain
placement near the aquifer base and suggest an alternate means of assessing drain flux with
the model.
Several other scenarios and combinations of remedial measures were simulated in the
original study. These included a shorter drain, a less penetrating drain, a smaller cap, a
shorter wall, and a drain at greater distance from the wall. The original study also focused on
discharge of contaminated groundwater to the seeps downgradient of the landfill. Seep
20-26
-------
discharge was obtained by summing fluxes to pertinent constant-head nodes. Individual nodal
discharges may be obtained by invoking the cell-by-cell flow option in the BCF package.
The original study also combined the numerical results with analytical results as a checking
procedure and to verify the validity of simplifying assumptions used in the numerical model.
An advective travel time was derived analytically from model results. A more sophisticated
method of assessing advective contaminant migration with a flow model is to use a particle
tracking module, such as MODPATH (Pollack, 1989).
20-27
-------
REFERENCES
Andersen, P.P., C.R. Faust, and J.W. Mercer, 1984. Analysis of conceptual designs for
remedial measures at Lipari Landfill, New Jersey. Ground Water Vol. 22, no. 2 pp
176-190.
Bear, J., 1972. Dynamics of fluids in porous media. American Elsevier, New York.
Bennett, D., Kontis, A.L. and Larson, S.P., 1982. Representation of multiaquifer well effects
in three-dimensional ground-water flow simulation, Ground Water v. 20, no. 3, pp.
334-341.
Bredehoeft, J.D., and G.F. Finder, 1970, Digital analysis of areal flow in multiaquifer
groundwater systems: a three-dimensional model. Water Resources Res., 6, pp 883-
888.
Faust, C.R., P.N. Sims, C.P. Spalding, P.P. Andersen, and D.E. Stephenson, 1989. FTWORK:
a three dimensional groundwater flow and solute transport code. Westinghouse
Savannah River Company WSRC-RP-89-1085.
Franke, O.L., Reilly, T.E., and Bennett, G.D., 1987. Definition of boundary and initial
conditions in the analysis of saturated ground-water flow systems — An introduction:
U.S. Geological Survey Techniques of Water-Resources Investigations, Book 3,
Chapter B5.
Freeze, R.A. and J. Cherry, 1979. Groundwater, Prentice Hall, Englewood Cliffs, NJ 604 pp.
GeoTrans, 1988. SEFTRAN: A simple and efficient two-dimensional groundwater flow and
transport model, Version 2.7. Software documentation.
Hantush, M.S., 1960. Modification of Theory of Leaky Aquifers, Journal of Geophysical
Research, 65 pp. 3713-3725.
McDonald, M.G., 1984. Development of a multi-aquifer well option for a modular ground-
water flow model. Proceedings of the FOCUS conference on practical applications of
groundwater models. NWWA.
McDonald, M.G. and Harbaugh, A.W., 1988. A modular three-dimensional finite-difference
ground-water flow model: U.S. Geological Survey Techniques of Water Resource
Investigations, Book 6 Chapter Al.
Moench, A.F. and T.A Prickett, 1972. Radial flow in an infinite aquifer undergoing
conversion from artesian to water table conditions. Water Resources Research. Vol 8,
No. 2 pp. 494-499.
R-l
-------
Papadopulos, I.S., 1965. "Nonsteady flow to a well in an infinite anisotropic aquifer," Symp.
Internn. Assoc, Sci. Hydrology, Dubrovnik.
Finder, G.F. and ID, Bredehoeft, 1968. Application of the digital computer for aquifer
evaluation. Water Resources Research, Vol. 4, pp. 1069-1093.
Finder, G.F. and E.O. Frind, 1972. Application of Galerkins procedure to aquifer analysis.
Water Resources Research, Vol. 8, pp 108-120.
Pollack, D.W., 1989. Documentation of computer programs to compute and display pathlines
using results from the USGS modular three dimensional finite difference groundwater
flow model, U.S. Geological Survey Open File Report 89-381.
Prickett, T.A., and C.G. Lonnquist, 1971. Selected computer techniques for groundwater
resource evaluation. Illinois State Water Survey, Bulletin 55, 62 pp.
Reilly, T.E., Franke, O.L., and Bennett, G.D., 1987. The principle of superposition and its
application in ground-water hydraulics: U.S. Geological Survey Techniques of Water-
Resources Investigations, Book 3, Chapter B6.
Rushton, K.R. and L.M. Tomlinson, 1977. Permissible mesh spacing in aquifer problems
solved by finite differences, Journal of Hydrology 34:63.76.
Theis, C.V. 1935. The relation between the lowering of the piezometrie surface and the rate
and duration of discharge of a well using groundwater storage. Trans. Amer.
Geophys. Union, 2, pp 519-524.
Trescott, P.C., G.F. Finder, S.P, Larson, 1976. Finite-difference model for aquifer simulation
in two dimensions with results of numerical experiments: U.S. Geological Survey
Techniques of Water-Resources Investigation, Book 7, Chapter Cl» 116 p.
van der Heijde, P.K.M. and R.A. Park, 1986. Report of findings and discussion of selected
groundwater modeling issues, Study conducted under cooperative agreement CR-
812603 with the U.S. Environmental Protection Agency, R.S. Kerr Environmental
Research Laboratory, Ada, OK 74820
R-2
-------
APPENDIX A
Abbreviated Input Instructions
These input instructions are intended as a quick reference for the experienced user. Most
explanations that are contained in the complete input instructions given in package
documentation have been omitted. The format of input fields is given only for those records
that contain fields that are not 10 characters wide. Each input item, for which format is not
given,, is identified as either a record or an array. For records, the fields contained in the
record are named. For arrays, only the array name is given. Input fields which contain codes
or flags are described. All other field and array descriptions have been dropped.
Array Input
The real two-dimensional array reader (U2DREL), the integer two-dimensional array reader
(U2DINT), and the real one-dimensional array reader (U1DREL) read one array-control record
and, optionally, a data array in a format
specified on the array-control record.
FOR REAL ARRAY READER (U2DREL or U1DREL)
Data: LOCAT CNSTNT FMTIN
Format; 110 F10.0 5A4
FOR INTEGER ARRAY READER (U2DINT)
Data: LOCAT ICONST FMTIN
Format: 110 110 5A4
IPRN
110
IPRN
110
IPRN-
is a flag indicating that the array being read should be printed and a code for
indicating the format that should be used. It is used only if LOCAT is not
equal to zero. The format codes are different for each of the three modules.
IPRN is set to zero when the specified value exceeds those defined in the
chart below. If IPRN is less than zero, the array will not be printed.
U2DREL
10G11.4
11G10.3
9G13.6
15F7.1
15F7.2
15F7.3 '
15F7.4
20F5.0
20F5.1
20F5.2
20F5.3
20F5.4
10G11.4
U2DINT
10111
6011
4012
3013
2514
2015
U1DREL
10G12.5
A-l
-------
LOCAT—indicates the location of the data which will be put in the array.
If LOCAT < 0, unit number for unformatted records.
If LOCAT = 0, all elements are set equal to CNSTNT or ICONST.
If LOCAT > 0, unit number for formatted records.
Basic Package Input
Input for the Basic (BAS) Package except for output control is read from unit 1 as specified
in the main program. If necessary, the unit number for BAS input can be changed to meet the
requirements of a particular computer. Input for the output control option is read from the
unit number specified in IUNIT(12).
FOR EACH SIMULATION
1. Record: HEADNG(32)
2. Record: HEADNG (continued)
3. Record: NLAY NROW NCOL NPER
ITMUNI
4. Data: IUNTT(24)
Format: 2413
8CF WEL DRN RIV EVT XXX GHB RCH SIP XXX SOR OC
1 2 3 4 5 6 7 8 9 10 11 12
5. Record: IAPART ISTRT
6. Array: IBOUND(NCOL,NROW)
(One array for each layer in the grid)
7. Record: HNOFLO
8. Array: Shead(NCOL,NROW)
(One array for each layer in the grid)
FOR EACH STRESS PERIOD
9. Data: PERLEN NSTP TSMULT
ITMUNI-is the time unit of model data.
0 - undefined 3 - hours
1 - seconds 4 - days
2 - minutes 5 - years
Consistent length and time units must be used for all model data. The user may
choose one length unit and one time unit to be used to specify all input data.
lUNTT-is a 24-element table of input units for use by all major options.
IAPART--indicat.es whether array BUFF is separate from array RHS.
If IAPART= 0, the arrays BUFF and RHS occupy the same space. This option
conserves space. This option should be used unless some other package
explicitly says otherwise.
If IAPART * 0, the arrays BUFF and RHS occupy different space.
A-2
-------
ISTRT—indicates whether starting heads are to be saved.
If ISTRT = 0, starting heads are not saved.
If ISTRT * 0, starting heads are saved.
IBOUND—is the boundary array.
If IBOUND(I,J,K) < 0, cell I,J,K has a constant head.
If IBOUND(I,J,K) = 0, cell IJ,K is inactive.
If IBOUND(I,J,K) > 0, cell IJ,K is active.
HNOFLO-is the value of head to be assigned to all inactive cells.
Shead-is head at the start of the simulation.
PERLEN-is the length of a stress period.
NSTP—is the number of time steps in a stress period.
TSMULT—is the multiplier for the length of successive time steps.
Output Control Input
Input to Output Control is read from the unit specified in IUNIT(12), All printer output goes to
unit 6 as specified in the main program. If necessary, the unit number for printer output can be
changed to meet the requirements of a particular computer.
FOR EACH 'SIMULATION
1. Record: IHEDFM IDDNFM IHEDUN IDDNUN
FOR EACH TIME STEP
2. Record: INCODE IHDDFL IBUDFL ICBCFL
3. Record: Hdpr Ddpr Hdsv Ddsv
(Record 3 is read 0, 1, or NLAY times, depending on the value of INCODE.)
IHEDFM-is a code for the format in which heads will be printed.
IDDNFM—is a code for the format in which drawdowns will be printed.
0-(10G11.4) 7-(20F5.0)
1-(11G10.3) 8-(20F5.1)
positive-wrap 2 - (9G13.6) 9 - (20F5.2) , <
3-U5F7.1) 10-(20F5.3) .. ' " J "''
negative-strip 4 - (15F7.2) 11 - (20F5.4)
5-(15F7.3) 12-(10G11.4) ,,
6 - (15F7.4) i—*
IHEDUN—is the unit number on which heads will be saved.
IDDNUN-is the unit number on which drawdowns will be saved.
INCODE-is the head/drawdown output code.
If INCODE < 0, layer-by-layer specifications from the last time steps are used. Input item
3 is not read. ,., ,
If INCODE = 0, all layers are treated the same way. Input item 3 will consist o
record. ,*~
IOFLG array will be read.
If INCODE > 0, input item 3 will consist of one record for each layer.
A-3
-------
IHDDFL-is a head and drawdown output flag.
If IHDDFL = 0, neither heads nor drawdowns will be printed or saved.
;_.•••• If IHDDFL * 0, heads and drawdowns will be printed or saved.
IBUDFL-is a budget print flag.
If IBUDFL = 0, overall volumetric budget will not be printed.
If IBUDFL * 0, overall volumetric budget will be printed.
ICBCFL-is a cell-by-cell flow-term flag.
If ICBCFL = 0, cell-by-cell flow terms are not saved or printed.
If ICBCFL * n, cell-by-cell flow terms are printed or recorded on disk depending on flags
set in the component of flow packages, i.e., IWELCB, IRCHCB, etc.
Hdpr--is the output flag for head printout.
If Hdpr = 0, head is not printed for the corresponding layer.
If Hdpr * 0, head is printed for the corresponding layer.
Ddpr—is the output flag for drawdown printout.
If Ddpr = 0, drawdown is not printed for the corresponding layer.
If Ddpr ?t 0, drawdown is printed for the corresponding layer.
Hdsv—is the output flag for head save.
If Hdsv = 0, head is not saved for the corresponding layer.
, If Hdsv ?t 0, head is saved for the corresponding layer.
Ddsv--is the output flag for drawdown save.
If Ddsv = 0, drawdown is not saved for the corresponding layer.
If Ddsv * 0, drawdown is saved for the corresponding layer.
Block-Centered Flow Package Input
Input for the BCF Package is read from the unit specified in IUNTT(1),
FOR EACH SIMULATION
1. Record: ISS IBCFCB
2, Data: LAYCON(NLAY) (maximum of 80 layers)
Format: 4012
(If there are 40 or fewer layers, use one record.)
. 3. Array: TRPY(NLAY)
4. Array: DELR(NCOL)
5. Array: DELC(NROW)
All of the arrays (items 6-12) for layer 1 are read first; then all of the arrays for layer 2, etc.
IF THE SIMULATION IS TRANSIENT
6. Array: sfl(NCOL,NROW)
IF THE LAYER TYPE CODE (LAYCON) IS ZERO OR TWO
7. Array: Tran(NCOL,NROW)
IF THE LAYER TYPE CODE (LAYCON) IS ONE OR THREE
8. Array: HY(NCOL,NROW)
9. Array: BOT(NCOL.NROW)
A-4
-------
IF THIS IS NOT THE BOTTOM LAYER L _.-L.
10. Array: Vcont(NCOL.NROW)
IF THE SIMULATION IS TRANSIENT AND THE LAYER TYPE CODE (LAYCON) is TWO
OR THREE ^ -.:__-.
11. Array: sf2(NCOL,NROW)
IF THE LAYER TYPE CODE IS TWO OR THREE
12. Array: TOP(NCOL,NROW) _..:LJ
ISS-is the steady-state flag.
If ISS * 0, the simulation is steady state.
If ISS = 0, the simulation is transient. .. .
IBCFCB-is a flag and a unit number. :
If EBCFCB > 0, cell-by-cell flow terms will be recorded if ICBCFL (see Output Control)
is set.
If IBCFCB. = 0, cell-by-cell flow terms will not be printed or recorded.
If IBCFCB < 0, print flow for constant-head cells if ICBCFL is set.
LAYCON-is the layer type table: 0 - confined, 1 - unconfmed,
2 - confmed/unconfined (T constant), and 3 - confmed/unconfined.
TRPY-is an anisotropy factor for each layer: T or K along a column to T or K along a row.
DELR—is the cell width along rows.
DELC—is the cell width along columns.
sfl--is the primary storage factor.
Tran-is the transmissivity along rows.
HY—is the hydraulic conductivity along rows.
BOT is the elevation of the aquifer bottom.
Vcont-is the vertical hydraulic conductivity divided by the thickness from a layer to the layer
beneath it.
sf2—is the secondary storage factor.
TOP--is the elevation of the aquifer top.
A-5
-------
River Package Input
Input to the River (RIV) Package is read from the unit specified in IUNIT(4),
FOR EACH SIMULATION
1. Record: MXRIVR IRIVCB
FOR EACH STRESS PERIOD
2. Record: ITMP
3. Record: Layer Row Column Stage Cond Rbot
(Input item 3 normally consists of one record for each river reach. If ITMP
, . is negative or zero, item 3 is not read.)
IRIVCB--is a flag and a unit number.
If IRIVCB > 0, cell-by-cell flow terms will be recorded.
If IRIVCB = 0, cell-by-cell flow terms will not be printed or recorded.
If IRIVCB < 0, river leakage will be printed if ICBCFL is set.
ITMP—is a flag and a counter.
If ITMP < 0, river data from the last stress period will be reused.
If ITMP > 0, ITMP will be the number of reaches active during the current stress period.
Recharge Package Input
Input to the Recharge (RCH) Package is read from the unit specified in IUNIT(8).
FOR EACH SIMULATION
1. Record: NRCHOP IRCHCB
FOR EACH STRESS PERIOD
2. Record: INRECH INIRCH
3. Array: RECH(NCOL,NROW) IF THE RECHARGE OPTION IS EQUAL TO 2
4. Array: IRCH(NCOL,NROW)
NRCHQP-is the recharge option code.
.o:lv Recharge is only to "the'top- grid layer. ;
2 - Vertical distribution of recharge is specified in array IRCH.
3 - Recharge is applied to the highest active cell in each vertical column.
IRCHCB-is a flag and a unit number.
If IRCHCB > 0, unit number for cell-by-cell flow terms.
If IRCHCB < 0, cell-by-cell flow terrhis will not be printed or recorded.
INRECH-is the RECH read flag. "
If INRECH'< 0, recharge fluxes from the preceding stress period are used
If INRECH > 0, an array of recharge fluxes, RECH (Lf1), is read.
INIRCH~is similar to INRECH. -
A-6
-------
Well Package Input
Input for the Well (WEL) Package is read from the unit specified in IUNIT(2),
FOR EACH SIMULATION '' . „ J,' f' \
1. Record: MXWELL IWELCB • ' .;\': "'
FOR EACH STRESS PERIOD -'; ;
2. Record: ITMP
3. Record: Layer Row Column Q
(Input item 3 normally consists of one record for each well. If ITMP is negative or zero,
item 3 is not read.) - —
MXWELL-is the maximum number of wells used at any time.
IWELCB—is a flag and a unit number.
If IWELCB > 0, unit number for cell-by-cell flow terms.
If IWELCB = 0, cell-by-cell flow terms will not be printed or recorded.
If IWELCB < 0, well recharge will be printed whenever ICBCFL is set. .-
ITMP—is a flag and a counter.
If ITMP < 0, well data from the last stress period will be reused.
If ITMP > 0, ITMP will be the number of wells active during the current stress period.
Drain Package Input
Input to the Drain (DRN) Package is read from the unit specified in IUNIT(3).
FOR EACH SIMULATION .< -.' :
1. Record: MXDRN IDRNCB .. :
FOR EACH STRESS PERIOD - . i
2. Record: ITMP
3. Record: Layer Row Col Elevation* Cond •_ ,:r.vi.i -.1; :''--J'J^i2
(Input item 3 normally consists of one record for each dratm If riMP is negative,® zero,
item 3 will not be read.) ., .,_ •, , •>•„•': ,•:-.'/ 1
MXPRN-is the maximum number of drain cells active at one time. . ; „ -jOl-r
IDRNCB—is a flag and a unit number. , , ; • >• ,;n i
If IDRNCB > 0, unit number for cell-by-cell flow terms. ' »/r 1
If IDRNCB = 0, cell-by-cell flow terms will not be printed or recorded, - jrj: ^--_-^'l_
If IDRNCB < 0, drain leakage for each cell will be printed whenever ICBCFL^is set^
ITMP-is a flag and a counter. , . r ... -_-"->c,;<[ L
If ITMP < 0, drain data from the last stress period will be reused.; ~ . , . ;• >;•
If ITMP > 0, ITMP will be the number of drains active during the current stress period.
A-7
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Evapotranspiration Package Input
Input tojthe Evapotranspiration (EVT) Package is read from the unit specified in IUNIT (5).
FOR EACH SIMULATION
1. Record: NEVTOP IEVTCB
FOR EACH STRESS PERIOD
2. Record: INSURF INEVTR INEXDP INIEVT
3. Array: SURF
4. Array: EVTR
5. Array: EXDP
IF THE ET OPTION IS EQUAL TO TWO
6. Array: IEVT
NEVTOP-is the evapotranspiration (ET) option code.
ul -JET is^ealculated only for cells in the top grid layer.
2 - The cell for each vertical column is specified by the user in array IEVT.
lEVTGB-ris a flag -and a unit number.
If IEVTCB > 0, unit number for cell-by-cell flow terms.
If IEVTCB < 0, cell-by-cell flow terms will not be printed or recorded.
INSURF-is the ET surface (SURF) read flag.
r; If. INSURF > 0, an array containing the ET surface elevation will be read.
If INSURF < 0, the ET surface from the preceding stress period will be reused.
INEVTR-is similar to INSURF.
INEXDP-is similar to INSURF.
INIEVT--is similamto INSURF^ ^ -
A-8 /
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General-Head Boundary Package Input
Input for the General-Head Boundary (GHB) Package is read from the unit specified in;
IUNIT(7).
FOR EACH SIMULATION >-:
1. Record: MXBND IGHBCB
FOR EACH STRESS PERIOD ' " '3 /K
2. Record: ITMP Boundary ^ -
3. Record: Layer Row Column Head Cond '- A
(Input item 3 normally consists of one record for each GHB. If ITMP is negative* -or zero,
item 3 is not read.)
MXBND— is the maximum number of general-head boundary cells at one time. •' -
IGHBCB-is a flag and a unit number.
If IGHBCB > 0, unit number for cell-by-cell flow terms.
If IGHBCB = 0, cell-by-cell flow terms will not be printed or recorded. .'. ____
If IGHBCB < 0, boundary leakage for each cell will be printed whenever ICBCFL is set.
ITMP-is a flag and a counter. . .^ T - -
If ITMP < 0, GHB data from the preceding stress period will be reused.
If ITMP > 0, ITMP is the number of general-head boundaries during the current stress.-
period.
Strongly Implicit Procedure Package Input
Input to the Strongly Implicit Procedure (SIP) Package is read from the unit specified in
IUNIT(9). , '• "
FOR EACH SIMULATION '. .L,; ! ; : ^
1. Record: MXITER NPARM
2. Record: ACCL HCLOSE IPCALC WSEED IPRSIP u ; -:l/l
IPCALC— is a flag indicating where the iteration parameter seed wifl/cdme>frora.' Eru- ^ -TVdi
0 - the seed will be entered by the user.
1 - the seed will be calculated at the start of the simulation from problem parameters.
IPRSIP-is the printout interval for SIP.
Slice-Successive Overrelaxation Package Input
Input to the Slice-Successive Overrelaxation (SOR) Package is read from the unit specified in
lUNIT(ll).
FOR EACH SIMULATION
1. Record: MXITER - #.WestJaL
2. Record: ACCL HCLOSE IPRSOR Chicago. «. 606043^
IPRSOR-is the printout interval for SOR.
A-9/; ,-
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