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PART I Simulating Flow in a One-Dimensional, Homogeneous System Consider the situation shown below where the conductivity of the homogeneous, isotropic, confined aquifer is (K = 0.5 m/day, thickness is b = 1.5 m, so transmissivity T = 0.5*1.5 = 0.75 m2/day). The heads on the left- and right-hand boundaries are constant at hA = 6.1 m and hB = 1.5 m, respectively. The mesh spading is Δx = 1 m and the initial head across the aquifer is h(x,t=0) = 6.1 m. The storage coefficient is S = 0.02.

Figure 1. Vertical cross-section through example aquifer.

The governing equation for flow in the aquifer is

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∂2h∂x2

=ST∂h∂t

A numerical approximation of the second derivative in space is

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∂2h∂x2 i

l

≈hi+1l − 2hi

l + hi−1l

Δx2

For the numerical approximation of the first derivative in time, we have two choices

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∂h∂t i

l≈hil+1 − hi

l

Δt (forward) or

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∂h∂t i

l≈hil − hi

l−1

Δt (backward)

Ground surface

Aquifer

x!y!z!

hB

Confining Layer

b

hA

Δx

i = 0 1 2 3 4 5 6 7 8 9 10

Node

Grid Cell

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Figure 2. Grid points used in a backward (explicit) finite-difference approximation in time Explicit method of solution In the “explicit” method of solution, we use all the information at the previous time step to compute the value at this time step. The method proceeds by calculating for the unknown head point-by-point across the domain (i = 1, …, n). This method can be unstable for large time steps. The finite-difference equation for the calculation is

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hi+1l − 2hi

l + hi−1l

Δx2 =SThil+1 − hi

l

Δt; i =1,...,n; l = 0,1,2,...

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TS

ΔtΔx2

(hi-1l − 2hi

l + hi+1l ) = hi

l+1 − hil

Define the parameter of the calculation

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r =TS

ΔtΔx2

so that

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hil+1 = hi

l + r hi-1l − 2hi

l + hi+1l( )

where the unknown head at time level l+1 is on the left-hand-side and all of the known information form the l time level is on the right-hand-side. Consider two slightly different values for the parameter r: r = 0.48, so that

i,l+1

i-1,l i,l i+1,l

i,l-1

t (l)

x (i) Δx

Δt

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Δt = rΔx2ST

= 0.0128d ≈18.5min

and r = 0.52, so that

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Δt = 0.0139d ≈ 20min When r = 0.48, we obtain the following results for the odd nodes in the mesh; however, when r = 0.52, we obtain the following results for the odd nodes in the mesh.

(a) r = 0.48

(b) r = 0.52 Figure 3. Results from solving the explicit approximation.

What’s going on here? At the initial time there is no flow across the aquifer, but at time t > 0, flow occurs. The water released from storage in a grid cell over time Δt is

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ΔV = SΔx(1)Δh Water flowing out of the cell over the time interval

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ΔV = TΔhΔx /2

Δt

So, this needs to be limited to be less than the total water available in the cell, or

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2T ΔhΔx

Δt ≤ SΔxΔh

or

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r =TS

ΔtΔx2

≤12

When r > 0.5 the time interval is too large and the cell doesn’t contain enough water. This causes an instable solution. Assignment: Water flows from left to right in the confined aquifer in the figure below. Use the following numerical parameters for the aquifer: 5 finite-difference cells; cell width (Δx) = 80 m; transmissivity (T) = 1000 m2/day; storage coefficient (S) = 0.002; pumping from first cell (Q1) = 3.96 m3/day/m2; pumping from fourth cell (Q4) = 6.04 m3/day/m2; left boundary head (h0) = 40 m; and right boundary head (h5) = 35 m. Solve for the head at each of the four unknown finite-difference nodes (i = 1, 2, 3, and 4) using an explicit method with a time step of Δt = 10 minutes. Is this a stable time step length for the explicit method? What is the r parameter?

Figure. Confined aquifer with no pumping from second and third cells (Q2 and Q3 = 0

m3/day/m2) Note: The initial conditions for the aquifer are

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h(x, t = 0) = h0l + h5

l − h0l

Lx = 40 − 1

80x m

The governing equation for one-dimensional flow in the aquifer is

Δ x Δ x Δ x Δ x Δ x i=0 i=1 i=2 i=3 i=4 i=5

Flow h0 h5

Q1 Q2 Q3 Q4

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∂∂x

T∂h∂x

#

$ %

&

' ( −Q = S

∂h∂t

The finite-difference approximation (explicit method) for the equation is

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hil+1 = hi

l + r(hi−1l − 2hi

l + hi+1l ) − Δt

SQi

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r =TS

ΔtΔx2

Qi [m3/day/m2 = m/day] is the pumping rate per unit area of aquifer. What to Turn in: Email to the instructor (daene@aol.com). Put your Team Members names in the email.

1. Spreadsheet showing calculations with graphs of the head at nodes i = 1, and 4 over time. 2. Answers to the questions: Is this a stable time step length for the explicit method? What

is the r parameter? 3. If the answer in (2) is “No”, then what is a stable time step?

PART II

Assignment: Use GrounwaterVistas to prepare the example model and run it as shown in class handout: Groundwater_Modeling_2.pptx (www.caee.utexas.edu/prof/mckinney/ce374l/Overheads/07_Groundwater_Modeling_2.pptx) What to Turn in: Email to the instructor (daene@aol.com). Put your Team Members names in the email.

1. GroundwaterVistas file for your model (*.gwv file). 2. MS Word document describing your results.

a. Color plots of heads in each of the three layers of your model. b. The Mass Balance page from the MODFLOW Output file (Model à

MODFLOW à View Output File) 1. Assignment Problem (see Figure below): Use GrounwaterVistas to prepare the

MODFLOW model for the problem described below1 • An alluvial valley contains a river that is in direct contact with an unconfined aquifer that,

in turn, overlies a semi-confined aquifer. • A grid of 3 layers, 11 rows (Δx = 200 m) and 8 columns (Δy = 300 m) is used to model

the aquifer. • A well pumps 1200 m3/day from cell (row, column, layer) = (5, 6, 3) • Recharge occurs at a rate of 200 mm/year

1takenfromR.J.Charbeneau,GroundwaterHydraulicsandPollutantTransport,PrenticeHall,2000

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• Heads along Row 1 = 22 m above the top of the confining bed • Heads along Row 11 = 18 m above the top of the confining bed • A river extends along Column 3 in Layer 1 of the model. The river is 10 m wide and 300

m long in each cell. The conductivity of the river bottom is Kriv = 0.1 m/day, and the thickness of the riverbed is = 0.2 m. The river stage (height of water in the river) and the river bed elevation vary according to the values in the following table.

o Layer 1: Kh = 5 m/day, Kv = 1 m/day, n = 0.35 o Layer 2: Kh = 0.1 m/day, Kv = 0.05 m/day, n = 0.40 o Layer 3: Kh = 10 m/day, Kv = 1 m/day, n = 0.30

Apply constant head boundary conditions to all cells on the ends in all three layers, EXCEPT for the cell containing the RIVER and make that cell a River Boundary Condition cell.

Row Col River Stage River Bottom 1 3 22.0 20.0 2 3 21.6 19.6 3 3 21.2 19.2 4 3 20.8 18.8 5 3 20.4 18.4 6 3 20.0 18.0 7 3 19.6 17.6 8 3 19.2 17.2 9 3 18.8 16.8 10 3 18.4 16.4 11 3 18.0 16.0

+20$m

0$m

&5$m

&15$m

Confining$bed

Semi&Confined$Aquifer

Unconfined$Aquifer

Well$(In$Layer$3)

Layer1

2

3

Column

1 81

11

Row

1600$m

River

Δy =$300$m Δx=200$mRiver&Cell&

River&Cell&

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Apply recharge to the TOP LAYER ONLY as shown in the following screen-shots. Select Modflow Options and select Recharge in Top Layer Only. Then select Recharge from the Props menu. Then select Property Values form the Props menu followed by Database. Then fill in the recharge value for zone 1. Then select Props Set Value or Zone à Window and select all the cells in the first layer and assign them to zone 1.

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