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3D Simulations-- the vertical dimension. Representation of hydrogeologic units in MODFLOW88. MODFLOW2000. In MODFLOW, vertical flow between layers Is governed by vertical conductance. Review. Horizontal Conductance. Vertical Conductance Terms. VCONT = K v /  z = leakance or

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Presentation Transcript
slide1

3D Simulations--

the vertical dimension

slide2

Representation of

hydrogeologic units

in MODFLOW88

slide4

In MODFLOW, vertical flow between layers

Is governed by vertical conductance.

slide5

Review

Horizontal Conductance

slide6

Vertical Conductance Terms

VCONT = Kv/z = leakance

or

VCONT = Vertical Conductance / (xy)

If Kv and z

are constants

slide7

Vertical Conductance Terms

VCONT = Kv/z = leakance

= Vertical Conductance / (xy)

MODFLOW uses special terms (VCONT) to calculate

vertical conductance between layers. VCONT

values are required in 3D problems.

Modflow.exe requires the user to supply values for

the VCONT arrays. GW Vistas gives you the option to

input the VCONT values, called leakance in GWV,

or will compute the values in the VCONT arrays for you.

slide8

Calculation of leakance for a simple example

3 layers

K= constant

=30 m/day

20 m

L=30/20

20 m

L=30/40

60 m

L=0

slide9

Leakance between layers where each

Layer belongs to a different hydrogeologic unit

See A&W, p. 81

slide11

“Quasi 3D” Models

The effects of confining layers are simulated

using leakance terms. The confining layers are

not physically represented as model layers.

Confining layer

Confining layer

slide12

Example

8 layer model

slide14

Leakance

“Quasi 3D” model

See A&W, p. 81

slide15

3D design of the Hubbertsville Problem

Raycine’s Problem

Three layers

1010

1000

1

1000

2

990

3

980

Leakance = Kv/ z

Layer 1 = 50/10

Layer 2 = 50/10

Layer 3 = 0

But leakance should be

calculated based on the

saturated thickness of the cell.

slide17

from GWV

Remember: In MODFLOW, a layer is considered “confined”

when the head in the cell is above the top of the cell. (If the top

layer is specified to be unconfined, it has no top elevation.)

slide18

3D design of the Hubbertsville Problem

Three layers with well nodes in each layer

Need to apportion total well discharge of 20,000 over

3 layers with 20,000/3 to each layer.

1000

1

2

3

But then, cells in the top layer go dry and MODFLOWshuts off the well. When this happens the total discharge

is less than 20,000.

slide19

3D design of the Hubbertsville Problem

Three layers with well nodes in two layers

1000

1

2

3

Apportion total well discharge of 20,000 to the

bottom 2 layers with 10,000 to each layer.

Now, heads in layers 2 and 3 are similar to heads in

the 2D (one layer) version of the problem.