3D Simulations--

the vertical dimension

Representation ofhydrogeologic unitsin MODFLOW88

MODFLOW2000

In MODFLOW, vertical flow between layersIs governed by vertical conductance.

Horizontal Conductance

Review

Vertical Conductance Terms

VCONT = Kv/z = leakance

or VCONT = Vertical Conductance / (xy) If Kv and z

are constants

Vertical Conductance Terms

VCONT = Kv/z = leakance

= Vertical Conductance / (xy)

MODFLOW uses special terms (VCONT) to calculatevertical conductance between layers. VCONTvalues are required in 3D problems.

Modflow.exe requires the user to supply values forthe VCONT arrays. GW Vistas gives you the option toinput the VCONT values, called leakance in GWV,or will compute the values in the VCONT arrays for you.

3 layersK= constant =30 m/day

20 m

60 m

20 m

L=30/20

L=0

L=30/40

Calculation of leakance for a simple example

Leakance between layers where eachLayer belongs to a different hydrogeologic unit

See A&W, p. 81

Leakance between layers where both layers belong to the samehydrogeologic unit

See A&W, p. 81

“Quasi 3D” ModelsThe effects of confining layers are simulatedusing leakance terms. The confining layers arenot physically represented as model layers.

Confining layer

Confining layer

8 layer model

Example

Quasi 3D design – 2 layer model

Leakance“Quasi 3D” model

See A&W, p. 81

980

1000

3D design of the Hubbertsville ProblemRaycine’s Problem

Three layers

1000

Leakance = Kv/ z

1

2

3

Layer 1 = 50/10Layer 2 = 50/10Layer 3 = 0

990

But leakance should becalculated based on thesaturated thickness of the cell.

1010

from GWV

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

3D design of the Hubbertsville ProblemThree layers with well nodes in each layer

1000 1

2

3

Need to apportion total well discharge of 20,000 over3 layers with 20,000/3 to each layer.

But then, cells in the top layer go dry and MODFLOWshuts off the well. When this happens the total dischargeis less than 20,000.

3D design of the Hubbertsville ProblemThree 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 inthe 2D (one layer) version of the problem.