기술자료 Strand7 Using Strand7 to model groundwater flow (seepage) problems

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Using Strand7 to model groundwater flow (seepage) problems

The Strand7 HEAT Flow solver solves the Laplace Equation. In addition to heat flow, this equation may also be used to describe groundwater flow (seepage). In the Laplace equation:

the variables have the following meaning:

Variable	Heat Conduction Problem	Seepage Problem
K	Thermal Conductivity	Permeability
f	Temperature at Nodes	Pressure Head at Nodes
Q	Internal Heat Generation	Flow due to a source or sink

To use the Strand7 Heat Solver for the analysis of seepage problems, we take the following steps:

Build the finite element model.
At model boundaries where there is no flow normal to the surface, do not apply any boundary conditions.
At regions of the model where there are flow sources and sinks, enter a Q value (equal to the flow rate per unit volume) for that element's property set. Note that if the source/sink is localised, we need a different property set for this element.
Define the pressure head in the model as follows:
1. Locate a level on the model to be used as the reference level. At this level set the nodal temperature to zero. (Zero nodal temperature indicates a pressure head value of zero).
2. Locate another level (or other levels) on the model where we have a known pressure head, relative to the zero level above. At these levels, apply nodal temperatures equal to the pressure head.
Enter the material data for the model. Here we enter the material permeability as the Conductivity entry in the element's Heat Property Data.
Solve the model using the Heat Solver.
Results can be plotted as follows:
1. Temperature contours are contours of pressure head.
2. Flux contours are contours of flow velocity.

NOTE

For transient seepage problems, we simply run the Transient Heat Solver (either linear or nonlinear depending upon the material properties).

EXAMPLE (Ref: "A Simple Guide to Finite Elements", Owen and Hinton)

The structure is an axisymmetric model of the groundwater flow beneath an inverted steel cylindrical vessel. There is no flow through the cylinder (only around it).

The excavated level is 3.0m below the ground level and the ground has a permeability of 0.864 m/day. Therefore there is a pressure head of 3.0 at the ground level relative to the excavated level.

The following is a representation of the physical situation: