This simple example provides an introduction to the kinds of problems that can be solved using Pipes. This particular example file, Pipe1.gsm, can be found in the Contaminant Transport Examples folder in your GoldSim directory (accessed by selecting File | Open Example... from the main menu).
Suppose that you wish to simulate a simple column experiment. In this experiment, water flows at a constant rate through a saturated cylindrical column filled with sand (porosity of 0.3 and bulk density of 1600 kg/m3). The column is 2 m long and has a cross-sectional area of 0.05m2. Water flows through the column at a rate of 1E-3 m3/day, and the dispersivity is assumed to be 10% of the column length. At one end of the column, two species (A and B) are injected at a rate of 1 mg/day. A does not partition onto the sand, but B does, with a partition coefficient of 2E-4 m3/kg. You wish to simulate the breakthrough of the two species out the end of the column.
To simulate this system in GoldSim, you would do the following:
1. Define two species (A and B);
2. Define two media (Water and Sand) and specify their properties;
3. Define a single Pipe to represent the column;
4. Specify the properties of the Pipe;
5. Define a sink Cell and create an advective mass flux link from the Pipe to the Cell.
6. Specify the boundary condition for the Pipe; and
7. Specify the simulation settings (i.e., duration and timestep) and run the model.
The output of this simulation, in the form of time histories of the mass rate of species A and B exiting the column, is shown below:
The mean unretarded travel time, t, through a Pipe can be computed as follows:
t = n A L / Q S
where n is the porosity of the infill, A is the cross-sectional area of the Pipe, L is the length of the Pipe, Q is the flow rate and S is the fluid saturation.
In this example,
t = (0.3)(0.05 m2)(2 m) / 1E-3 m3/day = 30 days
As can be seen from the figure, the unretarded species (A) reaches 50% of its steady state flux rate just befoe 30 days (due to the impact of boundary conditions, the solution to the equation is asymmetric and the 50% breakthrough value is expected to occur slightly before the calculated mean travel time). The spread around this point is due to dispersion in the pathway. Species B reaches 50% of its steady state flux at about 60 days. This is because the effective retardation factor, R, in this particular case can be computed as:
where ρb, Kd and n are the bulk density, partition coefficient and porosity, respectively, for the Solid infill through which the fluid is flowing.
Hence, in this example, the breakthrough curve was retarded by about a factor of two.