Freatic groundwater levels benchmark (Water Module)

This benchmark demonstrates a situation where a parcel of land is situated between two waterways, each with a stable but different water level. In combination with continuous rainfall, a characteristic curve will form over time, as shown in the image below.

Formulas

Due to continuous rainfall and two stable water levels left and right, a specific ground water table curve will form. Rainwater flows both left and right. Note that the time at which this balance occurs is dependent on the starting situation.

The following formula, taken from literature^[1], describes the curve of ground water levels when the ground water flow to the left and right have become stable:

$h^{2} (x) = h_{0}^{2} - (\frac{h_{0}^{2} - h_{L}^{2}}{L}) x + \frac{N}{k} (L - x) x$

where:

x

: distance of the left waterway edge (m)

h 0

: water level at x = 0 (m)

h L

: water level at x = L (m)

L

: distance between both waterways (m)

k

: horizontal infiltration speed (m / day)

N

: additional ground water (m/day).

Setup

The following setup has been taken from the use case Peilverschil over een strook grond (freatisch) at grondwaterformules.nl.

In the situation described there, the chosen Length L is set to 500 m. This is achieved with a grid size of 28 by 5, with a cell size of 20 m. Two stable groundwater levels are used: 11m (datum) on the left and 10m on the right. The rainfall is 0.8 mm per day, the horizontal infiltration speed is 3 m per day.

The terrain height is set to 12 for all cells and the ground bottom distance is also set to 12.

The terrain type's infiltration speed can be configured to 3 m / day.

For x = 1 the initial ground water level is set to 11 and for x = 26 the ground water level is set to 10. Initial groundwater levels between these points are linearly interpolated.

Additionally, two underground inlets are placed, one on x=1 and one on x=26, as an area over y = 1 to 3.

The simulation time is set to n days, with a rainfall of 0.8 mm per day. Configure the rain as: $[3600 \cdot 24 \cdot n, \frac{0.8}{1000 \cdot 3600 \cdot 24}]$ .