Bottom flow pankow benchmark (Water Module)

This testcase demonstrates a situation where a parcel of land is situated between two waterways with a stable water level. In combination with ground water flowing from deeper ground layers through the bottom boundary, a characteristic curve will form over time, as shown in the image below. It is described in a water balance research in 1968 of Pankow and Rijtema , and also contains the accompanying formulas that describe the curve. In this case, we use the case which does not take into account the additional water flow resistance of the waterway. Secondly, rain could also be taken into account, but we set that to 0 to exclusively benchmark the seepage mechanics. The continuous rainfall case is already tested in freatic groundwater levels benchmark.

A parcel of land situated between two waterways with ground bottom flow and optional continuous rainfall

Formulas

Due to ground bottom flow and two stable water levels left and right, a specific ground water table curve will form. Part of the water seeped in will flow left and part of the it will flow right. Note that the time at which this balance occurs is dependent on the starting situation.

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

$A={\frac {(h_{1}-h_{2})cosh{\frac {x_{1}}{\sqrt {KD\cdot c}}}}{{\frac {x_{2}^{2}}{2KD}}\cdot {\frac {2KD\cdot c}{x_{2}^{2}}}\cdot (cosh({\frac {x_{2}}{\sqrt {KD\cdot c}}})-1)-{\frac {x_{1}^{2}}{2KD}}\cdot {\frac {2KD\cdot c}{x_{1}^{2}}}\cdot (cosh({\frac {x_{1}}{\sqrt {KD\cdot c}}})-1)}}={\frac {(h_{1}-h_{2})cosh{\frac {x_{1}}{\sqrt {KD\cdot c}}}}{c\cdot (cosh({\frac {x_{2}}{\sqrt {KD\cdot c}}})-1)-c\cdot (cosh({\frac {x_{1}}{\sqrt {KD\cdot c}}})-1)}}$

$A=N+{\frac {h_{d}-h_{1}}{c}}$

Simplified without rain N:

$h_{1}=h_{d}-{\frac {(h_{1}-h_{2})cosh{\frac {x_{1}}{\sqrt {KD\cdot c}}}}{(cosh({\frac {x_{2}}{\sqrt {KD\cdot c}}})-1)-(cosh({\frac {x_{1}}{\sqrt {KD\cdot c}}})-1)}}$

To test the correctness of the ground bottom flow, the formula can test the following condition with the accepted error margin $\epsilon$:

$\left|h_{d}-{\frac {(h_{1}-h_{2})cosh{\frac {x_{1}}{\sqrt {KD\cdot c}}}}{(cosh({\frac {x_{2}}{\sqrt {KD\cdot c}}})-1)-(cosh({\frac {x_{1}}{\sqrt {KD\cdot c}}})-1)}}-h_{1}\right|<\epsilon$

where:

$x_{1}$: distance (m) of the point of measurement compared to the middle of the parcel.

$x_{2}$: distance (m) of the second point of measurement, which is always situated 3 meters from the edge of the waterway.

$h_{1}$: measured ground water level (m) at $x_{1}$

$h_{2}$: measured ground water level (m) at $x_{2}$

$h_{d}$: bottom head pressure

$KD$: measured horizontal transmissivity of the ground layer (m2/day)

$N$: additional ground water due to rainfall (m/day).

$\epsilon$: accepted error margin

Setup

We use the following setup. The grid size used is 53 by 5, with a configurable cell size of $dx$ in meters. There are two waterways, left and right, both with a stable water level of 3 meters.

One inlet is placed on the cells x = 1 and y = 1 to 3 and an other is placed on the cells x = 52 and y - 1 to 3, with the following setup to ensure a stable water level:

UPPER_THRESHOLD set to 3 m.

LOWER_THRESHOLD set to 3 m.

Inlet Q set to 0, such that is unlimited.

The terrain height is set to: ${\begin{cases}0,&{\text{if }}x<4{\text{ or }}x>48\\1.5,&{\text{if }}x=4{\text{ or }}x=48\\3.5,&{\text{if }}x=5{\text{ or }}x=47\\5,&{\text{otherwise}}\end{cases}}$

Depending on the specific test case, the Bottom head pressure $h_{p}$ is set to:

${\begin{cases}h_{p},&{\text{if }}x>5{\text{ and }}x<47\\none,&{\text{otherwise}}\end{cases}}$

Similarly, the Bottom resistance (in days) is set to:

${\begin{cases}c,&{\text{if }}x>5{\text{ and }}x<47\\none,&{\text{otherwise}}\end{cases}}$

An aquifer can be added to configure the horizontal infiltration speed. An important thing to note here, is that the KD value used in the report is derived from changes in water head over time, and does not mention the storage capacity of the soil. Therefor, in order to configure the Aquifer KD value in the Tygron Platform, the KD value has to be multiplied with the storage percentage.

The water level is configured to 3 m for all cells, resulting in an initial water level above and below ground of 3 meters.

The simulation time is set to n days, with rain set to 0 mm. To configure this, the rain set is set to $[3600\cdot 24\cdot n,0]$.

Results

Test case 1

cell size: 10 m;

$KD$: 17 m²/day;

Water storage fraction: 0.10;

Aquifer KD configured to 1.7 m²/day in the Tygron Platform;

Bottom resistance $c$: 20 days

Bottom head pressure $h_{d}$: 4.1

Simulation days $n$: 128 days

Test case 2

cell size: 10 m;

$KD$: 22 m²/day;

Water storage fraction: 0.25;

Aquifer KD configured to 5.5 m²/day in the Tygron Platform;

Bottom resistance $c$: 40 days

Bottom head pressure $h_{d}$: 4.1

Simulation days $n$: 128 days

Test case 3

cell size: 10 m;

$KD$: 22 m²/day;

Water storage fraction: 0.25;

Aquifer KD configured to 5.5 m²/day in the Tygron Platform;

Bottom resistance $c$: 40 days

Bottom head pressure $h_{d}$: 3.8

Simulation days $n$: 128 days

Test case 4

cell size: 10 m;

$KD$: 22 m²/day;

Water storage fraction: 0.25;

Aquifer KD configured to 5.5 m²/day in the Tygron Platform;

Bottom resistance $c$: 10 days

Bottom head pressure $h_{d}$: 4.8

Simulation days $n$: 128 days

Test case 5

cell size: 10 m;

$KD$: 17 m²/day;

Water storage fraction: 0.10;

Aquifer KD configured to 1.7 m²/day in the Tygron Platform;

Bottom resistance $c$: 20 days

Bottom head pressure $h_{d}$: 4.1 (but leads to $h_{d}=4.077$ for the comparisons with the middle value)

Simulation days $n$: 64 days

Test case 6

cell size: 10 m;

$KD$: 22 m²/day;

Water storage fraction: 0.33;

Aquifer KD configured to 7.26 m²/day in the Tygron Platform;

Bottom resistance $c$: 10 days

Bottom head pressure $h_{d}$: 4.5 (but leads to $h_{d}=4.499$ for the comparisons with the middle value)

Simulation days $n$: 64 days

Test case 7

cell size: 10 m;

$KD$: 220 m²/day;

Water storage fraction: 0.25;

Aquifer KD configured to 55 m²/day in the Tygron Platform;

Bottom resistance $c$: 20 days

Bottom head pressure $h_{d}$: 4.1

Simulation days $n$: 64 days

Test case 8

cell size: 10 m;

$KD$: 220 m²/day;

Water storage fraction: 0.10;

Aquifer KD configured to 55 m²/day in the Tygron Platform;

Bottom resistance $c$: 20 days

Bottom head pressure $h_{d}$: 4.1

Simulation days $n$: 64 days

Test case 9

cell size: 5 m;

$KD$: 22 m²/day;

Water storage fraction: 0.10;

Aquifer KD configured to 2.2 m²/day in the Tygron Platform;

Bottom resistance $c$: 20 days

Bottom head pressure $h_{d}$: 4.1

Simulation days $n$: 64 days

Test case 10

cell size: 2 m;

$KD$: 22 m²/day;

Water storage fraction: 0.10;

Aquifer KD configured to 2.2 m²/day in the Tygron Platform;

Bottom resistance $c$: 20 days

Bottom head pressure $h_{d}$: 4.1

Simulation days $n$: 64 days

Notes

• A higher KD value can lead to a lower stable ground water level than the head pressure.

References

Water Module

Tutorials

Overview

Theory

Benchmarks