Test 1: Uniform filling of a canal

*1°) Presentation of the test*

-Objectives of the test

The goal of this easy test is to check the behaviour of Mage in case of lateral flows. Consequently, we simulate the filling of a basin by lateral supplies regularly distributed on the lenght.

- Presentation of the canal

The basin is represented by a reach with a rectangular section, its lenght is 750 m, its width is 100 m and the depth equals 5 m.

- Singularity

To simulate a wall at the downstream of the canal, an inverted valve is inserted at the abscissa x=750 m. Its coefficient of flow rate equals 0 and its lenght is 100m.

- Friction

The chosen Strickler coefficient is K=50. In this test the friction has not any influence.

- Boundary conditions

- * To simulate the wall upstream we imposed a flow equals to 0.

* Dowstream, the level of the free surface is chosen equals to 1m.

At the beginning the flow is motionless. The level of water is Z = 2 m and there is no flow all along the basin.

- Lateral flows

A uniform lateral flow of 0.1 m3/s/m is imposed along the canal.

- Numerical data

- * Final time of the simulation: 15 min

* Step time: 60 s

* Space step: 10 m

We expected to observe a linear increase of the level of water with the time. That is what we can notice in the limnigram [ Z(t)] below:

For different times (t=0s in blue, t=0.1h in red and t=0.25h=15min in light blue)we can have the level of the free surface along the canal

More accurately, the height of water increases of 0.06 m / min.. Indeed the linear flow rate is 0.1m3/s/m and the width of the canal equals to 100 m.

The preservation of the flow equals to 0 in the basin can be observed in the hydrogram below:

Test 2: Back to the initial steady state

*1°) Presentation of the test*

- Objective of the test

The aim of this test is to validate easily the calculations obtained by Mage. Starting from an initial steady state, we make an unsteady calculation by varying boundary conditions with time. Back to the initial condition, the result obtained at the end of the calculation must be the sams as the initial steady state. Moreover, we check the mass conservation.

- Presentation of the canal

We modelise a 10km lenght canal, defined by three reaches as follows:

- Friction

The Strickler coefficient is defined on each reach:

- * K= 40 for 0<x<2000

* K= 30 for 2000<x<8000

* K= 40 for 8000<x<10000

Warning : if in a limnigram the last water level Z is the same as the first, Mage considers the downstream condition like a tide. To avoid this problem, we impose the level of water Z=3 m until the end of the simulation.

- Initial condition

Upstream Q = 100 m3/s and downstream Z = 3 m. FLUVIA gives a steady flow but we make a second calculation with Mage to eliminate differences between numerical schemes.

- Numerical data

- * Final time of the simulation: 1 jour

* Step time: 10 min

* Space step: 200 m

The final limnigram of the test is:

The most important point to notice in this graph is the downstream level of water. Z(tf)= 3m is the same as the initial condition. We have what we expected.

Moreover it is interesting to compare the initial and the final upstream
flow to be sure that the calculation give us right results.