The theoretical expression of waves' dissipation time is $ \displaystyle T = \frac{\rho a^2 L^2}{4 \mu g^2} $.

The influence of the parameters (viscosity $\mu$, tank's length L, initial water level h, and plan's acceleration a) will be studied to obtain the limits of this expression and to validate the numerical model.

The following drawing remind the configuration of the problem :

For the first simulations, the liquid phase is water but then we tested a liquid phase which comes up to a retardant substance. We take as hypothesis that the viscosity of a retardant is between the water viscosity and a thousand times the water viscosity. So we run four simulations for the four following cases with the same acceleration (a=1m/s²):

**Water level on the left wall function of time for the four viscosities **

For the four cases, we can see the oscillations obtained with Fluent and the theoretical envelope : $ y=A.e^{\frac{t}{T}} $

*$ \mu = 1000 \mu_{water} $*

*$ \mu = 750 \mu_{water} $*

*$ \mu = 250 \mu_{water} $*

**

*$ \mu = \mu_{water} $*

The results fit the theoretical dissipation time for the four viscosities

**Waves dissipation for the four viscosities (exponential envelopes)**

The more retardant there is, the shorter is the dissipation time and the briefer the sloshing is.

**Water level on the left wall for several tank's length **

*- L=1.5 m - L=2m*

**

*- L=2.5 m - L=3m*

*- L=3.5 m *

The results fit the theretical dissipation time for most of tank's lengths. However, for L=1.5m, Fluent results does not match the theory. For this case the length (L) and the height (H) are almost the same so the long waves hypothesis is no longer valid.

**Waves dissipation for several tank's length (exponential envelopes) **

Smaller is the tank, briefer is the sloshing.

**

Another parameter to take into account is the water/retardant intial level (or fill rate) in the tank.

The three cases in the middle of the graph (green, magenta and blue) have the same dissipation time. For the others two cases, the sloshing is briefer. During these two cases, there is a wetting, that is to say that the free surface touch the bottom.

So, without wetting, the dissipation time does not depend on the intial water level, and with wetting, the sloshing is brief and theoretical dissipation time is wrong.

A really important factor in this study is the acceleration of the plane. This is the more complicated factor to study because of high accelerations, with which wetting happens.

Firstly, the theoretical dissipation time is proportional to the squared acceleration over the squared gravity : $ T^2 \propto \frac{a^2}{g^2} $. From the plot, the dissipation time is larger for the high acceleration (blue curve). There is a contradiction with the theory.

In fact, $ T_{blue curve_ {th}} \approx 10^{4} s $ and $ T_{blue curve_{plot}} \approx 50 s $

But in this case (high acceleration), wetting happens. The following drawing shows the free surface form : .

We try to imagine a new case : we take the same tank turned of 90 degrees and we reverse the acceleration and the gravity :

We obtain a new dissipation time : $ T'_{blue curve_{th}} = 500 s $. This result is closer but it is not exact.

**$\Rightarrow$ To conclude this part about dissipation time, its theoretical expression is correct without wetting.**