A parametric study was performed with the software Fluent ANSYS in laminar flow.

The four following parameters varied :

- Fluid's viscosity : $\mu $ = {0.001, 0.25, 0.75, 1} Pa.s

- Tank's lenght : L = {1.5, 2, 2.5, 3, 3.5} m

- Inital water level : h = {0.2, 0.5, 0.7, 1.1} m

- Plan's acceleration : a = {1, g, 2.5g} m/s^{2}

The elevation of water level obtained with Fluent match the theoritical expression for all the cases.

Therefore, theoretical expressions and Fluent's results of dissipation time and natural frequency will be compared in the next parts.

**Meshing**

The first step of the numerical analysis is creating the mesh which will be our base to launch simulations. To do that, we used the software ICEM CFD, and we created a rectangular mesh representing the tank of water or retardant. The study is done in 2D.

Once the geometry created, we have to mesh it and to define the type of boundary conditions (only walls) before exporting it to FLUENT.

Initially, we worked with a rough mesh (5000 nodes) and then we refined it (up to 200 000 nodes), but the results were pretty much the same. So, to avoid longer calculation time, we opted for the rough mesh for all the simulations.

**Fluent - VOF Model**

The first step was to comprehend the diphasic mode on FLUENT, given the fact that we never used that mode before. There are different options in the diphasic model : Eulerian mode, mixture mode, and Volume Of Fluid mode. We chose to work with the VOF (Volume Of Fluid) option.

In the VOF model, the input are the following :

- number of phases
- volume fraction scheme (explicit or implicit)
- inclusion of the implicit body force formulation (optional)
- coupled level set with VOF (optional)
- inclusion of open channel flow
- inclusion of open channel wave boundary conditions
- inclusion of zonal discretization for applications such as diffused interface modeling in one zone and sharp interface modeling in another zone

Concerning our problem, we chose two phases, an explicit volume fraction scheme and the inclusion of the implicit body force formulation.

*The explicit scheme*

Explicit schemes can take on different forms :

used for hybrid meshes (containing twisted hexahedral cells)*time-dependent with the explicit interpolation scheme*:used if there is an interest in the time-accurate transient behavior of the VOF solution.*time-dependent with the geometric reconstruction interpolation scheme*:available only for quadrilateral and hexahedral meshes.*time-dependent with the donor-acceptor interpolation scheme*:gives interface sharpness of the same level as the geometric reconstruction scheme and is particularly suitable for flows with high viscosity ratio between the phase.*CICSAM scheme*:

In our case, we can use georeconstruct scheme or CICSAM scheme.

__Including body forces__

When large body forces (like gravity or surface tension forces) exist in multiphase flows, the body force and pressure gradient terms in the momentum equation are almoste in equilibrium, with the contributions of convective and viscous terms small in comparison. Segragated algorithms converge poorly unless partial equilibrium of pressure gradient and body forces is taken into account. That is why the optional "implicit body force" treatment has its interest in FLUENT. Indeed, this option account this effect and in this way make the solution more robust by achieving a realistic pressure field very early in the iterative process.

Then, after choosing the model and the options, we defined the phases : their material properties, interaction between them (here surface tension). In the VOF model, you can specify the primary and secondary phases as you prefer. In our case, we have air and water and it is preferable to specify the compressible ideal gas as the primary phase to improve solution stability. So the primary phase is air and the secondary is water.

If you have a difference between the densities of the two phases around 1000 or more of that, you have to choose, in the boundary conditions>operating conditions an operating density equel to 0 kg/m^{3}. This choice allow users to have a better convergence.

*Initialization*

- Initialize all the zone with a water volume fraction equal to 0
- Define a region of water (specifying the level of water) : >Adapt>Region and click on Mark
- Patch the water volume : Initialization>Patch. Choose, water volume function, value equals to 1, hewahedron-0. Click on Patch
- Check that the two phases are well defined : >Display>Contours>Phases>Water volum fraction (options : click on filled).

Several parameters have influence on the results of the simulations and can be adjusted :

- the accelereration of the plane $ a [m/s²] $
- the viscosity of the fluid (water or retaradant) $ \mu [Pa.s] $
- the length of the tank $ L [m] $
- the fill rate $ N=\frac{h}{L} $, or water level $ h [m] $ in the tank

By making vary all these parameters, we tested different cases ; herebelow is a recap chart of the different simulations :

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.**

Thanks to a parallel with the oscillating pendulum problem, we suppose that the waves' natural frequency has the following expression : $ f_0= \frac{1}{2\pi} \sqrt{\frac{g}{L}} $.

In a first time, we will check that the fluid's viscosity does not influence the natural frequency and we will verify the theoretical expression.

The natural frequency is the same for the four viscosities so the viscosity does not have any influence on the natural frequency.

The theoretical expression gives a natural frequency equal to 0,27 Hz but with Fluent results (plot) we obtain a natural frequency equal to 0,35 Hz. The expression is not right.

The expression of the natural frequency is modified supposing that it depends on all the dimensions of the problem. The expression becomes : $ \displaystyle f_0 = \frac{k}{2 \pi} \sqrt{\frac{g}{L*}} $ with L*=L*(L,h) a characteristic length and k a constant. We will find this characteristic length studying the influence of the initial level (h) and of the tank's length (L) on the natural frequency (f_{0}).

The Fast Fourier Transform of initial water level for five various initial water levels (first plot) allows to know that the initial water level has an influence on the natural frequency. Thanks to these Fast Fourier Transforms, the natural frequency can be extracted for all the cases and we obtain the second plot.

This plot represents the natural frequency (obtained after a fast fourier transform) for four initial water levels. Blue points are the Fluent results. A curve fitting returned a square root function (red points). So the natural frequency is proportional to the squared root of the initial water level : $f_0 \propto \sqrt{h} $. Therefore the characteristic length is proportional to the inverse of the initial level : $ L^* \propto \frac{1}{h} $. We can suppose that the characteristic length is proportional to the squared of tank's lenght over the intial water level : $ L^* \propto \frac{L^2}{h} $. We will verify this expression studying the influence of tank's lenght on the natural frequency.

The Fast Fourier Transform of water level for five various tank's length (first plot) allows to know that the length have an influence on the natural frequency. Thanks to these Fast Fourier Transforms, the natural frequency can be extracted for all the cases and we obtain the second plot.

This plot represents the natural frequency (obtained after a fast fourier transform) for five tank's length. Blue points are the Fluent results. A curve fitting returned an inverse function (red points). So the natural frequency is proportional to the inverse of the tank's length : $ f_0 \propto \frac{1}{L} $. Therefore the characteristic length is proportional to the squared of tank's length : $ L^* \propto L^2 $.

Thus the characteristic length is proportional to the squared of tank's lenght over the intial water level :

**$ \Rightarrow \displaystyle L^* \propto \frac{L^2}{h} $ **

All cases will be ploted in function of this characteristic length in the next part.

Thanks to the previous studies, the expression of the characteristic length is known as $ \displaystyle L^* = \frac{L^2}{h} $. All the studied cases are gathered on the following plot :

Blue points are Fluent Results. Red points are the mathematical approximation :

**$ \displaystyle f_0 = \frac{2.73}{2 \pi} \sqrt{\frac{g}{L*}} $**

Results and approximation match for almost all the cases. For the first one, there is a difference which can be explained by the non-validity of the long waves hypothesis for this case (L=1.5m).

The characteristic length and so the expression of the natural frequency were groped for but we find the waves celerity. In fact, with the long waves hypothesis, the waves' celerity is equal to : $ c = \sqrt{gh} $

So the new expression of the natural frequency is :

**$ \displaystyle f_0 = 0.53 \frac{c}{L} $**

$\Rightarrow$ **To conclude this part about the natural frequency, we found an expression function of a characteristic length and it does not depend on the fluid's viscosity.**

A solution to reduce sloshing in the tank is to compartmentalize the latter. Indeed, inserting walls into the tank can be a way to counter the sloshing of water or retardant, and consequently it can help to reduce the mechanical stress exerted on the plane. To realize that part of the study, we create a new geometry, including the new walls. Herebelow is a diagram of this new geometry in which we inserted four walls :

*Compartmentalized tank with four walls (same dimensions of the tank as before)*

Herebelow are two plots : water level function of time on the left wall and on the fourth inside wall, and fast fourier transform of water level for the same two walls :

We can notice that there is the same frequency on the two walls, but a phase difference appears. Plus, if we compare Fluent's results with the theory, we have :

Analysing this values, we see that the water level elevation is the same with Fluent and in theory. However, it is not the case for frequency and dissipation time. Indeed, Fluent results differs from theory for frequency and dissipation time. This can be due to the fact that the "long waves" hypothesis is no longer valid or to the fact that each part of the tank communicates with the others at the bottom.

Apart from that, the **compartmentalized tank solution obviously helps to reduce the sloshing **(elevation of water level much less important than before).