Preliminary calculation

Elevation e of the fluid in a tank subject to an acceleration a

In a first time, we theoretically calculate the level of water (or retardant) in a tank subject to an acceleration and gravity. Herebelow a diagram of the problem : 

                     

The blue line represent the interface between air and water and we have :

\[ \vec{a}=a.\vec{x} \]

\[\vec{g}=-g.\vec{y} \]

The forces' balance gives :

\[\rho \vec{a}=-\Delta P+\rho\vec{g}\]

After projection, we have : \[\frac{\partial P}{\partial x}=- \rho a \:  and \: \frac{\partial P}{\partial y}=- \rho g \]

On the surface, $ P= constant $ so : \[dP=0= \frac{\partial P}{\partial x} dx + \frac{\partial P}{\partial y} dy \]

Therefore, \[ \rho a.dx + \rho g.dy = 0 \]

 \[ \frac{dy}{dx} |_{surface} = - \frac{a}{g} \]

\[ y = - \frac{a}{g}.x + C \] and we know that $ \: y(x=0)=h \rightarrow C= h $ so $  \: y = - \frac{a}{g}x + h $ and now we can calculate $e$, the level of water : $ e + h = \left(- \frac{a}{g} \right)  . \left(- \frac{L}{2} \right) + h $ and finaly we get :

\[e = \frac{a.L}{2.g}\]

 

Dispersion time T of waves

This calculation is useful to know the time which is necessary for a wave to disperse when the acceleration moves from a value $ a $  to $ 0 $ . Yet, it is important to precise that the calculation is valid only if the value of the acceleration $ a $ is not too high (not over 2.5g). Herebelow is a diagram of the dimensions of the tank we need for this calculation :

 

Hypothesis :

\[\omega^2 = (gk + \frac{\sigma}{\rho}k^2) \tanh (kh)\]

Two hypothesis can be made to calculate the characteristic dispersion time :

• hypothesis of "deep water" : $ \lambda << h $,  $ \omega^2 = gk $,  $ c=\sqrt{\frac{g \lambda}{2 \pi}} $
• hypothesis of "long waves" : $ \lambda >> h $,  $ \omega^2 = gk^2h $,  $ c=\sqrt{gh} $

In our case, we choose to work with the hypothesis of "long waves" .

 

Calculation :

We have : \[\frac{\partial E_c}{\partial t} = D \thicksim 2\mu S_{ij}S_{ij} \thicksim 2\mu \frac{V^2}{L^2}\]

\[E_c \thicksim \rho V^2 \rightarrow \mu \frac{V^2}{L^2} \thicksim \frac{\rho V^2}{T}\]

\[\longrightarrow T = \frac{\rho V^2 L^2}{\mu V^2} \thicksim T=\frac{\rho L^2}{\mu}\]

Here, $ L = e $ (calculated just before), so we have : \[ T \thicksim \frac{\rho}{\mu}e^2\]

Thanks to the previous calculation, we know the expression of $ e $ :

\[e = \frac{aL}{2g}\] So we can write : 

\[T = \frac{\rho a^2 L^2}{4\mu g^2}\]

 

Natural frequency f₀

In the light of the framework of this problem, we can make the parallel with the problem of the oscillating pendulum and suppose, in a first time, that the natural frequency of a wave can be written as following : 

\[f_0 = \frac{1}{2\pi} \sqrt{\frac{g}{L}}\]