# Numerical Calculation of Slosh Dissipation
## 3 Relation between Energy Dissipation and Fluid Center of Mass Motion  
Considering the motion of the tank as a pure vertical translation, following the analysis in [12], the energy balance can be reshaped as:  
\[E_{K}(t)\,+\,m_{l}\,g\,y_{G}(t)\,+\,m_{l}\,\int_{t_{0}}^{t}\dot{y}_{G}(t)\,a_ {tunk}(t)\,dt\,=\,E_{diss}(t)\,,\quad y_{G}(t)\,:=\,\frac{\sum_{i}\,\rho_{i}\, V_{i}\,y_{i}}{m_{l}}\,, \tag{9}\]  
where \(a_{tunk}\) is the vertical acceleration of the tank and \(y_{G}\) the position of the fluid center of mass in the Ni-FoR. The latter is evaluated in a discrete form on the finite volume cells or on the SPH particles.  
The first equation of (9) highlights the role of the motion of the fluid center of mass in the Ni-FoR concerning the energy dissipated by the liquid sloshing. In particular, the phase lag between the velocity \(y_{G}\) and \(a_{tunk}(t)\) is a crucial aspect of the phenomenon. The third term of the left-hand side of the first Equation of (9) is the work completed by the non-inertial force acting on the liquid:  
\[W_{NF}(t)\,=\,\int_{t_{0}}^{t}\,\mathcal{P}_{NF}\,dt\,=\,\,-m_{l}\,\int_{t_{0 }}^{t}\dot{y}_{G}(t)\,a_{tunk}(t)\,dt \tag{10}\]  
In the problem studied in this article, \(W_{NF}\) is much larger than the kinetic and potential energy in the Ni-FoR, and therefore, as the time increases, \(W_{NF}\) becomes increasingly closer to the dissipated energy \(E_{diss}\). Considering the energy balance (9) at the time \(t_{f}=30\,\)T at the end of the experiment, when the fluid almost comes back to a rest condition, they are going to be very close to each other:  
\[W_{NF}(t_{f})\approx E_{diss}(t_{f})\]  
Following again [12], one can find that the acceleration of the center of mass \(\dot{y}_{G}\) is linked to the vertical forces \(F_{y}\) that the liquid exerts on the tank:  
\[m_{l}\,\ddot{y}_{G}(t)\,=\,-F_{y}(t)\,-\,m_{l}\,[g+a_{tunk}(t)] \tag{11}\]  
and the work completed by those forces of the above expression is:  
\[W_{ext}^{dyn}(t)\,=\,m_{l}\,\int_{t_{0}}^{t}\dot{y}_{G}(t)\,v_{tunk}(t)\,dt \tag{12}\]  
If the fluid is \"frozen\" (i.e., no liquid sloshing), the above external work performed by the tank walls on the fluid is zero since the center of mass \(y_{G}\) is a constant in the Ni-FoR. Considering that at \(t=t_{0}\), the tank is in a rest condition, using integration by parts, it follows that:  
\[W_{ext}^{dyn}(t)\,=\,m_{l}\,\dot{y}_{G}(t)\,v_{tunk}(t)\,+\,W_{NF}(t) \tag{13}\]  
where, as explained in [12], the first term on the right-hand side is linked to differences in the mechanical energy (i.e., \(E_{K}+E_{P}\)) evaluated in the I-FoR and in the Ni-FoR. Again,similarly to what was stated above, at the end of the experiment (i.e., \(t=t_{f}\)), the tank does not move anymore; hence, the two work terms \(W_{ext}^{dyn}\) and \(W_{NF}\), coincide and are practically both equal to the dissipated energy \(E_{diss}\) (see also Section 5.1).  
It is worth noting that the center of mass motion \(y_{G}(t)\) can be experimentally evaluated measuring \(F_{y}\) with a load cell, while the recording of \(y_{G}(t)\) from video-camera frames would require a more complex indirect measure, especially in a three-dimensional context [private communication, [PERSON], 2021][36]. Finally, all the relations written in this section are valid both for no-slip and free-slip conditions on the tank walls. In fact, those boundary conditions influence the fluid motion, which reflects directly on the center of mass motion \(y_{G}(t)\).