## 2 Theory
### Solution Strategy
#### 2.3.2 Power Dissipation  
An equation accounting for power dissipation via the Joule effect has been obtained [38]. This process gives rise to an increase in thermal energy and the corresponding fall in kinetic energy. In this case, volume fraction \(\phi_{l}\) and \(\phi_{S}\) must be incorporated. Revisiting the inviscid equation [39], Equation (23) considers the volume fraction of the liquid. The term at the left side in Equation (23) is the material derivative \(D/Dt=d/dt+(\mathbf{u}\cdot\
abla)\mathbf{u}\), with \(\mathbf{u}\) the velocity. \[\frac{D}{Dt}(\phi_{l}\rho_{l}\mathbf{u}_{l})=-\
abla p+\mathbf{J}\times\mathbf{B_{0}} \tag{23}\]  
Multiplied by \(\mathbf{u}_{l}\) (the fluid velocity), in the sense of a dot product, Equation (23) leads to Equation (24). \[\frac{D}{Dt}\Bigg{(}\phi_{l}\rho_{l}\frac{\mathbf{u}_{l}^{2}}{2}\Bigg{)}=(\mathbf{J} \times\mathbf{B_{0}})\cdot\mathbf{u}_{l}-\
abla\cdot(\rho\mathbf{u}_{l}) \tag{24}\]  
Rearranging terms allows us to write Equation (24) as follows. \[\frac{D}{Dt}\Bigg{(}\phi_{l}\rho_{l}\frac{\mathbf{u}_{l}^{2}}{2}\Bigg{)}=-\mathbf{J} \cdot(\mathbf{u}_{l}\times\mathbf{B_{0}})-\
abla\cdot(\rho\mathbf{u}_{l}) \tag{25}\]  
Using Equation (11) in the form of \(\mathbf{u}_{l}\times\mathbf{B_{0}}=\mathbf{J}/\sigma\) in Equation (25) and integrating over the entire volume leads to Equation (26). The volume integral of \(\rho_{l}u_{l}{}^{2}/2\) is the global kinetic energy (\(\Xi\)). \[\frac{D}{Dt}\int_{V}\rho_{l}\frac{\mathbf{u}_{l}^{2}}{2}dV=\frac{D}{Dt}\Xi=-\frac {1}{\sigma\phi_{l}}\int\mathbf{J}^{2}dV \tag{26}\]  
The kinetic energy can be approximated by solving Equation (27), where \(l_{\parallel}\) is the characteristic length scale for the flow (parallel to the magnetic field), and \(l_{min}\) is the smallest eddy size in a turbulent flow. The solution is expressed in Equation (28) [40]. This equation anticipates that, provided \(l_{min}\) and \(l_{\parallel}\) maintain the same order, the flow will be annihilated on a time scale \(\tau\). The dissipation is subject to the constraint that neither the creation nor the destruction of linear momentum is possible. This phenomenology has been treated in [41], where the discussion is centered in the properties of fluids moving in a uniform magnetic field, showing that the component parallel to the magnetic field is conserved while the normal components decay exponentially. \[\frac{D}{Dt}\Xi=-\Bigg{(}\frac{l_{min}}{l_{\parallel}}\Bigg{)}^{2}\frac{Dt}{\tau} \tag{27}\]\[\Xi(t)=\Xi_{0}e^{-\left(\frac{t_{min}}{t_{0}}\right)^{2}\frac{t}{t}} \tag{28}\]  
Denoting \(P\), the power dissipation via the Joule effect, as \(P=-\int\mathbf{J}^{2}dV/(\sigma\Phi_{l})\) and using \(J=-\sigma B_{0}w_{l}(r)\mathbf{\hat{e}_{\theta}}\) in Equation (28) (Section 2.3), the magnitude of the power dissipation can be found via Equation (29), meaning that the higher the conductivity, the higher the volume fraction of the liquid. Moreover, the higher the fluid velocity is, the higher the magnitude of the power dissipation through the Joule effect becomes. The volume can be expressed in terms of the generic \(r\)-coordinate, i.e., \(V=V(r)\), so the volume element \(dV\) is \(r\)-dependent. \[\left|P(t)\right|=\frac{\sigma B_{0}{}^{2}}{\Phi_{l}}\int_{V(r)}w_{l}(r,t)^{2}dV \tag{29}\]  
To use Equation (29), the domain corresponding to the portion of the flask where the fluid is circulating and the power is dissipated must be found. It can be obtained by defining the total volume \(V\) as the volume of a sphere of radius \(r\left(V_{0}=4\pi r^{3}/3\right)\) minus the volume of the top (\(V_{1}\)) and bottom (\(V_{2}\)) cap, defined by the water level and the geometry at the bottom of the flask, and minus the volume of the part of cylinder inside the flask (\(V_{3}\)) though which air is introduced. Thus, \(V=V_{0}-(V_{1}+V_{2}+V_{3})\), and the volume element \(dV\) can be expressed in terms of the \(r\)-coordinate.