## 2 Formulation of the Problem  
A horizontal magnetic field with strength \(\beta_{0}\) is influencing on an inclined trapezoidal domain heated by a triangular fin, as illustrated in Figure 1. The inclined walls of the containers are referred to with a low temperature \(T_{c}\) while the upper wall is insulated. The inclination angle is \(\gamma\) and the gravity and magnetic vectors are, respectively, \((-g\sin\gamma,\;-g\cos\gamma),\;(\beta_{0}\cos\gamma,\;-\beta_{0}\sin\gamma)\). The trapezoidal angle is \(\Phi\) and the geometry is filled by a variable properties porous medium. The non-equilibrium thermal model between the hybrid nanofluid and porous phases is valid. The mixture consists of alumina and copper nanoparticles suspended in a base fluid (water). Properties of the nanoparticles and water are given in Table 1. In addition, the porous elements are considered as a compacted spheres bed. Therefore, the model proposed by [PERSON] [43] for permeability is introduced. One of the features of this model is that the permeability depends on the particle size (particle diameter \(dp\)) and the porosity of the medium \(\varepsilon\). It is expressed as:  
\[K=\frac{d_{p}^{2}\varepsilon^{3}}{150{(1-\varepsilon)}^{2}} \tag{1}\]  
Figure 1: Schematic diagram of the current model. In addition, the variable porosity of the medium is taken as a function of the particle size \(dp\) and the smallest distance to the nearest wall \(|y|\). This relation is given as:  
\[(y)=\varepsilon_{\infty}\bigg{(}1+a_{1}\exp\bigg{[}-\frac{a_{2}|y|}{d_{p}}\bigg{]} \bigg{)} \tag{2}\]  
Values of the empirical constants \(a_{1}\) and \(a_{2}\) are introduced by [PERSON] and [PERSON] [44] as 1.4 and 5, respectively, and the value of \(\varepsilon_{\infty}\) is considered as 0.36. Using this correlation, the porosity near the bottom wall is equal to 0.864 and this value decreases exponentially as the distance from the bottom wall is increased. The \(Da\) is also considered a variable in this study. \[Da(y)=\frac{K(y)}{L^{2}}=\frac{d_{p}^{2}\varepsilon^{3}}{150(1-\varepsilon)^{ 2}}\frac{1}{L^{2}} \tag{3}\]  
When all these assumptions are taken into account, the two-energy model governing this physical case is given as:  
\[\frac{\partial u_{p}}{\partial x}+\frac{\partial v_{p}}{\partial y}=0 \tag{4}\]  
\[ \begin{array}{ll}\frac{\rho_{\text{ln}f}}{\varepsilon}\frac{\partial u_{p}} {\partial t}+\frac{\rho_{\text{ln}f}}{\varepsilon}&\left[u_{p}\frac{\partial \left(\frac{vp}{t}\right)}{\partial x}+v_{p}\frac{\partial\left(\frac{vp}{t} \right)}{\partial y}\right]\\ &=-\frac{\partial p}{\partial x}+\frac{v_{eff}}{\varepsilon}\left(\frac{ \partial^{2}u_{p}}{\partial x^{2}}+\frac{\partial^{2}u_{p}}{\partial y^{2}} \right)-\frac{v_{eff}}{K}u_{p}-\frac{v_{eff}}{\sqrt{K}}\sqrt{u_{p}^{2}+v_{p}^ {2}}u_{p}\\ &+(\rho\beta)_{\text{ln}f}\Big{(}T_{fp}-T_{c}\Big{)}g\sin\gamma-\frac{\alpha_{ \text{ln}f}\beta_{0}^{2}}{\varepsilon}\big{(}v_{p}\sin\gamma\cos\gamma+u_{p} \sin^{2}\gamma\big{)}\end{array}  \tag{5}\]  
\[ \begin{array}{ll}\frac{\rho_{\text{ln}f}}{\varepsilon}\frac{\partial v_{p}} {\partial t}+\frac{\rho_{\text{ln}f}}{\varepsilon}&\left[u_{p}\frac{\partial \left(\frac{vp}{t}\right)}{\partial x}+v_{p}\frac{\partial\left(\frac{vp}{t} \right)}{\partial y}\right]\\ &=-\frac{\partial p}{\partial y}+\frac{u_{eff}}{\varepsilon}\left(\frac{ \partial^{2}v_{p}}{\partial x^{2}}+\frac{\partial^{2}v_{p}}{\partial y^{2}} \right)-\frac{u_{eff}}{K}v_{p}-\frac{v_{eff}}{\sqrt{K}}\sqrt{u_{p}^{2}+v_{p}^ {2}}\;v_{p}\\ &+(\rho\beta)_{\text{ln}f}\Big{(}T_{fp}-T_{c}\Big{)}g\cos\gamma-\frac{\alpha_{ \text{ln}f}\beta_{0}^{2}}{\varepsilon}\big{(}u_{p}\sin\gamma\cos\gamma+v_{p} \cos^{2}\gamma\big{)}\end{array}  \tag{6}\]  
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