## 3 Dimensionless Groups of the Problem and Verification
### The Mechanical Problem  
Dimensionless variables are defined in the form  
\[x^{\prime}=\frac{x}{L};\qquad y^{\prime}=\frac{y}{H};\qquad h^{\prime}=\frac{h-h_{ 1}}{h_{2}-h_{1}} \tag{11}\]  
These variables are replaced in Equation (1):  
\[K_{x}\frac{\partial^{2}h^{\prime}(h_{2}-h_{1})}{\partial x^{\prime 2}L^{2}}+K_{y} \frac{\partial^{2}h^{\prime}(h_{2}-h_{1})}{\partial y^{\prime 2}H^{2}}=0 \tag{12}\]  
Averaging or approximating the derivative factors within the whole scenario to unity, the resulting coefficients of the addends of the above equation are \(\frac{K_{x}}{L^{2}}\) and \(\frac{K_{y}}{H^{2}}\). From them, a single dimensionless group is deduced:  
\[\pi_{1}=\frac{K_{x}}{L^{2}}\frac{H^{2}}{K_{y}} \tag{13}\]  
To this group, a discriminated group associated with the specific geometry of the problem must be added. It is generally denominated as the shape factor or aspect ratio, which defines the quotient between the horizontal extension of the river and that of the entire domain:  
\[\pi_{2}=\frac{l_{r}}{L} \tag{14}\]  
Thus, \(\pi_{1}\) and \(\pi_{2}\) necessarily control all the unknowns of the problem: dimensionless iso-potential (hydraulic) line patterns and dimensionless iso-streamline patterns. Moreover, such groups control the flow in its dimensionless form \(\frac{Q}{(h_{2}-h_{1})\sqrt{K_{x}K_{y}}}\)[34]. The pi theorem is applied:  
\[\frac{Q}{(h_{2}-h_{1})\sqrt{K_{x}K_{y}}}=f_{1}\left(\frac{K_{x}}{L^{2}}\frac{ H^{2}}{K_{y}},\frac{l_{r}}{L}\right) \tag{15}\]  
The dimensionless potentials field depends on the relative positions \(x/L\) and \(y/H\):  
\[\frac{h-h_{1}}{h_{2}-h_{1}}=f_{2}\left(\frac{K_{x}}{L^{2}}\frac{H^{2}}{K_{y}},\frac{l_{r}}{L},\frac{x}{L},\frac{y}{H}\right) \tag{16}\]  
It should be noted that discrimination, which is already applied in the very definition of the dimensionless variables (Equation (11)), leads to a more precise solution compared to that deduced from classical dimensionless considerations. From the latter, the group \(\frac{K_{x}}{L^{2}}\frac{H^{2}}{K_{y}}\) is separated into two monomials, \(\frac{K_{x}}{K_{y}}\) and \(\frac{H}{L}\), which, as is to be verified, are not independent dimensionless groups. For the values of \(\pi_{1}\) and \(\pi_{2}\), it should be noted that \(\frac{l_{r}}{L}\), which is not deduced from the governing equations but from the domain geometry, has a small value in general and has no physical meaning in terms of the balance of magnitudes, while \(\frac{K_{x}}{L^{2}}\frac{H^{2}}{K_{y}}\) balances the horizontal and vertical velocity components. When the order of magnitude of this group is greater than unity (\(K_{x}>>K_{y}\) or \(H>>L\), or both), the flow is confined to a region very close to the surface, and the pattern is independent of \(H\). If the value of this group is less than unity, the flow is distributed throughout the domain, creating a pattern clearly dependent on \(H\). As the boundary condition at the right of the domain implies a zero vertical velocity component (constant potential), very large domains would be similar to large regions where the vertical velocity component would be almost zero near the right edge. To verify Equation (15), the numerical solution of the set of cases shown in Table 1 is used. The illustrative scenarios (17 in total, introduced in Tables 1-3) are simulated witharbitrary values of geometric and hydraulic properties, which are clearly different from each other. With these, even if a parameter's value is changed from one case to another when the monomial does not change, the dimensionless result does not change either. This confirms that the dimensionless groups are correct; the relation between them, obtained using the strict formal protocol of the pi theorem and the type curves elaborated from these relations, leads to the expected solutions. For the scenarios in Table 1, the values of the monomials \(\pi_{1}\) and \(\pi_{2}\) are kept constant for all the scenarios, obtaining the same dimensionless flow rate \(\frac{\mathrm{Q}}{(\mathrm{h_{2}-h_{1}})\sqrt{\mathrm{K_{x}K_{y}}}}=0.215\).