## 3 Results and Discussion
### Effects of \(\vartheta\)  
Viscosity is assumed to be a function of pressure, as given by (29), while the pressure in the channel is given by (22), namely \(p(y)=p_{0}+(1-y)cos\vartheta\). From this equation, it can be observed that the pressure has its smallest value, \(p_{0}\), at the upper wall and increases linearly with decreasing \(y\) until it reaches its maximum \(p_{0}+cos\vartheta\) at the lower wall. Changes in pressure across the channel affect the viscosity distribution across the channel and result in a maximum viscosity at the lower wall and a minimum at the upper wall (in all but one case, when \(n=-1\)). The effects of \(\vartheta\) on \(u\), \(\mu\), \(Q\), \(-\frac{\mu u}{k}\), \(\mu\frac{d^{2}u}{dy^{2}}\), and \(\left(\frac{d\mu}{dy}\right)\left(\frac{du}{dy}\right)\) are illustrated in Figure 3.  
Figure 2: Visual display of parameters’ variations and their effects on the dimensionless volumetric flowrate in the channel, \(Q\).  
The inclinations of the flow domain used were \(\theta=15^{\circ}\), \(30^{\circ}\), \(45^{\circ}\), \(60^{\circ}\), \(75^{\circ}\), and \(90^{\circ}\), while all other parameters took on the values identified in the Reference Case given in Table 1. The variation of the angle has two distinct effects on the flow: the first is the variation of the pressure across the channel and thereby the variation of the viscosity; the second effect is the variation of the component of gravitational acceleration along the incline and therefore the driving force for the flow. When \(\vartheta=0^{\circ}\), the variation of the pressure across the channel is maximised but does not produce flow as there is no component of gravity acting along the incline. When the angle of inclination is \(90^{\circ}\), Equations (22) and (29) render \(p(y)=p_{0}\) and \(\mu(y)=\mu_{0}\), meaning the viscosity is constant across the channel.  
The plots of the velocity profiles, which can be found in Figure 3, show that for the case of \(\vartheta=90^{\circ}\), the velocity profile is symmetric with respect to the centre of the channel. This is a result of the uniform pressure, and therefore uniform viscosity, across the channel. As \(\vartheta\) decreases, the component of gravitational acceleration along the direction of flow decreases, thereby reducing the overall driving force behind the flow. Furthermore, the higher viscosity as at the bottom of the channel, illustrated in Figure 3, causes the velocity to be lower near the bottom of the channel, resulting in the asymmetry in the velocity profiles, which becomes more pronounced at smaller values of \(\vartheta\). Despite the lower velocity, Figure 3 shows that the Darcy friction is significant and nearly symmetric across the channel, with the low-speed regions corresponding to regions of high viscosity. As mentioned previously, when \(\vartheta=90^{\circ}\), there is no variation in pressure, and therefore viscosity, across the channel; so, \(\frac{d\mu}{dy}=0\), facilitating a symmetric velocity profile and therefore symmetric distributions of Darcy friction, and \(\mu\left(\frac{\vartheta^{2}u}{dy^{2}}\right)\). Examining the cases of \(\vartheta\
eq 90^{\circ}\), small asymmetries can be seen in all of the forces, and in the case of \(\left(\frac{du}{dy}\right)\left(\frac{du}{dy}\right)\), changes in direction of the force can be seen throughout the channel.