# Pressure gradient on flow on inclined plane

Is there any specific reason for pressure to be constant along the x axis, which is represented on this picture below.

If I think about an elemental volume of fluid with its base parallel to the x axis I see mainly 5 forces, which are the viscous shear forces on the upper and lower faces, two pressure forces on the right and left faces, and the x component of the force of gravity.

Now, apparently the two pressure forces cancel and the gravity alone will balance the shear forces, but how can we reach that conclusion?

It is easy for me to understand that a pressure gradient must exist on a horizontal flow with viscous stresses because it is what must balance those friction forces. But now the balance is made only due to gravity? Wouldn't it be plausible to assume that it would be a contribution from both gravity and pressure gradient?

Thank you!  This is a problem with a free surface. At the free surface, the pressure is equal to that of the air outside the fluid. The pressure is continuous across the free surface, so immediately below the free surface, the pressure is also atmospheric. This is true at all values of x. The fluid velocity normal to the incline (and normal to the free surface is zero). So,in the y direction, the pressure has to vary hydrostatically, with a pressure gradient in the y direction determined by $\rho g \sin{\theta}$. This happens at each value of x. So there is no pressure gradient in the x direction. All that is left to cause the fluid to flow (at constant thickness) down the incline is the gravitational component of force in the x direction.
• Pressure is continuous at the boundary between two materials (actually, normal stress and shear stress). If not, if you do a force balance on a segment of the boundary, it will have infinite acceleration (since the interface has zero mass). Regarding a random x section, the pressure will be a function of y (only), given by $$p=p_{atm}+(h-y)\rho g \cos{\theta}$$ (in my previous comment, I wrote sin when I should have written cos). So, the pressure will be a function of y, but not x. Oct 27, 2017 at 18:43
First we notice the notes [p.24 in https://projects.exeter.ac.uk/fluidflow/Courses/FluidDynamics3211-2/slides/lecture3-slides.pdf] for the figure in p.23 of the same webpage. Note 3 [No pressure gradient] is our final result for $$p$$ . It is important because it facilitates calculating the fluid velocity $$\pmb{u}$$ . The proof of $$\nabla p=\pmb{0}$$ is given in [Acheson, p.39, l.6--l.− 7]. §1.12.(A) in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/quantum-mechanics may help you read [Acheson].