I've stumbled upon the following question:

Consider a steady, incompressible and viscous flow on an inclined plane with an angle $\alpha$. The surface is in contact with air (which can be assumed to be non-viscous), where the air pressure is equal to $p_0$. Let us denote the flow axis by $x$ and the height from the bottom by $z$ (see the figure). The distance from the bottom to the surface is $h$. Write the Navier-Stokes equations for the pressure $p$ and velocity $v$, formulate the boundary conditions and solve for $p$ and $v$.

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In the solution they claim without proving that both $p$ and $\mathbf{v}$ depend only on $z$ and that $\mathbf{v}=v(z)\hat{\mathbf{x}}$. But how can this be shown? If we apply Bernoulli's principle to the two points $A$ and $B$ that lie along some streamline ($z=\mathrm{const}$), then we get a contradiction, because if $v_A=v_B$ and $p_A=p_B$ then

$$ \frac{\rho v_{A}^{2}}{2}+\rho gh_{A}+p_{A}=\frac{\rho v_{B}^{2}}{2}+\rho gh_{B}+p_{B}\Longrightarrow h_{A}=h_{B} $$

which is clearly incorrect. Perhaps Bernoulli's theorem isn't applicable in this case due to the fact that the fluid is viscous, i.e. there's friction between different layers, and since Bernoulli's principle stems from conservation of energy it must include some friction terms. Regardless, I'm not quite sure how to prove that $v$ and $p$ depend only on $z$.

  • $\begingroup$ You are correct that the viscosity renders Bernoulli irrelevent here. All the gravitational potential energy is being converted to heat by the friction, and not to kinetic energy as in the case of inviscid flow. That all quantities depend only on $z$ is just a consequence of it being a steady state flow. $\endgroup$
    – mike stone
    Feb 11 at 21:48
  • $\begingroup$ @mikestone - I don't see how the steady state condition implies that $v=v(z)$ and $p=p(z)$. Perhaps you meant the incompressibility condition? Assuming $\mathbf{v}$ has only the $x$ component, incompressibility implies $\partial_x v = 0$, i.e. $v=v(z)$. But what about the pressure? How do we show that $p=p(z)$ as opposed to $p=p(x,z)$? $\endgroup$
    – grjj3
    Feb 12 at 11:26

1 Answer 1


Yeah, bernoulli is not valid because there is viscous dissipation.

One way to "prove" that the velocity is dependent only on z is that the flow is one dimensional and incompressible, from the continuity equation:

$$\nabla \cdot v = 0; \frac{\partial v_x}{\partial x} =0 $$

This is also called "fully developed flow". When will you not consider fully developed flow for an steady flow? When entrance length effects are considerable, you can get a rough idea of this in this link: https://en.wikipedia.org/wiki/Entrance_length_(fluid_dynamics)

Basically, in steady flow, the velocity profile will change as it moves in the direction of the flow if there are sudden changes in boundary conditions. In the most common example, this happens right after a fluid enters a pipe, and the viscous forces make the velocity at the wall zero. Since this is a free surface flow moved only by gravity (and the fact that this is a river flowing, meaning the flow length will be much bigger than the surface height), its reasonable to say there are no changes in boundary conditions. So you can assume fully developed flow.

  • $\begingroup$ Thanks. Indeed, for a 1D flow incompressibility implies that the velocity profile is uniform in the direction of the flow. But what about the pressure? How can one show that $p=p(z)$ as opposed to $p=p(x,z)$? $\endgroup$
    – grjj3
    Feb 12 at 11:32
  • $\begingroup$ The flow is happening due to gravity. The local pressure is constant (atmospheric). The only pressure that changes is the hydrostatic pressure due to the height of the fluid. $\endgroup$
    – Klaus3
    Feb 12 at 14:03

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