We derive the Euler equation for inviscid flow. Then, for irrotational flow, we use the Euler equation to get Bernoulli equation (classic form) and show it holds in the irrotational region (still inviscid flow).
On the other hand, for a viscous flow, when it is irrotational, the viscous term disappears from Navier-Stockes equation and reduces to the Euler equation. This means the flow is irrotational (and viscous) and the governing equation is the Euler equation which we can again use it to get to Bernoulli equation. This means the Bernoulli equation holds for irrotational flow eventhough it is viscous?!
I know my reasoning has to be wrong, because Bernoulli equation (classic form) is conservation of energy which is not applicable to viscous flow. But I can't find where I am going off track in my reasoning, I appreciate your help.
2 Answers
Your question was a classical question I was used to ask at Fluid Mechanics course exams, BSc of Aerospace Engineering, since it gave me the chance to test the student on the whole dynamics of vorticity (quite important in Aerospace/Aeronautical applications).
Your reasoning is good so far, but you forget that the dynamics of vorticity is governed by the equation,
$$\frac{D \boldsymbol{\omega}}{Dt} = \left(\boldsymbol{\omega} \cdot \nabla \right) \mathbf{u} + \nu \Delta \boldsymbol{\omega} \ ,$$
i.e. the material time derivative of the vorticity field equals the sum of the vortex stretching-tilting term and the diffusion term.
If viscosity is zero, it's possible to prove that vorticity evolves as a material vector: so, if it's zero somewhere in the flow, it's zero in all the regions that are reached by the fluid starting from there.
If viscosity is not zero, diffusion creates/diffuses vorticity in viscous flows. If vorticity diffusion can't be neglected, the assumption about irrotational flow needed for this version of Bernoulli theorem fails.
-
$\begingroup$ Thanks for shedding light on another aspect of my question which I did not know. However, I will appreciate it if you can please clarify this: for a viscous irrotational flow, the viscous term disappears from Navier-Stocks equation and classic form of Bernoulli equation is the exact solution of the remaining equation. So, Bernoulli is governing a viscous (and irrotational) flow. How can that be correct when we know there is energy dissipation (energy loss) in viscous flow? Thanks again $\endgroup$– EbiCommented Jun 11 at 23:56
-
1$\begingroup$ If you assume that an incompressible flow is irrotational, thus the Laplacian of the velocity field vanishes since $\Delta \mathbf{u} = \nabla ( \nabla \cdot \mathbf{u} ) - \nabla \times \nabla \times \mathbf{u}$. But what you can assume from a mathematical point of view, must have some physical meaning to be useful. It's hard to assume that the whole flow remains irrotational for viscous fluids, even at high Reynolds number: if the fluid flows around a solid wall, boundary layer develops on the wall and this is a rotational region $\endgroup$– basicsCommented Jun 12 at 15:46
-
1$\begingroup$ These rotational regions usually develops and their thickness increases due to vorticity diffusion and are usually transported as free wakes behind streamlined bodies, $\endgroup$– basicsCommented Jun 12 at 15:48
If we take the dot product of the Euler equation and the velocity vector, we obtain the Bernoulli equation only if the vorticity is zero.
