fluid dynamics - Can vorticity be destroyed?

I have a professor that is fond of saying that vorticity cannot be destroyed. I see how this is true for inviscid flows, but is this also true for viscous flow? The vorticity equation is shown below for reference. From this equation, it looks as if vorticity only convects and diffuses. This would suggest that it can't be destroyed.

$$\frac{D\boldsymbol{\omega}}{Dt} = (\boldsymbol{\omega}\cdot\nabla)\boldsymbol{V} + \nu\nabla^2\boldsymbol{\omega}$$

However, consider this thought experiment:

Suppose we have a closed container filled with water with initial vorticity field $\boldsymbol{\omega}_0$ at time $t_0$. If the container is allowed to sit undisturbed, as $t\to\infty$ the water will become stationary ($\boldsymbol{V}\to 0$) with zero vorticity ($\boldsymbol{\omega}\to 0$).

This suggests vorticity can be destroyed.

My professor claims the boundary-layer vorticity at the sides of the container is equal and opposite in sign to the bulk vorticity. If this is the case, the vorticity cancels out after a long time resulting in the stationary fluid and vorticity is not destroy (just cancelled out).

EDIT: I'm looking for either a proof that the boundary-layer vorticity is equal and opposite to the bulk vorticity or a counter-explanation or proof. (I'm using proof in a very loose hand-wavy sense)

Answer

Your professor is correct, but I agree with you that the statement “vorticity can’t be destroyed or created” seems jarring - I would prefer to think of this as “vorticity is conserved” because the conservation of vorticity derives from the Navier-Stokes Eq and the conservation of angular momentum. I confess this is splitting terminology hairs (don’t push it with your professor) but I think it helped me.

So, I think, maybe I can understand this as an analogy with linear momentum, because linear momentum is conserved too. I remember the problem of a car of mass m, traveling toward the right at velocity v, and on the same road an identical car traveling to the left at velocity –v. They collide head-on and smash and stick together. Velocities after the crash – zero. Momentum after the crash – zero and of course, momentum is conserved. The total momentum of the system was zero before and after.

Let say your container filled with water is a long annulus with thick steel walls. The flow is initially a circular flow around the axis (i.e. 2D flow.) What is initial total angular momentum of the system? Eventually the fluid stops moving, so the final total angular momentum of the system must be zero. How do we show that the initial angular momentum is zero too?

At this point you need to recognize that the vorticity vector in the moving fluid is everywhere parallel to the axis of the container. And you need to use Stokes’ theorem to write an integral equation with a line integral on the LHS (the circulation) and a surface integral on the RHS (vorticity integrated over the container cross section. )

\begin{align*} \oint_{C} v \cdot dl = \int_{S} w \cdot dS\ \end{align*}

Take your integration path (the closed path C) entirely inside the steel wall of your container. The velocity inside the container wall is always zero, and so the circulation along the path is always zero, and so the total vorticity across the cross-sectional area (the area S) of the container and fluid is always zero too.

You can calculate for yourself that the boundary layer vorticity is equal and opposite to the bulk viscosity using a similar approach. Imagine spinning the container about its axis at a constant angular velocity. Eventually the entire viscous-fluid and container system will be rotating like a rigid body around the axis. Every point has the same angular velocity, and there is now a vortex located at the center. Compute the circulation for any closed path that includes the vortex inside it – this will be the strength of the vortex, and the magnitude and sign of vorticity in the bulk fluid. You can show yourself that the strength of this vortex is the total vorticity. Compute the circulation around any path that does not include the vortex, this will always come out to be zero. Pick a path near the fluid-container boundary, s’, so half of it is in fluid and half is inside the container wall, as long as the container and fluid are still rotating together the circulation around this path will be zero too. Now stop the container’s rotation. The fluid continues to move. Compute the circulation around the path s’ again, it is no longer zero and is the vorticity at the boundary layer. It’s sign is opposite that of the vortex at the center. Every point along the fluid-boundary can be associated with path like s’ and a small amount of boundary layer vorticity. Integrate around the entire boundary and the sum will be equal in magnitude and opposite in sign to the strength of the vortex at the center. Eventually, boundary layer vorticity will diffuse towards the center and annihilate the center vortex.

@Isopycnal_Oscillation is correct to point out that in 3D, and particularly near turbulent conditions, vorticity is not conserved. The second term on the RHS of your ‘transport equation’ says that the stretching and tilting of vortex tubes can change vorticity too. However, I expect that in the classes where your professor is fond of saying that “vorticity cannot be destroyed” turbulent flow is seldom if ever encountered.

Finally, assuming that the LHS of your ‘transport equation’ equals zero does not necessarily require that the fluid be inviscid or that the problem be 2D – you are assuming that the terms on the RHS happen to cancel exactly and the vorticity is fortuitously ‘steady-state.’ So yes, that is a very strong assumption to accept.