A simple problem on vorticity

When we "agitate" consciously the coffee in a cup or the rinsing water in a glass while washing dishes in the kitchen, we observe the so-called "vortex", i.e. a depression of the level of the surface of the liquid, the largest at the middle point. We have seen it a thousand times.

As part of my review of things long time ago forgotten, I want to go through calculations and confirm that, after proper handling of Navier-Stokes equations, this phenomena is well-predicted, as the free surface of a liquid with pure rotational motion is described by the equation of a parabola.

Statement: Find the surface of constant pressure (atmospheric pressure) of a uniformly rotating bucket with constant angular velocity, Ω,  full of an ideal fluid, under gravity, g, so the velocity components relative to fixed Cartesian system of reference set in the middle point of the cylindrical base of the bucket is u = (-Ωy, Ωx, 0) (Acheson, 1989).

One might be tempted to take the apparently gentle shortcut of Bernoulli's equation: the quantity, H, applied to any two points within the fluid must be constant, C:
H = p/ρ + 1/2 u^2 + gz = C                                                  (1)

The vector u squared is
u^2 =  Ω^2 (x^2 + y^2)                                                       (2)

According to the reference taken, at the point in the surface z = z0 and x = y = 0 (i.e., middle point), p = p0 (atmospheric pressure). This condition yields the value of constant C:
C = p0/ρ + gz0                                                                     (3)

When equations (2) and (3) are casted in equation (1), we obtain the pressure at any point:
p/ρ + Ω^2 (x^2 + y^2) + gz = p0/ρ + gz0                             (4)

To obtain the equation asked (free surface), we impose the condition that all points of such geometrical place (x,y,z) must have p = p0. By doing so:
z = z0 - (Ω^2/2g)*(x^2 + y^2)                                              (5)

Unfortunately, equation (5) does not make any sense. The last term in the RHS is always positive and so it appears that the highest point of the surface is the middle point, which is radically opposed to physical observations (a "surge" in the middle point, instead of a depression).

The root of the mistake lays in considering that Bernoulli's equation is applicable to any point within the fluid, and so to any point within the surface. This assertion is only valid when the fluid is irrotational (i.e. ω = curl u = 0). However, the flow given has vorticity, ω = (0, 0, 2Ω), which is not zero. Bernoulli's equation would be only applicable to points within the same streamline, which is completely different from two arbitrary points of the profile of the free surface.

To solve the problem, we need to write and apply Navier-Stokes equation to the problem. In this specific case, they are:
x - component:     - Ω^2 * x = - (1/ρ) * (δp/δx)                   (6)
y - component:     - Ω^2 * y = - (1/ρ) * (δp/δy)                   (7)
z - component:                   0 = - (1/ρ) * (δp/δz) - g              (8)

The three partial differential equations (PDE) can be integrated individually, without forgetting the integration functions as:
x - component:   p = Ω^2 * ρ * (x^2 / 2) + f(z,y)                 (9)
y - component:   p = Ω^2 * ρ * (y^2 / 2) + f'(z,x)                (10)
z - component:   p = - ρgz + f''(x,y)                                      (11)

A particular solution is a linear combination of (9) - (11), plus a constant. In this case, it is readily verifiable that equation (12) is a solution for (6) - (8):
p = ((Ω^2 * ρ) / 2) * (x^2 + y^2) - ρgz + C                        (12)

Now, re-using the boundary condition applied before to reach (3), C = p0 + ρgz0, and so, the general equation for the pressure of any point (x, y, z) within the liquid is:
p - p0 = ((Ω^2 * ρ) / 2) * (x^2 + y^2) + ρg (z0 - z)            (13)

Every point that belongs to the free surface must satisfy that p = p0. If this condition is applied to (13), we find the solution to our problem:
z - z0 = (Ω^2 / 2g) * (x^2 + y^2)                                        (14)

Now, the mathematical equation represents the correct parabola and reflects the depression of the middle point (lowest value).

The same result can be obtained if we work directly in cylindrical coordinates, with uniform and circular velocity, and only azimuthal component, uθ (Bird et al., edition 2007):
z - z0 = (Ω^2 / 2g) * (r^2)                                                  (15),
where, r is the inner radius of the bucket.

Exercise: Write and solve Navier-Stokes equations in cylindrical coordinates for this problem. Verify result (15).

Comments:

1) Acheson clearly states that vorticity does not mean circular motion. In this application, a pure circular (rotational) flow has vorticity, but take for instance, the simple 1-D flow between parallel planes of an ideal fluid with u = (βy, 0, 0), no rotation at all. It is straightforward to see that ω = (0, 0, -β), so the vorticity is not zero.

2) What it is normally called "vortex" appears in fluid mixing and other application where the rotational component of velocity is predominant over the other two components (radial and axial), and suggests that the mixing between adjacent layers of fluid will be poor. Intuitively, however, it seems to be just the opposite.

I can now recall my days in Vintage Pharmaceuticals, Huntsville, AL, with all those liquid-liquid mixture applications and that large number of Batch Product Records long having written instructions as: "set the impeller at such and such RPMs and wait for the vortex to be formed", and the long discussions with Y. K. about the possible partial inadequacy of it.

The thing is now remembered... And, flares and bitter fevers set aside, Y. K. mentioned it in his nice recommendation for me benefiting the admission in University College London.