According to Wikipedia, there is at least one solution to Laplace’s equation which could model fluid moving passed a cylinder. I’ve found an asymmetric flow (below) satisfying the necessary boundary conditions, with continuity. A spinning cylinder is asymmetric because it’s going one way, and not the other.
Suppose an infinitely long cylinder, with axis coinciding with z-axis, is in the flow of an inviscid, incompressible fluid, which would otherwise have velocity in the x-direction. I assume, like Wikipedia, that the velocity field is irrotational and free of divergence. Then there should exist a scalar velocity potential, whose gradient equals the velocity, and who satisfies Laplace’s equation in 2D. My velocity potential (scalar, units ) is given by
where is any real number are 2D polar coordinates and R (metres) is the cylinder’s radius. The far-away velocity is handled by the well-known “d’Alembert” solution, and I’ve added an asymmetry. I assume a potential may be discontinuous, since it’s not unique, but the unique real velocity must be continuous. The gradient
is the velocity field. The formula shows that at the cylinder’s radius, there is no radial component to velocity. Furthermore, the limit as is Therefore I’ve verified all boundary conditions are satisfied, and by inspection the velocity field is well-defined and continuous everywhere outside the cylinder. Furthermore, my asymmetry term in the scalar potential is the imaginary part of over the complex plane, so it must satisfy Laplace’s equation, except at the negative x-axis, where the velocity is still divergence and rotation-free, outside the cylinder. That is, my velocity field is correct.
Steady flow of U=1 m/s (3.6 km/h) current around cylinder of radius R=1 m, where I’ve assumed α (=a) is 1 m²/s. The asymmetry should be due to the cylinder spinning anticlockwise with frequency 1 radian/s ≈ 0.16 Hz ≈ 10 rpm, assuming coupling to the fluid. The fluid is dragged fast under the bottom. I’ve been trying to model inviscid drag. The image is from pplane8.
I have uncountably many partial solutions to Laplace’s equation, who satisfy all necessary boundary conditions, may model inviscid drag, but do result in discontinuity in velocity along the negative x axis. I can’t solve d’Alembert’s paradox, yet, which I encountered after pondering my reversible venturi experiment.
Update (2 Oct): The value of should be given by
where is the cylinder’s anticlockwise (scalar) angular velocity (radians/s). I base this on the case where – the fluid should move at the same rate as the cylinder, at its surface, when there is no external current. In the cylinder’s inertial frame of reference, when the current and cylinder have opposite non-zero velocities, the current should always win.
Update (3 Oct): Suppose there’s uniform density in a fluid with no external current. Then the fluid’s total angular momentum per unit length along its axis (Ns) diverges with This is plausible for an eternal rotating cylinder with fluid coupling. The kinetic energy per unit length (J/m) also diverges, except in proportion to The fluid’s velocity is just To obtain the cylinder’s rpm, multiply by