fluid dynamics - Terminal velocity of a settling sphere in 2D vs 3D

How can one analytically calculate the terminal velocity of a settling sphere in 2D? Actually it would be a circular disk. One cannot simple equate boyancy forces minus the drag right? As stated in Stokes law in 2-dimensions

the formula $\textbf{U}_s=2/9(\rho_p-\rho_f)\textbf{g}/9\mu$ [1] does not hold!

[1] https://www.aps.org/units/dfd/meetings/upload/APS_2005_Guazzelli_no_movie.pdf

Answer

Terminal velocity can be found from the balance of forces :

$$\text {buoyancy force + drag force = weight}$$ where the drag depends on the velocity $U$ of the cylinder through the fluid (or the velocity of the fluid past the cylinder) in some unknown way. Buoyancy cannot be neglected unless the density of the disk is much greater than that of the fluid. Weight and buoyancy are easily calculated. The issue is : What is the drag force in this situation?

For "slow" speeds (ie very low Reynolds number $Re=\frac{\rho L U}{\eta} \ll 1$, where $L$ is a characteristic length) drag is dominated by viscosity $\eta$ and is proportional to flow speed $U$. Stokes' Law gives a reasonable approximation whether the disk is perpendicular or parallel to the direction of flow. The exact formulae for a disk of radius $a$ and negligible thickness are [1]

$$F_D=k\eta a U$$ $$k_{\perp}=16, k_{||}=\frac{32}{3}$$

For large speeds such that $Re \gg 1$ drag is dominated by the inertia of the fluid (proportional to density $\rho$) which must be pushed out of the way, and is given by

$$F_D=\frac12 C_D \rho A U^2$$ where $A$ is the cross-sectional area presented to the flow. For a thin disk perpendicular to the flow $C_D\approx 1.17$. Parallel to the flow the inertial contribution is vanishingly small so viscous drag becomes important, but turbulence makes the Stokes' Law estimate unreliable.

Clift, Grace & Weber (1973) summarise numerical and experimental results for low to moderate $Re$ by the equations [2]

$$C_D=\frac{64}{\pi Re}(1+\frac{Re}{2\pi}) \text{ for } Re\le 0.01$$ $$C_D=\frac{64}{\pi Re}(1+10^x) \text{ for } 0.01 \le Re \le 1.5$$ $$\text { where } x=-0.883+0.906y-0.025y^2 \text{ and }y=\log_{10}Re$$ $$C_D=\frac{64}{\pi Re}(1+0.138Re^{0.792}) \text{ for } 1.5\le Re \le 133$$

Once wake shedding occurs, $C_D$ is insensitive to $Re$ and is constant at 1.17 for $Re\gt 1000$. There is some indication that $C_D$ passes through a minimum of 1.03 at $Re \approx 400$ but most data is correlated within 10% of the above with $C_D\approx 1.17$ for $Re \gt 133$.

As you have realised, you don't know the terminal velocity beforehand, so you do not know what $Re$ to use, because $Re$ depends on $U$. One way round this is to assume a value of $C_D$, calculate terminal velocity, then check that the value of $Re$ is consistent with the chosen $C_D$ according to the above equations. If not, adjust $C_D$ and repeat until values of $C_D$ and $Re$ are consistent. Alternatively, calculate the value of $U^2 C_D(U)$ required for terminal velocity, and then solve this transcendental equation by some numerical method.

Motion of a thin disk or even a finite cylinder would be unstable, even when oriented parallel to the flow. It would oscillate from side to side, and perhaps tumble, because of vortex-shedding. Oscillation changes the cross-sectional area and therefore the drag force. The oscillation problem doesn't affect a sphere because the cross-section is the same when it rotates.

Clift discusses motion of disks in free fall, along with oscillation and tumbling motions, on pp 148-149.

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[1] eqn 20-24, p 389 in chapter 20 Creeping Flow of Physics of Continuous Matter by Benny Lautrup. http://www.cns.gatech.edu/PHYS-4421/lautrup/book/creep.pdf

[2] eqns 6.1-6.4, p 145 in chapter 6 Non-sphericial rigid particles at higher Re of the book Bubbles, Drops and Particles by R Clift, J R Grace and M E Weber (Academic Press 1978)