# 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 $\textbf{U}_s=2/9(\rho_p-\rho_f)\textbf{g}/9\mu$  does not hold!

• I would be inclined to model this using the Blasius solution for drag flow parallel a flat plate;. – Chet Miller Aug 29 '17 at 20:24
• Could you elaborate more? – nabber Aug 29 '17 at 21:55
• Are you familiar with the Blasius solution? See this link:en.wikipedia.org/wiki/Blasius_boundary_layer – Chet Miller Aug 29 '17 at 22:03
• What context do you want to know the answer in? If you want to compare to an experimental system or a simulation, it probably isn't a true two-dimensional infinite plane - e.g. there is a finite system size, or some mechanical coupling to the third dimension. It turns out (see my answer) the answer can be subtle enough that it depends on these details! – AJK Aug 30 '17 at 1:02
• -1. Not clear. Are you asking about a flat disk and fluid which are both confined to 2D, or a thin disk in a 3D fluid? ... The 2D problem is the same as an infinite cylinder in 3D. Motion would be unstable, but that is a separate issue. Why do you think you cannot equate bouancy+drag with weight? – sammy gerbil Aug 30 '17 at 1:09

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

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

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.

 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

 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)

In principle, I believe you can just balance the buoyancy force and the drag. The problem is that the drag of a 2D disc is a surprisingly complex and potentially ill-defined problem.

Assuming that you are looking at the sedimentation of a circular disk in a true 2-dimensional system (i.e. no walls, not just a disc falling in a 3D fluid, infinite system size), there is no true solution to represent the drag on a 2D circle (i.e. an infinite cylinder) at zero Reynolds number: https://en.wikipedia.org/wiki/Stokes%27_paradox

I'm assuming you are looking at the low-Reynolds-number limit here based on your reference to Stokes law. For small, but nonzero Re, the answer turns out to be very, very complex, because there is no true "zero Re" solution (as long as you're assuming an infinite 2D system). The presence of a finite Reynolds number is a singular perturbation, fundamentally changing the nature of the problem. There are classical answers using ad-hoc approximations to the full Navier Stokes equations (the previous answer uses an Oseen approximation), but these turn out to not always be reliable.

This is discussed in detail in the review by Veysey and Goldenfeld, which goes through a lot of details and history, too. They also show the best approximations: https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.79.883 (free arxiv version here: https://arxiv.org/abs/physics/0609138)

• that paper is really informative...too much one might say for my case! Anyhow the link provided for the chart of Wolfram Science World for Cd for the infinitely long cylinder should work for my case, right? – nabber Aug 30 '17 at 9:00