Stokes flow and settling velocity

Stokes flow

Stokes (creeping) flow is flow at very low Reynolds number ($Re\ll1$, in practice $Re<1$) in which inertia is negligible compared with viscous forces. The Navier-Stokes equation reduces to the linear form

$$\nabla p=\mu\nabla^2\vec V,\qquad \nabla\cdot\vec V=0$$

Conditions: small particle or length scale, high viscosity, or very low velocity (e.g. micro-fluidics, sedimentation, lubrication films). The solution is linear and time-reversible.

For a sphere of diameter $d$ moving at velocity $V$ in an infinite fluid, Stokes' drag is

$$F_D=3\pi\mu dV,\qquad C_D=\frac{24}{Re}$$

Terminal velocity

At terminal velocity: weight $-$ buoyancy $=$ drag

$$\frac{\pi d^3}{6}(\rho_p-\rho_f)g=3\pi\mu dV_t\ \Rightarrow\ V_t=\frac{(\rho_p-\rho_f)\,g\,d^2}{18\mu}$$

Substituting ($d=1\times10^{-4}$ m):

$$V_t=\frac{(2650-998)\times9.81\times(1\times10^{-4})^2}{18\times1.002\times10^{-3}}=\frac{1652\times9.81\times10^{-8}}{0.018036}=8.99\times10^{-3}\ \text{m/s}$$

Check of validity

$$Re=\frac{\rho_fV_td}{\mu}=\frac{998\times8.99\times10^{-3}\times10^{-4}}{1.002\times10^{-3}}=0.895<1$$

The particle Reynolds number is below 1, so Stokes' law is acceptable (slightly at the upper limit).

The drag force is $F_D=3\pi\mu dV_t=8.5\times10^{-9}$ N, equal to the net weight.

Answer: $V_t=8.99$ mm/s (about 9 mm/s), $Re=0.89$, so Stokes' law is valid.