assume that a particle of radius $R_p$ is moving under influence of gravity $g$ in a fluid medium of density $\rho_l$ and viscosity $\mu_l$. then the Stokes settling velocity is given as
$$
\mathbf{v}_p= \frac{2}{9}\frac{\rho_l -\rho_p}{\mu_l}R^2_p\mathbf{g}
$$
which is found from balancing the buoyancy force with drag force. In this derivation the fluid itself is assumed stationary (at least far-field velocity and all the fluid velocity is generated because of spherical particle motion).
a) what is the time scale during which the settling takes place? is this timescale simply the viscous time scale $t_{scale} = R^2_p/\nu$ with $\nu$ being the kinematic viscosity
b)Now what should happen when the fluid itself is moving with a spatially non-uniform velocity field. Although the velocity field is non-uniform the Reynolds number is assumed to be small. Is the Stokes settling velocity still valid? In this case, the buoyancy force still remains the same but I am not sure about the drag force.
Any explanation will be highly appreciated.
Best Answer
a) The viscouse timescale is in play in the dynamics before reaching a steady-state situation.
Consider the equation of motion: $$m\frac{dv}{dt}=\left(m-m_w\right)g - 6\pi\mu R_p v$$ Clearly, a steady-state is reached once the derivative vanishes, this occurs as $v\rightarrow v_p$: $$0=\left(m-m_w\right)g - 6\pi\mu R_p v_p$$ Subtracting both equations yields: $$m\frac{dv}{dt}= - 6\pi\mu R_p \left(v-v_p\right)$$ which can be rearranged using $m=\frac{4}{3}\pi \rho R_p^3$ to: $$\frac{dv}{dt}= - \frac{9}{2}\pi\left(\frac{v-v_p}{\tau_v}\right)$$ where $\tau_v=R_p^2/\nu$ with $\nu=\mu/\rho$ which is the viscous timescale.
This equation dictatest that the velocity changes exponentially with a time-scale $\tau_v$ until a steady-state is reached ($v\rightarrow v_p$) because at that point $\frac{dv}{dt}\rightarrow 0$ and the rhs of the equation becomes $v-v_p \rightarrow 0$. As you see no dependence on the viscous time-scale once steady-state is reached.
So what other timescale is there to consider? Well the particle is now moving at a steady velocity, settling over a distance of height $H$ (i assume), so the equation of motion is now simply:
$$\frac{dv}{dt}=\frac{d^2y}{dt^2}=0$$
In other words:
$$ \frac{dy}{dt} = \frac{H}{t_s} = v_p $$
where $t_s$ is the settling timescale, i.e. $t_s=v_p/H$
b) As long as $\mathrm{Re}<1$, Stokes' flow is valid for non-uniform velocities however other effects may play a role.
Say that the non-uniform velocity is characterized by a shear rate of $\dot{\gamma}=\frac{U}{R_p}$ where $U$ is some characteristic velocity scale, then: $$\mathrm{Re} = \frac{UR_p}{\nu} = \frac{U/R_p}{\nu/R_p^2}$$ For $\mathrm{Re}<1$ it requires that $\dot{\gamma}<1$, i.e. the velocity change over the particle will be small (maybe negligible).
If the velocity change over the particle is small then you may average the velocity over the particle (or maybe several particle radii) and use that as a value for the far-field uniform velocity, $v_\infty$.
You haven't specified exactly how the velocity field is in relation to the motion of the particle, either it is normal or parallel to the motion:
As mentioned at the start other phenomena start to play a role, specifically 'lift' becomes dominant. First theoretically treated by Saffman, 'lift' occurs when the particle will start a rotation due to the gradient in velocity and causes it to move laterally depending on the orientation of the flow in relation to the motion of the particle.
I will not go into details here, for more information see e.g. this publication on Saffman lift phenomena, specifically ch. 2 'Lateral migration: lift in low-Reynolds-number flows'