Determine droplet velocity from equation of motion In multi-phase fluid mechanics the equation of motion for a drop can be described as follows:
$$
\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{f}{\tau_v}(u-v)+g
$$
$g$ is acceleration due to gravity, $u$ is the velocity of the medium the drop is travelling through, and $v$ is the velocity of the particle.
$\tau_v$ can be assumed to be a constant for a given drop parameters.
$f$ has various models as follows depending on Reynolds number:
for Re<1000
$$
f = 1+\frac{1}{6}\mathrm{Re}^\frac{2}{3}
$$
for Re>1000
$$
f = 0.0183\mathrm{Re}
$$
Reynolds number is defined as
$$
\mathrm{Re}=\frac{\rho|u-v|d}{\mu}
$$
How can I integrate the equation of motion for a given f model to determine $v$ as a function of $t$? I am interested in determining the velocity of a drop at a particular point in time. Similarly I would ideally want to be able to determine the distance travelled by the drop in a period of time. It is valid to assume all variables are constant except $v$ and $t$
Edit: note that $f$ cannot be treated as constant since it is a function of Re, therefore a function of $v$, for simplicity assume Re>1000 so the simpler equation can be used. For my conditions $u$ is always greater than $v$ as I have a high speed gas jet "accelerating" the drop.
 A: As you've said: all variables except $v$ and $t$ are constant, so I am safe in rewriting
$$
\frac{\mathrm{d}v}{\mathrm{d}t}=\frac{f}{\tau_v}(u-v)+g
$$
as $$
\frac{\mathrm{d}v}{\mathrm{d}t}=A-Bv
$$
with $A\equiv g+\frac{f}{\tau_v}u$ and $B\equiv \frac{f}{\tau_v}$.
Now just separating variables,
$$
\frac{\mathrm{d}v}{A-Bv} = dt
$$
and integrating;
$$
-\frac{\ln(A-Bv)}{B} = t + C
$$
solving for $v=v(t)$,
$$
v(t)=-(e^{-B(t+C)}-A)/B.
$$
We can solve for $C$ with our initial condition $v(0)=v_0$,
$$v(0)=-(e^{-(BC)}-A)/B=v_0$$
or
$$ C= -\ln(-Bv_0+A)/B $$
and feel free to substitute in our definitions of $A,B$. (it's just a bit messy)
A: The main difficulty here is that $f$ is a function of $Re$, and it is conceivable that your initial conditions could have $Re>1000$ but that you would slow down to $Re<1000$ at terminal velocity. We also don't know if the initial velocity, $v_0$, is greater or less than $u$. This means that in general there is not an analytic solution to the problem.
By definition, $Re>0$ and therefore so is $f$, whether the flow is laminar or turbulent. Setting $dv/dt=0$, we find the terminal velocity to be
$$
v_\infty = u+\frac{g\tau}{f}.
$$
As $f>0$, this means that $v_\infty>u$ and so the velocity difference $v_\infty-u>0$. This means that if initially $v_0<u$, we will need to keep the absolute value for the velocity difference in the definition of Reynolds number and we cannot solve the equation analytically. However, if $v_0>u$ we can define
$$
Re=\frac{\rho(v-u)d}{\mu}
$$
as we know that $v-u>0\,\forall\,t>0$.
To help solving the equation, we can define $w=v-u$. Substitution into the differential equation gives
$$
\frac{dw}{dt}=-\frac{fw}{\tau}+g.
$$
Considering $Re<1000$ first, the differential equation becomes
$$
\frac{dw}{dt}=-\frac{w}{\tau}-\frac{1}{6\tau}\left(\frac{\rho d}{\mu}\right)^{2/3}w^{5/3}+g,
$$
which doesn't have a nice analytic solution. Considering $Re>1000$ instead, the equation becomes
$$
\frac{dw}{dt}=-\frac{0.0183\rho dw^2}{\tau\mu}+g,
$$
which has solution
$$
w=w_\infty \tanh\left(\frac{g(c_1+t)}{w_\infty}\right),
$$
where we have used the fact that the terminal velocity difference is
$$
w_\infty=\left(\frac{g\tau\mu}{0.0183\rho d}\right)^{1/2}.
$$
At $t=0$, $w=w_0$, therefore
$$
c_1=\frac{w_\infty}{g}\tanh^{-1}\left(\frac{w_0}{w_\infty}\right).
$$
Putting everything together and substituting back for $w=v-u$, we get
$$
v=u+(v_\infty-u)\tanh\left(\tanh^{-1}\left(\frac{v_0-u}{v_\infty-u}\right)+\frac{gt}{v_\infty-u}\right).
$$
We can see that as $t\to\infty$, $v\to v_\infty$ as required. The solution can also be integrated analytically to find the distance.
The above solution requires that $Re>1000$ throughout, and that the initial velocity $v_0>u$. In all other situations, the equation will have to be solved numerically.
