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.

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    $\begingroup$ If $v$ is not constant, $Re$ is not constant ... which will complicate the integration somewhat. $\endgroup$ Mar 31, 2021 at 19:04
  • $\begingroup$ Something to consider is that when Re>1 and depending on fluid properties (see Morton number), the shape of the droplets may start to deform and cause a wake which would affect the analysis. $\endgroup$
    – nluigi
    Apr 1, 2021 at 8:33

2 Answers 2


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.

  • $\begingroup$ This is based on an experiment, and it is not possible for v0 (drop velocity) to be greater than u (gas velocity), as it is surrounded by a high speed air stream travelling in the same direction as the drop. u-v is the relative velocity of the drop to the air. I am inclined to believe that this needs to be solved numerically as you state, due to these factors. $\endgroup$ Apr 1, 2021 at 19:42

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)

  • $\begingroup$ This is a good answer. This comment is merely adding to the analysis the observation that the speed as time grows $t\rightarrow+\infty$ is a constant which is referred to as the terminal velocity. $\endgroup$
    – kbakshi314
    Mar 31, 2021 at 23:43
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    $\begingroup$ Yes indeed, @kb314. Furthermore, the terminal velocity is simply $A/B$, which is easy to see upon taking that limit in my equation for $v(t)$. In the original notation this is just $\frac{\tau_v}{f}g + u$. This can also be seen by letting $dv/dt=0$ in the original EOM (drag and grav forces are equal in magnitude at terminal velocity). Please upvote if you like the answer! $\endgroup$
    – Jacob A
    Apr 1, 2021 at 1:04
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    $\begingroup$ The above answer is correct only if $f$ is a constant but it isn't. Instead it depends on $Re$, which complicates things and means that your result is not correct. $\endgroup$
    – Nick
    Apr 1, 2021 at 7:43
  • $\begingroup$ Nick, check OP's last sentence. $\endgroup$
    – Jacob A
    Apr 1, 2021 at 10:52
  • $\begingroup$ As Nick states f is not a variable but a function of Re, therefore is a function of v and hence cannot be treated as constant. This is where the issue arises, otherwise the above solution is valid. $\endgroup$ Apr 1, 2021 at 19:13

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