# 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.

• If $v$ is not constant, $Re$ is not constant ... which will complicate the integration somewhat. Mar 31, 2021 at 19:04
• 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. Apr 1, 2021 at 8:33

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 $$v_0, we will need to keep the absolute value for the velocity difference in the definition of Reynolds number and we cannot solve the equation analytically. However, if $$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.

• 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. 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)

• 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. Mar 31, 2021 at 23:43
• 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! Apr 1, 2021 at 1:04
• 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.
– Nick
Apr 1, 2021 at 7:43
• Nick, check OP's last sentence. Apr 1, 2021 at 10:52
• 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. Apr 1, 2021 at 19:13