# Viscosity Displacement Velocity Time relation

1. Sep 26, 2015

### SataSata

If I were to drop a ball down a fluid with high viscosity, the ball decelerates over time and then reaches close to terminal velocity.

How does displacement and velocity relates with time in terms of mathematical equation?
Am I right to say:
v(t) = Vterminal(1 - exp(-bt))
d(t) = Vterminal*t(1 - exp(-bt))
whereby b is proportional to the viscosity of the fluid.
The velocity-time equation looks about right but I'm not sure about displacement.

2. Sep 26, 2015

Well it won't start off decelerating unless you give it some initial speed greater than terminal velocity. There are also other effects that are important, like buoyancy and gravity.

It would probably be best to write a force balance and go from there. Also, your velocity formula is not the time derivative of your distance formula.

3. Sep 26, 2015

### SataSata

From my data, the ball did decelerate, hence I believe that the initial speed come from me dropping the ball from a certain height.
Would the equation be valid if buoyancy and gravity is accounted for in 'b' in the equation? Since both affect the rate at which the ball reaches terminal velocity?
If my ball is extremely light with just a mass of 0.9grams and the viscosity of the fluid is 1.4Pa-s, would the time to reach terminal velocity be very short to notice?

Last edited: Sep 26, 2015
4. Sep 26, 2015

### Nidum

Stokes Law

5. Sep 26, 2015

### Staff: Mentor

You need to follow Boneh3ad's suggestion, and first draw a free body diagram identifying the forces acting on the ball. Then write the 2nd law force balance equation for the acceleration.

6. Sep 26, 2015

Unless you are dropping the ball from very, very high (as in so high that the density of the air changes appreciably), then I can't imagine it ever slowing down. If it is starting from rest, then its velocity should increase monotonically until it reaches terminal velocity (or rather asymptotically approaches it). If it slowed down at any point, that means it would have had to start at a velocity higher than the terminal velocity.

No. Like I said, velocity should be the time derivative of displacement and yours is not, so that alone would imply that your equations are wrong. Like I said before, your best bet is to start with a free body diagram and a force balance and then solve it from there.

It could be. There are other factors that are important, though, such as the density of the fluid and the velocity of the ball. Maybe you ought to describe your experiment a bit more.

That is only relevant if $Re \ll 1$ and it is falling either just through open air or a tube that is large enough that there are no side wall effects.

7. Sep 26, 2015

### SataSata

$F=ma$
$mg-b-f =m\frac{dv}{dt}$
Whereby m is mass of the ball, g is 9.81, b is buoyancy and f is fluid resistance.
$\frac{dv}{dt}=\frac{mg-b-f}{m}$
Then I'll integrate from here and assuming Stoke's Law is valid, $f=6\pi\mu rv$ Where by $\mu$ is viscosity of the fluid.
I then simplify it and got:
$v=\frac{t(mg-b)}{m+6\pi\mu rt}$
Then again I integrate this and got:
$x=\frac{(mg-b)(6\pi\mu rt-mln(m+6\pi\mu rt))}{(6\pi\mu r)^2}$
Am I doing this right?

If I were to take my previous equation:
$v(t) = Vterminal(1 - exp(-kt))$
and I integrate this and would get:
$x(t) = Vterminal(t + \frac{exp(-kt)}{k})$
I thought this is right because as time increases, v(t) will approach Vterminal. And the only thing affecting the rate of it is the constant 'k' which should change base on different factors like mass of ball, viscosity of fluid and buoyancy.

The experiment is done such that I dropped a tiny ball with radius of around 2mm down a tube of fluid of radius of 1.75cm. I then record the time and its corresponding displacement. A graph is plotted and it looks linear hence I assume the time taken to reach terminal velocity would be very very short and unnoticeable. I thought it decelerate because I dropped it about 3cm above the fluid.

Last edited: Sep 26, 2015
8. Sep 26, 2015

Are you familiar with the concept of Reynolds number? How large is your ball, through what fluid is it falling, and about what do you expect to be the terminal velocity?

9. Sep 26, 2015

### SataSata

The ball has a radius of 2mm and mass of 0.07g. The fluid is undiluted detergent. If I were to look at the gradient of the displacement time graph, the terminal velocity would be around 0.15cm/s. And if I use $x(t)=Vterminal(t+\frac{exp(−kt)}{k})$ to fit the graph, the program derive Vterminal as also around 0.15cm/s.

I'm not familiar with Reynolds number, but judging from the size of the ball and the size of the tube wall effect should be minimum and Stoke's Law should apply here.

10. Sep 27, 2015

So the Reynolds number,
$$Re = \dfrac{\rho v \ell}{\mu},$$
is a dimensionless number representing the relative importance of inertial effects and viscous effects, where $\rho$ is density, $v$ is velocity, $\ell$ is a characteristic length (here, diameter), and $\mu$ is dynamic viscosity.

Now, I don't know what kind of detergent you are using, so I will use some data on dish detergent that I got from Dow Chemical. Its density is approximately 1060 kg/m3 and its viscosity is 0.35 Pa⋅s. Therefore, $Re \approx 10^{-2} \ll 1$, so Stokes' Law is reasonable here.

Buoyancy is simply archimedes principle, so,
$$F_b = \rho_{\mathrm{fluid}} V_{\mathrm{sphere}} g.$$

The force of gravity is, of course,
$$F_g = m_{\mathrm{sphere}} g.$$

Stokes' Law is, in terms of diameter,
$$F_d = -3\pi \mu D v$$
(the negative is because the drag force must opposed the direction of velocity, so if the velocity is negative, as in freefall, the drag is upward).

So your equation is essentially correct, though let's choose another sign convention so that it represents freefall:
$$m\dfrac{dv}{dt} = -F_g + F_b + F_d = -mg + \dfrac{1}{6} \rho_f \pi D^2 g - 3\pi \mu D v,$$
or
$$\dfrac{dv}{dt} = \dfrac{\rho_f\pi D^2 g}{6m} - g - \dfrac{3\pi\mu D}{m} v.$$
That's a separable, first-order ODE (albeit a messy one), so its solution is
$$v(t) = \dfrac{\rho_f\pi D^2 g - 6mg}{18\pi\mu D} + C_1\exp\left[- \left( \dfrac{3\pi\mu D}{m} \right)t \right].$$
So if we say $v(0) = v_0$ is some initial velocity, then
$$C_1 = v_0 - \dfrac{\rho_f\pi D^2 g - 6mg}{18\pi\mu D},$$
and
$$\boxed{v(t) = \dfrac{\rho_f\pi D^2 g - 6mg}{18\pi\mu D} + \left( v_0 - \dfrac{\rho_f\pi D^2 g - 6mg}{18\pi\mu D} \right)\exp\left[- \left( \dfrac{3\pi\mu D}{m} \right)t \right]}.$$

That, of course, can be framed in terms of some positions, $y$, so
$$\dfrac{dy}{dt} = \dfrac{\rho_f\pi D^2 g - 6mg}{18\pi\mu D} + \left( v_0 - \dfrac{\rho_f\pi D^2 g - 6mg}{18\pi\mu D} \right)\exp\left[- \left( \dfrac{3\pi\mu D}{m} \right)t \right],$$
so
$$y(t) = \dfrac{\rho_f\pi D^2 g - 6mg}{18\pi\mu D}t - \left( \dfrac{mv_0}{3\pi\mu D} - \dfrac{\rho_f m\pi D^2 g - 6m^2g}{54\pi^2\mu^2 D^2} \right)\exp\left[- \left( \dfrac{3\pi\mu D}{m} \right)t \right] + C_2.$$
So, letting $y(0) = y_0$ leaves
$$C_2 = y_0 + \left( \dfrac{mv_0}{3\pi\mu D} - \dfrac{\rho_f m\pi D^2 g - 6m^2g}{54\pi^2\mu^2 D^2} \right),$$
and
$$\boxed{y(t) = y_0 + \dfrac{\rho_f\pi D^2 g - 6mg}{18\pi\mu D}t - \left( \dfrac{mv_0}{3\pi\mu D} - \dfrac{\rho_f m\pi D^2 g - 6m^2g}{54\pi^2\mu^2 D^2} \right)\left\{ 1 - \exp\left[- \left( \dfrac{3\pi\mu D}{m} \right)t \right] \right\}}.$$

You can make the equations less unruly by recognizing that the terminal velocity is
$$v(t\to\infty) = v_{\infty} = \dfrac{\rho_f\pi D^2 g - 6mg}{18\pi\mu D}.$$
So then,
$$\boxed{v(t) = v_{\infty} + \left( v_0 -v_{\infty} \right)\exp\left[- \left( \dfrac{3\pi\mu D}{m} \right)t \right]},$$
and
$$\boxed{y(t) = y_0 + v_{\infty}t - \dfrac{m\left( v_0 - v_{\infty} \right)}{3\pi\mu D}\left\{ 1 - \exp\left[- \left( \dfrac{3\pi\mu D}{m} \right)t \right] \right\}}.$$

That equation changes if the Reynolds number increases such that Stokes' Law is no longer valid, however, as the drag becomes proportional to $v^2$ instead of $v$, and that is sort of a pain in the butt.

So the bottom line is that your insight on the form of the solution was close, but not quite right (assuming I didn't make any mistakes. It is late and that was a lot of algebra).

As a fun exercise, you could play with the value of $v_0$ (which is negative if you were to throw it downward) and look at how the velocity evolves with time. Essentially, in the velocity equation, the exponential term represents a correction factor that will tend to bring any initial velocity back toward the terminal velocity, and the rate at which it does so is related to how far the current velocity is from terminal and the Stokes' drag on the ball. Given that this is how it should behave, I believe I went through this error free.

Last edited: Sep 27, 2015
11. Sep 27, 2015

### SataSata

Thank you boneh3ad. That is very detailed and is what I'm looking for.
Looking at v(t), is it right to say that terminal velocity in this case would be: $\frac{\rho_f\pi D^2 g-6mg}{18\pi\mu D}$ ?

12. Sep 27, 2015

Yes I added that in as an edit. I had gotten excited and clicked submit before I meant to.

13. Sep 27, 2015

### Staff: Mentor

Hi bh.

In this last equation, as best I can tell, v represents the upward velocity. So, in free fall, your v is negative. Your math is correct, but, for what it's worth, I think it would have been clearer for the OP to choose the downward direction as positive.

One correction is, in the buoyant term, that should be a D3, not a D2.

Chet

14. Sep 27, 2015

### Staff: Mentor

Actually, since $m=\frac{πD^3}{6}ρ$, this equation can be simplified to:
$$v_∞=\frac{D^2g}{3μ}(ρ_f-ρ)$$
where ρ is the density of the sphere.