# Are there any cases where Stokes law does not apply in viscous fluids?

A friend of mine and I are conducting an experiment to find the relationship between terminal velocity and radius of a sphere (i.e trying to confirm Stokes law). We are using spheres ranging from 1.5mm radius to 3mm, and dropping them in a large beaker (Of radius 7cm) filled with canola oil, using a tracker software to measure the terminal velocity. Terminal velocity is indeed reached for all balls, and we get consistent values of Vt for each repition of the experiment.

Our results, however, are inconclusive. We have been trying to figure out why for the last week or so to no avail. Our teacher has no idea either.

The problem: We assumed Stokes law would apply to this scenario (perhaps a mistake?), and that Terminal velocity would be proportional to r^2 yet our results show overwise. (And Terminal velocity plotted over Radius seems to be almost exactly linear) Here is our plot of V over R^2:

Sorry for the unfinished formatting; I quickly entered it in plot.ly because I thought google sheets wasn't scaling properly. Since the trendline couldn't fit within the uncertainty bars assumed there must be something else happening to cause the mismatch with what was expected.

Does anyone have any ideas as to why this is? We have double checked everything and are 100% sure we have made no measurement errors. Perhaps it is simply because we do not have enough data points? Wet tried calculating the terminal velocity from the Stokes formula, and it is consistently higher than what we measured in the experiment. We have absolutely no idea what could be causing this. All help is appreciated.

Thanks :)

• The Stokes formula applies only in the limit of low Reynolds Number. Have you calculated the Reynolds Number for the situation you are encountering. See this: grc.nasa.gov/www/k-12/airplane/dragsphere.html – Chet Miller Nov 14 '18 at 4:12
• @ChesterMiller Yes we have - sorry for not mentioning that. From memory, it was ~0.3, so it should apply. – Manu Nov 14 '18 at 6:19
• What was the mass of each sphere, and how did the mass vary with radius? – Chet Miller Nov 14 '18 at 12:33
• Are the spheres solid or hollow? – Chet Miller Nov 14 '18 at 13:41
• Did your calculations take into account buoyancy? – Chet Miller Nov 14 '18 at 13:43

Here the obvious error is due to the fact that the point (0,0) is related to the number of data. Meanwhile, in the case of the Stokes law with a small radius, the ratio $$v/r^2$$ tends to a constant. Therefore, you can add another point, for example (0.01,0.28272), repeating the previous data, but not zero. Theoretically, if there is a dependence on the initial velocity, the measurement result should fall on the curve $$\frac {v}{r^2}=a+(b/r^2-a)\exp(-c/r^2)$$. Using the data we find the constants, construct the curve and plot the data.

We give an explanation.The resistance force acting on a spherical particle in a case of the Stokes law is $$F=6\pi \mu r u+\frac {2}{3}\pi r^3\rho \frac {du}{dt}+...$$. The equation of motion, taking into account the force of gravity and the force of Archimedes is $$\frac {du}{dt}=g-ku$$, $$g=g_0(\rho _s-\rho )/(\rho_s+\rho /2)$$, $$k=\frac {9\mu }{2(\rho _s+\rho /2)r^2}$$. We set $$u(0)=u_0$$, then the general solution of the equation can be written as

$$ku(t)=g+(ku_0-g)e^{-kt}$$

We assume that the speed measurement is carried out at a fixed time $$t=t_0$$, then we put $$k=k_0/r^2, a=g/k_0, b=u_0, c=k_0t_0,v=u(t_0)$$, as a result we find $$\frac {v}{r^2}=a+(b/r^2-a)\exp(-c/r^2)$$. Using the data, we find constants $$a=0.2828, b=18.46, c=15.63$$

• What is the thought behind that correlation? – Eashaan Godbole Dec 28 '18 at 20:29
• Can you clarify the question? – Alex Trounev Dec 29 '18 at 21:45
• You used $\dfrac{v}{r^2}=a+\dfrac{1}{r^2}(b\exp({-\dfrac{c}{r^2}}))$. Where did you get this function from? – Eashaan Godbole Dec 31 '18 at 7:37
• @EashaanGodbole I updated the post. – Alex Trounev Jan 2 at 21:25

Stokes law is a corollary of the Navier-Stokes law.

It is not applicable in cases where

• Reynolds number is too high: For particles, the regime in which Stokes law is applicable - the Stokes regime - is when $$Re=\dfrac{d_p\cdot v\cdot \rho}{\mu}<<1$$. If Re is too high, then drag forces will cause deviations from your expectations.
• Non-Newtonian fluids are involved: The viscosity of these fluids changes on application of shear stress and you won't get a linear relationship as expected for Newtonian fluids.

The Navier – Stokes equation describes not only the laminar regime, but also the turbulent one. but the turbulent solution is complex. In addition, there is a connection between the solution of the Navier – Stokes equation and the solution of the Schrödinger equation see file 3. See attached files where there is a solution for the sphere in file 2

file 1 Values of large dimensionless unknown functions (for example, a large Reynolds number) can be found out as solutions of non-linear partial differential equation. In this case these equations can be brought to some number of non-linear ordinary differential equations. Turbulent solutions corresponding to large values of unknown function are complex. Transition from real solution to complex turbulent solution is realized through infi nity of the right parts of ordinary differential equation system to which Navier – Stokes equations are brought. Thus, real solution of Navier – Stokes equation for turbulent mode yields function going to infi nity. At the same time, complex solution for the turbulent mode is fi nite. Fluid fl ow resistance coeffi cient is calculated for round pipeline with different pipeline walls roughness.

file 2 The problem of turbulent fluid motion description has not been solved yet. It creates difficulties when oil, gas pipelines design calculation is performed. Besides, there are no theoretical methods for description of bodies motion in turbulent environment. These methods would be necessary for description of motion of aircrafts, submarines or above-water ships in the turbulent mode. Without simulation of bodies motion in wind tunnels or water basins, design of the bodies moving in the viscous environment is impossible.

file 3 In this part of the article is described the physical meaning of complex velocity. The relationship between the Schrödinger equation and the Navier – Stokes equation obtain. Schrödinger equation and the Navier – Stokes equations are, in general, have a countable number of turbulent energy and solutions.

1. YAKUBOVSKIY, EG. "STUDY OF NAVIER-STOKES EQUATION SOLUTION I. THE GENERAL SOLUTION OF NONLINEAR ORDINARY DIFFERENTIAL EQUATION." EUROPEAN JOURNAL OF NATURAL HISTORY 3 (2016): 60-66. https://world-science.ru/pdf/2016/3/14.pdf
2. YAKUBOVSKIY, EG. "STUDY OF NAVIER-STOKES EQUATION SOLUTION II. THE USE OF LAMINAR SOLUTIONS." EUROPEAN JOURNAL OF NATURAL HISTORY 3 (2016): 67-83.https://world-science.ru/pdf/2016/3/15.pdf
3. YAKUBOVSKIY, E. G. "STUDY OF NAVIER–STOKES EQUATION SOLUTION III. THE PHYSICAL SENSE OF THE COMPLEX VELOCITY AND CONCLUSIONS." EUROPEAN JOURNAL OF NATURAL HISTORY 3 (2016): 84-87. https://www.world-science.ru/pdf/2016/3/16.pdf
• Please summarize the essential points of these papers so as to make your answer self-contained. – ZeroTheHero Dec 29 '18 at 0:16