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I am measuring the viscosity in water using a falling ball viscometer. In my system, I believe that exists two forces:

The force difference between the weight and buoyancy of the sphere:

$F_g = m'g = \frac{4}{3} \pi g r^3 ( \rho _s - \rho_l) $

where $m'$ is the effective mass of the sphere, $ \rho _1 $ the density of the sphere and $ \rho _l $ the density of the liquid

A frictional force given by the Stokes law:

$ F_f = 6 \pi r \eta v$

where $ r $ is the radio of the sphere, $ \eta $ the viscosity of the fluid, $v$ the velocity of the sphere

Equaling the two forces I obtain:

$ \eta = \frac{2}{9} \frac{r^2 g ( \rho _1 - \rho_2 )}{v_t} $

where $ v_t $ is now the terminal velocity

I believe that exists another force due the walls of the tube, since when I drop balls with similar densities but with very little diameter differences, the balls achieve quite different terminal velocities. But I would like to know how can I describe this force based in the geometry of the tube

I believe this is true because if I use the last equation, I obtain viscosities very different from the literature

For example the water density is $ 1000 kg/m^3 $, the density of the ball is $ 2400 kg/m^3 $, the diameter of the sphere is 0.01581 m, and the diameter of the tube is 0.016 m, so:

$ \eta = \frac{2}{9} \frac{(0.01581 m/2)^2 (9.8 m/s^2) ( 2400 kg/m^3 - 1000 kg/m^3 )}{0.000802 m/s} = 237.55 pa*s $

Considering that the water was at room temperature the answer should be 0.00089 pa*s . So there is a lot of difference

I am not even sure if this law is correct, if I consider the correct answer, then:

$ v_t= \frac{2}{9} \frac{(0.01581 m/2)^2 (9.8 m/s^2) ( 2400 kg/m^3 - 1000 kg/m^3 )}{0.00089 pa*S} = 213.79 m/s $

That is a very high velocity in my opinion.

I would like to deduce the full equation, but I am not sure if I am making the correct assumptions or if I missing something important, I would appreciate any suggestions, thanks in advance

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  • $\begingroup$ Minor correction: $r^3$ should be in the first formula for $F_g$. But your final equation is correct. $\endgroup$ – Deep Aug 27 '16 at 6:00
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    $\begingroup$ Since this is a low Reynolds number flow, effect of ball's motion on fluid extends to many multiples of ball diameter. Given that the tube in only a little bigger than the ball, walls are very much affecting the ball's motion. Can't you work with a container at least say 100 times larger than ball diameter? $\endgroup$ – Deep Aug 27 '16 at 6:15
  • $\begingroup$ Well the idea was to use a commercial viscometer. What is strange to me is that any of the technical manuals talk about this wall effect brookfieldengineering.com/download/files/Falling_Ball_B.pdf . And yes it was true, I forgot to write $r^3$, I am going to fix that. thanks $\endgroup$ – Delfin Aug 27 '16 at 15:04
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Deducing the full equation is difficult.
For such classical mechanics problems I often refer to "The General Properties of Matter" by F H Newman & V H L Searle first published in 1929 with the latest edition in 1957. Searle ran the undergraduate experimental laboratory at the Cavendish Laboratory in Cambridge and many of the experiments that he developed are still used today.
The book is old fashioned with no photographs and has no colour, however it has lots of derivations, that now have gone out of fashion, and reference to scientific papers.

In case you cannot easily access a copy (Fifth edition pages 229-230) I will give you some information as to how improve the accuracy of your experiment.

In there you will find that the length of the column should be divided into three and you should measure the terminal speed (and check that it is indeed the terminal speed, in the middle third of the column.
To correct for the wall effect:

$$v_\infty = v \left ( 1+2.4 \frac r R\right) $$ and for the end effect

$$v_\infty = v \left ( 1+3.6 \frac r h\right) $$

where $r$ is the radius of the sphere, $v$ is the measured terminal speed, $v_\infty$ is the speed for an infinite expanse of liquid, $R$ is the radius of the column of liquid and $h$ is the height of the column of liquid.

The text gives references for these derivation and also nots that there is a more accurate correct for the effect of the wall which is really just a polynomial in $\frac r R$.

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  • $\begingroup$ Very interesting answer, I am going to check that book, there is only one copy in the library of my university, although I don't know what edition is. Maybe I need to take more terms for the polynomial in r/R, since when I use $ v_{\infty} = 0.000802 m/s(1+2.4 \frac{0.01581 m}{0.016m}) = 0.0027 m/s $ and I put it in my equation $ \eta = \frac{2}{9} \frac{(0.01581 m/2)^2 (9.8 m/s^2) ( 2400 kg/m^3 - 1000 kg/m^3 )}{0.0027 m/s} = 70.56 m/s pa*s $ I still get a very high viscosity different from literature $\endgroup$ – Delfin Aug 27 '16 at 14:50
  • $\begingroup$ Do not forget that viscosity is very temperature dependent as shown by this table kayelaby.npl.co.uk/general_physics/2_2/2_2_3.html $\endgroup$ – Farcher Aug 28 '16 at 9:06

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