# Determine viscosity using falling sphere (Stokes Law, Ladenburg correction)

## Introduction

I am trying to determine the viscosity of a fluid. Therefore, I let a sphere of known mass m and radius r fall (sink) into the fluid. I then measure the time it takes to sink/travel a given height.

Then there are the following forces having a effect on the sphere:

$F_g = mg$ and

$F_a = \frac{4 \rho \pi r^3 g}{3}$ (buoyant force)

with $g = 9.81 \ ms^{-2}$ and $\rho$ being the density of the fluid.

The result is a motion downwards (sinking), hence I get friction $F_r$.

Assuming this friction gets that strong to have no resulting force on the sphere (i.e. its velocity remains constant), I can use the following equation:

$F_g - F_a - F_r = 0$

According to Stokes Law, the friction of a moving sphere in a fluid with viscosity n is:

$F_r = 6 \pi r n v$ where v is the constant velocity.

Using these equations, I can determine the viscosity:

$n = \frac{g}{6 \pi r} (m - \rho \frac{4}{3} \pi r^3)\frac{1}{v}$

with $v = \frac{s}{t}$.

The fluid is inside of a cylinder, so I cannot use Stokes Law "as it is" because the walls of the container add more friction.

Hence I have to apply the so called Ladenburg correction with a correction factor

$f = (1 + \frac{(2.1 r)}{R}) (1 + \frac{(3.3 r)}{H})$

When / where do I have to apply this factor?

As I am doing this with spheres of different radiuses, I get multiple values for the viscosity:

radius/mm | viscosity/(kg/(m*s))
--------------------------------
1.0       | 0.594
2.0       | 0.608
2.5       | 0.631


Extrapolation to $r = 0 \ mm$ gives a value $n_0 = 0.575$.

I do not understand: Why do I need this extrapolation?

• I am not familiar with the Ladenburg correction, but judging by the form of the equation (going to 1 for infinite cylinder radius and height) I think you just have to multiply $F_r$ with it Mar 10 '13 at 15:19
• I also do not understand the extrapolation. I would assume that if the calculations are correct, you would get viscosities which are independent of the radii of the spheres Mar 10 '13 at 15:20
• Yep...the viscosity should come out same for all radii. Mar 10 '13 at 23:12
• Well, by fitting a straight line to the viscosity values, I get a non null slope ( > 0). It seems as if I should apply the correction factor to the extrapolated value n_0 at some radius r, which gives my a corrected value for this radius. Example: r = 10 mm, R = 100 mm, H = 100 mm. So f = 1.21 * 1.33. Hence i would use (0mm|n_0) as a first point, and (10mm|n_0 * f) as second point. Between thos points, I have to draw the corrected (less slope, but not null) line/graph. That is what I have to do, but I don't understand it :( Mar 10 '13 at 23:18

The correction needs to be applied to the terminal velocity you obtain :- $v$.$$v_{corrected} = v_{measured} L$$

, where $L$ is Ladenburg correction

Here is the link. Though the correction expression is different due to different conditions, but the concept is same.

• R = 23.25 mm and H = 800 mm. Mar 11 '13 at 10:26