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 $v = \frac{s}{t}$.
Question about Ladenburg correction
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?
Question about extrapolation
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?
Best Answer
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.
Also, after applying the Ladenburg correction, your viscosity should become constant for all radii.