Stokes' law and falling sphere method

  • #1
xenoidmaster
2
0
Homework Statement
I need help at calculating the viscosity of a fluid. I did an experiment of dropping spherical steel balls through a liquid. The diameter includes 5 mm, 10 mm, 15 mm, 20 mm. What makes me confuse is that the viscocity is difference for each diamater, isnt it suppose to be the same? viscocitty shouldnt be affected by diameter of the balls. As i know only terminal velocity will be affected. And so how to calculate the viscocity of the liquid, to get similar answers/small difference between each diameter.
Relevant Equations
η = 2gr^2(d'– d)/9v
where:
v is the particles' terminal velocity velocity (m/s),
r is the radius of the sphere,
g is the gravitational acceleration,
d' is the density of the falling sphere,
d is the density of the liquid,
and η is the viscosity.
In dire need of help, someone please explain the correct method for this, if its not possible what should i write in the conclusion for this?
 
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  • #2
You say you did an experiment. What did you measure? Did you plot your results to see if the viscosity is (or is not) constant?
 
  • #3
Did you loose your last thread on the topic?

https://www.physicsforums.com/threads/viscosity-by-falling-sphere-equations.1058374/#post-6978437

Your ball is not falling through the liquid (water) at terminal velocity. You have a transient (time varying) velocity to contend with (in your measurements). By solving (with help if necessary) the ODE for position vs time in the other thread you can examine what the data would "look like" if the conditions were actually met in the experiment. It is almost certain (steel ball falling in water) that the flow around the ball is not laminar. This is (likely) an unmet requirement for the equation wish to use.

You need to post the results of the experiments so it can be discussed.
 
Last edited:
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  • #4
xenoidmaster said:
Homework Statement: I need help at calculating the viscosity of a fluid. I did an experiment of dropping spherical steel balls through a liquid. The diameter includes 5 mm, 10 mm, 15 mm, 20 mm. What makes me confuse is that the viscocity is difference for each diamater, isnt it suppose to be the same? viscocitty shouldnt be affected by diameter of the balls. As i know only terminal velocity will be affected. And so how to calculate the viscocity of the liquid, to get similar answers/small difference between each diameter.
Relevant Equations: η = 2gr^2(d'– d)/9v
where:
v is the particles' terminal velocity velocity (m/s),
r is the radius of the sphere,
g is the gravitational acceleration,
d' is the density of the falling sphere,
d is the density of the liquid,
and η is the viscosity.

In dire need of help, someone please explain the correct method for this, if its not possible what should i write in the conclusion for this?
Let's see your calculations. How far from the walls of the container were the balls?
 

What is Stokes' law?

Stokes' law is a formula that describes the force of viscosity acting on a small sphere moving through a viscous fluid at a low Reynolds number.

How is Stokes' law used in the falling sphere method?

In the falling sphere method, Stokes' law is used to calculate the viscosity of a fluid by measuring the terminal velocity of a sphere falling through the fluid.

What factors affect the accuracy of the falling sphere method in measuring viscosity?

The accuracy of the falling sphere method in measuring viscosity can be affected by factors such as the size and shape of the sphere, the temperature of the fluid, and the presence of any impurities in the fluid.

What are the limitations of Stokes' law in the falling sphere method?

Stokes' law assumes that the fluid is incompressible and that the sphere is small and moving at a low Reynolds number. These assumptions may not always hold true in practical applications, leading to limitations in the accuracy of the method.

How can the falling sphere method be improved to overcome the limitations of Stokes' law?

To improve the accuracy of the falling sphere method, researchers can use correction factors to account for deviations from the ideal conditions assumed by Stokes' law. Additionally, using more sophisticated equipment and techniques can help reduce errors in the measurements.

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