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.
 
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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?
 

FAQ: Stokes' law and falling sphere method

What is Stokes' Law?

Stokes' Law describes the force of viscosity on a small sphere moving through a viscous fluid. It states that the drag force acting on the sphere is directly proportional to the sphere's radius, the fluid's viscosity, and the sphere's velocity. Mathematically, it is given by F = 6πrηv, where F is the drag force, r is the radius of the sphere, η is the dynamic viscosity of the fluid, and v is the velocity of the sphere.

How is the falling sphere method used to measure viscosity?

The falling sphere method involves dropping a sphere of known size and density into a fluid and measuring the time it takes to fall a certain distance. By applying Stokes' Law, the viscosity of the fluid can be calculated. The terminal velocity of the sphere is determined, and using the known values of the sphere's radius and density, as well as the density of the fluid, the viscosity can be computed.

What are the assumptions made in Stokes' Law?

Stokes' Law assumes that the flow around the sphere is laminar, the sphere is rigid and smooth, the fluid is incompressible and Newtonian, and the sphere is small enough that its Reynolds number is less than 1. These conditions ensure that the drag force is linearly proportional to the sphere's velocity.

What is the Reynolds number and why is it important in Stokes' Law?

The Reynolds number (Re) is a dimensionless quantity that predicts the flow regime around an object in a fluid. It is defined as Re = ρvr/η, where ρ is the fluid density, v is the velocity of the sphere, r is the radius of the sphere, and η is the dynamic viscosity of the fluid. In Stokes' Law, it is crucial that Re < 1 to ensure laminar flow, which is a key assumption for the law to hold true.

What are some practical applications of Stokes' Law and the falling sphere method?

Stokes' Law and the falling sphere method are used in various fields such as material science, biology, and engineering. They are applied in measuring the viscosity of fluids, characterizing the behavior of suspensions, determining sedimentation rates in natural waters, and even in medical diagnostics to analyze blood viscosity. These techniques are valuable for quality control, research, and development in industries dealing with fluids and particulate materials.

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