Science Buddies: "Ask an Expert"

Hello,
I am trying to find the dynamic viscosity of water and I am using the formula
"Viscosity = (2(ΔP)ga^2)/9v
where:

Viscosity is in newton-seconds per meter squared (Nsec/m2).
Delta (Δ) P is the difference in density between the sphere and the liquid, and is in kilograms per meter cubed (kg/m3).
g is the acceleration due to gravity and equals 9.81 meters per second squared (m/s2).
a is the radius of the sphere in meters (m).
v is the average velocity, defined as the distance the sphere falls, divided by the time it takes to fall in meters per second (m/s)." (from this site)
I have found ΔP to be 330 kg/m^3, a to be 0.012 m, and v to be .25m/1.2 sec. This comes to be some viscosity around .8; however i then checked the true viscosity of water in pa*s (same as Nsec/m2) and saw that it was about 8.9 x10^-4. Which is not even cloes to what I got. I don't understand what I am doing wrong. Please help.

Hello Soulous ,

I'm not sure why you are getting the wrong answer, but here are a few comments.

1.2 sec is a very short time to measure by eye and hand. Could you use a longer column of water and a less dense ball to make the fall time longer?

The formula assumes that you measure the falling velocity after the ball has accelerated to it's final speed. Made-for-purpose falling-ball viscometers measure the falling time between two points that are below the liquid surface so the ball has time to get up to speed. For more viscous liquids, the ball gets to terminal speed quickly, so the error is small. In water and with so short a falling distance, maybe it's not so small.

While these are issues affecting the accuracy of your result, they don't seem big enough to account for the three orders of magnitude error.

One other consideration involves the “Reynolds number.” For any fluid-flow situation, this number determines whether the fluid moves as “laminar flow” or “turbulent flow.” See http://en.wikipedia.org/wiki/Reynolds_number.

I calculate that your data indicates a high Reynolds number, and thus indicates that you have turbulent flow around the ball. Unfortunately, the equation you are using is for laminar flow.

I have no experience in measuring viscosity, just what I learned in school. I will try to find someone else who is more knowledgeable than I. Until then, you could try performing the falling-ball experiment in a more viscous fluid where the Reynolds number will be low enough to make the result valid.

Sorry to give you an uncertain answer, WW

I see, I came to the same conclusion during further research? Do you know of any other way to calculate viscosity without having to resort to specific scientfic tools? I used a graduated cylinder during this test which as around 30 cm tall, so I don't think I'm going to be able to find something larger than that. Falling ball viscometers were too expensive for my budget and with the experiment already done, I would rather not redo everything. Thanks for the response.

Soulous,

I don't know of any way to compute viscosity from the data you have. The only idea I have is to try to find a ball that would give a meaningful result in your 30 cm cylinder.

The Reynolds number is given by

R = (ρ V L) / μ,

where ρ is the density of the liquid, V is the velocity of the ball, L is the diameter of the ball, and μ is the dynamic viscosity.

If you could find a very small ball to reduce L and density close to water so that V is reduced considerably, you might be able to get a useable result. No guarantees.

If it meets the requirements of your project, you could change the liquid to one that is very viscous, say motor oil, and get a useable result that way.

Good luck, WW