Analysis of Stokes law?

[donsrno1]

Hi,
I am currently involved in an investigation in which I am looking at stokes law for measuring the viscosity of different liquids. I realise that stokes law only applies with infinite boundaries (i.e. a tube infinately wide and long) however what I wanted to invvestigate was what sort of effect on the reading for viscosity there was when these conditions were not applied and the sphere used was not very small compared to the diameter of the tube (with the largest ball bearing used it's diamter was half that of the tube.)

Would anybody be able to tell me if it is possible to get an accurate relationship between the diameter of the ball bearing and the value attained for viscosity?

So far my results appear to show that there is perhaps a relationship that the diameter of the ball bearing squared is roughly proportional to the reading I got for viscosity.

Any resonses would be much appreiciated or any link to other source of information.

Thanks.

 

[rcooper]

Are you sure you are working in the laminar region? What is the Reynolds number (measured velocity*diameter/kinematic viscosity). If it is greater than 1 then the flow is transitional or turbulent and Stoke's law does not apply.

 

[25362]

Let's start by saying that the terminal velocity of the falling ball is that velocity when the gravitational force equals the frictional "drag" force.

([π]/6)d[sup]3[/sup].[Δ][ρ].g = f ([π]/4)d[sup]2[/sup].[ρ].V[sub]t[/sub][sup]2[/sup]/2​

The generally-accepted validity of the Stokes' law is for 0.00001<Re<2 and f = 24/Re

Where the terminal velocity, V[sub]t[/sub], m/s

V[sub]t[/sub] = 0.0556.d[sup]2[/sup].[&Delta;][&rho;].g/[&mu;]​

d, diameter of the ball, m
[&Delta;][&rho;], density difference, [&rho;][sub]ball[/sub]-[&rho;], kg/m[sup]3[/sup]
[&rho;], the density of the liquid, kg/m[sup]3[/sup]
[&mu;], absolute viscosity, kg/(m.s)
g, acceleration of gravity, m/s[sup]2[/sup]

For 2<Re<500 (f = 18.5/Re[sup]0.5[/sup]), the formula would be:

V[sub]t[/sub] = [0.072.d[sup]1.6[/sup].[&Delta;][&rho;].g/([&mu;][sup]0.6[/sup].[&rho;][sup]0.4[/sup])][sup]0.714[/sup]​

In which the viscosity and the ball diameter are differently related.

If you do a Google search on falling ball viscometry
you may find more answers to your queries.

 

[donsrno1]

Thanks for the help but I'm afraid my knowledge on the topic is not great so I was hoping you might be able to expand a bit on where the different equations and figures you used came from. If you had any suggestion on what viewpoint I would be best to base my investigation that would also be much appreciated because there may be a better way to look at what I was trying to study.

I have taken several reading for the terminal velocity of different diameters of ball bearings in three different liquids and my initial plan was to look at what the effect on the results I got for viscosity were as the diameter increased.

Thanks.

 

[donsrno1]

can anyone help?

 

[25362]

If you read again my post you'll see that by substituting the value of the friction factor in the gravitation/drag forces equilibrium equation you'd get the Stokes'law as expressed above.

 

[jmw]

The Stokes law method for viscosity depends on finding the terminal viscosity which is not as easy to impliment in a practical viscometer as would be liked.
Instead, most fallling ball viscometers use elapsed time: the time taken for the ball to fall a fixed distance from rest. Many such instruments do not not use a perpendicular tube but an inclined tube.
I would suggest this as a practical way forward (why re-invent the wheel?) and suggest you can read something more in thread124-55699 or you can do a search for "Hoffler" "Falling ball viscometers" or similar searches. You could also visit the Kittiwake website. They produce a falling ball viscometer for fuel oils and provide an excellent instrunctional CD.

JMW

http://www.ViscoAnalyser.com