Stoke's Law - Viscosity

[SpykerC8Aileron]

For a piece of GCSE physics coursework, my science teacher has decided that we must all work out the viscosity of some fluids using Stoke's Law, despite there being no mention of it anywhere in the syllabus. Apparently, we are doing it to 'impress the examiner'.

 

The trouble is that I cannot find any websites which provide Stoke's Law in the format necessary to calculate viscosity. I'm aware that there are internet-calculators that can do this for me, however my teacher quickly got wise of these and is now forcing us to write down every piece of working.

 

I'm not asking you to do the calculations for me - I'm not that lazy - but it would really help if someone could provide the formula in the format for working out viscosity.

 

Thanks in advance!

 

-Spyker

Edited February 18, 2014 by SpykerC8Aileron

[studiot]

I am having real trouble reconciling GCSE with first year university, which is where a student would normally meet the Stokes falling ball viscometer.

 

Google provides many hits on stokes visometer here is a university lab from one.

 

http://www.engr.uky.edu/~egr101/ml/ML3.pdf

 

This gives the fully corrected formula.

 

A simpler one is

viscoscity, f =[math]\frac{{2{r^2}g}}{{9\nu }}({\rho _s} - {\rho _L})[/math]

p_s and p_L are the sphere and liquid densities respectively

r is the sphere radius

v is the terminal velocity.

 

I can't see that you could be expected to produce a derivation of even this simplified equation for GCSE

Edited February 18, 2014 by studiot

[SpykerC8Aileron]

I know. I promise that I'm not making this up. My teacher said himself that it was A-level, and as far as I can tell I can still get full marks without it, but to put it in his own words: 'I'd be very disappointed if you didn't include Stoke's Law'. Thanks for the speedy response!

Edited February 18, 2014 by SpykerC8Aileron

[CharonY]

Depends on what information you have. If you have the drag force and the dimension of a sphere, it is quite straightforward to derive the dynamic viscosity (it is, after all, part of Stokes's law).

 

Edit: cross-posted, but assuming you have settling velocity as measured parameter, deriving the equation is actually relatively straightforward. You just have to start with balancing the forces acting on a particle (drag, buoyancy and gravitational force).

Since you have not posted the experimental setup to derive viscosity the question could also relate to what you would have to measure to derive viscosity. In the end it all boils down to get the drag force (and all the hints have already been shown here).

Edited February 18, 2014 by CharonY

[SpykerC8Aileron]

I don't have the drag force - only the masses and lengths, and therefore the densities.

Edited February 18, 2014 by SpykerC8Aileron

[studiot]

 

work out the viscosity of some fluids using Stoke's Law,

 

Just out of interest what do you mean by this?

 

To balance the forces (is force balance on the GCSE? syllabus) in an infinite ocean it is a relatively simple calculation as CharonY suggests, although you must know a good deal of underlying physics to do this.

 

But to measure the viscosity using stokes law requires some quite sophisticated corrections for real world apparatus.

[SpykerC8Aileron]

I'm not sure if I fully understand it either. I'm just doing what I've been told to do, and I hate coursework anyway.

[studiot]

g in the formula is the acceleration due to gravity.

 

If you multiply the mass of an object by g you get the weight, which is the force of attraction between the Earth and the body.

 

So the forces acting on the falling ball are the weight w=mg, the bouyancy force due to the displaced fluid and the viscous frictional drag against the motion.

 

The mass of the ball obviously equals its volume times the density, and the mass of displaced fluid is similar, hence the subtracted quantities in brackets.

 

It is difficult to help without knowing what you are expected to produce.

[SpykerC8Aileron]

I'm probably going to make myself sound incredibly stupid now, but is the acceleration due to gravity constant then?

Also, is there any way to insert a picture into this thread?

 

Once again, thanks for all the help so far!

 

EDIT: I think that I've worked out how to do it, as it says on 'Bitesize' that 1kg = 10N, so I take it that the force of attraction is 10... I think.

Edited February 23, 2014 by SpykerC8Aileron

[studiot]

No you are definitely not stupid.

 

If you use the more reply options (button towards bottom right) you will find a way of uploading files to your post.

 

Click the cursor where you want it, after uploading, and it will be displayed there.

 

 

g is regarded as constant for schoolwork, even at A level.

It actually gets smaller and smaller the further from the centre of the Earth you get, so that has to be taken into account for rockets etc.

It is normally taken as 9.81 Newtons per kilogram of mass.

 

So a mass of 1kg weighs 9.81 Newtons force. (10 is often a good figure in rough calcs)

Edited February 23, 2014 by studiot

[SpykerC8Aileron]

I thought that that was V.

 

EDIT: Ignore what I just wrote there; V is 9.8, of course.

I think that I'm ready: {[(2X2.5^2)9.81]/9(9.8)}(0.35-1.00652)

 

EDIT (2): I got a minus-number for an answer. That can't be right, can it?

 

I got -0.9127637, followed by many more numbers

Edited February 23, 2014 by SpykerC8Aileron

[studiot]

One of the most important lessons to learn in GCSE is to be consistent in your units.

 

This is why it is recommended to always state them in full.

 

Let us look at your figures.

 

 

{[(2X2.5^2)9.81]/9(9.8)}(0.35-1.00652)

 

So do you have a 2.5 metre ball?

 

you say the density of the ball is 0.35 whereas the density of the liquid is 1 and a bit.

 

So why does the ball fall at all?

[SpykerC8Aileron]

The marble (ball) is 1.6cm in diameter and weighs 5.92g. Therefore, its volume is 17.15728468cm^3. The measuring-cylinder that I dropped the marble in is 25.5cm tall and has a radius of 2.5cm. The water weighed 503.26g in this case. The density of the measuring-cylinder is therefore in kg/cm^3.

 

So do you have a 2.5 metre ball?

 

Is 'r' the radius of the ball, or the radius of the measuring-cylinder? '2.5cm' is the radius of the measuring-cylinder.

 

2.5 metre

 

I thought that it was in cm already. I can't quite see where I've put something in metres. If everything is in cm, then surely the units are consistent. Both densities and all radii are in centimetres.

Edited February 23, 2014 by SpykerC8Aileron

[studiot]

You should be using the MKS system.

 

Length is measured in metres, so bring all lengths to metres (and powers of 10)

 

Force is measured in Newtons

 

Mass is measured in kilograms

 

density is measured in kg per cubic metre.

 

So if your radius is in cm, and therefore your volume is cubic centimetres, your mass will be wrong, and thus your forces will be wrong.

 

You can work in other units, but it is just so easy to make a mistake and get things 10, 100, 1000 times too big or small.

That is why I said be consistent.

 

Have we cured the negative sign now?

[SpykerC8Aileron]

Yes, hoorah! It's a whole number. Thanks for all the help (and patience). One final question: what are the units for this; I know that it can vary, so it would be great if you could tell me the specific unit for the one that you've shown me. Then, I promise, that's it.

Edited February 23, 2014 by SpykerC8Aileron

[SpykerC8Aileron]

Fine, don't answer me.

[studiot]

 

Fine, don't answer me.

 

 

Pardon?

 

The units of viscosity are fairy easy to look up on the internet eg this pdf

 

http://www.hydramotion.com/pdf/Website_Viscosity_Units_V2.pdf

 

You should be using the SI units.

 

You will note that they have quoted two types of viscosity.

 

Dynamic viscosity, which is the one I have given you as f

 

Formulae in fluid dynamics are so often divided by the density that a second form of viscosity is often given called the

 

Kinematic viscosity, which is simply the dynamic viscosity divided by the density.

[math]h = \frac{f}{p}[/math]

 

 

This has SI units of metres squared per second or m² s^-1.

 

To see how the units of dynamic viscosity are derived we use Newton's law of viscosity which says that the shear stress in a fluid is proportional to the velocity gradient.

[math]Shearstress = \frac{{Force}}{{Area}} \propto velocitygradient = f\frac{{dv}}{{dz}}[/math]

 

Where the dynamic viscosity, f, is the constant of proportionality

 

 

[math]f = \frac{F}{{A\frac{{dv}}{{dz}}}}[/math]

 

If you are familiar with the following it is called dimensional analysis. we can represent a formula by inserting M for mass, L for length and T for time as the basis of mechanical units

 

[math]\frac{{ML{T^{ - 2}}}}{{{L^2}L\frac{{{T^{ - 1}}}}{L}}} = M{L^{ - 1}}{T^{ - 1}}[/math]

 

So the units work out to mass divided by (length x time) or

 

kg m^-1 s^-1

 

However this is normally expressed in as Newton seconds per metre squared, (N-s m^-1) which is equivalent (you should check this for yourself).

 

Edited February 26, 2014 by studiot