Find the viscosity coeffienct of glycerol, using stokes law

[RoryScanlan]

Homework Statement

I have been given an experiment to perform, whereby I drop steel ball bearings into a glass measuring cylinder, and time how long it takes for the ball bearing to fall a set distance.
From this I should be able to work out the viscosity coeffienct of the glycerol.

I have been given this sum to work with:
[itex]2g/9(p-o)r^2/v [/itex]
Whereby :
p = the density of the ball bearing
o = the density of the glycerol(given to me as 1260kg/m³
r = radius of sphere
v = average velocity

I have crunched some number but I think they represent the reistants the sphere feels in the fluid, not the actual viscosity of the fluid. This is because the number I am receiving from the sum is raising in line with the radius of the sphere. If this is the case how to can i make the connection between this ristance and the coeffient viscosity of the fluid?

This is not a physics course and I do not study physics so if u can try to keep it simple it would be nice :)
Any links or help would be greeat!

Homework Equations

The Attempt at a Solution

Here is one of my test, does the final number represent the viscosity coeffient of the fluid or does it represent the resistance the object feels? or does it represent nothing? :P

Ball Bearing: 4mm. Mass: 0.00026kg. Distance traveled 0.20m.
Radius (m) : Diametre(mm)/2 = 4/2 = 2 ÷1000 = 0.002
Volume (m3) : V= 4/3 πr³ = 4/3 ∙ 3.141592 ∙ 2³ = 33.51031469 ×〖10〗^(-9) = 3.351031469 × 〖10〗^(-8)
Density (kg/m3) : Mass/Volume = 0.00026/(3.351031469 × 〖10〗^(-8) ) = 7758.805085

Time Taken Test 1 (s) : 4.65
Time Taken Test 2 (s) : 4.80
Time Taken Test 3 (s) : 4.79

Average Time (s) : 4.75
Average Velocity (m/s) 0.20/4.75=0.04
Viscosity Coefficient of glycerol according to this test: Viscosity = 2(g/9) × (∆P) × (r^2/v)=
2(9.81/9) × (7758.81-1260) × (〖0.002〗^2/0.04) = 1.41675584

 

[RoryScanlan]

Can any clear this up for me coefficient viscosity of a fluid should be constant no matter the radius of the ojects ? or would it increase as the radius of the object increased?

 

[Chestermiller]

Hi Rory,

The expression should read:

[tex]μ=\frac{2}{9}\frac{(ρ_b-ρ_g)}{v}gR^2[/tex]

Is this the form of the equation you were using (it's hard to tell from how you have written it)?

The equation was derived originally using the approach that you suggested. The drag force for stokes flow is given by:
[tex]F = 6πμRv[/tex]
You should be able to use your analysis to derive the equation for the falling ball viscosity.

Chet

 

[RoryScanlan]

Hi Chet,

Thanks for the reply yes the is the formulae I am using, but i am getting some strange results, for instance this is my equation for a 12 mm ball.
[tex]
{\frac{2}{9}}*{\frac{(7718.78−1260)}{0.27}}*9.81*0.006^2 = 1.87735
[/tex]
and this is my equation for a 14mm ball.
[tex]
{\frac{2}{9}}*{\frac{(7718.78−1260)}{0.30}}*9.81*0.007^2 = 2.29975
[/tex]
The densities are in kg/m3 the velocity is ms-1 and the radius is in meters

Any ideas why these numbers are so different? surely if I am using exactly the same liquid and only changing the ball. The viscosity coeffience should be similar?

 

[RoryScanlan]

Hi Chet,

Thanks for the reply yes the is the formulae I am using, but i am getting some strange results, for instance this is my equation for a 12 mm ball.
[tex]
{\frac{2}{9}}*{\frac{(7718.78−1260)}{0.27}}*9.81*0.006^2 = 1.87735
[/tex]
and this is my equation for a 14mm ball.
[tex]
{\frac{2}{9}}*{\frac{(7718.78−1260)}{0.30}}*9.81*0.007^2 = 2.29975
[/tex]
The densities are in kg/m3 the velocity is ms-1 and the radius is in meters

Any ideas why these numbers are so different? surely if I am using exactly the same liquid and only changing the ball. The viscosity coeffience should be similar?

Hmmm I am analayzin my results and the only way that this is possible if the coeffienct of viscosity is dependant on radius and is not constant just like with velocity.
This should really be called the coeffienct of drag on an object as it travels through a liquid under certain conditions. ?:P

 

[TheAustrian]

This may or may not help to clear issues, but the whole experiment is also dependent on temperature. If you do this experiment in the Summer, and in the Winter, it'll be a bit different.

 

[RoryScanlan]

All information helps at the moment, thanks :)

Can any clear up this for me though, is it accurate to say.
The viscosity of a fluid is constant at a constant tempreture.
But the coeffecient viscosity is not constant and is dependant on the objects radius and velocity(and thus force) that it is applying to the liquid.
EI the coeffecient viscosity is restance produced by a liquid as an object applies a force?
This is where I am getting most confused.

 

[Chestermiller]

All information helps at the moment, thanks :)

Can any clear up this for me though, is it accurate to say.
The viscosity of a fluid is constant at a constant tempreture.
But the coeffecient viscosity is not constant and is dependant on the objects radius and velocity(and thus force) that it is applying to the liquid.
EI the coeffecient viscosity is restance produced by a liquid as an object applies a force?
This is where I am getting most confused.

The viscosity is a property of the liquid, and is independent of the objects that are used. Did you wait for the balls to reach terminal velocity before you measured the velocity? How did the diameters of the balls compare with the diameter of the cylinder? (The equation you used applies to the balls falling through an infinite ocean of the fluid. If the ball diameter is on the same order as the cylinder diameter, the relationship you used will not be as accurate.)

On another note, what were the units of the viscosities you calculated.

Chet

 

[RoryScanlan]

The viscosity is a property of the liquid, and is independent of the objects that are used. Did you wait for the balls to reach terminal velocity before you measured the velocity? How did the diameters of the balls compare with the diameter of the cylinder? (The equation you used applies to the balls falling through an infinite ocean of the fluid. If the ball diameter is on the same order as the cylinder diameter, the relationship you used will not be as accurate.)

On another note, what were the units of the viscosities you calculated.

Chet

I think the problem was with the length of my tube, and the heavier balls where not reaching terminal velocity. The tube was much larger than the balls in diameter so i don't think the walls would have affected the results as drastically as this.

I tested ball bearings ranging from 4mm-14mm. In the lower ranges 4-8mm the results are all quite constant(around 1.4) but at 10mm+ the results begin ro rise incrimentally(10mm=1.7,12mm=1.8,14mm=2.2). The tube was around 3inches wide so i don't think it would be that? The length of the tube was only 20cm.

This brings me to believe the higher mass caused a slower acceleration not allowing them to reach terminal velocity in 20cm thus distorting the results?

I am not told what unit i am attempting to measure, I was simple given an experiment sheet with the Aim to:-
Determine the coefficent of viscosity of glycerol.

The method is as described.

Then Finally i am told to plot a velocity over r² graph, and then to determine the viscosity of glycerol(η) from the equation desribed above.

In my equation i converted:-
Radius from mm to m.
Volume is in mm3.
Mass is in grams.
I used [tex]\frac{Mass}{Volume} X 1000000[/tex] to give Density in kg/m3
Velocity ms^-1

^ This was in the attempt to achieve kg/ms²

Thanks again for the help.

 

[Chestermiller]

I think the problem was with the length of my tube, and the heavier balls where not reaching terminal velocity. The tube was much larger than the balls in diameter so i don't think the walls would have affected the results as drastically as this.

I tested ball bearings ranging from 4mm-14mm. In the lower ranges 4-8mm the results are all quite constant(around 1.4) but at 10mm+ the results begin ro rise incrimentally(10mm=1.7,12mm=1.8,14mm=2.2). The tube was around 3inches wide so i don't think it would be that? The length of the tube was only 20cm.

This brings me to believe the higher mass caused a slower acceleration not allowing them to reach terminal velocity in 20cm thus distorting the results?

I am not told what unit i am attempting to measure, I was simple given an experiment sheet with the Aim to:-
Determine the coefficent of viscosity of glycerol.

The method is as described.

Then Finally i am told to plot a velocity over r² graph, and then to determine the viscosity of glycerol(η) from the equation desribed above.

In my equation i converted:-
Radius from mm to m.
Volume is in mm3.
Mass is in grams.
I used [tex]\frac{Mass}{Volume} X 1000000[/tex] to give Density in kg/m3
Velocity ms^-1

^ This was in the attempt to achieve kg/ms²

Thanks again for the help.

The units of viscosity should be kg/ms. Please double check.

Chet

 

[RoryScanlan]

Sorry yes I just check that 1 kg/ms = 1 Pa.s which is the unit of dynamic viscosity, while looking at this I also read a published value for glycerol viscosity to be 1.412Pa.s

Which is close to my first 3 results, and make's me believe the above post(about why my results are distored) could be true. Do you mind if i post my first result(which i believe to be correct) in proper tex so u can point out any mistakes (if you can see any).

Just so i don't burn any more time on this theory i have created in the above post.

Thanks :)

Rory

 

[Chestermiller]

Sorry yes I just check that 1 kg/ms = 1 Pa.s which is the unit of dynamic viscosity, while looking at this I also read a published value for glycerol viscosity to be 1.412Pa.s

Which is close to my first 3 results, and make's me believe the above post(about why my results are distored) could be true. Do you mind if i post my first result(which i believe to be correct) in proper tex so u can point out any mistakes (if you can see any).

No need. Your sample calculations show that you had it right.
Regarding your measured value compared to the literature value, your data may have been taken at a temperature a few degrees below where the literature value was measured. This could easily account for the difference.

Chet

 

[RoryScanlan]

I cannot thank you enough, I was completely confused at the beginning of this analysis I now have an idea what's going on and I have nearly 3 pages of calculations to review and show off, plus i have two more equations that i can play with(terminal velocity, and drag force) more brownie points!
Thanks Once again! :)