# Futher Viscosity Help

1. Feb 26, 2005

### tntcoder

Hi,

Im doing an experiment to determine the viscosity of glycerol this is my collected data:

Data:
Sphere mass: 16.3 g
Sphere voume: 2.14 cm -³
Sphere density: 7.62 g cm -³

Glycerol density 1.26 g cm -³

Sphere Terminal Velocity: 23 cm/s
Light Gate Seperation: 20cm

Given value for viscoisty of glycerol in data book: 9420 10^-4 N sm-2

My Calculations:

By Stokes law:

V = (2gr²)(d1-d2)/9µ
µ = ((2gr²)(d1-d2)/V) / 9

Where:

V = velocity of fall (cm sec-¹),
g = acceleration of gravity (cm sec-²),
r = "equivalent" radius of particle (cm),
dl = density of particle (g cm -³),
d2 = density of medium (g cm-³), and
µ = viscosity of medium (dyne sec cm-²).

µ = ((2gr²)(d1-d2)/V) / 9

µ = ((2*980*0.8²)(7.62-1.26) / 23) / 9

µ = 38.54 Dynes
µ [DATA]: 0.942 N sm-2

im not great at physics, but something looks very wrong with that value :surprised cany anyone put me right, are my calculations correct and i have very in accurate data? Or have i used the wrong equation?

Ive seen lots of equations like:
F= 6*pi*µ*r*v
&
mg - u = 6*pi*µ*r*v

I have no idea if the above equation is correct? It seems the other two take account of viscous drag? Weight & upthrust, should i not be taking these into account?

And how can i use the above equations if i dont know the Forces involved?

Thanks very much for any help, and sorry if ive made a load of mistakes.

Jack

Last edited: Feb 27, 2005
2. Feb 26, 2005

### Integral

Staff Emeritus
verify the value you used for g.

3. Feb 27, 2005

### tntcoder

Thanks integral, ive updated g to 980cm sec-², and now i get an even stranger result :( any ideas?

4. Feb 27, 2005

### Integral

Staff Emeritus
Why did multiplying your result by 100 change the digits?

5. Feb 27, 2005

### tntcoder

I dont think it did? Ive double checked the calculation, 0.29 without and 29.21 with g as 980.

6. Feb 27, 2005

### Integral

Staff Emeritus
Double check the density of the sphere, I do not get the same thing you show. The results will be in Dynes, not Newtons as you show.

7. Feb 27, 2005

### Integral

Staff Emeritus
Use your given values to determine an expected terminal velocity, compare it to the velocity you give. What can you conclude? You may also want to do a careful unit conversion of your data value to the units you are using, or convert all of your measurements to the same units as your data value. It does not matter which way you do it, just get them all in the same units.

8. Feb 27, 2005

### Integral

Staff Emeritus
Humm... I thought I had seen a 1.xx last night, may just be my memory.

9. Feb 27, 2005

### tntcoder

Thanks integral, i made a mistake in the sphere volume, ive updated that and the density and the final equation and the units changed to Dynes.

Could you tell me how i calculate an expected terminal velocity please? Could i do it with suvat equations.

Last edited: Feb 27, 2005
10. Feb 27, 2005

### Integral

Staff Emeritus
Take your original equation (before solving for viscosity), use the data value of the viscosity, to compute the terminal velocity. That is the V in your first equation. (I get something over 30cm/s)

OPPS! I just did a re-compute I now have 94 cm/s as the expected terminal velocity.

There are some things that I cannot check but you need to verify, is .8 indeed the radius and not the diameter of the sphere?

If you still have access to the experiment try positioning your sensors closer together and then at different depths. If the velocity between 10 and 15 cm is different from that between 15 and 20 then you have verified that you have not yet reached terminal velocity.

I have to go now will not be back till later tonight (10 - 12 hrs from now) Good luck, perhaps someone else will take over.

Last edited: Feb 27, 2005
11. Feb 27, 2005

### tntcoder

Thanks very much, I have checked everything and the radius of the sphere is 0.8 cm, I will be continuing with my experiment tomorrow, so i will move my light gates up and down the tube to ensure i am measuring terminal velocity. Please could you tell me how i can convert from dyne sec cm-² to Newtons sm -2 in order to calculate the expected terminal viscosity.

12. Feb 28, 2005

### Integral

Staff Emeritus
Understanding units and unit conversion is essential to doing physics. In many ways it is Physics, if you have no idea where you are going, but are able get the correct units, chances are you will end up pretty close to where you wanted to be.

You need to find (Google is your friend) A conversion factor for Newton to Dyne and also the one for meter to centimeter. Then do the following computation:

$$.942 \frac {N s} {m^2} \times \mmbox{ y }\frac {Dyne} N \times \mmbox{z}\frac {m^2} {cm^2}$$

If you look at the units in the equation as just another algebraic expression, you will see that you have a N(ewton) in both the numerator and denominator, thus they cancel, likewise for the m2. The units that remain are the ones you want.

You need to find the numeric values of x and y i

Last edited: Feb 28, 2005