Viscosity Help: Calculating Glycerol Viscosity w/ Data

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The discussion revolves around calculating the viscosity of glycerol using experimental data and Stokes' law. The original calculations yielded an unexpected viscosity value, prompting questions about the accuracy of the data and the equations used. Participants suggested verifying the sphere's dimensions, ensuring consistent units, and calculating expected terminal velocity to compare with the measured value. There was also emphasis on understanding unit conversions between dyne and Newton, as well as meter and centimeter, to ensure accurate results. The importance of correctly identifying the radius versus diameter of the sphere was highlighted as a potential source of error.
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Hi,

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

Data:
Sphere radius: 0.8 cm
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 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 don't know the Forces involved?

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

Jack
 
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verify the value you used for g.
 
Thanks integral, I've updated g to 980cm sec-², and now i get an even stranger result :( any ideas?
 
Why did multiplying your result by 100 change the digits?
 
Integral said:
Why did multiplying your result by 100 change the digits?

I don't think it did? I've double checked the calculation, 0.29 without and 29.21 with g as 980.
 
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.
 
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.
 
tntcoder said:
I don't think it did? I've double checked the calculation, 0.29 without and 29.21 with g as 980.
Humm... I thought I had seen a 1.xx last night, may just be my memory.
 
Thanks integral, i made a mistake in the sphere volume, I've 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.

Thanks for all your help.
 
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  • #10
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.
 
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  • #11
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
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
 
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