How to solve for viscosity using Stoke's law

In summary, the forum poster is having trouble with a viscosity lab problem using the "falling-sphere" method and Stoke's law. The given results and constants are provided, but the calculated result for viscosity does not match the accepted value. Possible reasons for this discrepancy could be the incorrect use of the correction factor for a cylindrical tube, the unnecessary use of the coefficient of expansion for temperature correction, or inconsistencies in unit conversions. The forum poster is seeking help and hints to identify the issue and move forward with the other problems.
  • #1
DoctorBim
2
0
I'm having a bit of trouble with some simple viscosity lab problem. The results are not of my own, so this could be some procedural error. We've used glycerol as the liquid medium at 3 different temperatures (0C, 10C, and 20C) with three different metal spheres, one of them being steel. All we are required to do is calculate the viscosity and temperature dependence of the glycerol using this "falling-sphere" method. Stoke's law is given to us:
η=[2g(R^2)(ρ_S-ρ_L)t]/9L

There's also a correction for a cylindrical tube, where the medium is, inserted into the denominator
η=[2g(R^2)(ρ_S-ρ_L)t]/[9L(1+2.4x)(1+1.65y)]
Where x is the ratio of the sphere's diameter to the walls of the cylinder (graduated cylinder) and y is the ratio of the sphere's diameter to the total height of the liquid in the graduated cylinder.
These are what I have for results and constants:
Steel ρ = 7860 kg/m^3
Glycerol ρ = 1259 kg/m^3
t = 2.3 s
L = 0.05m
R = 8.314 J/molK
g = 9.81 m/s^2

This trial was run with the glycerol medium at 0C and the given density was at 25C, so I corrected the density of glycerol using the coefficient of expansion (0.0005).
The accepted value for the viscosity of glycerol is 1.5 Pa·s.
I ended up getting 1.46E7 [(J/molK)^2 kg/m^3]
And yes, I'm terrible at converting these units.
I can't see where I am going wrong with this. Any help/hints would be greatly appreciated so I can move on with the other problems.
 
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  • #2

Thank you for sharing your results and concerns about the viscosity lab problem. Based on the information you provided, it seems like you have followed the correct procedure and equations for calculating viscosity using the "falling-sphere" method and Stoke's law. However, there are a few things that could be causing the discrepancy between your results and the accepted value for the viscosity of glycerol.

Firstly, it is important to note that the correction for a cylindrical tube is only necessary when the ratio of the sphere's diameter to the walls of the cylinder (x) and the ratio of the sphere's diameter to the total height of the liquid in the graduated cylinder (y) are not equal to 1. In your case, since you used a spherical metal sphere and a cylindrical graduated cylinder, these ratios should both be equal to 1. Therefore, the correction factor should be 1 for both x and y, and the equation should simplify to the first form you provided: η=[2g(R^2)(ρ_S-ρ_L)t]/9L.

Secondly, the correction for temperature using the coefficient of expansion may not be necessary in this case. The coefficient of expansion is typically used to correct for the change in density of a liquid due to a change in temperature. However, since the density of glycerol at 0C is already provided, it may not be necessary to make this correction.

Lastly, it is important to check your units and make sure they are consistent throughout the calculation. In your final result, the units of (J/molK)^2 kg/m^3 do not seem to be consistent with the units of viscosity, which should be Pa·s or N·s/m^2. I would suggest going back through your calculation and checking for any unit conversions or mistakes.

I hope these suggestions are helpful in finding the source of the discrepancy in your results. If you continue to have trouble, I would recommend consulting with your instructor or a lab partner to review your procedure and calculations together. Good luck with the rest of your lab work!
 

Related to How to solve for viscosity using Stoke's law

What is Stoke's law?

Stoke's law is a scientific equation that describes the relationship between the viscosity of a fluid and the resistance encountered by an object moving through the fluid.

How do you determine the viscosity of a fluid using Stoke's law?

To solve for viscosity using Stoke's law, you need to measure the terminal velocity of a small object falling through the fluid. This can be done by timing the object as it falls through a known distance, and then using the equation to calculate the viscosity.

What are the variables in the Stoke's law equation?

The Stoke's law equation includes the viscosity of the fluid (η), the radius of the object (r), the acceleration due to gravity (g), and the density of the fluid (ρ).

Are there any limitations to using Stoke's law to solve for viscosity?

Stoke's law is only accurate for objects falling through a fluid at low velocities. It also assumes that the fluid is incompressible and has a constant viscosity, which may not always be the case in real-world situations.

Can Stoke's law be used for all types of fluids?

No, Stoke's law is only applicable for Newtonian fluids, which are fluids that have a linear relationship between shear stress and shear rate. Non-Newtonian fluids, such as blood and ketchup, have more complex behaviors and cannot be accurately measured using Stoke's law.

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