Viscosity of a Fluid measured by a viscometer

[DylanW]

Homework Statement

The viscosity of a fluid is to be measured by a viscometer constructed of two 75 cm long concentric cylinders. The outer diameter of the inner cylinder is 15 cm and the gap between the two cylinders is 1 mm. With the outer cylinder fixed the inner cylinder is rotated at 300 RPM, and the torque is measure by a spring gauge to be 0.8 Nm. Determine the viscosity.

Homework Equations

Viscosity = Shear Stress/Shear Rate

The Attempt at a Solution

I am continually getting a value of 0 for the Shear Rate, and a value of 757.51 for the Sheer Rate. I think the problem may be within my value for the Radius at location of calc as I am not entirely sure what is meant by this. Any advice would be appreciated

 

[mishek]

I am continually getting a value of 0 for the Shear Rate, and a value of 757.51 for the Sheer Rate.

Could You please elaborate this result?

 

[DylanW]

Could You please elaborate this result?

I've made some progress since then using the formula Viscosity = (T.h)/(2.Pi.R^3.w.L)
Which works out to Viscosity = (0.8 * 0.001)/(2Pi*(0.075)^3.(300*(2Pi/60))*0.75) which then equates to 0.0128 N.s/m^3

I hope I'm right :D

 

[sankalpmittal]

I've made some progress since then using the formula Viscosity = (T.h)/(2.Pi.R^3.w.L)
Which works out to Viscosity = (0.8 * 0.001)/(2Pi*(0.075)^3.(300*(2Pi/60))*0.75) which then equates to 0.0128 N.s/m^3

I hope I'm right :D

Seems to be the correct formula and the work looks good to me.If you have put there 2 because torque is given when one wheel is fixed, then you are correct.

 

[DylanW]

Thanks very much Sankal :) To be honest I found a perfect example of the question I was attempting in a textbook so I just adapted that for my question - 16.66% of my assignment locked in :D

 

[Chestermiller]

I've made some progress since then using the formula Viscosity = (T.h)/(2.Pi.R^3.w.L)
Which works out to Viscosity = (0.8 * 0.001)/(2Pi*(0.075)^3.(300*(2Pi/60))*0.75) which then equates to 0.0128 N.s/m^3

I hope I'm right :D

Your units are incorrect. They should be N.s/m^2 = Pa.s

What did you get for the shear rate? I got 2356 s^-1

What did you get for the shear stress? I got 30.2 Pa

I want to strongly encourage you not to use formulas like this to solve problems. Some day, you may not have the formula handy, and you may need to solve the problem. It would be much better if you worked from fundamentals. For example, in this situation, the gap between the cylinders is very small compared to the radii of the cylinders. So the situation is closely equivalent to shear between two infinite parallel plates. The gap between the plates is 0.1 cm, and the relative velocity of the cylinders (plates) is ωr, where ω is 10π radians per second and r = 7.5 cm. From this you can get the shear rate. The torque is equal to the shear stress times the radius times the surface area of the cylinders. From this you can get the shear stress.

 

[DylanW]

I typed the units into the forums incorrectly but thankfully I had N.s/m^2 for my units in my assignment.
I didn't end up successfully calculating shear rate and shear stress. I would ideally always like to solve problems from first principles/fundamentals although I am just getting back into physics so my intuitive understanding still has plenty of holes in it. Thanks for your explanation I think I know have a slightly clearer understanding of how the system of a viscometer works which should make my future assignments easier :)