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Concentric Cylinder Viscometer

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data

    A viscometer consists of two concentric cylinders, 10.20cm and 10.60 cm in diameter. A liquid fills the space between them to a depth of 12.0cm. The outer cylinder is fixed, and a torque of 0.024 N*m keeps the inner cylinder turning at a steady rotational speed of 57 rev/min. What is the viscosity of the fluid?

    2. Relevant equations

    F = ηA[itex]\frac{v}{l}[/itex]

    3. The attempt at a solution

    The torque needed to move the cylinder at constant angular velocity must be equal to the total torque exerted by the viscosity of the fluid. Around the outer, vertical edge of the cylinder, this torque is equal to F*R = ηA[itex]\frac{v}{l}[/itex]*R.

    Where R = [itex]\frac{10.20}{2}[/itex]cm, A = 2[itex]\pi[/itex]R*h (h=12cm - l), v = ωR = [itex]\frac{57*2\pi}{60}[/itex]*R and l = [itex]\frac{10.60-10.20}{2}[/itex]

    So that

    F*R = [itex]\frac{η(2\pi*Rh)ωR}{l}[/itex]*R = [itex]\frac{η2\pi*hωR^3}{l}[/itex]

    But this doesn't account for the torque acting on the bottom of the cylinder, which varies with radius. At a distance from the center r, the torque at the bottom is equal to

    F*r = [itex]\frac{ηAv}{l}[/itex]*r = [itex]\frac{η(2\pi*r*dr)ω*r}{l}[/itex]*r

    So the total torque on the bottom is the sum of all torques from radius=0 to the radius R found above, or:


    = [itex]\frac{η\piωR^{4}}{2l}[/itex]

    So the TOTAL torque on the cylinder from the viscoity of the liquid is

    T = [itex]\frac{η2\pi*hωR^3}{l}[/itex] + [itex]\frac{η\piωR^{4}}{2l}[/itex]

    And solving for η gives

    η = [itex]\frac{2Tl}{4\pi*hωR^3 + \piωR^{4}}[/itex]

    Wow, that was a bunch of annoying latex. I'm sure I've got a typo or two in there somewhere, but I don't see any right now. Anyway, after plugging in value, I get an answer that's close to, but not quite what's in the book. I end up with 0.074 Pa*s and the book gives 0.079 Pa*s. I figure I'm probably making a mistake in finding the torque on the bottom of the cylinder, but I don't know what it would be.
    Last edited: Feb 16, 2012
  2. jcsd
  3. Feb 19, 2012 #2
    Any help would be appreciated.
  4. Feb 20, 2012 #3


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    Staff Emeritus
    Science Advisor
    Homework Helper

    From the wording of the problem statement, I would assume that the bottom of the cylinder does not have any fluid in contact with it, and does not contribute anything towards the torque.

    (If the bottom were in contact with fluid, they would have to tell us the gap between the rotating surface and the fixed surface below it, and they did not tell us this.)
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