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Fluid Mechanics: Rotating cylinder viscometer

  1. Jan 25, 2015 #1
    1. The problem statement, all variables and given/known data
    1_6.png

    2. Relevant equations
    τ = μ (du/dy)

    3. The attempt at a solution

    For part B)

    dM = τr dA

    τ = μ (rΩ/Δr)

    ∫ (μrΩr2πr / Δr)dr = μR4πΩ / 2Δr

    pretty sure that's correct.

    I'm confused for part A though.

    I want to set it up as

    dM = τR dA = τR (2πRdL)

    and

    τ = μ (RΩ/L)

    substitute and integrate over 0 to L. But this doesn't yield the correct result.
     
  2. jcsd
  3. Jan 25, 2015 #2
    Part A is done incorrectly. The velocity gradient in the gap is ##\frac{ΩR}{Δr}##. So the shear stress is ##μ\frac{ΩR}{Δr}##

    Chet
     
  4. Jan 25, 2015 #3
    Ahh of course. It's so obvious now. Thank you, Sir.

    Would that mean that part A becomes:

    dM = τR dA = τR (2πRdL)

    M = ∫ R(2πRdL) (ΩR/Δr) = 2πR3ΩLμ / Δr

    correct?
     
  5. Jan 25, 2015 #4
    Yes.
     
  6. Jan 25, 2015 #5
    Thanks!
     
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