Fluid Mechanics: Rotating cylinder viscometer

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Homework Help Overview

The discussion revolves around a fluid mechanics problem involving a rotating cylinder viscometer. Participants are examining the relationships between shear stress, velocity gradients, and the resulting moments in the context of the problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the setup of equations related to shear stress and moment calculations. There is confusion regarding the correct application of the velocity gradient and its impact on the results for different parts of the problem.

Discussion Status

Some participants have provided feedback on the setup of the equations, suggesting corrections to the original poster's approach. There is an ongoing exploration of the implications of these corrections, but no consensus has been reached on the final formulation.

Contextual Notes

There appears to be confusion regarding the integration limits and the definitions of variables such as shear stress and velocity gradient. The original poster's attempts at integration have not yielded the expected results, indicating potential misunderstandings in the problem setup.

Feodalherren
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Homework Statement


1_6.png


Homework Equations


τ = μ (du/dy)

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.
 
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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
 
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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?
 
Feodalherren said:
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?
Yes.
 
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Thanks!
 

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