Fluid Mechanics: Rotating cylinder viscometer

  • #1
603
6

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.
 

Answers and Replies

  • #2
20,875
4,548
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
 
  • Like
Likes Feodalherren
  • #3
603
6
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?
 
  • #4
20,875
4,548
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.
 
  • Like
Likes Feodalherren
  • #5
603
6
Thanks!
 

Related Threads on Fluid Mechanics: Rotating cylinder viscometer

Replies
6
Views
2K
  • Last Post
Replies
2
Views
16K
Replies
6
Views
5K
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
2K
Replies
6
Views
991
Replies
9
Views
4K
Replies
1
Views
597
  • Last Post
Replies
2
Views
124
Top