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Fluid Mechanics~ Wall shear stress

  • #1
3,003
2

Homework Statement


Picture1-31.png





My main problem here is that I do not understand what they are asking. What is the wall shear stress? Do they mean the stress at the "floor" (or whatever you want to call it)?

If so, I am assuming I use Newton's Law of Viscosity [itex]\tau=\mu\frac{du}{dy}[/itex] since this is 1-dimensional flow.

Would the wall shear stress [itex]\tau_w[/itex] be given by

[tex]\tau_w=\int_{y=0}^\delta \tau(y)\, dy[/tex]

and then the problem reduces to finding an expression for [itex]\tau(y)[/itex]

Or am I way off here? thanks :smile:
 

Answers and Replies

  • #2
3,003
2
I just found http://www.cfd-online.com/Wiki/Wall_shear_stress" [Broken]. So I guess my interpretation was wrong.

So, according to this definition, I should have:

[tex]\tau_w=\mu*\frac{d}{dy}[U\sin(\frac{\pi y}{2\delta})]|_{y=0}[/tex]


Is that all?
 
Last edited by a moderator:
  • #3
3,003
2
I just found http://www.cfd-online.com/Wiki/Wall_shear_stress" [Broken]. So I guess my interpretation was wrong.

So, according to this definition, I should have:

[tex]\tau_w=\mu*\frac{d}{dy}[U\sin(\frac{\pi y}{2\delta})]|_{y=0}[/tex]


Is that all?
Not that anyone will respond to this (since no one ever looks in this forum), but I am assuming that if the above is correct, than part (b) is as simple as solving

[tex]\frac{1}{2}\tau_w=\mu*\frac{d}{dy}[U\sin(\frac{\pi y}{2\delta})][/tex]

for y.

Sound good? Good :smile:

So I have:

[tex]\frac{\tau_w}{2}=\mu U\cos(\frac{\pi y}{2\delta})*\frac{\pi}{2\delta}[/tex]

[tex]\Rightarrow \frac{\tau_w\delta}{\mu U\pi}=\cos(\frac{\pi y}{2\delta})[/tex]
 
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  • #4
3,003
2
71 views, no responses. Yesssss I am just going to keep chatting it up with myself.

Maybe I can set a new record?

Has there ever been a thread locked where there was only one speaker?

Oops! Don't answer that :smile:

I need help. (The kind PF cannot offer)
 
  • #5
1
0
I'm gonna reply to this thread, just to break your monologue. =)
 
  • #6
1
0
I agree with Saladsamurai.

τw = μ×[itex]\frac{d}{dy}[/itex][Usin([itex]\frac{\pi y}{2δ}[/itex])][itex]|y=0[/itex]

so that, τw = 180[itex]\pi[/itex]μ

and τ = μ×[itex]\frac{d}{dy}[/itex][Usin([itex]\frac{\pi y}{2δ}[/itex])] when τ = τw/2

thus, 90[itex]\pi[/itex]μ = μ×[itex]\frac{d}{dy}[/itex][Usin([itex]\frac{\pi y}{2δ}[/itex])]

[itex]\frac{1}{2}[/itex] = cos([itex]\frac{\pi y}{2δ}[/itex])

[itex]\frac{\pi}{3}[/itex] = [itex]\frac{\pi y}{2δ}[/itex]

y = 0.02 m
 
  • #7
19,799
4,048
I just found This. So I guess my interpretation was wrong.

So, according to this definition, I should have:

[tex]\tau_w=\mu*[U\sin(\frac{\pi y}{2\delta})]|_{y=0}[/tex]


Is that all?
Yes. If you actually apply the above formula, you get

τw=(μUπ)/(2δ)

Is that what you got?
 

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