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Homework Help: Fluid Mechanics~ Wall shear stress

  1. May 7, 2009 #1
    1. The problem statement, all variables and given/known data

    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:
  2. jcsd
  3. May 7, 2009 #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: May 4, 2017
  4. May 7, 2009 #3
    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]
    Last edited by a moderator: May 4, 2017
  5. May 8, 2009 #4
    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)
  6. Dec 21, 2010 #5
    I'm gonna reply to this thread, just to break your monologue. =)
  7. Nov 7, 2012 #6
    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
  8. Nov 8, 2012 #7
    Yes. If you actually apply the above formula, you get


    Is that what you got?
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