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I Second derivative of friction force question

  1. Apr 29, 2017 #1
    I'm studying boundary layers. I am confused by what I am reading in this book.

    The book says the friction force (F) per unit volume = $$\frac{dF}{dy}=\mu\frac{d^2U}{dy^2}$$
    They say $$\frac{dU}{dy}=\frac{U_\infty}{\delta}$$

    This makes sense to me, delta is the thickness in the y direction, and the velocity is 0 at the airfoil (or whatever object) and it equals the freestream velocity after the boundary thickness delta. so du/dy being proportional to the free stream velocity divided by the thickness makes sense.
    But then they say this : $$\frac{d^2U}{dy^2} =\mu\frac{U_\infty}{\delta^2}$$

    Can someone explain why the second derivative is proportional to delta^2 ?
    Is it that, delta represents the thickness y, and so the second derivative of 1/y = -1/y^2, and we just ignore the negative?

    Attached Files:

  2. jcsd
  3. Apr 29, 2017 #2


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    Well for one, friction force per unit volume is not ##dF/dy## as you have written it. It is ##\partial \tau/\partial y##. The difference is important, because ##F## is a force, and ##\tau## is a stress. The units don't work out if you use ##F##.

    So then the friction force per unit volume is
    [tex]\dfrac{\partial \tau}{\partial y} = \mu\dfrac{\partial^2 u}{\partial y^2}.[/tex]
    So, if you've already come to terms with the fact that
    [tex]\dfrac{\partial u}{\partial y} \sim \dfrac{U_{\infty}}{\delta},[/tex]
    Then the same logic leads to
    [tex]\dfrac{\partial \tau}{\partial y} = \mu\dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial y}\right) \sim \mu\dfrac{\partial}{\partial y}\left(\dfrac{U_{\infty}}{\delta}\right) \sim \mu\dfrac{U_{\infty}}{\delta^2}.[/tex]
  4. May 3, 2017 #3
    Hi thank you for replying.

    Ah okay, I think I figured it out.
    I see how $$\frac{\partial u}{\partial y} = \frac{U\infty}{\delta}$$ because... it U=0 at the boundary and U_infty after a distance delta, so it should be proportional.

    So I think :
    $$\frac{\partial^2 u}{\partial y^2} = \frac{\frac{\partial u}{\partial y}_\infty}{\delta} = \frac{U\infty}{\delta^2}$$ Or basically du/dt/delta = u_infty/delta^2
    Is this the correct interpretation?
  5. May 3, 2017 #4


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    You got it.
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