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Homework Help: Shering stress

  1. May 8, 2012 #1
    1. The problem statement, all variables and given/known data
    The velocity distribution for the floe of a newtonian fluid between 2 wide parallel plates is given by the equation u=3V/2[1-(y/h)^2] where V is the mean velocity. The fluid has a viscosity of 1.915Ns/m^2. When V=0.61m/s and h=5mm, determine:
    A) the shearing stress acting on the bottom wall.
    B) the shearing stress acting on a plane parallel to the wall.
    C) the shearing stress at the centerline.

    2. Relevant equations
    ζ=μ du/dy

    3. The attempt at a solution
    For C) I worked out u=0.915[1-y^2/2.5X10^-3]
    From that I worked out ζ=1.915 X 0.915[1-y^2/2.5X10^-3] X 1/dy.

    I have no idea how to work out B).

    I think A) =0 as there is no velocity there? I could be very wrong saying that though.
    Last edited: May 8, 2012
  2. jcsd
  3. May 8, 2012 #2
    Where is the origin of h? Bottom plate, top plate, center?
  4. May 9, 2012 #3
    Either bottom or top plate, doesn't make a diff in this question. It's like a radius if you will. It says the flow is symetrical.
  5. May 10, 2012 #4
    I hit the h key instead of the y key. Typo. Where is the origin of y?
  6. May 10, 2012 #5
    "For C) I worked out u=0.915[1-y^2/2.5X10^-3]
    From that I worked out ζ=1.915 X 0.915[1-y^2/2.5X10^-3] X 1/dy."

    How can this be when ζ=μ du/dy?
  7. May 10, 2012 #6
    Origin of y is the centerline.

    I dunno, I'm really confused. :/
  8. May 11, 2012 #7
    Here is some more help. The profile is parabolic with the velocity function given by u=V/2[1-(y/h)^2] where V is the average velocity.

    The shear stress is given by:
    S = mu*du/dy which is the viscosity multiplied by the velocity gradient in the direction perpendicular to the flow. If you plot the velocity profile you will note that when y is zero (centerline), the velocity is maximum. Furthermore the fluid velocity is zero at the walls where y= +h or -h. To determine shear, you take the derivative with respect to y. V, the average velocity, is constant.

    So all you need to do for part C is take the derivative and plug in the value of y=h.
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