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Viscosity (Fluid mechanics)

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Attached.

    I am working on part (i). What is the resistive force?

    Shear stress= μ*(du/dy) = μU/l = F/A

    Since the force is up and down, it causes a shear stress, thus area is parallel to the force and thus it is the 4 faces B*L

    Thus, A=4*B*L

    → F=μ*U*A/l = 4*μ*U*L*B/l

    Where = viscosity
    l=clearance gap
    L=length of block
    F=Resistive force
    A=Area

    I'm not sure if that is the correct method. Can someone please verify?
     

    Attached Files:

  2. jcsd
  3. Aug 28, 2012 #2
    Looks good. The fluid in the gap exerts a vertical shear force on the 4 vertical sides of the block. Looks like the analysis is right on target.
     
  4. Aug 29, 2012 #3
    Okay thanks, the second part asks me to find the "terminal velocity"

    This is what I did

    F=4*μ*U*L*B/l=ma
    m=ρV
    Where ρ is the density of the solid, not the fluid
    and V is the volume of the block = (B^2)*L

    Plugging that in and simplifying I got U=ρ*B*g*l / (4*μ)

    I'm a bit hesitant on that answer because it did not involve the density of the fluid. Is this the correct method?
     
  5. Aug 29, 2012 #4
    No. There is an upward (buoyant) force from the surrounding liquid acting on the block, equal to weight of the displaced liquid ρliqVg. So the net downward force from the weight of the block combined with the upward (buoyant) force from the surrounding liquid is equal to....?
     
  6. Aug 29, 2012 #5
    Oh, I think I'm supposed to ignore that. The problem stated to neglect buoyancy. Is that still necessary?
     
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