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Bingham Plastic vs Gravity

  1. Sep 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Acrylic latex can be described by a Bingham Plastic model where the yield stress is 11.2 N/m2, a limiting viscosity, m0, of 80 cp and a density of 0.95 g/cm3. What is the maximum thickness of this paint that can be applied to a vertical wall without running.


    2. Relevant equations
    Shear stress = [itex]\tau[/itex]xz = -[itex]\mu[/itex]*dvz/dx

    Where positive z direction is down (with gravity) and positive x direction is to the right (wall on the left, paint on the right of the wall)

    Unfortunately, that is all I know of relevant to this problem. I would also say that a chart of shear rate vs shear stress depicting the Bingham Plastic model would be relevant. And perhaps the shear stress in the above equation should actually be differential (as it goes from 0 to the yield stress), but I am not entirely sure about that.

    3. The attempt at a solution
    I'm not quite sure where to start with this as I am not at all familiar with non-Newtonian fluids. I can only imagine that I should treat it as a Newtonian fluid at it's yield stress and solve for x in some manner and at that point it would flow.

    I would try to partially solve the above equation leading to [itex]\tau[/itex]xz*∫(from x = 0 to x = x) dx = -[itex]\mu[/itex]*∫(from vz = 0 to vz = vz) dvz
    I feel this is a good start but I'm a bit confused on how to deal with the velocity integral. Ideally I would think you want to integrate from the velocity is 0 to when it is infinitesimally small, which would essentially be 0 making the equation unsolvable (unless x = 0 which does not work).

    I also can't help but notice that unused density number which I can only assume plays some part.

    Help is much appreciated.
     
  2. jcsd
  3. Sep 12, 2013 #2
    Hi again. This is much simpler than you think. Hint: The paint won't run if the shear stress does not exceed the yield stress. The shear stress throughout the layer will be constant if the fluid is not deforming. Do a vertical force balance, and determine the shear stress as a function of the layer thickness.
     
  4. Sep 12, 2013 #3
    I was thinking force might be involved, but how do I go about a vertical force balance. The force, I assume, would be directly a result of gravity so it should = m*g? There doesn't seem to be an m to calculate with but I don't remember much of basic physics unfortunately so I may be a bit off on the force balance.

    As for the second part I assume once I can figure out the force, I take the shear stress = F/A, but again I'm a bit confused what constitutes the area.


    I'm sure you're right, that this is much easier than I'm making it out to be. But I really do appreciate all the help, and thanks again for answering that previous question for me.
     
    Last edited: Sep 12, 2013
  5. Sep 12, 2013 #4
    Focus on a small section of the coating between z and z + Δz. The thickness of the coating is h. Let L be the width of the section of coating. What is the volume of this section of coating? What is the weight of this section of the coating? From a force balance, what vertical shear force must the wall exert on the section of coating to support its weight. What is this force divided by the contact area LΔz? This is the shear stress.

    Chet
     
  6. Sep 12, 2013 #5
    I was just in the middle of writing this up as I thought about it more when you posted your reply:

    Thinking about it further, I think that doing the vertical force balance would be a good time to make use of that density. So knowing the density and assuming a thickness of x, I can make up a cross-sectional area of paint that is z high by L long by x thick. Multiplying my volume (x*L*z) by my density would give me a mass m. Then I can get the vertical force pushing down on that area.

    To get shear stress would I then use that force and divide by the area L*z? This would give me shear stress = [itex]\rho[/itex]*x*g

    From there I think I would simply set the shear stress equal to yield stress to solve for x and give me the thickness at which the shear stress overcomes (≥) the yield stress.
     
  7. Sep 12, 2013 #6
    Nicely done!!!
     
  8. Sep 12, 2013 #7
    Just one last question for you. Where does the viscosity play into all of this, if at all?
     
  9. Sep 12, 2013 #8
    It doesn't unless the thickness is large enough for the coating to start slipping. Incidentally, I made an incorrect statement in you first response. The shear stress will not be constant throughout the layer. At the air interface, of course, the shear stress will be zero. So the shear stress will be increasing linearly with distance from the air interface.
     
  10. Sep 12, 2013 #9
    Ok that makes sense. And the shear stress still follows a linear profile, decreasing with increasing thickness (x)? I'm curious though what the profile would look like as the thickness is increased past the distance for yield stress. For the shear stress less than yield stress, I'm guessing it's just a linear line that looks like a forward slash, /[\b] (where the wall is on the left side), based on what you said. But after? Maybe a parabolic curve of sorts, not entirely sure?
     
  11. Sep 13, 2013 #10


    If you do a differential force balance (with respect to the thickness direction), you will find that the shear stress profile is linear irrespective of whether the yield stress is exceeded. The real question is, what does the velocity profile look like? To get an idea about that, first solve the problem for a Newtonian fluid.
     
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