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High viscous fluid, how fast will a object sink?

  1. Apr 12, 2013 #1
    1. The problem statement, all variables and given/known data
    I am currently looking for a (simplified) model to see how far/fast a object will sink in a high viscous fluid.
    I found alot of information about sinking spheres and the maximum contact angle with the fuild because of the surface tension before it sinks. And a example explanation of a sinking sphere in a fluid (to calculate the speed it sinks), but this one did not start at the surface:
    "Let's combine all these things together for a sphere falling in a fluid. Weight goes down, buoyancy goes up, drag goes up. After awhile, the sphere will fall with constant velocity. When it does, all these forces cancel. When a sphere is falling through a fluid it is completely submerged, so there is only one volume to talk about — the volume of a sphere."
    Here they calculate the sinking speed when the sphere is fully emerged in the water.

    F.buoyance + F.viscousdrag = F.massa ;ρ*g*V + 6πηrv = ρ*g*V

    But in my model I need to know how fast a object (in my case a rectangle form with low height, a small flate in the order of 2 mm in length) sinks when put on the surface area of a fluid (glass) with high viscosity. In the beginning not fully emerged into the fluid.
    Is it possible to give an esstimate of how fast the object will sink. I need this info because then I can calculate for how long the glass needs to be in that viscous state before the object sinks in too deep into the glass.
    the object needs to stay on top of the glass but can sink a little bit (but never be fully emerged in the fluid) before changing the glass viscosity back (by lowering the temperature).


    2. Relevant equations
    Is the surface tension neglectiable when the viscosity is very high?
    Because then the object will only sink very slowly, and no high contact angle will be made?
    What physics law's do I need to use for my problem (can be simplified)?

    3. The attempt at a solution
    Only valid with a sphere fully emerged into the water
    F.buoyance + F.viscousdrag = F.massa ;ρ*g*V + 6πηrv = ρ*g*V
    V = 2* ∇ρ*g*r^2 / ( 9*viscosity )

    The first attempt was with the law of archimedes, but that is only to see if it would sink.
    With his law I could calculate how far it would sink because of its Volume and desity. But not how fast it would sink.

    - Massa of the object is known
    - volume is known
    - viscosity at given temperature is known
    - desity are known

    I hope you can help me to give me the right direction to look into, to solve my problem.

    Greetings,
    spitskool
     
  2. jcsd
  3. Apr 12, 2013 #2

    haruspex

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    You need to develop an equation (DE) of motion. At a given depth of immersion, you can calculate the buoyancy. How viscous drag changes during immersion I've no idea. A first approximation might be to assume it's the same as if fully immersed. If surface tension is significant then this will make the equation quite nasty, as the angle of contact will decrease from 90 degrees to some minimum.
     
  4. Apr 13, 2013 #3
    How accurately do you need to know the answer? Would an upper bound or a lower bound to the distance vs time be adequate?
     
  5. Apr 15, 2013 #4
    Thank you for the replies,
    the accuracy does not need to me high, but a nice esstimate will be good. A simplified model will do.

    Density.object = 2330 kg/m^3
    Area.object = 1.7*0.8 = 1.36 mm^2
    Volume.object = 1.7*0.8*0.1 = 0.136 mm^3
    Massa object = 0.31688 mg

    Density.fluid = 2300 kg/m^3, at 500 degrees Celsius
    Viscosity.fluid = 500-1000 Poise (is this 50 kg/ms?)

    Fg = m*g = 0.31688*9.81 = 0.00311 mN

    m.fluid = m.object Archimedes?
    V.fluid = m.object / ρ.fluid = 0.31688 mg / 2300 = 0.13777 mm^3
    Height = Volume / area = 0.13777 / 1.36 mm = 0.101 mm

    Thus the object will sink 0.101 mm in the fluid???
    The object height is 0.1 mm, so (0.1 - 0.101 =) -0.001 mm is the object under the fluid level.

    It sinks untill fully emerged into the fluid. But I need to know a which time the object is fully emerged. because before this time I can turn down the Temperature so the viscosity of the fluid will increase and the object will not sink anymore (not become fully emerged into the fluid). I need to find the maximum time bofere the object will emmerge into the fluid.

    I hope I explained my problem correctly, a simple model will be enought. But it would be nice to know a which given time the object has sink too deep.

    Greetings
     
    Last edited: Apr 15, 2013
  6. Apr 17, 2013 #5
    Can I assume that because of the wetting between the object and molten fluid (glass) that the surface tension can be neglected?

    Are the only forces I can work with now is the Buoyancy force and the viscous drag force?
    So at the given temperature for example:

    Stokes Law:
    v = (2*∇ρ*g*r^2) / (9*η) = (2*(2330-2300)*9.81*r^2) / (9*500). what is r now?

    but this formula is only for a sphere, stokes law gives F.drag = 6*pi*r*η*v. I think a rectangle prism got a higher drag resistance then a sphere.
    How can I rewrite his equation for a rectangle form with a small height?
    cuboid.gif

    If I am going in the wqrong direction please tell me, I really need help with this.
    Are there any other possibilities to tell how far it will sink in a given time?
     
  7. Apr 17, 2013 #6
    You can get a lower bound on how far the object sinks by considering the drag on a rectangular prism (of your desired shape) entirely immersed in a uniform fluid flow. In this case, the drag force will be higher than in your situation, where, presumably, the upper portion of the prism has not yet submerged. See if you can find something on the internet on fluid dynamic drag on a prism, or creeping flow past a prism.

    As far as surface tension is concerned, the upward force cannot be higher than the surface tension times the wetted perimeter. If molten glass is non-wetting, then the downward force cannot be higher than the surface tension times the wetted perimeter.

    Try to think of other ways of bounding the answer.
     
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