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Fluid dynamics, friction and pressure

  1. Sep 17, 2012 #1
    Hello,
    this is not homework, I am trying to derive some physics results using intuition, I am currently looking at some fluid dynamics problems.

    Consider a water droplet on a frictionless horizontal plane, subject to gravity. The cross-section of the droplet would look something like this (water volume at any given position along the cross-section discretized into water columns):

    2weawyv.png

    Now because the surface on which the droplet is resting is frictionless, the water droplet will spread out forever until it covers the plane completely and uniformly. This is logical and can be tested easily with a material with a very low friction coefficient, such as glass.

    This happens because at the bottom of each water column, there is some amount of pressure, depending on the height of the column, as:
    $$P = \rho gh$$
    Water being incompressible, fluid at the bottom of the column will be "pushed" aside by the column's pressure, and will attempt to enter the neighboring columns. Because the plane is frictionless, nothing prevents it to do so, so it does and the droplet spreads out.

    Well, actually, water will only go from higher-height columns to lower-height columns, because the column "receiving" the water will oppose a pressure force dependent on its current height; clearly, if the "receiving" column has a higher height, its pressure force will exceed the force pushing the water towards it, and the water will not move. But from the argument that the maximum height of the columns has to be bounded, it follows that the system's equilibrium is a uniform water distribution (possibly of infinitesimal height, if the plane has an infinite surface area).

    Now suppose the plane has some amount of friction. We would expect the droplet to enter a more interesting state of equilibrium, because there is now one force opposing water transfer between columns: friction. This means that as soon as the friction force exceeds the pressure force (pressure multiplied by the column's surface area) the water cannot move from that column, it is "held back" by friction. We know that friction depends on the normal force on the surface, conveniently this is the pressure force multiplied by the friction coefficient:
    $$F_{\mathrm{friction}} = \mu N = \mu P A = A \mu \rho gh$$
    The force exerted by the water column on the water at the bottom of said column is more complicated to calculate. Basically, the column receiving the water will oppose a pressure force dependent on its current height. Let h' denote this height, then the opposing force is:
    $$F_{\mathrm{opposing}} = A \rho g h'$$
    Since the columns have the same area. Then, the net force without friction on the water to transfer is:
    $$F_{\mathrm{net}} = A \rho g h - A \rho g h' = A \rho g(h - h')$$
    So the water will be transferred if:
    $$F_{\mathrm{net}} > F_{\mathrm{friction}}$$
    $$A \rho g(h - h') > A \mu \rho gh$$
    $$h - h' > \mu h$$
    $$(1 - \mu) h > h'$$
    Which yields:
    $$\displaystyle h > \frac{h'}{1 - \mu}$$

    This conclusion agrees with our previous observations: with a frictionless surface (μ = 0) we get h > h' which is what we found at the beginning of the post (water will only be transferred if the receiving column has less water than the "donor" column. Note I ignored the fact that the water spreads to more than one column (probably four) since we are in two dimensions, but it should just be a constant factor on the (h - h') term - but I haven't checked. With this relation and some integration, it should be possible to compute the steady-state height and water distribution of the droplet depending on its original height, its initial water distribution and the surface friction.

    Now my question - is this actually correct? It seems to make sense to me, and the results seem to be coherent and consistent with experiment, but can anyone check through my work and see if they agree with it? Does it make sense, is this what is actually happening physically or are there extra things to consider?

    I also have an extra question: this analysis ignores surface tension as the droplet's surface area changes - how important are those effects and what would be the first step to introducing them into the equation? I am unsure how to calculate the effects of surface tension - my first idea was to consider the change in (h - h') after the water has moved, which is what is causing the droplet's area to change.

    Thanks in advance :)
     
  2. jcsd
  3. Sep 17, 2012 #2
    Friction is not an issue with a fluid because of the well-known no-slip boundary condition at a solid surface. This is a fluid mechanics problem, and viscous forces play a major role. For this particular problem, surface tension effects are also very important. Surface tension affects the pressure distribution within the fluid, which feeds into the viscous flow. Another factor consider is the contact angle at the interface between the solid, liquid, and gas, which is determined by the chemical nature of the three materials (primarily the liquid and solid). The contact angle is typically not zero. Non-stick surfaces have lower contact angles.

    So, if you are going to start modeling this problem properly, you need to forget about friction, but include viscous flow, surface tension, non-slip boundary condition, and contact angle.
     
  4. Sep 17, 2012 #3
    In this problem, because of the very thin nature of the fluid film on the surface and the slow velocity of flow, fluid inertia is not significant, and the behavior is dominated by the creeping viscous flow equations. The Euler equations do not apply to this situation.
     
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