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Simplified derivation of viscosity equation

  1. Jan 31, 2014 #1
    1. The problem statement, all variables and given/known data
    Think of a fluid between two plates at a distance d from each other in the z-direction; then slide the top plate at a constant speed v_0 in the x-direction while holding the bottom plate still. The fluid pushes back, so you need to apply a force F in order to keep the plate moving at constant speed. Near the top plate, friction makes the fluid move with speed v0, whereas near the bottom plate the fluid is stationary. This causes a velocity gradient v_x(z), with v_x(d) = v0 and v_x(0) = 0.
    Take a slab of gas at some value z. Assume all particles have the same magnitude of z-velocity, |v_z| (but have a gradient in their v_x velocity as explained above). Half move up and half move down. Also assume they all collide for the last time exactly λ distance (the mean free path) before passing through the slab and in the process acquire the x-velocity v_x(d − λ) if they collide below the slab and go above and v_x(d+λ) if they collide above the slab and go below. Their z-velocity does not change during this collision (Can you explain why this is reasonable assumption?).
    Knowing that the density of particles in the fluid is ρ = N/V , give the number of particles (∆N) per unit area that pass through the horizontal slab from below in a period of time ∆t. Assume ∆t|v_z| ≫ λ. How many cross from above?

    3. The attempt at a solution
    The assumption that the z-velocity doesn't change is reasonable if we move the plate slowly enough.

    The particles pass through an horizontal slab, so the result is proportional to the z-component of the velocity. Also, the number of particles passing must be proportional to the particle density. We thus get
    \begin{equation*}
    \text{\# of particles per unit time per unit area} =\dfrac{\Delta N}{A\Delta t}= \rho |v_z |
    \end{equation*}
    I'm just not sure on how to figure out how many come from above or below. Also, I am not supposed to use the Maxwell distribution.
     
  2. jcsd
  3. Jan 31, 2014 #2
    No worries I found the solution. Silly me.
     
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