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Use viscosity/shear stress relations to find torque needed to rotate disk

  1. Sep 20, 2014 #1
    The problem asks for an approximation for the torque needed to rotate a disk that is separated from a stationary boundary by a viscous fluid, given that [itex]\tau = \mu \frac{U}{H}[/itex].

    I did it first using like this:
    [itex]T = F R[/itex]
    [itex]F = \tau A = \mu \frac{U}{H} \pi R^2[/itex] where [itex]U = \Omega R[/itex]
    And thus [itex]T = \mu \frac{\Omega \pi R^4}{H}[/itex]

    Now making the following assumptions...
    Flow is steady
    Flow is a function of height (z) only
    Film is thin enough that the pressure gradient is negligible
    ...I started with momentum:

    [itex]\rho \frac{du}{dt} = -\nabla p + \mu \nabla^2 u[/itex] which reduces to [itex]0 = 0 + \mu \nabla^2 u(z)[/itex]
    Where [itex]u(z) = \frac{C_1 z}{\mu} + C_2[/itex]
    Using the boundary conditions
    [itex]u(0) = 0[/itex] and [itex]u(H) = U[/itex] yields [itex]C_1 = \mu \frac{U}{H} = \tau[/itex] and [itex]C_2 = 0[/itex] so that
    [itex]u(z) = \tau \frac{z}{\mu} = \frac{U z}{H} = \frac{R \Omega z}{H}[/itex]

    Here is where I have trouble; a classmate told me to square the velocity and multiply it with mass
    [itex]T = M u(z)^2 = M \frac{\Omega^2 R^2 z^2}{H^2}[/itex] and let z = R and [itex]\mu = M \frac{\Omega}{H}[/itex] so that the equation becomes
    [itex]T = \mu \frac{\Omega R^4}{H}[/itex]
    Which is the same as I got before, but without the constant [itex]\pi[/itex]
    I'm note very comfortable with this because it takes the same form as kinetic energy [itex](\frac{1}{2}Mv^2)[/itex] but it does seem to yield the correct answer. Is this true? Another problem I have is keeping track of the vectors vs. scalars but I'll get into that later on. Any help is greatly appreciated.
     
  2. jcsd
  3. Sep 21, 2014 #2
    This is not done correctly, because the shear stress on the disk is a function of radial location. The calculation needs to involve an integration with respect to r, over annular increments.

    Chet
     
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