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Homework Help: Fluid Dynamics Forces

  1. Apr 27, 2007 #1
    I would really appreciate any help with this problem. I'm pretty stuck.

    1. The problem statement, all variables and given/known data

    An incompressible viscous fluid occupies the region between two parallel plates. The lower plate is stationary and the upper plate moves with a velocity u. Find the velocity of a steady, pressure-driven flow between the plates. Sketch the velocity profice and consider separately the cases where the pressure gradient is positive or negative. Compute the force per unit area on the upper and lower plates.

    2. Relevant equations

    most likely navier stokes for incompressible newtonian fluids found at http://en.wikipedia.org/wiki/Navier-Stokes_equations

    and continuity equation for fluid dynamics, found at http://en.wikipedia.org/wiki/Continuity_equation

    3. The attempt at a solution

    If the upper plate is taken to be moving in the x-direction, then (velocity) Vy = Vz =0.

    With the continuity equation, d(rho)/dt becomes zero as the fluid is incompressible so density never changes. WHere U is fluid velocity, Continuity becomes del(dot)(rho*u) = (del(rho))(dot)u + rho(del(dot)u). (del(rho))(dot)u is zero due to incompressibility as well, so this gives dUx/dx + dUz/dz = 0.

    Vz is 0 at the bottom and top plate and thus everywhere else. so Vz=0, dUz/dz=0, dUx/dx=0 by symmetry

    so Ux is a function of z. Ux=Ux(z)

    WIth the navier stokes equation becomes
    integrating both sides gives
    dUx/dz=1/μ *dP/dx *z + C
    Ux(z) = 1/2μ * dP/dx Z^2 + C(1)z + C(2)

    Applying the boundary conditions: Ux(0)=0 -> C(2)=0
    Ux(h) = V0 -> C(1)=V0/h-1/2μ*dP/dx*h

    After this I'm not sure about the pressure gradients part or the forcer per unit area on the plates. I'm not even completely sure I did the flow velocity (Ux(z))) part correctly. Any help would be really appreciated.

    I also hope you can understand what I wrote, it's quite hard to type these formulas.

  2. jcsd
  3. Apr 28, 2007 #2


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    Have your read the https://www.physicsforums.com/showthread.php?t=8997" for this forum? It's an easy way to display mathematical formulas.

    Seems correct. Now, rearrange it to get,

    [tex] U_x(z) = \frac{z(dP/dx)(z-h)}{2 \mu} + V_0z/h [/tex]

    How will the plot of U_x(z) vs z, look when dP/dx is positive or negative? If you want, you can put numbers and check the plots.

    Now that you've got the velocity profile, and you know that the shear stress is [tex] \mu \frac{dV}{dz}[/tex], can you calculate the forcer per unit area on the plates?
    Last edited by a moderator: Apr 22, 2017
  4. Apr 28, 2007 #3
    Thanks a lot!

    So the force per unit area on the upper plate would be

    and on the lower plate, it would be

    If i haven't made a mistake

    However, I still have a question about the velocity profile. When plugging in arbitrary numbers, the profile changes drastically depending on the μ value in the positive dP/dx case. With a small μ, the velocity can suddenly become negative between the plates. With the negative dP/dx case, the flow increases in the Z direction as expected, but then suddenly slows when z=h at the top plate. If the top plate is causing the movement of the fluid, then the flow velocity should be greatest at the top plate (z=h). The physical reasoning velocity profile is of a 0 fluid velocity at the bottom plate (Z=0) increasing linearly to a velocity V=U at the top plate (Z=h). Neither of these cases match. The only thing I can think of is if my original velocity equation is wrong.
  5. Apr 28, 2007 #4


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    How did you get that? For example, if you want to find the stress at the lower plate, evaluate [tex]dU_x/dz[/tex] at z=0, and multiply it with [tex]\mu[/tex]. Can you take it from here?

    Interesting, isn't it? And it's true!

    Consider the equation,
    [tex]v_z = \frac{z(dP/dx)(z-h)}{2 \mu} + V_0z/h [/tex]

    The first term, [tex] \frac{z(dP/dx)(z-h)}{2 \mu}[/tex], is a parabola, and the second term, [tex] V_0z/h[/tex] is a straight line. When dP/dx is positive, notice that they have opposite signs. The final answer will be a superposition of these two.

    No, that's not right.

    The linear increase will take place only when the pressure gradient is 0.
    Last edited: Apr 28, 2007
  6. Apr 28, 2007 #5
    haha, sorry, I actually did this but I made a stupid mistake in derivation mixed with messing up how to use "Latex". I have an answer that makes more sense now.

    I think I understand the velocity profile now too.

    Thanks a lot for your help, I really appreciate it.
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