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Navier-Stokes and why DP/Dz is constant

  1. Jan 19, 2015 #1
    So in trying to solve for Poiseuille's Law using the continuty eq and Navier-Stokes, the differential EQ becomes DP/dz = some function of R only. The professor says that because DP/dz is ONLY a function of Z, and the left side is ONLY a function of R, the only way they can be equal is if both are constants.

    What's the intuition or reasoning behind this?
  2. jcsd
  3. Jan 19, 2015 #2
    If the flow is fully developed, then the fluid velocity does not depend on z. Also, the flow velocity is purely axial. Are you comfortable with this?

  4. Jan 19, 2015 #3
    Oh sorry, I should have labeled. the P means pressure of flow and z is the direction.
    I have found a similar explanation on this video which may give you a better understanding of what I'm asking. This is at time 1:50.

    Sorry, the explanation still isn't working for me!
  5. Jan 19, 2015 #4
    What I was trying to do is to explain why your professor is correct. For fully developed flow in a horizontal pipe, the velocity vector is given by:
    where ##\vec{i}_z## is the unit vector in the axial direction. This relationship satisfies the continuity equation exactly and, if you substitute it into the z component of the momentum equation in cylindrical coordinates, you end up with your professor's equation. The right hand side of this equation is independent of z. That means that dp/dz is independent of z. Right now, I don't see how to prove that dp/dz is not a function of r, but I'll think about it. I know that it is.

  6. Jan 20, 2015 #5
    After thinking about it some more, I realized that dp/dz not being a function of r comes from the components of the NS equation in the other two directions. For these other two directions, the pressure variations are hydrostatic.

  7. Jan 20, 2015 #6
    I just reread my question and it's more vague than I thought it was, sorry about that!
    I do agree with all the equations in the video and what you've posted, but here's my real question:

    I'm curious as to why, at time 1:50 in the video, a function of Z only being equal to a function of R only means that both of them have to be constants for this to be true. It doesn't haveWhat is the logic behind this?
  8. Jan 21, 2015 #7
    If f(r) = g(z), give me a specific example of a case in which f and g are not constants.

  9. Jan 25, 2015 #8


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    Homework Helper

    Start with: [tex]
    \frac{\partial p}{\partial z} = f(r).
    [/tex] Using the given information, [tex]
    \frac{\partial f}{\partial r} = \frac{\partial}{\partial r}\left(\frac{\partial p}{\partial z}\right) = 0[/tex] and [tex]
    \frac{\partial}{\partial z}\left(\frac{\partial p}{\partial z}\right) =\frac{\partial f}{\partial z} = 0.[/tex] Hence [itex]f(r) = \frac{\partial p}{\partial z}[/itex] is constant.
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