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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?

    Chet
     
  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:
    [tex]\vec{V}=v(r)\vec{i}_z[/tex]
    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.

    Chet
     
  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.

    Chet
     
  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.

    Chet
     
  9. Jan 25, 2015 #8

    pasmith

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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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