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Help with Hagen-Poiseuille Flow

  1. Jan 6, 2010 #1
    Hi

    I'm trying to derive the velocity profile for Hagen-Poiseuille flow through a pipe.
    Using cylindrical coordinates (z direction horizontal), I began by applying the Navier-Stokes equations to each coordinate.

    For z, I got: [tex]\frac{1}{\eta}[/tex] [tex]\frac{\partial p}{\partial z} = \frac{1}{\rho}[/tex] [tex]\frac{\partial}{\partial\rho}[/tex] [tex](\rho \frac{\partial v_{z}}{\partial\rho})[/tex] and from this equation I got the result that [tex]v_{z} = \frac{1}{4\eta} \frac{\partial p}{\partial z} (\rho^{2} - R^{2})[/tex] where R is the radius of the pipe

    The Navier-Stokes equations for the [tex]\rho[/tex] and [tex]\phi[/tex] directions give [tex]\frac{\partial p}{\partial \rho} = \mu g_{\rho}[/tex] and [tex]\frac{\partial p}{\partial \phi} = \mu g_{\phi}[/tex] respectively, where g is gravity in each direction and [tex]\mu[/tex] is the density of the fluid.

    I know that when deriving the velocity profile for flow between parallel plates, you need to use the [tex]\phi[/tex] and [tex]\rho[/tex] equations to show what is a function of what, and what is independent of what.
    However, I've managed to get the velocity profile without using any information from the other two coordinate equations. Have I missed something? Do I need to use these two equations for something?

    Many thanks
    teeeeee
     
  2. jcsd
  3. Jan 6, 2010 #2

    minger

    User Avatar
    Science Advisor

    Let me walk you from the beginning. If we assume fully-developed flow, then the velocity becomes purely axial, and varies only with the lateral coordinates, that is:
    [tex]
    \begin{equation}
    \begin{split}
    v=w=0 \\
    u=u(y,z)
    \end{split}
    \end{equation}
    [/tex]
    The continuity and momentum equations can then reduce to:
    [tex]
    \begin{equation}
    \begin{split}
    \frac{\partial u}{\partial x}=0 \\
    -\frac{\partial \hat{p}}{\partial x} + \mu\left(\frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\right) = 0 \\
    -\frac{\partial \hat{p}}{\partial y} = -\frac{\partial \hat{p}}{\partial y}
    \end{split}
    \end{equation}
    [/tex]
    These indicate that the total pressure is a function only of x. Since u does not vary with x, we can say that the gradient dp*/dx must be a negative constant. Then, we can combine to form the basic equation as:
    [tex]
    \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = \frac{1}{\mu}\frac{d\hat{p}}{dx} = \mbox{const}
    [/tex]
    We can then non-dimensionalize as:
    [tex]
    \begin{equation}
    \begin{split}
    y* &= \frac{y}{h} \\
    z* &= \frac{z}{h} \\
    u* &= \frac{\mu u}{h^2(-d\hat{p}/dx)}
    \end{split}
    \end{equation}
    [/tex]
    Where h is a characteristic duct width. OK, now that we have those variables defined, we can rewrite the general equation as:
    [tex]
    \nabla^{*2}(u*) = -1
    [/tex]

    OK, we're just about there. Now, for a Hagen-Poiseuille Fow, we have a circular duct, so the single variable is of course r. Non-dimensionally, we can write r*=r/r_o where r_o is the pipe radius. The Laplacian operator in cylindrical coordiantes reduces to:
    [tex]
    \nabla^2 = \frac{1}{r}\frac{d}{dr}\left(r\frac{d}{dr}\right)
    [/tex]

    From here you can plug your operator into the general equation using the non-dimensional terms. If you need more help, come back with a good attempt and we can get you the rest of the way.
     
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