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Laminar flow in a pipe (Poiseuille flow)

  1. May 13, 2014 #1
    1. The problem statement, all variables and given/known data

    I am trying to derive Poiseuille's equation using an infinitesimal volume element of the pipe in cylindrical coordinates as seen below. I am using x for the flow direction, u for the velocity in the flow direction, and μ for the viscosity constant.

    https://www.physicsforums.com/attachment.php?attachmentid=69754&stc=1&d=1400000671

    2. Relevant equations

    I know the viscous stress varies radially by τ = μ(du/dr)

    This means that the viscous force on one face of my infinitesimal volume element is μ(du/dr)r r δ[itex]\phi[/itex] δx

    And I think the viscous force on the opposite face is μ(du/dr)r+δr (r+δr) δ[itex]\phi[/itex] δx

    Now I want to subtract one of these forces from the other to get the net viscous force on the infinitesimal element.

    3. The attempt at a solution

    When I attempt to do this, I am left with two radial derivatives evaluated at different points. I get:

    μ ( (du/dr)r+δr r δ[itex]\phi[/itex] δx + (du/dr)r+δr δr δ[itex]\phi[/itex] δx - (du/dr)r r δ[itex]\phi[/itex] δx)

    Now I'm not exactly sure how to use Taylor's Theorem to get all of the derivatives evaluated at the same point and combine terms. Thanks.
     
  2. jcsd
  3. May 13, 2014 #2
    You're very close to having the right answer. There are two methods you can use.

    Method 1:
    μ(du/dr)r+δr (r+δr) δϕ δx-μ(du/dr)r r δϕ δx=[itex]μ[(r+δr)(\frac{du}{dr})_{r+δr}-r(\frac{du}{dr})_{r}]δϕ δx=μ\frac{[(r+δr)(\frac{du}{dr})_{r+δr}-r(\frac{du}{dr})_{r}]}{δr}δrδϕ δx[/itex]

    This is equal to
    [tex]μ\frac{d}{dr}\left(r\frac{du}{dr}\right)δrδϕ δx[/tex]

    Method 2:

    μ ( (du/dr)r+δr r δϕ δx + (du/dr)r+δr δr δϕ δx - (du/dr)r r δϕ δx)=[itex]μ(r(\frac{du}{dr})_{r+δr}-r(\frac{du}{dr})_{r}+(\frac{du}{dr})_{r+δr})δr δϕ δx[/itex]
    This is equal to:[tex]μ(r\frac{(\frac{du}{dr})_{r+δr}-(\frac{du}{dr})_{r}}{δr})δr δϕ δx+μ((\frac{du}{dr})_{r}+\frac{d^2u}{dr^2}δr)δr δϕ δx=μ(r\frac{d^2u}{dr^2})δr δϕ δx+μ(\frac{du}{dr})δr δϕ δx+μ\frac{d^2u}{dr^2}(δr)^2 δϕ δx[/tex]
    This is equal to:
    [tex]μ(r\frac{d^2u}{dr^2}+\frac{du}{dr})δr δϕ δx+μ\frac{d^2u}{dr^2}(δr)^2 δϕ δx[/tex]
    The last term has 4 differentials in it compared to only 3 from the previous term, so it can be neglected. We are left with:

    [tex]μ(r\frac{d^2u}{dr^2}+\frac{du}{dr})δr δϕ δx[/tex]

    Chet
     
  4. May 14, 2014 #3
    Ah, I see thank you. I should have recognized the definition of a derivative.

    My only question is that in method 1 it doesn't look like you are making any approximations because you're just applying the definition of a derivative, while in method 2 it looks like you use an approximation because you neglect the term with 4 differentials. But you end up with the same result both times. Why is this the case?
     
    Last edited: May 14, 2014
  5. May 14, 2014 #4
    They're just different ways of doing the approximation, but the final result has to come out the same either way.
    Chet
     
    Last edited: May 14, 2014
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