Laminar flow in a pipe (Poiseuille flow)

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SUMMARY

This discussion focuses on deriving Poiseuille's equation for laminar flow in a pipe using cylindrical coordinates. The participants detail two methods for calculating the net viscous force on an infinitesimal volume element, employing the viscosity constant (μ) and the velocity gradient (du/dr). Both methods yield the same result, μ(r(d²u/dr²) + (du/dr))δr δϕ δx, despite one method involving an approximation by neglecting higher-order terms. The discussion emphasizes the importance of understanding the definition of a derivative in fluid dynamics.

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rmc240
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Homework Statement



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

Homework 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\phi δx

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

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

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\phi δx + (du/dr)r+δr δr δ\phi δx - (du/dr)r r δ\phi 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.
 
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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=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

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

Method 2:

μ ( (du/dr)r+δr r δϕ δx + (du/dr)r+δr δr δϕ δx - (du/dr)r r δϕ δx)=μ(r(\frac{du}{dr})_{r+δr}-r(\frac{du}{dr})_{r}+(\frac{du}{dr})_{r+δr})δr δϕ δx
This is equal to: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
This is equal to:
μ(r\frac{d^2u}{dr^2}+\frac{du}{dr})δr δϕ δx+μ\frac{d^2u}{dr^2}(δr)^2 δϕ δx
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:

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

Chet
 
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:
rmc240 said:
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?

They're just different ways of doing the approximation, but the final result has to come out the same either way.
Chet
 
Last edited:

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