Laminar flow in a pipe (Poiseuille flow)

[rmc240]

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.

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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.

 

[Chestermiller]

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

 

[rmc240]

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?

 

[Chestermiller]

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

 

[paranormal]

Hello, it seems like you are on the right track with your derivation. In order to use Taylor's Theorem to evaluate the derivatives at the same point, you can expand the terms in your equation to include higher order terms. This will allow you to use Taylor's Theorem to approximate the derivatives at the same point. Additionally, you may need to use the chain rule to evaluate the derivatives in terms of the velocity and position variables. Keep in mind that the net viscous force on the infinitesimal element should be equal to the change in momentum over time, which can be represented by the Navier-Stokes equation. I hope this helps. Good luck with your derivation!