# Laminar flow in a pipe (Poiseuille flow)

1. May 13, 2014

### rmc240

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$\phi$ 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.

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$\phi$ 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.

2. May 13, 2014

### Staff: Mentor

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

3. May 14, 2014

### 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?

Last edited: May 14, 2014
4. May 14, 2014

### Staff: Mentor

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