# Homework Help: Justifying steps in the Navier-Stokes calculation for simple circular pipe flow.

1. Feb 28, 2012

### Peeter

1. The problem statement, all variables and given/known data

In [1], problem 2.3 (ii), we are asked to show that for a pipe with circular cross section $r = a$ and constant pressure gradient $P = -dp/dz$ one has

$$u_z = \frac{P}{4 \mu}\left( a^2 - r^2 \right)$$
$$u_r = 0$$
$$u_\theta = 0$$

References:

[1] D.J. Acheson. <em>Elementary fluid dynamics</em>. Oxford University Press, USA, 1990.
2. Relevant equations

If one does assume that $u_r = u_\theta = 0$, then the Navier-stokes equations for an incompressible steady state flow takes the form

\begin{align}u_z \frac{\partial {u_z}}{\partial {z}} = \frac{P}{\rho} + \nu \left(\frac{1}{{r}} \frac{\partial {}}{\partial {r}}\left(r \frac{\partial {u_z}}{\partial {r}} \right) + \frac{\partial^2 {{u_z}}}{\partial {{\theta}}^2} + \frac{\partial^2 {{u_z}}}{\partial {{z}}^2}\right)\end{align}
\begin{align}0 = \frac{\partial {p}}{\partial {r}}\end{align}
\begin{align}0 = \frac{\partial {p}}{\partial {\theta}}\end{align}

3. The attempt at a solution

I can actually solve this problem completely, but have some trouble justifying some of the steps.

a) One of them is the assumption that $u_r = u_\theta = 0$, which I use as a starting point to express NS above. This intuitively makes sense, but are there other physical principles (other than intuition) that lead to this conclusion?

b) In NS above we can kill of the $\partial_{\theta\theta}$ portion of the Laplacian, again by intuition. However, it's not too hard to imagine that you could have a flow where one could have an angular dependence in the flow (ie. imagine a really big pipe where fluid is being injected harder at the base of the pipe than at the top). Something like that would probably have a different pressure dependence. It isn't clear to me what sort of mathematical argument that one could use to show that for this constant pressure gradient scenerio one must have no $\theta$ dependence in the velocity.

c) Finally, once we make assumption (b) so that $\partial_\theta u_z = 0$, NS takes the form (with $w = u_z$)

\begin{align}w \frac{\partial {w}}{\partial {z}} = \frac{P}{\rho} + \nu \left(\frac{1}{{r}} \frac{\partial {}}{\partial {r}}\left(r \frac{\partial {w}}{\partial {r}} \right) + \frac{\partial^2 {{w}}}{\partial {{z}}^2}\right).\end{align}

Attempting separation of variables with $w = Z(z) R(r)$, this appears to be inseparable due to the non-linearity of the $(\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u}$ term on the LHS. The fact that this is inseparable doesn't seem like it's a good justificaion for requiring that $w = w(r)$ only and not $w = w(r,z)$. I could imagine that it is still possible to find solutions of the form $w = w(r,z)$. Is there some other argument that can be made to show that there is no $z$ dependence in this remaining component of the fluid's velocity?

Last edited: Feb 28, 2012
2. Feb 28, 2012

### Jolb

I think all of your questions basically have the same answer. Just think of NS as a statement of Newton's second law and it should become obvious.
$$\mathbf{F}=m\mathbf{a}$$
becomes
$$-\nabla p=\rho\frac{D\boldsymbol{u}}{Dt}$$ where the capital D's denote the advective derivative.

(I've written it with no externally applied forces because your question seems to imply that's the case. I don't think the question would make any sense without that assumption.)

Therefore, you should be able to see very clearly that since the pressure gradient (force) only has a z component, there shouldn't be any change in the flow (accelleration) in the radial or angular directions. The pressure field is a constant so the flow is steady--i.e., you don't have to worry about residual currents or changes in the flow as if the question were "at t=0 the pressure field is turned on" or anything like that.

Let me know if this makes sense to you.

Last edited: Feb 28, 2012
3. Feb 29, 2012

### Peeter

I think that answers why we'd have no $u_r$, and $u_\theta$ terms, but we have one more (the Laplacian) term in this version of F=ma:

with that term it still seems to me that we can have functional dependence on z and $\theta$, even without components directed in the radial or angular directions.

Yes, in this problem, no external forces are being considered.

4. Feb 29, 2012

### Peeter

Okay, I see why there's no z-dependence in u_z. The statement of the problem should have included one more equation. This was Navier-Stokes for incompressible fluids, so must be augmented by

$$\nabla \cdot \mathbf{u} = \frac{1}{r}\partial_r( r u_r) + \frac{1}{r} \partial_\theta u_\theta + \partial_z u_z = 0$$

so if $u_r = u_\theta = 0$, we can have no z dependence in u_z.