# Symmetry Arguments: Flow Through Cylinder

Why can for symmetry reasons a steady, viscous, incompressible flow, obaying the N.S equation:

$$\rho(v \nabla)v = -\nabla p + \eta \Delta v$$

That flows through a cylindrical(very long) pipe not have a component in x or y direction and the velocity can not depend on z? (setting aside any end of pipe effects) I just can not find any good argument that is actually mathematically sound and not just "sounds right".

• It can easily have velocity components in $x$ and $y$ direction, depending on the boundary conditions on the inlet and outlet. (just imagine a point source on the left end and a point sink on the right end). This way you can even easily create a flow that is not rotation symmetric around the $z$-axis. The symmetry is rather an asymptotic behaviour (for long pipes). – Sebastian Riese Mar 17 '16 at 15:56
• sorry to be confusing, there is always something i forget. infinite length and a no slippage condition seem to beimpied in this by the setup and the viscosity i think. Also it has to be incompressible to obay this version of the NS, i added that also. – pindakaas Mar 17 '16 at 16:30

I remember from my bachelor studies that my teachers would often just neglect terms based on experience which often wasn't obvious to me and left me with the same questions. Not until later when i learned about advanced dimensional analysis topics like scaling did i fully understand when and why certain terms may be neglected. A really good book on this is 'Analysis of Transport Phenomena' by Deen, i really recommend it to get a good grasp on these topics.

The answer to your question for a pipe flow isn't actually answered using the Navier-Stokes equations but using the continuity equation (in radial coordinates):

$$\boldsymbol{\nabla}\cdot \boldsymbol{u}=\partial_{z}u_{z}+\frac{1}{r}\partial_{r}\left(ru_{r}\right)=0$$

which is an equation my teachers also tended to forget about precisely because they immediatly assumed there were no velocity fluctuation in the radial direction, i.e. $u_r\left(r,z\right)\equiv0$, and the velocity in the axial direction only depended on the radial direction, i.e. $u_z\left(r,z\right)=u_z\left(r\right)$. Substituting this in, you can see this exactly satisfies the continuity equation and you can subsequently forget about it (just like my teachers did) and focus on solving the Navier-Stokes equations (usually leading to Hagen-Poiseuille flow, Couette flow, etc.).

Now lets assume that we are a bit naive and don't immediately see that in an infinitely long pipe without entrance/exit effects we can make these assumptions. Then we can turn to dimensional analysis to give us a mathematical basis for which to arrive at these assumptions. To do so we introduce the following dimensionless quantities:

$$\tilde{z}=\frac{z}{L}\quad\tilde{r}=\frac{r}{R}\quad\tilde{u}_z=\frac{u_z}{U}\quad\tilde{u}_r=\frac{u_r}{V}$$

where $L$ is the length scale in the axial direction (pipe length), $R$ is the length scale in the radial direction (pipe radius), $U$ is some velocity scale for the axial direction and $V$ is some velocity scale for the radial direction (usually assumed to be the maximum velocities in each respective direction). Now let's non-dimensionalize the continuity equation:

$$\frac{U}{L}\partial_{\tilde{z}}\tilde{u}_{z}+\frac{V}{R}\frac{1}{\tilde{r}}\partial_{\tilde{r}}\left(\tilde{r}\tilde{u}_{r}\right)=0$$

In scaling equations we are looking to reduce all terms in equation to $O\left(1\right)$. Assuming our scales are correctly chosen, the dimensionless quantities are at most $O\left(1\right)$ which implies for all terms in the dimensionless continuity equation to be $O\left(1\right)$ we require:

$$\frac{U}{L}\sim\frac{V}{R}$$

This then simplifies the continuity equation to:

$$\partial_{\tilde{z}}\tilde{u}_{z}+\frac{1}{\tilde{r}}\partial_{\tilde{r}}\left(\tilde{r}\tilde{u}_{r}\right)=0$$

and we have new information on the relative size of the velocity scales since:

$$\frac{V}{U}\sim\frac{R}{L}$$

For an infinitely long cylinder ($R/L\ll1$), we can then say that $V/U\ll1$ and that the velocity scale in the radial direction is much smaller than the velocity scale in the axial direction. The next step in the simplification is then simply saying that there isn't any velocity in the radial direction at all which leads us back to the assumptions made earlier.

If from the continuity equation you can simply neglect a velocity component, then subsequently you don't have to consider the respective component of the Navier-Stokes equatins either.

• hey thanks for the answer, sounds very reasonable. What happend to the angular derivative ${1 \over r} {\partial F \over \partial \theta}$ though? I guess since the problem is perfectly symmetric in $\theta$ there can be no change along this variable and thus the derivative must be zero aswell? – pindakaas Mar 20 '16 at 9:36
• @pindakaas - yeah I am a bit of a hypocrit :) i assumed the velocity fluctuation in the angular coordinate was insignificant because the geometry chosen was a perfectly straight pipe. The same analysis applies ofcourse, where showing that if the radius of curvature is negligible to the pipe length (sort of like looking at a straight horizon while knowing the earth is round), the angular component of the velocity is negligible and can be set identically to zero. – nluigi Mar 20 '16 at 9:56