5
$\begingroup$

I'm taking a first course on fluid dynamics, and I have this (sort of) conceptual question that's been nagging me for a moment now. I can completely follow the mathematics behind the derivation of the time-independent plane Poiseuille flow, it's the symmetry considerations at the beginning that are giving me a headache.

Briefly, the "plane Poiseuille flow" is the steady pressure-driven laminar flow of a Newtonian fluid between two fixed parallel walls of infinite extent separated by a distance d. Most books that I've read begin by saying something in the lines of "because of the translational symmetry", the flow "cannot depend on the longitudinal coordinate". In fact, it is true that the problem looks the same if one shifts the origin an arbitrary distance along a line parallel to the walls.

What I cannot understand is how this last observation can be consistent with the fact that the pressure field does depend on that same coordinate. I know its gradient does not.

My question is: do these (so-called) symmetry considerations only apply to the velocity field? If that's the case, I cannot understand why the velocity field and the pressure field are treated differently.

I'm looking for an answer to this problem that can be extrapolated to other laminar viscous flows (such as plane and circular Couette's, etc.). I'm also interested in answers pointing to a formalization of these symmetry considerations. I've already browsed through Cantwell's Introduction to Symmetry Analysis, but right now it seems like an overkill for this problem.

$\endgroup$
  • 1
    $\begingroup$ One thing to think about is whether you are talking about compressible or incompressible flow. Traditional Poiseuille flow is found assuming incompressibility -- what is the relationship between the velocity and pressure in incompressible flow? What does that relationship tell you about why the two fields can be treated differently? $\endgroup$ – tpg2114 Jul 6 '15 at 0:51
  • $\begingroup$ Thanks for your answer. Can you elaborate a little more? I would like to make sure I understand what you mean. Thanks again for your quick answer. $\endgroup$ – Just Asking Jul 6 '15 at 1:00
  • $\begingroup$ Well, I left it intentionally vague (and as a comment because of that) because I want you to think more about the assumptions that go into the problem. By the time you get to the "Because of the translational symmetry" point of the derivation, several very important assumptions have been made. These assumptions have an influence on what happens next. $\endgroup$ – tpg2114 Jul 6 '15 at 1:05
  • $\begingroup$ Thanks for your help, although I'm still lost on this issue. I know you meant well by motivating me to think on this and that your answer is kind of a hint, but I've spent so much time thinking on this problem that I'm really confused and I'm starting to think I do not understand laminar viscous flows at all. Thanks again, tpg2114. $\endgroup$ – Just Asking Jul 6 '15 at 1:18
  • $\begingroup$ Incidentally, does your answer/hint means that I was right on thinking that these "symmetry considerations" apply only to the velocity field (although i do not yet see why)? $\endgroup$ – Just Asking Jul 6 '15 at 1:22
1
$\begingroup$

The solution lies inside the reduction of the Navier-Stokes equations for this particular problem and the assumptions of plane Poiseuille flow. The general 2-dimensional, incompressible, and constant property Navier-Stokes equations take the form,

$$ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 0 $$ $$ \frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y}= -\frac{1}{\rho} \frac{\partial p}{\partial x} + \nu \left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right)$$ $$ \frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y}= -\frac{1}{\rho} \frac{\partial p}{\partial y} + \nu \left(\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} \right)$$

For steady flow we have,

$$ \frac{\partial u}{\partial t} = \frac{\partial v}{\partial t} = 0 $$

Similarly, because the flow is confined between two non-porous parallel plates, and we are interested in laminar flow solutions, we automatically deduce,

$$ u = u(y) \quad \text{or} \quad \frac{\partial u}{\partial x} = 0$$ $$ v = 0 $$

Reducing the general two-dimensional, incompressible, and constant property Navier-Stokes equations yields,

$$ \frac{\partial u}{\partial x} = 0 $$ $$ \frac{1}{\rho} \frac{\partial p}{\partial x} = \nu \frac{\partial^2 u}{\partial y^2} $$ $$ \frac{\partial p}{\partial y} = 0 $$

Focusing on the $x$-momentum equation and using the results of $\partial u/ \partial x = 0$ and $\partial p/ \partial y = 0$, we are left with a linear ordinary differential equation,

$$ \frac{1}{\rho} \frac{d p}{d x} = \nu \frac{d^2 u}{d y^2} $$

The important part here is that $dp/dx$ is a constant for the derivation of plane Poiseuille flow. This is because plane Poiseuille flow is concerned with fully developed laminar flow between two parallel plates. The plane Poiseuille flow has a generalized solution assuming the walls are $\pm W$ away from the centerline,

$$ u(y) = - \frac{W^2}{2\mu} \frac{dp}{dx} \left[1-\left(\frac{y}{W}\right)^2\right]$$

The fully devloped assumption is a major distinction that most textbooks fail to emphasize regarding plane Poiseuille flow or even Hagen-Poiseuille pipe flow. Below is a schematic for flow in a pipe, but the illustration looks the same for plane flow between parallel plates. Pipe flow and pressure profile.

Notice, only in the fully developed region does the profile $u(y)$ and the driving pressure gradient $dp/dx$ become indepedent of the $x$ location between the plates or inside the pipe. This is the basis of the symmetry arugument for $u$ and $dp/dx$. However, you seem to be hung up at the physical value of $p$ along $x$ instead of $dp/dx$. You just need to realize that the driving function for this particualr flow is not $p = p(x)$, but simply $dp/dx$. Lastly, a final comment is that this flow is resticted to fully develped laminar flow, and given a long enough distance, the Reynolds number will become high enough that transition to fully develped turbulent flow will take place. In which case, the Poiseuille flow solution or Hagen-Poiseuille are no longer applicable.

| cite | improve this answer | |
$\endgroup$
1
+100
$\begingroup$

I will offer two comments, noting that these answers partially rely on the material already presented by TRF:

  1. For incompressible flow, it is only the pressure gradient that matters, and pressure is only ever determined up to an arbitrary constant. In the full system of momentum and mass conservation equations (presented in the first answer), only the gradient of the pressure appears. This should be sufficient to explain why a pressure that varies in the streamwise direction does not break the proposed symmetry, as long as its gradient is constant. Perhaps the short answer to the question of why "we can assume translational symmetry when the pressure is not symmetric under translation" is that the pressure itself does not appear in the equations. Only its gradient does, and this is translationally invariant. Notice that we do not need to assume a constant pressure gradient from the outset: The assumption of a translationally invariant, one-dimensional velocity profile is sufficient to derive this result.
  2. As an additional comment on these symmetry "assumptions", it is perhaps helpful to understand that the original problem (incompressible Navier-Stokes flow between infinite parallel plates) is not fully formulated, and the question for its solution is thus ill-posed from a mathematical perspective. The issue is that no initial and not all boundary conditions are specified. One way to look at the way this particular problem is formulated is to rephrase it as follows: Find all steady solutions (if any) of the Navier-Stokes equations in the given geometry that can be characterized by a single velocity component parallel to the plates, varying only in the coordinate normal to these plates. It turns out that such solutions do indeed exist, and the answer to the question is a family of parabolic velocity profiles associated with a constant pressure gradient of an appropriate size. Of course, there are infinitely many other solutions that do not satisfy the given conditions, but nevertheless represent viable, physical flows. Examples are unsteady (laminar or turbulent) flows, and various kinds of steady developing flows. Thus, a second answer to the extended question of why "we can assume translational symmetry when the pressure is not symmetric under translation" might be that we don't need any specific justification. We just state these assumptions and see if we can find a solution that satisfies them. It turns out that this is indeed the case, which can serve to justify the assumption after the fact.

Update: In a comment to this answer, a question regarding the physical interpretation of the above came up. Here is my answer to this: It is important to realize that the "pressure" that appears in the incompressible Navier-Stokes equations is unphysical in important ways. In incompressible fluid mechanics, pressure is not a thermodynamic quantity. As I had said above, it is a Lagrange multiplier to ensure incompressibility. This means that in some sense the model of "incompressible flow" is unphysical in some important ways as well. For example, signals will travel at infinite speed in a theoretical incompressible fluid. And, in this model, the absolute value of pressure has no meaning.

The model of incompressible flow has value because it turns out that, as long as the speed of the flow is significantly below the speed of sound, the solutions it produces represent well-behaved perturbations of physical solutions to the full, compressible flow equations, or at least so we believe. It bears mentioning that the question of whether indeed this assumption is true is closely linked to the answer to one of the Millenium Problems of the Clay Institute, which has remained elusive so far.

Finally, yes, if we drop the assumption of incompressibility, then the solution of the compressible Navier-Stokes equations will not retain translational invariance of the pressure gradient or the velocity field.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I admire your mathematical description of fluid dynamics. I have a question regarding the statement "notice that we do not need to assume a constant pressure gradient". For this class of flows, namely laminar parallel flows between two plates, we deduce that $v = 0$ and $\partial u / \partial x = 0$. This implies that $u = u(y)$ and cannot be dependent on $x$. However, if our pressure gradient $dp/dx$ is not constant, then we obtain a solution such that $u = u(x,y)$, which violates are reduction of $u = u(y)$. Is it correct to suggest $dp/dx$ can be arbitrary for this class of flows? $\endgroup$ – TRF Dec 10 '16 at 15:08
  • $\begingroup$ If you look at the structure of the system of partial differential equations describing incompressible flow (known as the "Navier-Stokes equations" by which I refer to the system of equations consisting of the continuity and momentum equations), you can see that it has an interesting structure. Mathematically speaking the pressure that appears in the momentum equations acts as a Lagrange multiplier which is required to ensure that the divergence-free condition can be met. For example, if you take the divergence of the momentum equations, you obtain a Poisson equation for the pressure. Contd. $\endgroup$ – Pirx Dec 10 '16 at 16:34
  • $\begingroup$ In the case of one-dimensional flow we are discussing here, this Poisson equation becomes a Laplace equation, so for the 1-D case we get $\partial^2 p/\partial x^2=0$, meaning pressure must be a linear function, and its gradient is constant. So what I was saying is that the linearity of pressure follows from our assumptions on the velocity distribution, and does not need to be assumed. In the general case $u=u(x,y)$ the pressure gradient is still not arbitrary, since pressure needs to satisfy the Poisson equation I alluded to above. $\endgroup$ – Pirx Dec 10 '16 at 16:36
  • $\begingroup$ @Pirx Two points: Concerning your first sentence - the bounty was put on this question before TRF's answer (I didn't mean to put 'The current answers do not contain enough detail.') Secondly you say 'pressure is only ever determined up to an arbitrary constant.' Pressure has a well defined physical meaning (namely the average normal force per unit area on 3 mutually perpendicular surfaces) so how can it only be defined up to a constant? $\endgroup$ – Quantum spaghettification Dec 11 '16 at 6:12
  • $\begingroup$ Excellent question. I'll add to my answer. $\endgroup$ – Pirx Dec 11 '16 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.