4
$\begingroup$

I have a question about non-dimensionalization of the incompressible Navier-Stokes (NS) equations. My understanding is that the purpose of non-dimensionalization is to "collapse" solutions onto one curve so that the solution space can be explored with fewer parameters. It can also give insight into the physics. Consider the dimensional incompressible NS equation:

$\partial_t u_i + u_j \partial_j u_i = -\rho^{-1}\partial_i p + \nu \partial_{jj} u_i, \qquad \partial_j u_j = 0.$

Non-dimensionalizing using the scales $ [x,y,z] = L, [u] = U, [p] = \rho U^2$. Plus

$[t] = L/U, \qquad \qquad \text{convective time}$

$[t] = L^2/\nu, \qquad \qquad \text{diffusive time}$

Results in two different equations:

$\partial_t u_i + u_j \partial_j u_i = -\partial_i p + Re^{-1} \partial_{jj} u_i, \qquad \qquad \qquad \text{using convective time scale}$

$Re^{-1}\partial_t u_i + u_j \partial_j u_i = - \partial_i p + Re^{-1}\partial_{jj} u_i, \qquad \qquad \text{using diffusive time scale}$

Here $Re = \frac{UL}{\nu}$ is the Reynolds number.

Questions

1) Why is it that these two equations are not the same given the fact that $Re$ is defined exactly the same between the two equations?

2) Can the solutions from each equation be related somehow?

3) Is the span of both parameter spaces the same for both equations?

I'm looking for a methodical view of answering these questions since I'm interested in how understanding the answers extend to a more complicated problem.

I appreciate any help!

$\endgroup$
1
$\begingroup$

The two equations only differ by a trivial rescaling of the time coordinate, and are therefore equivalent. If $u(x,t)$ is a solution of the first equation, then $u(x,{\it Re}\, t)$ is a solution of the second. Re is physically relevant because it governs the relative importance of the advective and dissipative terms. Multiplying $\partial_t u$ by a constant is just a rescaling, and has no physical significance.

$\endgroup$
  • $\begingroup$ I agree with everything you've said, but I'd like to ask about your 2nd and last sentences. I think your 2nd sentence is not very general. Suppose more terms are in the NS equation. I'm not clear how your notation of comparing the solutions $u(x,t)$ and $u(x,Re t)$ would work with comparing, e.g., the advection term and a electromagnetic Lorentz force. The last sentence you wrote does seem to be a more general approach, but how can you re-scale a single term in an equation while maintaining the same parameters (Re, e.g.)? $\endgroup$ – Charles Jun 3 '16 at 6:31
  • $\begingroup$ @Charlie If there is only one $\partial_t$ term in the the PDE, then multiplying that terms by a constant just changes the unit of time, and therefore has no physical significance. $\endgroup$ – Thomas Jun 3 '16 at 12:46
  • $\begingroup$ If you add other terms to the Navier-Stokes equation, related to heat flow, MHD terms, etc, then of course additional dimensionless ratios appear (wikipedia has a whole table, en.wikipedia.org/wiki/Dimensionless_numbers_in_fluid_mechanics ) but that has nothing to do with your ability to rescale t. $\endgroup$ – Thomas Jun 3 '16 at 12:51
  • $\begingroup$ By rescaling the time coordinate you will rescale the 'convective' time scale to a 'diffusive' time scale so while the equations are mathematically equivalent, physically they are not equivalent. I tried explaining this in my answer, but perhaps this wasn't clear. @Charlie - an example of how to rescale a single term is shown in my answer where the pressure scale was rescaled. $\endgroup$ – nluigi Jun 3 '16 at 15:04
  • $\begingroup$ How can changing the unit of time from [t]=1 second to [t]=17 min change anything about the solution? $\endgroup$ – Thomas Jun 3 '16 at 15:39
-1
$\begingroup$

... the purpose of non-dimensionalization is to "collapse" solutions onto one curve so that the solution space can be explored with fewer parameters.

That's one 'purpose' of non-dimensionalization; two others are the identification of the characteristic length, velocity, pressure, etc. scales and the analysis of the equation under different regimes, in this case $\mathrm{Re}\ll1$ (viscosity dominated) and $\mathrm{Re}\gg1$ (convection dominated).

Characteristic length scales are the scales of the system which uniquely characterize the dynamics such that all terms in the equations (including initial/boundary conditions) are $O(1)$ or smaller. You have chosen a set of scales with which you non-dimensionalize the equation, but are they characteristic scales? Well let's put it to the test.

First, consider the convectively dominated regime for which you determine:

$$\partial_t u_i + u_j \partial_j u_i = -\partial_i p + \mathrm{Re}^{-1} \partial_{jj} u_i, \qquad \qquad \qquad \text{using convective time scale}$$

In this regime, $\mathrm{Re}\gg1$ and assuming the scaling was done correctly, i.e. all non-dimensional variables are $O(1)$, then we see that all terms are $O(1)$ or smaller (since $O(\mathrm{Re}^{-1})\ll O(1)$). This indicates that the scaling was properly done and that the scales are the characteristic scales in this regime. Since viscous term is neglibly small compared to the convective term, we are allowed to completely disregard it. Note that the pressure term is of the same order as the convective term; the pressure gradient must always be of the same order as the dominant term in the equation in order to balance the terms.

Let's now consider the viscous regime where $\mathrm{Re}\ll1$, you find (after dividing by $\mathrm{Re}^{-1}$):

$$\partial_t u_i + \mathrm{Re} u_j \partial_j u_i = - \mathrm{Re} \partial_i p + \partial_{jj} u_i, \qquad \qquad \text{using diffusive time scale}$$

We see here that again all terms are $O(1)$ or smaller (since $O(\mathrm{Re})\ll O(1)$). In this regime, we can neglect the convective terms as the viscous terms are dominant. However, apparantly the pressure gradient is also negligible which should in fact be of the same order as the viscous terms. This indicates that the proposed scaling in this regime is not correct.

To resolve this problem we require a rescaling of the pressure scale. Let's rescale $[p]^* = \mathrm{Re}^{\alpha}[p]$ where $\alpha$ is some constant to be determined. Substituting into the equations yields:

$$\partial_t u_i + \mathrm{Re} u_j \partial_j u_i = - \mathrm{Re}^{1+\alpha} \partial_i p^* + \partial_{jj} u_i, \qquad \qquad \text{using diffusive time scale}$$

For the pressure gradient to be of the same order as the viscous terms, we require $1+\alpha=0$ or $\alpha=-1$. We then get:

$$\partial_t u_i + \mathrm{Re} u_j \partial_j u_i = - \partial_i p^* + \partial_{jj} u_i, \qquad \qquad \text{using diffusive time scale}$$

and the pressure scale was rescaled to $[p]^* = \mathrm{Re}^{-1}[p] = \frac{\mu U}{L}$ which is clearly a viscous pressure scale as evident from the presence of the viscosity $\mu$. This is in contrast with the original pressure scale $[p] = \rho U^2$ which clearly doesn't contain a viscosity and was only appropriate in the convective regime $\mathrm{Re}\gg1$. We can therefor refer to it as a convective pressure scale.

To conclude with answering your questions:

  1. Because the equations are valid only for two different regimes as set by your choice of scales, $\mathrm{Re}\ll1$ and $\mathrm{Re}\gg1$.
  2. The only 'related' solution will be found for $\mathrm{Re}=1$, otherwise the solutions will be completely different. In fact analytical solutions are generally only possible for $\mathrm{Re}\ll1$ because the equations become linear. For $\mathrm{Re}\gg1$, the equations are highly non-linear making it only numerically possible to obtain solutions.
  3. No, see answer to 2.
$\endgroup$
  • $\begingroup$ I appreciate the semi-methodical approach (using an equation $1+\alpha = 0$ to determine an appropriate scaling). However, this statement: "... the pressure gradient must always be of the same order as the dominant term in the equation in order to balance the terms." is not, in general, true. A counter example is Stokes' first problem, there is no pressure gradient. The other problem that I have is that you seem to suggest that a solution is incorrect if the wrong scales are used. This does not sit well with me. As @Thomas pointed out, both solutions can be related for this simple case. $\endgroup$ – Charles Jun 3 '16 at 16:01
  • $\begingroup$ @Charlie - If the pressure gradient is what is driving the flow (be it viscous or inertial), then always the pressure gradient necessarily needs to be of the same order as the viscous or inertial terms respectively. Clearly, this doesn't apply to Stokes' first problem. Regardless of the choice of scales, the two equations will give the same solutions in dimensional terms since it is just a rescaling of the time coordinate. However, if your aim is to determine the scales which dominate then clearly you will not get the correct solution. $\endgroup$ – nluigi Aug 10 '16 at 15:08

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.