So I am trying to do some further tests on a 2nd-order code Incompressible Navier Stokes equations, by studying transition to turbulence in a Poiseuille flow. Specifically, I'm interested to see whether the code is energy-conserving by modelling a 3D TS wave in a Poiseuille flow. Note that I have tested the code for a 2d inviscid Taylor vortex and the code holds. Now, I have calculated the perturbation velocities via the Orr-Sommerfeld/Squire equations, and I get the Eigenspectrum and corresponding Eigenfunctions (velocities/wall-normal vorticity) for the following Poiseuille flow:

$U^0_1 = 1-y^2,\,\, -1 \le y \le 1$

With a Reynolds number based on the centreline velocity and the channel's half-height


From here, the driving pressure gradient is just:

$\Pi_1 = -\frac{2}{Re}$

For the linear perturbation I choose a streamwise and spanwise wavenumbers of $\alpha = 1.12, \beta= 2.1$, respectively. For the eigenvectors, the most unstable mode has been chosen (that where Real(Lambda) is the closest to zero). All the results are the following:


Eigenvectors for <span class=$\lambda \approx -0.07 - 0.36i$">

I have imposed the velocity/vorticity perturbations into my code, without success (meaning not getting the transition), so clearly there's something I'm not quite getting right. Now, my questions are:

  1. Orr-Sommerfeld Modelling: Are my results obtained (meaning perturbation velocities/vorticity) reasonable? Are these correct?
  2. Equations to solve for temporal transition modelling: In my opinion, the equations to be solved when studying spatio-temporal transitions are the incompressible Navier-Stokes equations (INSE) for the perturbation velocities, that is:

    $ u_{i,t} +u_j u_{i,j} + u_j U^0_{i,j} + U^0_j u_{i,j} = -p_{,i}+\frac{1}{Re}u_{i,kk}+\Pi_i$

    Which results from the splitting of the instantaneous fields into base and perturbation fields, $\widetilde{\phi_i} = \phi^0_i + \phi_i$ in the INS, and then substracting the INSE of the base flow. Now my questions are: Is the previous equation correct? I'm having doubts about the pressure gradient $\Pi_i$. Second: Should be these the equations I should be solving for simulating temporal transitions? Or, Should I go just with the classic INSE for the poiseuille flow and superposed fluctuations?

  • $\begingroup$ You can be fairly certain you have a typo in your perturbation equation as currently written, since you have a bare $j$ in your $-p_{,j}$. Dummy indices can only appear in pairs. $\endgroup$ – origimbo Mar 27 at 12:10
  • $\begingroup$ Great. The only other notational point I'd make is that you've been slightly inconsistent in writing $U^0$ and $\Pi$ as scalars versus vectors, which may explain why you're concerned about your pressure gradient term (which you can eat by defining $p' = p-\frac{2x}{Re}$ ). $\endgroup$ – origimbo Mar 27 at 14:22
  • $\begingroup$ @origimbo True! But I think my doubt is beyond such details. That is: when I make the difference between perturbation+baseflow and base-flow INSE, does the $\Pi$ term remain in the final perturbation-only INSE? I mean, should the base-flow INSE include the bodyforce $\Pi$? In that case $\Pi$ dissapears from the perturbation-only INSE... BTW, I fixed the inconsistencies, I think. $\endgroup$ – Kbzon Mar 27 at 14:32
  • $\begingroup$ Your pressure gradient isn't precisely a body force. It's the pressure gradient required to maintain incompressibility and a steady state for Poiseuille (i.e. laminar) flow in the state you're perturbing around. Either something must do the work on the fluid, or else the fluid will decelerate as momentum is lost though viscous shear. $\endgroup$ – origimbo Mar 27 at 14:37
  • $\begingroup$ @origimbo Again, I agree that in a mathematical sense the pressure is a lagrange multiplier, but when I'm going to do a pressure gradient-driven Poiseuille flow (temporal boundary layer), numerically, you impose the 'steady' (calculated using basic formulae) pressure gradient as a bodyforce under the assumption that you are in an accelerating frame of reference. Then, the 'transient' pressure becomes the lagrange multiplier, if you may. $\endgroup$ – Kbzon Mar 27 at 14:50

Your Answer

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

Browse other questions tagged or ask your own question.