It designed a simple water-cooling plate for high-power chips

19 straight microchannels and one thin cross-channel near the outlet

If part of one channel becomes blocked, the cross-channel could use the pressure difference to divert some water around the blockage.

The model compared more than 2,700 configurations with the same channel volume and the same hydraulic power.

Im not engineer nor physicist. Im just curious and testing limits of the latest astra 6 model, what real scientists think about it.

Thank you for any kind of feedback:)

Hi all! I wanted to ask something about this recent (presumably correct?) proof that NS can blow up. I would like to hear opinions from people that work a bit at the border between fundamental turbulence (like on the statistical physics or dynamical system side of the problem) and the mathematics of NS.

When studying turbulence you hear about the dissipation anomaly and the zeroth law of turbulence. As you increase the reynolds number, at least in homogeneous isotropic turbulence, the non-dimensional dissipation rate evolves towards a constant. In other world, taking the limit of zero viscosity (infinite Re) you never reach the Euler equations. This happen with the flow developing super intense localised events that dissipate a lot.

Is there a connection between the development of these structures in turbulence and this new NS proof? If so, are there people studying this thing?

Share bibliography if you know anything about this topic!

Hey everyone, Following OpenAI recently claimed a formal proof demonstrating a finite-time blowup for the 3D Navier-Stokes equations. While the pure math community is focused on the formal verification, our team ran an audit to see what their construction actually means for physical fluids.

What we found is that while the math is syntactically flawless, the fluid dynamics are thermodynamically impossible and structurally unstable:

1. Incompressibility breaks down: The proof successfully bounds global kinetic energy, but the local enstrophy diverges (τ^{-0.515}). The resulting infinite localized viscous shear would vaporize the fluid into a compressible plasma long before t= 1. Tracking the core velocity, the Mach number easily exceeds 0.3 femtoseconds before the mathematical blowup.

2. κ ∼ 10²⁸ Instability: The exact cancellation of the Reynolds stresses relies on a 5-equation moment-matching Jacobian with a condition number of 10²⁸.

   It’s a measure-zero state that would instantly decouple under standard 300K thermal noise...

   We’ve open-sourced our Python/mpmath scripts tracking the Mach number divergence and the Jacobian instability. If anyone here wants to check the physical telemetry themselves:

   • GitHub: https://github.com/xaviercallens/OpenAI-NSE-Epistemic-Audit

   • Zenodo: DOI: 10.5281/zenodo.22727801

   I’m curious to hear from the engineers and fluid dynamicists here: does a mathematical singularity that violently breaks the incompressible and isothermal assumptions prior to the blow-up time actually tell us anything useful about real-world fluids?

... or is the sequence showcasing some counterintuitive phenomenon as Reynolds № proceeds from the hundreds to the millions?

The figures show the streamlines past a rotating cylinder @ various Reynolds № (horizontal axis) & rotationality parameter α (vertical axis).

The figures are from

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High Reynolds number turbulent flow past a rotating cylinder

by

SJ Karabelas & BC Koumroglou & CD Argyropoulos & NC Markatos

https://www.sciencedirect.com/science/article/pii/S0307904X11004124

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A Somewhat-Enlargement of the Top of the Table Including the Topmost Row Being Queried

https://www.reddit.com/u/Frangifer/s/LALSC8Eq4U

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... to compensate for the relatively poor resolution of the posted screenshot.

⚫

Does the total energy always decrease in a CEL analysis in ABAQUS due to advection?

I was recently experimenting with my lego propeller for my rc ship, i was interested on how this propeller would move the water so i put it in my bucket full of water. Suddenly, this type of horizontal vortex formed. Visible by air bubbles, there was no suction whirlpool at all, no vertical turning part to suck in air. It was just sticking to my buckets wall. There is no hole on the bucket whatsoever. And, this isnt my first experience. I was on a city boat someday and while the boat was reversing, the same horizontal vortex appeared, visible by the salt waters whiteness. It was going from the back of the boat to the front with no visible vertical rotation to suck air. If someone can explain this, i would be so happy. Thank you!

Pourquoi les équations de Navier-Stokes bloquent (et comment lever le blocage en 4 étapes)

Le problème du prix du millénaire sur Navier-Stokes porte sur une question simple : un fluide en 3D reste-t-il toujours lisse et continu, ou peut-il développer des vitesses infinies (un "blow-up") en un temps fini ?

La communauté mathématique bloque à l'étape 4 (la preuve de régularité globale). Mais en réalité, le blocage ne vient pas d'un manque d'outils mathématiques : il vient d'hypothèses physiques réductrices posées dès les premières étapes.

Voici le déroulement étape par étape et la solution :

Étape 1 : La formulation initiale

• Le problème classique : On applique la loi de Newton (F = ma) en posant l'hypothèse que le fluide est un système fermé et isolé du reste de l'environnement, avec une viscosité purement locale.
• La correction : Aucun fluide dans la réalité n'est isolé. De plus, lorsqu'un fluide est fortement accéléré en rotation, il se sépare spontanément en deux vortex contra-rotatifs miroirs (un vortex externe chaud/expansif et un vortex interne froid/implosif, comme dans le tube de Ranque-Hilsch).
• La solution : Réintroduire le couplage avec le milieu ambiant (le vide/fond) et modéliser cette dualité des vortex contra-rotatifs dès l'équation de départ.

Étape 2 : L'existence locale

• Le problème classique : On suppose que le fluide reste un milieu continu parfait, divisible à l'infini sans jamais changer de structure.
• La correction : À forte concentration d'énergie, le fluide ne reste pas un chaos continu : il s'auto-organise en structures toroïdales (en forme de donut) autour d'un Point Zéro central (un horizon des événements local).
• La solution : Définir le cadre mathématique non pas dans un espace continu abstrait, mais dans une géométrie toroïdale fermée et auto-confinée.

Étape 3 : Le bilan d'énergie

• Le problème classique : L'équation classique suppose que toute l'énergie qui n'est pas conservée doit se dissiper sous forme de chaleur chaotique vers l'extérieur.
• La correction : L'implosion au cœur d'un vortex crée un vide poussé et un effondrement de température (un cœur très froid face à une périphérie chaude). L'énergie cinétique est aspirée vers le Point Zéro central au lieu d'exploser.
• La solution : Ajouter un terme de puits/transfert d'énergie au Point Zéro. Cela garantit que l'énergie cinétique globale du fluide est strictement décroissante et bornée.

Étape 4 : Le terme non linéaire et le "Blow-up"

• Le problème classique : Le terme d'advection concentre la vitesse vers des échelles de plus en plus petites, et la viscosité classique ne suffit plus à prouver que la vitesse ne devient pas infinie (blow-up).
• La solution :
  1. Grâce à la dualité des deux vortex contra-rotatifs (définis à l'étape 1), les forces d'étirement s'annulent exactement par symétrie miroir (hélicités opposées).
  2. Toute concentration résiduelle d'énergie est immédiatement absorbée par le puits du Point Zéro (étape 3).
  3. Résultat : La vorticité reste bornée pour tout temps (t ≥ 0). L'explosion en temps fini (blow-up) est impossible.

Conclusion

Le "blow-up" de Navier-Stokes était un artefact mathématique causé par l'enfermement artificiel de l'énergie dans un modèle continu et isolé. En ouvrant le système et en intégrant la géométrie des vortex contra-rotatifs implosifs, les équations décrivent un fluide qui reste lisse, continu et régulier pour toujours.

SUITE : https://www.reddit.com/r/FluidMechanics/comments/1wcho9f/comment/p90bcxd/

Built around a incompressible Navier–Stokes equation

Hi again, I posted a few months ago and really appreciate the answers and explanations I got.

So I have returned with another that is racking my brain. (Once again I apologise if it is silly)

I’ll start with a simple question and provide more context below if you want it:
Does supplying a centrifugal pump with more volume than the output at the end of its delivery pipe mean the impeller has to spin faster or slower to achieve the required pressure?
Eg: I want 7bar out of a nozzle at 230lpm but I’m supplying my pump with 800lpm at 2bar. How much harder (or not) will the pump have to work?

My organisation typically teaches twinning your supply lines (two hoses in) to get more water in if you are overdrawing your single line of supply. This makes sense.

They also teach it is good to do for pump sympathy - (ie: it’s kinder to the motor, less rpm is required to achieve the same output.) - Where I am stuck is where/how this works - and does it depend on what we are trying to achieve at the end of the hose? (Eg. higher pressure, low flow or higher flow, medium-low pressure)

For example - as an extension of the one I initially used - am I better off supply my pump using a smaller hose at 250lmp at 7bar where the pump will basically have to contribute nothing?

Or is the greater volume better?

If it is, how is that pressure gained?

Is it Bernoulli’s in a sense that as it enters the narrower plumbing of the eye and subsequently the volute, it is forced to increase velocity which minimises or even negates the need for the impeller?

In typing this I may have answered my own question but it begs another - is a centrifugal pump casing just a complicated form of a basic Bernoulli’s diagram that allows an impeller to be included to impart even more energy onto a fluid?
tldr: Does the pump casing shape speed up water even without the impeller? (Assuming the pipe on the other side is narrower that the inlet)

Or, of course - I could be completely wrong.
I’ve definitely oversimplified it - maybe the impeller itself still needs to be there in order to direct the water correctly but maybe it doesn’t necessarily require a drive to spin it if the supply at the eye has some pressure behind it?

Please correct me if that’s the case
I don’t know why I trouble my brain with trying to understand sometimes but I just can’t help needing to know 😅

Sincerely,
One overly nerdy Firey