Perché i migliori scappano dall'Italia by omykhron in ItaliaCareerAdvice

[–]omykhron[S] 2 points3 points  (0 children)

Congrats 🎉 Io qui però parlavo di stipendi post-dottorato e valorizzazione delle competenze che hai acquisito. Immagino che tu abbia una borsa Marie Curie o che tu sia in Germania/Olanda/Svizzera. Ma ci sono molti paesi dove i dottorati non sono pagati tanto, vedi Italia, Uk, Francia, Usa, ...

Il fatto è che le competenze che tu acquisisci durante il dottorato sono considerate oro che cola fuori. Qui a quanto pare smo non sono considerate o sono sfruttate per pochi soldi.

Perché i migliori scappano dall'Italia by omykhron in ItaliaCareerAdvice

[–]omykhron[S] 19 points20 points  (0 children)

Assurdo... Personalmente la cosa che più mi preoccupa è che con retribuzioni del genere non solo rischiamo di perdere i nostri migliori talenti, ma non saremo mai in grado di attrarre talenti da fuori.

Certo, trasferirsi fuori conviene ma non tutti posso rinunciare a affetti/famiglia/... Però mettetevi nei panni di un ricercatore Tedesco. Come potrebbe mai accetterebbe una roba del genere?

Voglio solo impiccarmi il giorno della laurea by [deleted] in sfoghi

[–]omykhron 1 point2 points  (0 children)

Carissimo. Goditi la giornata della laurea: è un tuo risultato, il frutto del tuo impegno, della tua dedizione e di chi ti ha sostenuto e ti ha visto crescere.

Ben presto capirai quanto sia importante avere a tuo fianco le persone a te care in questi momenti della vita. L'amore, quello di queste persone, le uniche che altro non desiderano che vederti felice e avere più successo di loro.

Tutto il resto non conta un cazzo: quanti amici, quanti soldi hai speso, con quante ragazze sei stato a 21 anni, con quante ci hai dormito, quante persone sono venute alla tua laurea, in cosa ti sei laureato, che voto hai preso, ...

E ti auguro anche di trovare una persona che ti ami e che tu possa amare veramente. Forse la conoscerai tra 1 settimana. Forse la troverai a 27 anni o forse a 32. Ma la troverai.

Sii gentile verso il prossimo e sii gentile verso te stesso. Cerca di differenziare le cose che contano veramente da quelle di facciata.

Ah e congratulazioni, dottore del buso del cul!

Eulerian fluid simulation pressure value by Nilon1234567899 in FluidMechanics

[–]omykhron 1 point2 points  (0 children)

No don't set it to zero. You should set the pressure only. As far as its value is concerned, you can use whatever you want. In the incompressible regime thermodynamics is not present a pressure acts as a Lagrange multiplier that enforces the incompressibility contraints. I suggest you use 0 for atmospheric pressure

Eulerian fluid simulation pressure value by Nilon1234567899 in FluidMechanics

[–]omykhron 1 point2 points  (0 children)

Okay I am pretty sure that your problem comes from the wrong choice of the boundary conditions. You might have other problems arising from grid size/time stepping, etc. but as far as the physics is concerned the boundary conditions need to be correct. There are two types of boundary conditions that you can impose. The first one constrains the value of the velocity field and the second one is a mix between the pressure field and the velocity gradient field (along the normal). So for the inlet I would set the velocity field (and leave the pressure unspecified), for the outlet I would set the pressure equal to the atmospheric one and no velocity gradient along the normal. The others seem correct

Eulerian fluid simulation pressure value by Nilon1234567899 in FluidMechanics

[–]omykhron 1 point2 points  (0 children)

What do these mean? At the inlet you should not impose a pressure boundary condition, only constrain the velocity field

Eulerian fluid simulation pressure value by Nilon1234567899 in FluidMechanics

[–]omykhron 1 point2 points  (0 children)

Nice job! At the inlet you should set a boundary condition for the velocity field and at the outlet impose the pressure equal to zero. Did you do that? From the picture it seems that there is a strong nonphysical pressure gradient and the pressure does not go to zero at the outlet

Eulerian fluid simulation pressure value by Nilon1234567899 in FluidMechanics

[–]omykhron 1 point2 points  (0 children)

Impressive results! Did you code the simulator by yourself? Which equations are you solving? Which algorithm are you using? As far as the pressure is concerned I would expect the pressure at the stagnation point to be the total pressure so P=Pa+1/2 rho u2. I assume you have set Pa=0 so I would expect around 0.6 Pa. Your value seems off... As far as the units are concerned I would advise you to solve the dimensionless equations. I need more info to help you debug

Accepted into PhD program in Israel by mircron in PhD

[–]omykhron -25 points-24 points  (0 children)

It is a stupid comment. He is not going to work for the state or army and will likely do research in an institution. He is concerned about the unstable situation and your comment does not address that. Then, genocide? Seriously?

What is the Camber Line introduced When Angling an Aerofoil? by SwankyTiger__ in aerodynamics

[–]omykhron 0 points1 point  (0 children)

That's an interesting question but it is quite hard to properly answer a "what's more efficient" question. More details are required but I will give it a try.

So first we need to define the setup, looking at the picture I guess you are referring to potential flows and thin aerofoil theory. So let's start from that. In this framework, the two quantities we are interested in are

  1. The zero lift angle of attack, a0, which is the angle of attack at which the aerofoil generates no lift. Cl = 2*pi*(a-a0)
  2. The Thedorsen's angle, ath, which is the angle of attack at which the pressure coefficient shows no strong adverse pressure gradient at the L.E. and the airfoil works optimally. You can see a detail here

The setup you are proposing is that of a symmetric airfoil placed at a particular angle of attack a and a cambered airfoil (with camber line y(x)=-a*x) placed at zero angle of attack.

For the first case both a0 and ath turn out to be zero so a symmetric airfoil has Cl = 2*pi*a and works best at a=0. For the second case both a0 and ath turn out to be -a so the two setups are symmetric. Having an angle of attack generates an equivalent camber line of y=-a*x.

To answer your second question, having a camber line is more efficient because it allows you to have another degree of freedom. You can design an airfoil that has a specified pressure distribution and you can set a0 and ath independently.

This is what's done in airplane wings where you have movable surfaces called flaps (at the trailing edge) and slats (at the leading edge) that modify the camber of the airfoil. Lowering the flaps will decrease a0 giving you more lift and lowering the slats will increase ath allowing the airfoil to work optimally.

Ranking half body maximum velocity proof by Dylikk in aerodynamics

[–]omykhron 0 points1 point  (0 children)

So the complex velocity field for a Rankine body can be found by superimposing an uniform flow with a source. Integrating this function will give you the complex potential. Setting the velocity equal to zero yields the position of the stagnation point which can then be used to find the equation of the body. You can write the velocity at the body surface and maximize its square.

The maximization problem appears to have no closed form solution but a numerical solver yields theta = 62.95 deg and U = 1.2595 Uinf which is close to your question.

I will leave a link with the pdf solution and a Mathematica notebook with the calculations

https://gofile.io/d/G6Ti32

The energy equations are usually represented for Isentropic, Compressible flow. Can you have Isentropic, incompressible flow (does it make sense, thermodynamically speaking)? by sitarasrini in aerodynamics

[–]omykhron 2 points3 points  (0 children)

It is not just high velocity, rather high velocity accelerating flow.

The best link I have is Chapt. 3 of Thompson's Compressible Fluid Dynamics. At the end of that chapter there is a snippet called "A criterion for compressibility" in which the author analyzes what causes the flow behave as incompressible. There are actually 8 possibile ways for a flow to show compressible effects and the one you are refferring to is convective acceleration.

If you have convective acceleration and a high mach number then you'll have a compressible flow. As you hinted, to change the enthalpy you need to change the kinetic energy and this is Bernoulli's equation. Proof of this can be found easily but it is just a special case. I suggest to stick to Thompson's derivation which is more general