Talk:Euler equations
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I would assume that the first equation expresses the conservation of mass and is the equation of continuity, the second one is the conservation of momentum and is the true Euler equation. The third is the conservation of enthalpy, and is the enthalpy or energy equation. None of them incorporates outer force field effects, wich would be nice tough.
should I put stuff about Rankine-Hugoniot conditions here or under shock waves?
Mo ena ene test lor eulers la. Ki pou fou????
Is it possible for someone to define all the variables in this article? Rtdrury 10:59, 19 December 2005 (UTC)
As an answer to : "in particular, it is not intuitively clear why this equation is correct and :
is incorrect" : the reason is simple enough, Newton's second law states Force=mass*(derivative of u) not Force=derivative of (mass*u), where u is the velocity, therefore it is clear that the formula given above cannot be correct if the mass density is not constant (like in the flow of a compressible medium). This is a common misconception of Newton law, which for example leads to incorrect results for the eqution of motion of rockets. Gringo.ch 10:44, 3 October 2006 (UTC)
Incorrect! If you read Principia, Newton's law states that F = dp/dt, where p is the momentum. This must be the correct form because it can be shown that it is equivalent to both Lagrangian and Hamiltonian dynamics, for generalized coordinate q and its canonical momentum p. --GringoNegro
Yes, you are right, sorry I wasn't clear enough. My point was that, when you have systems where the mass is not constant in time (say, a droplet falling while losing mass to evaporation) then you have to be a bit careful on how to apply Newton's 2nd law. In that example, blindly writing F=dp/dt => mg=d/dt(mv)=dm/dt v+ m dv/dt would yield the acceleration of the droplet as a=dv/dt=g-v/m*dm/dt which is not the correct answer a=g, whereas just writing F=m(t)*dv/dt gives you directly the right solution (isotropic evaporation assumed here, no air friction). To get the right answer with the former approach one should account also for the evaporated material. Gringo.ch 17:01, 28 February 2007 (UTC)
It doesn't give the right answer because, as you say, there are other quantities that must be accounted for as forces on the left hand side--that's one of the main difficulties of Newtonian mechanics: expressing all the forces acting on a body. That's also why Lagrangian dynamics is so appealing. As we both agree that F = dp/dt (as long as F is the correct F) is the correct equation, we still haven't answered why "it is not intuitively clear why this equation is correct and : is incorrect" --GringoNegro
What about this: we are following a mass element dm in the flow, which has a volume dV(t) which depends on time. The density is then r=dm/dV, which changes in time, but dm is constant. Then we have that F=dp/dt. The force per volume is -grad p, hence F = -grad p *dV(t)=d/dt(dm*v(t))=dm*dv/dt, which can be written as -grad p= r* dv/dt. Transformation from the streaming system to the laboratory system (d/dt goes into d/dt + v grad ) delivers the Euler equation. So the confusion seems to arise in the step going from the force F to the force per volume, right? Gringo.ch 11:27, 1 March 2007 (UTC)
While it could be argued that the non-conservation form of the Euler equations is less intuitive, it remains the standard form used in analytic solution of the Euler equations. If there are no objections, I will do some rewording to place the conservation and non-conservation forms of the equations on a more equal footing. On another note, could someone explain "Although the Euler equations formally reduce to potential flow in the limit of vanishing Mach number, this is not helpful in practice, essentially because the approximation of incompressibility is almost invariably very close."? It seems to me that if the approximation of incompressibility is very close, an assumption of incompressibility would be a very good one to make. Cheers. Chrisjohnson 22:24, 16 March 2007 (UTC)
- Should the note about the other equation not being true dropped? Also, the appended comment of both being "correct"? (what does that exactly mean? I can see both are correct only if the fluid is incompressible) --Daniel (talk) 12:51, 27 April 2008 (UTC)
I would like to object to the remark about the use of the ideal gas law as p = ρ(γ − 1)e which is only true for caloric perfect gases and relatively cold mixtures of caloric perfect gases (i.e. gas mixtures around 300K). In fact, p = ρ(γ − 1)e is only true if e = cvT, which is only relatively true for caloric perfect gases, and nR = Cp − Cv which is not true for mixtures of ideal gases that are allowed to react with each other. While a mixture of ideal gases is not ideal it is still possible to use the ideal gas law in it's original formulation if no other assumptions are made.
There are thus two problems with the rewriting of the ideal gas law as p = ρ(γ − 1)e:
1) cv is a function of temperature and is defined as 
. If cv is assumed to be constant the equation e = cvT is valid, but only then.
2) For a mixture of ideal gases the number of moles n changes with temperaure and pressure. Since Cp and Cv are defined as differentials of the internal energy nR = Cp − Cv will not always be true. It is, however, still possible to use the ideal gas law in it's original formulation for such a gas.
While it's a good approximation for many gases it is unnecessary to refer to a version of the ideal gas law that makes assumptions that are not always true. The best choice would be to use pV = nRT since this equation is always valid, even for non-ideal reacting gas-mixtures where n changes with temperature and pressure. (I know that the ideal gas law isn't correct in itself but if one wants to use the ideal gas law, one should use it correctly or have to make the assumptions, and check their validity, themself.)
Per Öberg - Nov 15 2007 (With minor changes Nov 21)
—Preceding unsigned comment added by 130.236.50.246 (talk) 09:23, 15 November 2007 (UTC)
