Talk:Euler equations (fluid dynamics)

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Old discussions and comments

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)[reply]

As an answer to : "in particular, it is not intuitively clear why this equation is correct and : $\left(\partial /{\partial t}+{\mathbf {u} }\cdot \nabla \right)(\rho {\mathbf {u} })+\nabla p=0$ 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)[reply]

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)[reply]

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 : $\left(\partial /{\partial t}+{\mathbf {u} }\cdot \nabla \right)(\rho {\mathbf {u} })+\nabla p=0$ 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)[reply]

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)[reply]

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)[reply]

I would like to object to the remark about the use of the ideal gas law as $p=\rho (\gamma -1)e$ which is only true for caloric perfect gases and relatively cold mixtures of caloric perfect gases (i.e. gas mixtures around 300K). In fact, $p=\rho (\gamma -1)e$ is only true if $e=c_{v}T$ , which is only relatively true for caloric perfect gases, and $nR=C_{p}-C_{v}$ 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 $p=\rho (\gamma -1)e$ :

1) $c_{v}$ is a function of temperature and is defined as $\left(\partial e/\partial T\right)_{p}$ . If $c_{v}$ is assumed to be constant the equation $e=c_{v}T$ is valid, but only then.

2) For a mixture of ideal gases the number of moles $n$ changes with temperaure and pressure. Since $C_{p}$ and $C_{v}$ are defined as differentials of the internal energy $nR=C_{p}-C_{v}$ 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)[reply]

Stiff equation of state

The article says "The well-known Bernoulli's equation can be derived by integrating Euler's equation along a streamline, under the assumption of constant density and a sufficiently stiff equation of state."

Would it be correct to say that an assumption of sufficiently stiff equation of state is equivalent to saying "sufficiently incompressible"?

If so, I would suggest that we say

Integration of Euler's equation for incompressible flow, (specifically an assumption of constant density and a sufficiently stiff equation of state), yields the well-known Bernoulli's equation.

--Hroðulf (or Hrothulf) (Talk) 14:51, 23 March 2010 (UTC)[reply]

"Sufficiently stiff equation of state" is not equivalent to "sufficiently incompressible". Neither depends on the other. Constant density, however, already implies "suffuciently incompressible", indeed it means "not compressible at all". So I would agree with substituting "incompressible flow" for "constant density", but the content of the rest of the sentence should not be changed. Cs32en Talk to me 18:45, 23 March 2010 (UTC)[reply]

Thanks. You corrected a couple of my misunderstandings.

Could you explain the precise meaning of "stiffness of the equation of state"? My only recollection of the term stiff equations is from the field of numerical analysis, and the top hits for stiff equations of state in Google and Google Books are advanced pieces in astrophysics and nuclear physics (interesting but unusual fluids.)

Landau and Lifshitz (Vol 6 Ch 1) integrate Euler's momentum equation to get forms of Bernoulli's equation (twice, once for constant density as does Wikipedia, and once for the general case.) If I understand the calculation correctly, (and there is every chance I don't), they do so by examining steady flow, and by constraining thermal conductivity and viscosity to be unimportant. Is the formulation in the article a way of saying the same thing, or another way of getting to the adiabatic and isentropic constraints that L&L use?

--Hroðulf (or Hrothulf) (Talk) 16:39, 24 March 2010 (UTC)[reply]

the gravity term (!)

I think there should a gravity term rho * g be included in the momentum equation since dm * Dv/Dt = dF = -dP*dA + dm*g ... rho * Dv/Dt = -grad(P) + rho * g. —Preceding unsigned comment added by 193.190.253.144 (talk) 22:33, 5 October 2010 (UTC)[reply]

Body forces, of which gravity is an important one, are not included in the presented form of the Euler equations. -- Crowsnest (talk) 11:10, 6 October 2010 (UTC)[reply]

Removed comment

I removed these comments from the conservation and component section:

in particular, it is not intuitively clear why this equation is correct and $\left(\partial /{\partial t}+{\mathbf {u} }\cdot \nabla \right)(\rho {\mathbf {u} })+\nabla p=0$ is incorrect) -- Both forms are "correct"

FR - This note represents the author's opinion of what he/she believes is more intuitive. As opposed the this author, I would say that it is the non-conservation form that is more directly linked to the classic representation of Newton's second law F = m du/dt when a fixed mass is followed in time. For Fluids the non-conservation expression is just: mass/volume du/dt = F/Volume, where du/dt is the derivative of the velocity of the fluid particles as they pass through a certain point in the flow field (= substantial derivative, typically written as Du/Dt to remind the reader that an Eulerian frame of reference is used). The non-conservation form maps directly to the interpretation of Newton's 2nd law for fixed mass particles. What is more, the conservation form can be easily reworked from the non-conservation form by means of the continuity equation. That is using additional information (conservation of mass) thereby obscuring Newton's 2nd law. Of course, the conservation form can also be directly derived from principles of momentum changes, representing an alternative interpretation of Newton's 2nd law. This is then just a limiting case of the Reynolds Transport Theorem for which the control volume size is shrunk to an infinitesimal one. But, whether this shows the "direct connection" to Newton's 2nd law can be argued as I have done. Therefore I would suggest to delete the opinions from this page. Best Regards, FR

PAR (talk) 06:27, 22 February 2012 (UTC)[reply]

What about external forces?

What about external forces, such as gravity? Are they not supposed to be included in these equations? Or maybe this fluid isn't affected by gravity??

At least, the section about the Bernoulli equation for unsteady potential flow states that the equation is obtained by integrating the momentum equations of the Euler equations, and that equation does contain gravity. —Kri (talk) 16:29, 9 March 2012 (UTC)[reply]

External forces can off course be added to the momentum equations, and might also appear in the energy equation if they perform work. -- Crowsnest (talk) 14:01, 10 March 2012 (UTC)[reply]

Well, you say that they can off course be added, but they are not present in the equations in their current form in the article, which has to be the most important thing. Why are they not there? They are obviously supposed to be there. Do you agree with me? —Kri (talk) 10:55, 12 March 2012 (UTC)[reply]

Sure I agree with you. There are however not many textbooks of renowned fluid-dynamics scientists including them in the equations. Emphasis is often on how to present/derive them in a conservative form. And further, in aerodynamics (where they are mostly used) gravity is often of minor importance. So a reliable source including the external (body) forces, like gravity, has to be found. -- Crowsnest (talk) 18:04, 12 March 2012 (UTC)[reply]

What do you mean by that a reliable source including the external (body) forces has to be found? —Kri (talk) 23:15, 15 March 2012 (UTC)[reply]

Ah, never mind. I realized that you where talking about a literary source and not a source that should be included in the equations. —Kri (talk) 23:17, 15 March 2012 (UTC)[reply]

Merger of Streamline curvature theorem

There is hardly anything to find on "streamline curvature theorem": 4 papers in Japanese scientific journals. For a fundamental theorem in fluid dynamics this seems very unlikely, one would expect tons of search results. The same is the case for the term "Euler's normal equation".

What can be found is several references to the Euler equations in "streamline coordinates" or "path coordinates". These have a component normal to the streamline, the subject of the article proposed to merge here, and along the streamline one recovers Bernoulli's equation.

See:

Graebel, W. P. (2001). Engineering Fluid Mechanics. Taylor & Francis. pp. 168–170. ISBN 9781560327110.
Paterson, Andrew Robert (1983). A First Course in Fluid Dynamics. Cambridge University Press. pp. 183–184. ISBN 9780521274241.
Fay, James A. (1994). Introduction to Fluid Mechanics. MIT Press. pp. 150–152. ISBN 9780262061650.{{cite book}}: CS1 maint: postscript (link) This book is the reference in the article, and talks about "Euler's equation in streamline coordinates".

See also this Google search for the combination "Euler", "equations", "streamline" & "coordinates".

So my suggestion would be to merge Streamline curvature theorem into a section in this article, e.g. "Steady flow in streamline coordinates". -- Crowsnest (talk) 14:15, 11 May 2012 (UTC)[reply]

A section ”Steady flow in streamline coordinates" is added according to the proposal. If no problems in this section, I will merge Streamline curvature theorem into this section. くま兄やん (talk) 11:28, 17 May 2012 (UTC)[reply]

That looks really very good, くま兄やん, to my opinion. I changed Streamline curvature theorem into a redirect to this article in order to finish the merger. -- Crowsnest (talk) 12:54, 17 May 2012 (UTC)[reply]

Thank you, Crowsnest, for merging. くま兄やん (talk) 13:09, 17 May 2012 (UTC)[reply]

Minus signs

Energy equation in convective form is given as

${De \over Dt}={\frac {p}{\rho }}\nabla \cdot \mathbf {u}$

Should not it be

${De \over Dt}=-{\frac {p}{\rho }}\nabla \cdot \mathbf {u}$ ?

See, e.g., http://www.mathematik.uni-dortmund.de/~kuzmin/afc2.pdf , beginning of section 2, which has $\nabla (\rho H\mathbf {v} )$ , with $\rho H=\rho E+p$ .

Also the section of "Enthalpy conservation" in the article seems to have lost a minus in ${De \over Dt}={Dh \over Dt}-{\frac {1}{\rho }}\left({Dp \over Dt}+{\frac {p}{\rho }}{D\rho \over Dt}\right)$

It should be ${De \over Dt}={Dh \over Dt}-{\frac {1}{\rho }}\left({Dp \over Dt}-{\frac {p}{\rho }}{D\rho \over Dt}\right)$ 192.198.151.43 (talk) 13:56, 29 July 2015 (UTC)[reply]

The above sign errors were corrected on 30 July 2015. There is another sign error in the 3rd equation of Sec. 4.2 on Enthalpy Conservation. It was:

${De \over Dt}={Dh \over Dt}+{\frac {1}{\rho }}\left(p\nabla \cdot \mathbf {u} -{Dp \over Dt}\right)$

and I am now changing it to:

${De \over Dt}={Dh \over Dt}-{\frac {1}{\rho }}\left(p\nabla \cdot \mathbf {u} +{Dp \over Dt}\right)$

Plasma g (talk) 16:42, 17 August 2015 (UTC)[reply]

Error?

(I give up using wikipedia math expressions:)

Section: "Incompressible Euler equations with constant and uniform density"

Contain the equation: du/dt + u . grad u = - grad w + g

the u . grad u term is not very clear. Clearer to write (u.grad) u. see [1] p.1-2

This is made worse by the equation: du_j/dt+Sum_i u_j d(u_i + w e_i)/dr_i == 0 where to my eyes the second term is plainly wrong.

should be: du_j/dt +sum_i u_i d u_j/dr_i + dw/dr_j == 0 see [2] p.1 — Preceding unsigned comment added by 194.47.168.94 (talk) 09:01, 10 May 2017 (UTC)[reply]

Deduction of the form of sound speed valid for ideal gases

Firstly, in my opinion, the second equation in the deduction of the sound speed for ideal gases in chapter Euler equations (fluid dynamics)#Ideal gas is wrong. There should be a minus sign instead of a plus sign. Can somebody confirm that?

Second, division of the second equation by ρ^−γ dρ does not yield the third equation, because ρ^−γ-1 divided by ρ^−γ yields ρ⁻¹ instead of ρ.

Belgariath (talk) 12:08, 12 November 2019 (UTC)[reply]

singularity with a computer-aided demonstration

A preprint, posted online in October 2022, states that, under certain boundary conditions, the Euler equations fall in a mathematical singularity. It was the first time in which a singularity was discovered through a computer-aided demonstration. (source: Wired). 95.237.45.179 (talk) 20:59, 29 December 2022 (UTC)[reply]

Spontaneous stochasticity

This article notes that the eqns are ill-posed, without also specifying eqn of state. Additional refs for this are:

C. De Lellis and L. Szekelyhidi, On admissibility criteria for weak solutions of the Euler equations, Archive for Rational Mechanics and analysis 195, 225 (2010).
C. De Lellis and H. Kwon, On non-uniqueness of Holder continuous globally dissipative Euler flows, arXiv:2006.06482 (2020)
S. Daneri, E. Runa, and L. Szekelyhidi, Non-uniqueness for the Euler equations up to Onsager’s critical exponent, Annals of PDE 7, 1 (2021)

At high Reynolds numbers, this apparently leads to "Eulerian spontaneous stochasticity" during turbulent fluid motion:

A. A. Mailybaev, Spontaneously stochastic solutions in one-dimensional inviscid systems, Nonlinearity 29, 2238 (2016).
S. Thalabard, J. Bec, and A. A. Mailybaev, From the butterfly effect to spontaneous stochasticity in singular shear flows, Communications Physics 3, 1 (2020).

I'd like to read more about how/why the eqns are ill-posed without the eqn of state, and a description of spontaneous stochasticity. 67.198.37.16 (talk) 21:11, 20 March 2024 (UTC)[reply]

Error in Second Euler Equation For Compressible Fluid

The equation is supposed to represent conservation of momentum for the general case where rho can be a function of space and time. In that case, the equation should be D/Dt (rho * u) = -grad p + g. NotSqwerty (talk) 06:21, 5 May 2024 (UTC)[reply]