Talk:Potential flow

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[edit] Hi!

Hello everybody.

I have been planning to edit this page for some time...there was nothing about complex analysis at all.

So I added a pointer to potential flow in 2d. I did this (rather than put the stuff in potential flow per se), for several reasons:

(1) potential flow is a *physical* set of assumptions while potential flow in 2d is basically a set of *mathematical* concepts

(2) potential flow is 3d and potential flow in 2d is a strict subset of this

(3) potential flow in 2d includes a whole shed-load of complex analysis that is not directly relevat to the physics of potential flow (nor is it interesting to someone who wants to know the physics).

(4) a pointer from complex numbers to potential flow is nonsense...the correct pointer should be to potential flow in 2d.

Anyway, someone changed potential flow in 2d into a redirect (to potential flow), which I think is wrong.

How do other wiki people see this issue?

[edit] Confusing article

This article is a bit confusing to me... and I'm an Aeronautical Engineer. The material presented is a highly mathematical approach, which is hard to understand for most non-mathematicians. I'l have a go at re-writing it when I have time. If anyone else is thinking of re-writing it, it'd help to include an example or two, mention actual uses (lifting cylinders, Joukouski transforms)...

(Halabut 23:40, 9 November 2005 (UTC))

[edit] Inviscid condition

Potential flow does not have to be inviscid. I am removing these references. The existing external link confirms this. Josh Quinnell 02:47, 2 December 2005 (UTC)

Hi. I don't see that potential flow can have viscous effects. Could you elaborate, rather than just deleting what appears to be good-faith material? Robinh 15:38, 2 December 2005 (UTC)

How is it good faith material when the link provided in the article states otherwise?
proof:
Per the conditions of potential flow:

\nabla \cdot \mathbf{v} = 0 (1)
\nabla \times \mathbf{v} = 0 (2)
(2) is satisfied by

\mathbf{v} = \nabla \Phi (3)

Substituted into the continuity equation yields

\nabla^2 \Phi = 0 (4)

Now lets go to the viscous term

\mu\cdot\nabla^2\mathbf{v} = \mu\nabla\nabla^2\Phi = 0 (5)

The above is zero regardless of the value of μ. μ does not matter. Josh Quinnell 20:51, 2 December 2005 (UTC)


Hi again. Well I buy the maths above, but am still confused. Surely a free vortex is potential flow (the div and the curl are zero). But ramp the visosity up to 10^13 times that of water and the whole thing will grind to a halt. So physically there is a dependence. How to square these two appararently contradictory approaches?

best wishes, Robinh 22:32, 2 December 2005 (UTC)

The viscous forces are 0. Josh Quinnell 01:35, 3 December 2005 (UTC)

I think you are looking at the problem from the wrong point of view. An actual flow is a very complex process. The complete Navier-Stokes equations, or better, its computational Reynolds averaged version, still require a cluster to compute a solution for any simple geometry. And still, the fluid is assumed to be continuous and homogeneous, therefore neglecting a lot of effects easily observable in actual fluids, such as a termodynamic effect as the phase change (important for cavitation analysis on boat propellers, for example). When we take the Navier-Stokes frame-work to model a fluid, it becomes evident that the rotation of fluid particles is propagated from particle to particle due to shear stresses, i.e. viscous forces. So, if you take the Navier-Stokes to start with, you'll see that the rate of change of vorticity is (see chapter 2 of "low-speed aerodynamics" from Joseph Katz and Allen Plotkin, for example)

{D\zeta }/{Dt} = \zeta \cdot \nabla \mathbf{v} + \mu \nabla^2 \zeta

where \zeta = \nabla \times \mathbf{v} is the vorticity. So, as you can see, talk about viscosity is something that only makes sense when you consider your fluid to be (capable of being) rotational. If you impose it to be irrotational, implicitly you're saying "there are no shear stresses between fluid particles", therefore the whole viscosity concept loses its theoretical ground. By taking the Potential flow as a starting point and then try to conclude about viscosity... well, sounds the same as trying to see the relativity effects of the speed of light by starting with the Newton laws... know what I mean? Joaoliveira 16:00, 23 August 2006 (UTC)

[edit] To Mr Hi!

Potential flow is by definition a (almost) 2d-phenomenon therefore I think it's wrong to have both an article for 2d-potential flow and one for potential flow. If you wish to add something I think it's best to do so in the actual potential flow article.

This is mainly because people look stuff up here when they don't understand it. Having both 2d-potential flow and potential flow articles might suggest that (the concept of) 3d-potential flow exists.

Update===forget the stuff above. Here one can find material about 3d-potential flow: http://www.ocp.tudelft.nl/mt/journee/Files/Lectures/OffshoreHydromechanics.pdf (page 288 and beyond) As my exams are coming up I don't have time to study and write on 3d-potential flow but I think it should be explained in this article.

By the way, I added two lines about "gradient" and the note that "gradient" is different from "divergence" in the mathematical introduction. This is just because I used to mix them up sometimes. I am a wannabe mechenical engineer so I'm not as good at math as physicists so I tend to mix up stuff which can look quite elementary for someone more experienced in the subject.

What often happens is that people explain "curl" and "divergence" and then just throw the "gradient" concept in ones face without any explanation at all, which I think is confusing.

[edit] Comment

It is not obvious if/when potential flow satisfies the momentum equation. I think this should be addressed on this page. To clarify my point: The momentum equation has 3 components. Usually the velocity has 3 components, so we have 3 equations for 3 variables. But for potential flow the velocity only has 1 independent variable. So we seem to have 3 equations and only 1 variable. In general one might expect that no solution exits. The best way to answer this question would be to write down a 1-component momentum equation involving the velocity potential.

[edit] Merger proposal

I suggest this article to be merged with the article on popential flow, since pretty much all the content of this article is already covered there and the rest would well fit in. —Preceding unsigned comment added by 128.196.169.66 (talk • contribs) 07:15, 31 January 2008

The above was copied from Talk:Potential flow in two dimensions. — Crowsnest (talk) 19:15, 23 April 2008 (UTC)

[edit] Equations

they're a real mess. need consistently sized equations. why oh why the wikipedia web designers did not set one size for the equations (as done in every textbook i've seen, where they're the same height as the bodytext) escapes me... —Preceding unsigned comment added by 141.213.220.186 (talk) 07:00, 24 April 2008 (UTC)