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Old   June 29, 2006, 15:46
Default transform navier-stokes eq. to euler-eq.
  #1
pxyz
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hello,

i want to transform the navier-stokes-eq for incompressible flows into the none-viscous formulation of euler. Unfortunately I have no idea how I can combine the momentum-eq with rot(c) = 0.

Thanks for help.
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Old   June 29, 2006, 16:43
Default Re: transform navier-stokes eq. to euler-eq.
  #2
Adrin Gharakhani
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what's rot(c)?

anyway, you've already answered your own question: "non-viscous formulation".

So, what does non-anything mean in mathematical form? Look at the navier stokes equation again and try to make it "non-viscous"

Adrin Gharakhani
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Old   June 29, 2006, 16:50
Default Re: transform navier-stokes eq. to euler-eq.
  #3
Anonymous
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I think that by rot(c) = 0 you are trying to make the equations irrotational. The Euler Eqns are NOT irrotational, the are just inviscid (non-viscous).

To make the equations inviscid, just drop off the viscous terms, it is pretty simple!!

If you only want to model an irrotational flow, you only need to solve the potential equation.

Any basic fluid mechanics bok will explain this to you.

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Old   June 29, 2006, 23:52
Default Re: transform navier-stokes eq. to euler-eq.
  #4
pxyz
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[dv/dz - dw/dy] rot(c) = [dw/dx - du/dz] = 0

[du/dy - dv/dx]

1) You are right, I want to eliminate or integrate the rotation in the NS-eq. But I don't see where these terms occur. I even could not find it in my books. That's why I am asking. 2) How do viscosity and rotation play together? Isn't a requirement?

NS-eq:

d(rho*u_i)/dt + d(rho*u_i*u_j)/dx_j = -dp/dx_i -2/3*eta (d^2u_j/dx_j*dx_i) + eta*(d^2u_i/dx_j^2 - d^2u_j/dx_j*dx_i)

Thanks!
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Old   June 30, 2006, 02:25
Default Re: transform navier-stokes eq. to euler-eq.
  #5
Luca
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Yes you're rigth...but I have a doubt: how to treat the wall boundary condition? will this be automatically accounted as an Euler solver does? I mean: by neglecting viscosity I'll automatically assure the normal wall velocity to be zero and the tangential velocity to be found by the solver?
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Old   June 30, 2006, 03:58
Default to potential flow
  #6
pxyz
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Hi,

you are right! What I asked was wrong. Of course you can eleminate the viscous terms in the NSE if mue=0.

So, I have to formulate the question in another way:

1) Which terms (in the NSE) are zero, when I introduice the condition rot(c)=0? (This will lead to a potential flow).

2) Can you give any example of a flow, which would not work without rotation?

3) So, rotation and viscosity are NOT connected with each other?

Thanks for your help.
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Old   June 30, 2006, 04:16
Default to potential flow
  #7
pxyz
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Hi,

you are right! What I asked was wrong. Of course you can eleminate the viscous terms in the NSE if mue=0.

So, I have to formulate the question in another way:

1) Which terms (in the NSE) are zero, when I introduice the condition rot(c)=0? (This will lead to a potential flow).

2) Can you give any example of a flow, which would not work without rotation?

3) So, rotation and viscosity are NOT connected with each other?

Thanks for your help.
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Old   June 30, 2006, 04:31
Default Re: to potential flow
  #8
Dominic
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Hi,

Taking the CURL of the momentum equation, you end up in the Vorticity equation, basically the equation that governs 'rotation'. Although you consider a fluid with Nu=0, you still can have an evolution equation for Rotation. If this is the case, then vorticity will only get convected from t=0. If there is no Vorticity at t=0, there will never be at all for any t>0 if Nu=0. (Unless you have curved shock waves).

-Dominic
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Old   June 30, 2006, 04:37
Default Re: to potential flow
  #9
Tom
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Ignoring boundary conditions any irrotational flow will solve the full Navier-Stokes equations automatically (Actually this is true for any constant vorticity inviscid solution as well).

More specifically if w=curl(u) is the vorticity then the visucous term in the Navier-Stokes equations for an incompressible flow is just

-nu.curl(w),

where nu is the kinematic viscosity.

Have a look at the recent papers by D.D. Joseph in JFM.
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Old   June 30, 2006, 09:20
Default Re: to potential flow
  #10
Anonymous
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Hi Dominic,

Take another look at your vorticity equation ... you will get vorticity generated in an inviscid flow if you have a gradient of density that is not parallel to a gradient of pressure ... i.e. in an a stratified flow.
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Old   June 30, 2006, 10:52
Default Re: to potential flow
  #11
Dominic
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Yes, I agree with you..about the baroclinic generation of vorticity
: grad p x grad rho

Also, the case that this term is still Zero where there are significant density variations cannot be overuled. ie..
:For a stratified fluid, this term can still be zero. It just requires the gradient vectors be parallel.

-Dominic
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Old   June 30, 2006, 14:31
Default Re: to potential flow
  #12
Anonymous
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You are absolutely right Dominic, that is why I wrote

" ... if you have a gradient of density that is not parallel to a gradient of pressure ... "
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Old   June 30, 2006, 15:14
Default Re: to potential flow
  #13
Adrin Gharakhani
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Any velocity vector field can be decomposed (Hodge, Helmholtz) into a vortical component and a potential component:

u = u_omega + u_potential

The potential component can be written in terms of the gradient of a scalar (potential) field:

u_potential = grad(phi)

The vortical component is solenoidal, which means its divergence is zero. So, now, if you apply continuity (in an incompressible flow setting) you'll get:

Div(u) = Div(grad(phi)) = Laplacian(phi) = 0

So, you see that a potential flow (above) is a manifestation of continuity. A potential flow does not account for the dynamics of the flow: it just follows the geometric shape of the domain (the kinematics)

At least in an incompressible flow case, rotation (vorticity) and viscosity are "related" only at solid boundaries. This (vorticity generation mechanism) is actually a matter of great debate. Anyway, viscosity does not directly generate vorticity; it only acts to diffuse (spread) it.

Also, note from the decomposition of velocity that:

vorticity = curl(u) = curl(u_omega)

because curl(grad(phi)) = 0. Therefore, you see clearly the decoupling of vorticity (vortical flow) from the potential flow component. The two velocity components are coupled only at the boundaries. This should hopefully help out with the confusion between potential and euler flow

Adrin Gharakhani
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Old   June 30, 2006, 20:12
Default Re: transform navier-stokes eq. to euler-eq.
  #14
Tian_FB
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OK,I just talk about 2D. As Adrin Gharakhani said above,you make {miu}=0,youcan get the none-viscous formulation of euler.if you want to get the irrotational,you can take cirl on the both sides of the eqn.,and get D(w)/Dt=0,it's vortex conservertion!And if you want to get the integral for,you can make F=(nabla)(phi),and p=p(rho) act.,and you can get it!
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Old   June 30, 2006, 21:05
Default Re: transform navier-stokes eq. to euler-eq.
  #15
Adrin Gharakhani
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>>.if you want to get the irrotational,you can take cirl on the both sides of the eqn.,and get D(w)/Dt=0,it's vortex conservertion!

You correctly state that Dw/Dt=0 is conservation of vorticity in 2D. However, this is _not_ the formulation for irrotational flow! The equation says that vorticity is constant along flow particle trajectories. It does not say vorticity is zero everywhere, which is the condition for potential flow. The confusion may be arising from the modeling of vortex wakes behind streamlined objects in the literature using an "unsteady potential flow" model (which is strictly speaking not a potential flow)

Adrin Gharakhani
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Old   July 3, 2006, 04:05
Default Re: to potential flow
  #16
Tom
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"At least in an incompressible flow case, rotation (vorticity) and viscosity are "related" only at solid boundaries. This (vorticity generation mechanism) is actually a matter of great debate. Anyway, viscosity does not directly generate vorticity; it only acts to diffuse (spread) it."

In what sense is it a subject of "great debate" - Prandtl's boundary layer theory essentially explains this (and just about every mathematician/boundary layer theorist takes this view).

Your statement about potential flow is also a bit misleading since you appear to be ignoring the application of potential flow to water waves & the motion of vortex sheets. You should also remember that a steady vortical flow also has the property you describe as kinematic via Bernoulli; i.e. these are both dynamics.
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Old   July 3, 2006, 15:32
Default Re: to potential flow
  #17
Adrin Gharakhani
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>>Your statement about potential flow is also a bit misleading since you appear to be ignoring the application of potential flow to water waves & the motion of vortex sheets.

Tom,

I'd already mentioned your example, in an earlier response (duplicated below):

>>The confusion may be arising from the modeling of vortex wakes behind streamlined objects in the literature using an "unsteady potential flow" model (which is strictly speaking not a potential flow)

Vortex flow is not potential flow; I'm sure you know. The potential flow due to the wake has in it a number of modeling assumptions, which leads to an equation that _looks_ like an "unsteady" potential flow (an unsteady potential flow is a misnomer too). At the very least, potential flow is prescribed by a Laplace equation for a _scalar_ variable. Vortex flow (even in the simplified flow problem of invscid wakes) is the result of a _vectorial_ Poisson equation that has the vorticity distribution as it source. These are two different things/problems and the only time they are one and the same is when vorticity is zero (which is the original definition of potential flow anyway).

The key word in your statement is the "application" of potential flow to waves, wakes etc. Yes, the potential flow "model" is applied to these cases, but these flows are _not_ potential; thus my own orginal comment "the confusion arises ..."

As for the issue related to "great debate", the full sentence in my original messages was:

>>This (vorticity generation mechanism) is actually a matter of great debate.

Prandtl theory or any other that I know of, say nothing about vorticity generation, but its evolution once it's already generated. HOW vorticity is generated is not obvious, and that has been a source of problem/confusion especially from the perspective of applying vorticity boundary condition for vortex-based methods.

Luigi Morino has a number of papers on boundary integral equations related to these "potential flow" problems (80s and 90s) and provides information on this "debate".

Very briefly, the question is whether vorticity generation is an inviscid process or a viscous process. Do you need to have viscosity to generate vorticity at solid walls or is viscosity the mechanism/medium that diffuses the vorticity from the wall into the flow once it's already been generated inviscidly? There are strong arguments in support of both positions.

Adrin Gharakhani
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Old   July 3, 2006, 22:22
Default Re: to potential flow - a clarification
  #18
Adrin Gharakhani
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A bit more on potential flow vs. vortical flow.

If we consider the Navier Stokes equation in its vorticity transport formulation we have the continuity eqution, which is simply a statement about the solenoidality of velocity (in incompressible flow), and the transport of vorticity, which is the balance between convection, stretch and diffusion of vorticity.

Now, if we agree with the premise that what separates potential flow from inviscid flow is vorticity - potential flow being the case with zero vorticity, then it's clear from the vorticity transport equation that the "momentum" equation produces nothing - we have zero vorticity on the left and right hand sides of the equation. So, the only remaining equation is that of continuity, in which case when we apply the Helmholtz decomposition to the velocity field, we end up with a Laplace equation for a potential field (this is what I meant by the potential flow being an expression of continuity and thus kinematics)

However, as we know we can apply the "potential equation" to the flow of waves and _vortex_ sheets; so the question is "what gives"? This is a very interesting (beautiful) consequence of the boundary integral formulation of the (vector) Poisson equation for vorticity source terms. The boundary integral equation equivalent of the latter for a distribution of vorticity sources gives us surface integrals (in 3D) of vortex sheets and sources on solid boundaries and volume integrals of vortices. It just so happens that if we make these volumes infinitesimally thin we end up with surface integrals which look identical (in form) to the former integrals for boundary sheets. So, in a sense, the fluid vorticity (in the zero thickness limit) is like a collection of "fluid" walls that extends from the solid walls.

Even in a purely potential flow, we end up with a potential distribution on the solids, which can be rewritten to look as a collection of vortex sheets; i.e., the slip velocities at the potential flow walls are equal in size to vortex sheets (the same amount of vorticity that has to be generated to satisfy no-slip boundary condition). So, there is a close link between "potential" and "vortical" flows, and that link is right at the boundaries (be they solid walls or fluid walls). When we solve a "potetial" flow with a complex unsteady boundary condition, what we've essentially done is to shift the vorticity dynamics onto the potential flow as a boundary condition. Now, does this mathematical trick and coincidence make this flow potential? I'd say not.

Adrin Gharakhani

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Old   July 4, 2006, 04:00
Default Re: to potential flow - a clarification
  #19
Tom
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"Now, does this mathematical trick and coincidence make this flow potential? I'd say not."

The fact that there is a velocity potential that defines the flow everywhere means that it is potential - a potential flow is one whose velocity can be determined from a "velocity potential".

You also have to be careful in talking about vorticity the way you are above - the problem with vorticity boundary conditions arises because vorticity is a "derived" rather than a physical quantity (it was originally introduced to exploit the relationship of the solenoidal condition to the Biot-Savart law). The physical quantities (the velocities) have boundary conditions but the derived quantities require derived boundary conditions which have to be determined by consistency with the physical ones!

For Prandtl's boundary layer theory and vorticity generation just differentiate the boundary layer equations with respect to the wall normal to get the vorticity equation; i.e. vorticity generation is nothing more than the reduction in streamwise velocity to satisfy the no-slip condition. More precisely if y is the normal direction from the wall then the vorticity is approximately (u tangential to the wall) u_y=du/dy which is just the shear. Now as the freestream is approached u_y->0 (but u is nonzero) while at y=0 u=0. This implies that u_y must be nonzero in the boundary layer => vorticity generation mechanism.
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Old   July 4, 2006, 12:20
Default Re: transform navier-stokes eq. to euler-eq.
  #20
Anonymous
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Maybe I am too simple minded, but to me vorticity is just a rotational flow. Anything that creates a velocity gradient will lead to some vorticity.

In a boundary layer, viscosity leads to a velocity gradient due to the no slip condition which in turn leads to some vorticity. The same thing will happen in a shear layer.

In the case of a baroclinic torque, some of the flow has a different density than other portions (i.e. different inertia), and when subjected to a pressure gradient will move with a different velocity, this velocity gradient will lead to vorticity.

Like I said, maybe I am too simple minded. However, I do sometimes think that people get too caught up in mathematical formalism and forget that we are dealing with a physical system. It sometimes seems like people talk of vorticity like this weird magical quantity ... it is not, it is just rotational flow!!
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