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Physics of an incompressible fluid

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Old   March 20, 2015, 10:32
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Martin Hegedus
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Hey, that's nice how Jonas didn't acknowledge my contributions. I guess they were not insightful.
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Old   March 20, 2015, 10:45
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The velocity term in the compressible Navier Stokes equations are non-dimensionalized by the speed of sound. There is a difference between Mach number approaching zero and the Mach number actually being zero. The flow for the Mach number actually being zero is boring.
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Old   March 20, 2015, 11:03
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Oh, and if one views my 1-D bar of accelerating steal as a fluid I gather the reason I have a dp/dx is that there is a dp/dt at the boundary. Like I said, Dp/Dt = V*dp/dx + dp/dt = 0.
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Old   March 20, 2015, 11:23
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Thank you too, Martin.
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Old   March 20, 2015, 11:41
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OK, so does your original statement/theory agree or disagree with the fact that a 1-D accelerating incompressible bar of steal has a dp/dx? And a 1-D bar of steal can be modeled by the incompressible 1-D Euler equations. If you can not show that, then your mathematical analysis has a hole in it.
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Old   March 20, 2015, 13:22
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well, however 1D case implies du/dx = 0 from the divergence-free constraint?

Otherwise we are speaking of the classical Burgers equation (zero pressure gradient) which model compressibility effects?

I am not understanding
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Old   March 20, 2015, 13:33
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V of steal coming into control volume equals V of steal coming out of control volume, so yes, du/dx is zero. So yes.

However, if the bar of steal is accelerating, i.e. dV/dt, then the Euler equation gives a dp/dx. I showed that above.
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Old   March 20, 2015, 14:03
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but I don't think that the equation du/dt + dp/dx = 0 can be considered as counterpart of Euler incompressible equation for a fluid
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Old   March 21, 2015, 00:24
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The incompressible Euler's momentum equation states -dp/dx = rho*DV/dt. It is just F=ma. It is not a counterpart. Or maybe I don't understand what you mean by a counterpart. If DV/dt is not zero, i.e. acceleration or deacceleration, then there must be a dp/dx. Or, if the pressure field is constant everywhere then acceleration is zero. For anything interesting to occur there must be a pressure gradient for the incompressible Euler equations. Sorry, I know the sentences above are basically repeating themselves.

Yet, it seems as if people are saying that the pressure gradient is zero. That can only be true if there is no acceleration.

So I'm really confused about what people are writing. At this point, I sure don't think any evidence has been provided that the Euler equations need to be adjusted.
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Old   March 21, 2015, 00:31
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The incompressible Euler equations are shown here.

http://en.wikipedia.org/wiki/Euler_e...id_dynamics%29

I'm surprised I feel I need to make reference to this. Which makes me wonder if I really understand what is being said in this discussion thread.
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Old   March 21, 2015, 04:18
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if your example is well understood, you are considering the 1D case with du/dx=0, therefore the pressure gradient balance only the Eulerian acceleration being the convective term udu/dx zero.
That is not a general Euler equation where the non linear part is fundamental in the fluid dynamics and the pressure gradient contributes to balance also that
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Old   March 21, 2015, 04:24
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Quote:
Originally Posted by Martin Hegedus View Post
The incompressible Euler's momentum equation states -dp/dx = rho*DV/dt. It is just F=ma. It is not a counterpart. Or maybe I don't understand what you mean by a counterpart. If DV/dt is not zero, i.e. acceleration or deacceleration, then there must be a dp/dx. Or, if the pressure field is constant everywhere then acceleration is zero. For anything interesting to occur there must be a pressure gradient for the incompressible Euler equations. Sorry, I know the sentences above are basically repeating themselves.

Yet, it seems as if people are saying that the pressure gradient is zero. That can only be true if there is no acceleration.

So I'm really confused about what people are writing. At this point, I sure don't think any evidence has been provided that the Euler equations need to be adjusted.
If you consider your case with a steel bar then I don't see why you must have a pressure difference. You have no friction that works against the motion in the 1d equation so you will have continuous acceleration if you have that pressure gradient. (as you state)

So, If you go back and consider a trivial case of 1d flow in a frictionless pipe then, if you prescribe a velocity inlet (dirichlet) the continuity equation prevents a continuous acceleration which means that you can't have a pressure gradient. This example would represent your steel bar at free flight in vacuum I guess (after the solution of the Poisson equation).

Basically: if you prescribe the pressure gradient then you have a pressure gradient and if you set the inlet velocity then you have no pressure gradient. (in your 1d frictionless case)
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Old   March 21, 2015, 10:38
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"Basically: if you prescribe the pressure gradient then you have a pressure gradient and if you set the inlet velocity then you have no pressure gradient. (in your 1d frictionless case"

I'm referring to non-steady case flow, i.e. dV/dt is not zero. dV/dt is set at the boundary and since dV/dx is zero throughout the fluid by incompressibility then dV/dt is constant throughout the fluid. And since dV/dt is constant throughout the fluid then so is dp/dx.
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Old   March 21, 2015, 11:50
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Filippo:

I gather your point is that my example is too simple. And yes, I too feel that it is simple.

OK, I should include u*du/dx

Du/dt = u*du/dx + du/dt and, in 2-D, by conservation of mass du/dx = -dv/dy.

So Du/dt = -u*dv/dy + du/dt.

The -u*dv/dy term is centripetal acceleration. Therefore the Euler momentum equation says dp/dx is a function of centripetal acceleration and linear acceleration.

Centripetal acceleration brings up the subject of vorticity and the incompressible vorticity transport equation says vorticity comes from the boundary too.

Acceleration, in one form or another, for an an incompressible fluid must come from the boundary. And without acceleration dp/dx is zero.
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Old   March 21, 2015, 11:53
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By the way this (irrotationality and incompressibility) brings us back to potential flow, which Jonas blew off earlier.
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Old   March 21, 2015, 12:13
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Quote:
Originally Posted by Martin Hegedus View Post
Filippo:

I gather your point is that my example is too simple. And yes, I too feel that it is simple.

OK, I should include u*du/dx

Du/dt = u*du/dx + du/dt and, in 2-D, by conservation of mass du/dx = -dv/dy.

So Du/dt = -u*dv/dy + du/dt.

The -u*dv/dy term is centripetal acceleration. Therefore the Euler momentum equation says dp/dx is a function of centripetal acceleration and linear acceleration.

Centripetal acceleration brings up the subject of vorticity and the incompressible vorticity transport equation says vorticity comes from the boundary too.

Acceleration, in one form or another, for an an incompressible fluid must come from the boundary. And without acceleration dp/dx is zero.


no...because if you use a 2d example then

Du/dt = du/dt + udu/dx+vdu/dy
Dv/dt = dv/dt + udv/dx+vdv/dy

and you have to take into account also dp/dy in the Euler system
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Old   March 21, 2015, 12:26
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It's an acceleration due to the change in the direction of the flow.

Are people saying that an incompressible flow can not change direction?
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Old   March 21, 2015, 12:27
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1D Euler + divergence-free contraint (du/dx=0) drive to:

d rho/dt + u(t) d rho/dx = 0

du/dt + (1/rho) dp/dx = 0

dp/dt + u(t) dp/dx = 0


therefore, rho and p can vary in time and space but they must be constant along the path-line dx/dt = u(t).

If you add the constraint that the fluid is really incompressible, then rho is constant both in time and in space as well as the pressure
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Old   March 21, 2015, 12:40
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So, you are saying that Euler equations do not boil down to the Bernoulli equation. I believe they do. But, unfortunately, I just don't have the time to prove it or dig up a reference. I'll have to leave it at this.

And we both agree that the velocity profile can be found using potential theory, correct?
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Old   March 21, 2015, 12:48
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In regards to this:

dp/dt + u(t) dp/dx = 0

dp/dt comes from the boundary. As I mentioned before.
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