[Jonas Holdeman]

It is useful to consider the necessary properties of a hypothetical incompressible fluid in the same sense that one considers an inviscid fluid as the limiting case of vanishing viscosity. Consider a small volume of fluid. It is usually assumed that the effect of the surrounding fluid can be represented in terms of surface (contact) forces or stresses. For compressible fluids, the assumption of a normal surface force (due to pressure) correctly leads to a gradient term in the momentum balance equation, which is the basis of the Navier-Stokes equation. It is conventionally assumed that the same holds true in the incompressible case.

But by the Helmholtz-Hodge decomposition, the incompressible Navier-Stokes equation is composite, separable into a pressure-free equation for the divergence-free velocity and an equation for P as a function of U. So we see a contradiction. The assumption of a normal force leads to a grad P term, while the Helmholtz decomposition excludes such a term. We must conclude that an incompressible fluid does not support the assumption of a normal surface force. If not a surface or contact force, what force keeps the velocity divergence-free? We invoke the classical concept of action-at-a-distance, resulting in a body force.

In general, the advective/convective term (-U dot grad U) is not divergence free. When the divergence of (-U dot grad U) is positive, this 'virtual' fluid source acts like a repulsive force, pushing adjacent fluid away. When negative, there is a 'virtual' sink, resulting in an attractive force as adjacent fluid is pulled in.

We consider the form of the force on fluid particle at r due to a nonzero divergence at r’. Assume the strength of this force is proportional to the divergence of (-U dot grad U), the force acting along the line joining r and r’, and decreasing with distance inversely as the distance squared in 3-D and inversely as the distance in 2D. Consequently the force on the particle at r due to the divergence of the advective term at r’ would be f(r,r’)=rho Cd (div (-U dot grad U)) (r-r’)/|r-r’|^d , where Cd is a constant to be determined.

Summing over all the distant positions r’ gives the force on the fluid particle at r. With a(r)=f(r)/rho, a(r)=integral(Cd (div (-U dot grad U)) (r-r’)/|r-r’|^d dr’). We require the divergence of ( -U dot grad U+a(r) ) to be equal to zero. In free space, the integral can be evaluated to find the coefficient Cd.

So the governing equation for incompressible velocity U is an integro-differential equation. There is an elegant way to solve the variational/weak form of the pressureless incompressible equation using finite elements.

Question: Can integro-differential equations like this be solved using the methods of finite differences?

[FMDenaro]

Hello Jonas,
this statement is not clear to me:
The assumption of a normal force leads to a grad P term, while the Helmholtz decomposition excludes such a term. We must conclude that an incompressible fluid does not support the assumption of a normal surface force.


However, FD can be used as way to discretize any differential operator (provided regularity of the function is ensured), of course when an integral needs to be discretized you have to complete the FD formulation with some integration rules such as trapezoidal, Simpson..

[Jonas Holdeman]

Originally Posted by FMDenaro

Hello Jonas,
this statement is not clear to me:
The assumption of a normal force leads to a grad P term, while the Helmholtz decomposition excludes such a term. We must conclude that an incompressible fluid does not support the assumption of a normal surface force.

I remarked:
(1) The assumption (existence) of a normal surface force (pressure) leads to a grad P term (force). True?
(2) The Helmholtz decomposition excludes such a (grad P) term. True?

These statements are incompatible: ((1) there is ... or (2) there isn't ... a grad p term. They can't both be true). True?
How can we make them compatible except excluding some assumption about incompressible fluids? What other assumption is there than the existence of a normal surface force? It took me a long time to accept this conclusion and its consequences. I am still working on this.

However, FD can be used as way to discretize any differential operator (provided regularity of the function is ensured), of course when an integral needs to be discretized you have to complete the FD formulation with some integration rules such as trapezoidal, Simpson..

I guess I am thinking generally of FD as (discrete) operations performed on values at nodes of a mesh, and those operations could include sums (like trapezoid, Simpson) as well as differences. I think there is a beauty in the FEM missing in the Spartan FD. There has been a lot of trial and error that has gone into making FD work. I wonder if there might be some unifying principles that might come out of the "integro-" part. I am less than a beginner in FD, and at my age I feel I must stay focused on just a few things.

[FMDenaro]

I try to clarify my doubt...Consider the momentum equation written in terms of accelerations:

a + Grad p = a*

where
a is divergence-free
Grad p = Div (I p) is curl-free
a* is convection + diffusion

therefore the decomposition of a* is already present in the equation.
You see also that that action of the pressure is by the isotropic part.
I dont see contradiction...

PS: yes, FEM, FV and FD could be seen in a unified framework of projective methods along suitable shape functions.

[Martin Hegedus]

I'm not sure I understand the question/topic, but for steady incompressible flow, it is conservation of mass which keeps the flow divergence free.

[Martin Hegedus]

Oops, let me change my statement. Conservation of mass keeps incompressible flow divergence free regardless of whether the flow is steady or not since drho/dt is neglected.

[Martin Hegedus]

BTW, what happens in real life is that an acoustic wave goes out and sets up this divergence free flow. And for incompressible flow it moves infinitely fast. It is the acoustic wave which provides the "action at a distance." But the acoustic wave is a pressure wave.

[FMDenaro]

yes, no matter steady/unsteady, the mass equation is alway Dv v = 0, for steady flow the decomposition writes as

Grad p = a*

the incompressible model tell us that a small variation of density (d rho) causes istantaneusly an infinite variation of pressure in isoentropic condition, such that

dp/d rho -> Infinity

Of course, that is a model, not the real physics in which the compressibility is always present

[Martin Hegedus]

And, I gather that the original posters "P as a function of U" is Bernoulli's equation.

[Martin Hegedus]

Oh, I guess the other assumption that the original poster is missing is that the pressure wave moves very fast relative to the motions of interest. Another way to view the incompressibility assumption is that the speed of sound is much much larger than the velocity of the object and/or flow.

[FMDenaro]

Originally Posted by Martin Hegedus

And, I gather that the original posters "P as a function of U" is Bernoulli's equation.

well, Bernouilli is valid only for the case of non-dissipative condition

[Jonas Holdeman]

I posted this thread hoping to prompt more discussion about the physical properties one might find for a hypothetical incompressible fluid (in contrast to incompressible flow), which might be used as a model for incompressible fluid flow. One thinks of a hypothetical inviscid fluid without viscosity. That concept is very useful, but has limitations near boundaries, where viscosity, however small, dominates. My tact was to model the force between fluid particles in terms which would lead to the incompressible Navier-Stokes equation.

In the classical mechanics of the 19th century, forces were viewed as due to contact, or by action-at-a-distance. Examples of the latter were gravitational (Newton), electric (Coulomb), and magnetic (Biot-Savart). Maxwell united the latter two with the concept of an electromagnetic field as an intermediary between charged particles. The interactions of Coulomb and Biot-Savart were instantaneous (effects propagated at an infinite velocity, if you wish). Einstein’s theory of relativity, with a limiting velocity of interactions, changed that. Although there were theories of direct interaction using ‘retarded’ potentials, field theories gained acceptance over action-at-a-distance. For fluids this means fluid particles interact via pressure fields, interactions limited by the speed of sound in the fluid.
For incompressible fluids, the speed of sound would be infinite. It would seem that to accommodate this fact, we need to resurrect the classical concept of direct interaction-at-a-distance. The form I described for the force in the post is just a ‘physical’ interpretation of the terms in a Green’s function for the projection.

In response to Martin Hegedus: (1) A basic assumption is that the velocity of pressure waves (if they exist in this fluid) would be infinite. (2) Incompressibility (or div U = 0) is treated as a conservation law, not the agency of a thermodynamic pressure. (3) Bernoulli’s equation is just a statement that the incompressible part of (- U dot grad U) is invariant along the stream line. (4) The solenoidal projection of the Navier-Stokes equation does not contain a pressure, and divergence free flow can be computed without reference to a pressure, so what evidence exists that a pressure exists in an incompressible fluid? Isn’t it used as just another name for a subtractive term in the solenoidal projection?

In response to Filippo Denaro: As you note, P^S (-U dot grad U - grad P) = (-U dot grad U - grad P’) . The form is conserved but grad P is not necessarily the same as grad P’. By P one might mean pressure, but P’ would be something else. This argument worries me too because it seems a little circular. The result comes from writing P^S=(1-P^I), where P^S and P^I are the solenoidal and irrotational projection operators. But we could formally write P^S=(curl((curl curl)^(-1))curl) and P^I=1-P^S instead of P^I=grad(Delta^(-1))div, in which case a grad P term would never appear. Of course we would have to find an expression for the Green’s function of the (curl curl) operator. This is much closer to the modified Hermite finite element method I use.

Still unresolved in this discussion is how to handle the influence of boundaries, and the force an incompressible fluid might exert on a boundary.

[Martin Hegedus]

Sorry, a little lost here but this sounds like a discussion about potential flow (Panel methods, etc.)

[Martin Hegedus]

BTW, neither Bernoulli's equation or potential take into account viscosity, but, I assume, that if this discussion applies to the N.S. equations, it also applies to the Euler equations.

[Jonas Holdeman]

Originally Posted by Martin Hegedus

Sorry, a little lost here but this sounds like a discussion about potential flow (Panel methods, etc.)

It has nothing to do with potential flow. Further short description with publication references can be found on the CFD-online\wiki\source code archive\educational, along with MATLAB code for computing pressureless flow using primitive variables.

[Martin Hegedus]

An incompressible, irrotational, inviscid flow = potential flow.

Whatever you are talking about should boil down to potential flow. I guess.

And potential flow is "action-at-a-distance"

[H0T_S0UP]

so what evidence exists that a pressure exists in an incompressible fluid?

Actually, wouldn't an incompressible fluid have infinite pressure for a 0 amount of time?

Consider the equation for the speed of sounds

c^2 = dP/drho

which describes how a wave (ie a brief density gradient) propagates with a change in pressure. So, provided there is no change in density, it is possible to have a pressure gradient, but in this case c^2 should go to infinity, and the pressure felt in response to a change in density (c^2 drho = dP), should also approach infinity. Of course, if the speed is sound is infinite then you have to have felt this force for t = 0 seconds (or your favorite unit). Im at work so I cant dig into this deeper.

[Martin Hegedus]

A truly incompressible fluid means that the fluid is actually at rest. So, no pressure or velocity gradients.

[Martin Hegedus]

http://www.engineeringtoolbox.com/so...ids-d_713.html

[FMDenaro]

Well, "incompressible fluid" is a model and as a model has approximations and implications...
What we call pressure has no thermodinamic meaning (p=rho*R*T in the compressible formulation but if we would use rho=constant and T= constant, we would get dp=0).
That sounds strange but the momentum equation is mathematically a well posed problem when the divergence-free is enforced. What we call "pressure" comes from the continuity equation and is simply a scalar function having gradients such to ensure continuity. We should simply denote as "phi" function and it contains any scalar function acting as gradient.
It can be shown that it acts as a lagrangian multiplier.

As the BC.s are concerned, the pressure problem is well posed with one boundary condition, that is the projection of the Hodge decomposition along the normal direction to the boundary. That ensures to fulfill the compatibility condition and have a solution for the problem