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Junseok Kim September 1, 2004 23:09

meaning of pressure
 
What is meaning of pressure term in the incompressible viscous Navier-Stokes equation?


noName September 2, 2004 12:22

Re: meaning of pressure
 
An isotropic forcing potential?

Jonas Holdeman September 2, 2004 16:54

Re: meaning of pressure
 
The pressure in incompressible flow can be divided into two parts, hydrostatic and hydrodynamic. The hydrostatic pressure is that which balances conservative forces such as gravitational.

By the Helmholtz decomposition, the incompressible Navier-Stokes (momentum) equation can be separated into two parts, a solenoidal part which becomes a pressureless governing equation for the velocity (and depends on any nonconservative forces), and an irrotational part for the pressure (gradient) as a function of the velocity.

The two equations are integro-differential equations, which is one reason you usually don't see them in this form. When the two equations are combined, the result is a partial differential equation (actually an differential algebraic equation (DAE)in time). The dependence of the pressure on the velocity remains in that there is no explicit mechanism for independently advancing the pressure in time.

For computation, the equations may be multiplied by solenoidal test or weight functions for the velocity and irrotational test functions for the pressure to form and equivalent formulation. In this way, the test functions serve as projection operators and the integro-differential form is not evident. This "weak" form is the basis for the finite element method, and is the form used by mathematicians to "prove" the existence and uniqueness of solutions of the NSE. Actually, that existence has not been proved in 3D, and there is a million dollar prize offered to anyone who can prove it.

Normally, the sum of solutions of two equations is not equal to the solution of the sum of the equations. However in this case it is true because the two are orthogonal; the velocity is the curl of a stream function or vector potential and the pressure gradient is the gradient of a scalar potential.

In this view, the incompressible pressure is that quantity which is consistent with the velocity in the NSE, and is an approximation to the physical velocity in compressible constant-density flow.

Junseok Kim September 3, 2004 02:47

Re: meaning of pressure
 
What does mean by the sentence in your paper, Governing Equation for Incompressible Flow; Revisiting the Navier-Stokes Equation.

What kind of adjustments you have in mind?

The problem with the classical derivation of the incompressible NSE is irrefutable, and appropriate adjustments should be made in the physics curriculum.

Thanks

Tom September 3, 2004 07:51

Re: meaning of pressure
 
Pressure is simply the macroscopic manifestation of the inter-molecular force which resists compression.

For an inviscid fluid it can be interpreted as the Lagrange muliplier for mass conservation in the varaitional principle - hence it's usual interpretation of enforcing incompressibility. ( In Lagrangian mechanics this is actually how the pressure enters the problem with the Lagrange multiplier being interpreted as the pressure after application of the Euler-Lagrange equations to the variational principle).

Tom.

Jonas Holdeman September 3, 2004 11:43

Re: meaning of pressure
 
Re: Kim JunSeok,

The usual derivation of the governing equations for compressible flow are OK. The same derivation of the governing equation for incompressible flow is flawed in that the assumption of a pressure gradient across a control volume cannot be supported because an assumed dynamic pressure gradient would be instantly equilibrated because pressure waves propagate at infinite speed in an incompressible fluid. One just cannot create a dynamic pressure gradient in such a fluid. A false assumption usually leads to a false conclusion.

Look at this another way. Suppose one expands the pressure around the control volume in a Fourier series. Then for each sinusoidal component, the wavelength is given by: wavelength = (speed of sound)/frequency. Now if we let the speed of sound go to infinity, the wavelength of each component goes to infinity. A simple graphical picture will convince that each component is constant across the control volume. The sum of the constant components is constant, so the gradient of the pressure must be zero.

However, one often argues that the incompresible equations can be reached as a limiting case of the compressible equations. But this limit is a singular one (contrary to opinion stated in this forum). As evidence, the nature of the pressure changes from a thermodynamic variable to something else. The pressure is a state variable in the compressible case but not in the incompressible case. Pressure boundary conditions are needed in the compressible case but are not necessary in the incompressible case. I think the astute student is left confused by all this but is intimidated into accepting the usual conflicting arguments and is left feeling intellectually inadequate, and ultimately accepts the contradictions as a matter of faith.

One can derive the pressureless governing equation for the velocity and the pressure as a function of the velocity algebraically using projection operators. The projection operators can be written in terms of the Green's function for the Laplace equation subject to the same velocity boundary conditions as the fluid problem. This is what leads to the integro-differential form of the governing equations.

But integro-differential equations (IDE) are justifiably not taught to undergraduates, so how could one teach incompressible fluid mechanics? There is a way of deriving all the classical analytical incompressible flow solutions without IDEs using what I have called "projection by inspection." I have examined enough of the known closed-form solutions to conclude that such an approach is entirely feasible. But this approach would require further development into a consistent curriculum.

The easy way is to continue to ignore the contradictions. But this will continue to lead to the type of confusion that lead you to ask your original equation.

To Tom:

When one sets up the variational integral for the Stokes equation, one must specify (or imply) over which functional space one is varying the velocity. If the variation is over only the space of solenoidal vector functions, the Lagrange multiplier is not necessary nor is it defined because (lambda div U) is zero for all values of lambda. If this lambda is interpreted as the pressure, then the pressure is obviously not defined by this.

If one varies the velocity over all vector functions, then the Lagrange multiplier is necessary to restrict the acceptable solutions to solenoidal functions. This leads to an equation with the (non-solenoidal) gradient of lambda used to subtract or project out the non-solenoidal part of the trial solution for U. The equation looks like the pressure is involved, but to quote one author, "not everything called P is a pressure."

With appropriate boundary conditions, div U = 0 implies U = curl Psi, where Psi is called the stream function in 2D and the vector potential in 3D. Then there is no continuity equation to deal with. One associated method is the stream function-vorticity formulation in which the velocity is found with no explicit pressure.

I think that it is unacceptable to say that sometimes (with some methods) the incompressible pressure enforces continuity and sometimes it doesn't.

Tom September 4, 2004 06:09

Re: meaning of pressure
 
I think you've misinterpreted my above comment since I didn't say anything about the Stokes equations - I was talking about the Euler equations of an ideal fluid (in which case when you perform the variation all you do is restrict yourself to the condition that the mass should not be varied). Incompressibility then arises through the condition that variations in the volume must vanish.

The Stokes equations are a bit strange since they are derived on the assumption of strong viscosity and end up being time reversible. The restrictions on the variations appear because, as opposed to the usual case of starting with a variational principle and then deriving the equations, you start knowing the equations and reconstruct the variational principle.

Another difference, because of the above point, is that in the case of Stokes flow the equations will be in Eulerian form whereas in the ideal fluid case the equations appear in Lagrangian form.

Tom.


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