[armyou]

Is the pressure term in the N-S equation (momentum) mechanical pressure or thermodynamic pressure? It seems mechanical pressure is not the same as thermodynamic pressure.

Any answer is appreciated.

[tas38]

Originally Posted by armyou

Is the pressure term in the N-S equation (momentum) mechanical pressure or thermodynamic pressure? It seems mechanical pressure is not the same as thermodynamic pressure.

Any answer is appreciated.

In the Navier-Stokes equations, Stokes' hypothesis amounts to the assumption that
the mechanical and thermodynamic pressures are the same. In the case of most fluids,
the thermodynamic and mechanical pressures do not differ appreciably. Only for highly
ionized fluids, shock waves, etc. wherein additional energy modes of the fluid molecules
are exicted do the thermodynamic and mechanical pressures begin to diverge.

[FMDenaro]

pm=p -lambda*Div v

they are the same provided either the Stokes assumption lambda= 0 or a divergence-free velocity field apply in the flow field

[Jonas Holdeman]

One may question whether the p-term in the INSE is a pressure at all in the usual sense. In a compressible fluid, one fluid particle interacts with another through the medium of a pressure field. A fluid particle acts and some time later (the time = particle_distance/wave_velocity) the second fluid particle responds. Momentum and energy are conserved because the pressure field carries the energy and momentum from the first particle in the interval between the action and response. Note that Newton's third law does not apply between particles in the interim. For an incompressible medium, the response to the action is immediate, and Newton's third law can be applied. There is no finite time interval in which a pressure field carries momentum or energy. The incompressible "pressure field" is not material, and if it carries no momentum or energy, is it in any sense detectable so it can be said to exist? It would seem that incompressible fluid particles must interact instantaneously through "direct-action-at-a-distance", like Coulomb's law in electrostatics or Newton's law of gravity.

[armyou]

Thank everyone for your reply in this thread. It really helps.

[FMDenaro]

Originally Posted by Jonas Holdeman

One may question whether the p-term in the INSE is a pressure at all in the usual sense. In a compressible fluid, one fluid particle interacts with another through the medium of a pressure field. A fluid particle acts and some time later (the time = particle_distance/wave_velocity) the second fluid particle responds. Momentum and energy are conserved because the pressure field carries the energy and momentum from the first particle in the interval between the action and response. Note that Newton's third law does not apply between particles in the interim. For an incompressible medium, the response to the action is immediate, and Newton's third law can be applied. There is no finite time interval in which a pressure field carries momentum or energy. The incompressible "pressure field" is not material, and if it carries no momentum or energy, is it in any sense detectable so it can be said to exist? It would seem that incompressible fluid particles must interact instantaneously through "direct-action-at-a-distance", like Coulomb's law in electrostatics or Newton's law of gravity.

yes, if Div v = 0 is the assumption then p acts only by means of its gradient and is not what we rigorously say thermodinamic pressure but is a scalar function like a sort of potential.

However, in using a numerical method, we can somehow reconstruct a pressure field. For example, the well-known second order fractional step of Kim and Moin, uses the pressure-free projection method. The pressure is never really computed at any time step, only a scalar function phi at each time step is computed. However, Phi has an analytical functional relation with p (an elliptic-type equation) that can be solved if we know a value for the thermodinamic pressure.

Still more complex appears the reconstruction if a turbulence model is used. That is due to the fact that the isotropic part of the modelled tensor is merged into the Phi function giving sourse to a new scalar function.

[armyou]

I have clarified with the differences between the thermodynamic pressure and mechanical pressure.

But still have a question: Although stokes assumed P_thermodynamic = P_mechanical, strictly speaking, should the pressure in the momentum conservation equation be the thermodynamic pressure or mechanical pressure or neither?

[FMDenaro]

Originally Posted by armyou

I have clarified with the differences between the thermodynamic pressure and mechanical pressure.

But still have a question: Although stokes assumed P_thermodynamic = P_mechanical, strictly speaking, should the pressure in the momentum conservation equation be the thermodynamic pressure or mechanical pressure or neither?

if Div v = 0, what we call "p" is nothing else but a scalar function that acts by means of gradient to ensure the continuity constraint. It is not rigorously the pressure.

[LuckyTran]

Originally Posted by armyou

I have clarified with the differences between the thermodynamic pressure and mechanical pressure.

But still have a question: Although stokes assumed P_thermodynamic = P_mechanical, strictly speaking, should the pressure in the momentum conservation equation be the thermodynamic pressure or mechanical pressure or neither?

I feel the pressure in the N-S equations should be the mechanical pressure, as it is derived from a force-momentum balance and not derived from thermodynamic principles. And mechanical pressure seems sufficient, and another pressure is not needed? I would appreciate a correction if I'm wrong.

[FMDenaro]

Originally Posted by LuckyTran

I feel the pressure in the N-S equations should be the mechanical pressure, as it is derived from a force-momentum balance and not derived from thermodynamic principles. And mechanical pressure seems sufficient, and another pressure is not needed? I would appreciate a correction if I'm wrong.

As I wrote above, if the working assumption is that the velocity field must be divergence-free, the grad p term in the momentum has a different role and "p" is not the pressure. You can call the scalar function in any other way

[Jonas Holdeman]

Originally Posted by LuckyTran

I feel the pressure in the N-S equations should be the mechanical pressure, as it is derived from a force-momentum balance and not derived from thermodynamic principles. And mechanical pressure seems sufficient, and another pressure is not needed? I would appreciate a correction if I'm wrong.

The problem is that the assumption that there is a normal force term (pressure) acting on a fluid surface in the case of incompressible fluid flow is fundamentally flawed. The assumption is correct in the case of compressible fluids, and normal-force balance correctly leads to the pressure gradient term. However, if the flow is incompressible, if the divergence of the flow is zero, then a mathematical property of vector fields (the Helmholtz theorem/decomposition - derivable by simple integration by parts) precludes a force which is the gradient of a potential from influencing the divergence-free flow. The resolution of this dilemma seems to be that there can be no normal force (no pressure) in the first place. Usually the first thing methods do when solving the INSE is to eliminate the "pressure". For example the stream function-vorticity method finds the incompressible flow without using a "pressure", showing that the "pressure" gradient term is not necessary. A code for another "pressureless" method is given in the CFD-Online wiki pages under educational codes. The conclusion is that the "grad p" term in the INSE is not a pressure. It results from a particular application of the Helmholtz decomposition. But this is likely a fertile area for future research.

[FMDenaro]

Originally Posted by Jonas Holdeman

The problem is that the assumption that there is a normal force term (pressure) acting on a fluid surface in the case of incompressible fluid flow is fundamentally flawed. The assumption is correct in the case of compressible fluids, and normal-force balance correctly leads to the pressure gradient term. However, if the flow is incompressible, if the divergence of the flow is zero, then a mathematical property of vector fields (the Helmholtz theorem/decomposition - derivable by simple integration by parts) precludes a force which is the gradient of a potential from influencing the divergence-free flow. The resolution of this dilemma seems to be that there can be no normal force (no pressure) in the first place. Usually the first thing methods do when solving the INSE is to eliminate the "pressure". For example the stream function-vorticity method finds the incompressible flow without using a "pressure", showing that the "pressure" gradient term is not necessary. A code for another "pressureless" method is given in the CFD-Online wiki pages under educational codes. The conclusion is that the "grad p" term in the INSE is not a pressure. It results from a particular application of the Helmholtz decomposition. But this is likely a fertile area for future research.

I agree, a classical way to see that, is using the weak form you get by using the velocity as test function. This way you get (let me yet call "p")

Int (v.Grad p) dx = Int ( Div (vp)) dx - Int (p Div v) dx
= Int (n.vp) dS

the last integral vanishing under suitable conditions. Therefore, p does not enter into the solution of v.
The property of orthogonality in the Hodge decomposition is not always fulfilled by the BC.s but one can work on
If you like this topic, some years ago I published a paper about that

[Jonas Holdeman]

Originally Posted by FMDenaro

I agree, a classical way to see that, is using the weak form you get by using the velocity as test function. This way you get (let me yet call "p")

Int (v.Grad p) dx = Int ( Div (vp)) dx - Int (p Div v) dx
= Int (n.vp) dS

the last integral vanishing under suitable conditions. Therefore, p does not enter into the solution of v.
The property of orthogonality in the Hodge decomposition is not always fulfilled by the BC.s but one can work on
If you like this topic, some years ago I published a paper about that

Yes, I have read your paper, but I should review it again.

One can formally write an operator to project out the divergence-free part of a vector as (1-grad Delta^-1 div), where Delta^-1 is the inverse of the Laplacian operator (expressible in functional form as a Green's function). Then the second term applied to (u dot grad u) gives the (-grad p) term. But one can also formally write the solenoidal projection operator as (curl (curl curl)^-1 curl), where (curl curl)^-1 is the inverse of the (curl curl) operator. Applied to (u dot grad u) the (-grad p) term never appears. This latter form is most closely related to the "p"-less equation for u.

[FMDenaro]

Originally Posted by Jonas Holdeman

Yes, I have read your paper, but I should review it again.

One can formally write an operator to project out the divergence-free part of a vector as (1-grad Delta^-1 div), where Delta^-1 is the inverse of the Laplacian operator (expressible in functional form as a Green's function). Then the second term applied to (u dot grad u) gives the (-grad p) term. But one can also formally write the solenoidal projection operator as (curl (curl curl)^-1 curl), where (curl curl)^-1 is the inverse of the (curl curl) operator. Applied to (u dot grad u) the (-grad p) term never appears. This latter form is most closely related to the "p"-less equation for u.

yes, using the curl-based projection I think you adopt the second choice of the HH decomposition, suited for the stream-function/vorticity formulation, isn't it?

[Martin Hegedus]

Originally Posted by Jonas Holdeman

The problem is that the assumption that there is a normal force term (pressure) acting on a fluid surface in the case of incompressible fluid flow is fundamentally flawed.

I believe this was discussed before.

For an incompressible fluid, dp/dx (or dp/dt) comes from the boundary conditions.