| Author | Message | ||
     David Shelby (daveshel) New member Username: daveshel Post Number: 1 Registered: 10-2007 |
Hi, I've read through a number of postings related to incompressibility and there has been alot of helpful information. Nevertheless, I am still a bit confused on a few of the issues and had some follow-up questions. I was hoping you could help me. 1. Do you have a more detailed description of how the "div(grad(p))=M*div(U)" equation was derived in the example file "viscous.pde"? I can't seem to arrive at the same equation using the equation of state described. Was the grad(p) term included to help with smoothing? 2. A previous posting used the direct penalty substitution of p=c*div(U) into the Navier-Stokes equation (as opposed to the mixed approach with interpolation on p). Does one approach work better than the other in FlexPDE? 3. I'm still confused as to how pressure boundary conditions can even be defined for incompressible fluids (such as in "viscous.pde") . My understanding was that p was arbitrary, and that only the total stresses (e.g. p+Txx, p+Tyy, etc. where Txx and Tyy are the deviatoric stresses) were known on the boundaries. Thanks in advance for your help. David | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 964 Registered: 06-2003 |
1. As the notes to "viscous.pde" say, "The compressible form of the continuity equation is dt(dens) + div(dens*U) = 0 which, together with the equation of state p(dens) = p0 + L*(dens-dens0) yields dt(p) = -L*dens0*div(U) " Our Tech Note "Smoothing Operators in PDE's" makes the argument that a smoothed representation of p can be achieved by replacing dt(p) by dt(p)-eps*Div(grad(p)), where eps is 2*d*c/pi, d is the spatial wavelength of oscillations to be smoothed, and c is a typical signal propagation speed. In steady state, we drop the dt(p), resulting in the equation div(grad(p)) = (L/eps)*dens0*div(p) The choice of (L/eps) is somewhat arbitrary, as it encapsulates ideas of smoothing distance, stiffness of material, etc. 2. The direct penalty substitution is a more common application of the penalty pressure idea. However, we find it to be sometimes rather harsh, and the diffused version to allow smoother solution. This is not always true, and I don't know a general rule for how to choose. Try it both ways and see what happens. 3. The "pressure" equation is really a device for imposing div(v)=0. Since only derivatives of P appear in the equations, the absolute value of P is arbitrary. Imposing value BC's on P is really just a means of imposing an overall pressure drop, or average gradient. In the equation div(grad(p))=..., FlexPDE will integrate the div(), resulting in surface terms whose value must be supplied by a Natural BC. Natural(p)=0 implies that pressure contours meet the boundary at right-angles, which may or may not be true. P. Gresho makes an argument that you can differentiate and combine the momentum equations and get a pressure equation and boundary condition. I think this is a tautology, and adds no new information. It is also destabilizing. In short, I don't know an answer to the question. | ||
| David Shelby (daveshel) New member Username: daveshel Post Number: 2 Registered: 10-2007 |
Thanks Robert, Your response has been very helpful. I will try the various solution strategies and see how they ultimately affect the results. Rgds, David | ||
| Adam Burbidge (skippington) Member Username: skippington Post Number: 6 Registered: 10-2007 |
Dear David, for incompressible fluids the pressure is not really arbitrary just because the equation of state is degenerate. In fact in 3D you have a system of 4 equations (3 NS, 1 for each velocity component and 1 incompressibility relation between the 3 velocity components (div u vanishes). If there were no pressure then we would have 4 equations to determine 3 dependent variables (the velocity components) and the system would be over specified. The consequence of this is that a part of the pressure is kinematically determined and not really arbitrary as such. Generally speaking the pressure is traditionally defined as -1/3 of the trace of the total stress tensor so that it coincides with the inviscid definition where positive pressures are compressive and not tensile as the normal sign convention requires. Of course you can add an arbitrary offset to your pressure variable and this will not affect the velocity field (unless you have a free surface) such that using gauge or absolute pressures makes no difference. I guess in that sense you could say that it is arbitrary to some extent, although perhaps it would be clearer to say that the pressure field is fully determined but for and unknown constant. To solve the incompressible NS equations you should therefore specify at least a single point pressure value in the same way as you would when integrating any other 1st order (wrt pressure) differential equation. Hope this helps to clarify a bit Cheers Adam | ||
| Adam Burbidge (skippington) Member Username: skippington Post Number: 7 Registered: 10-2007 |
Dear David, for incompressible fluids the pressure is not really arbitrary just because the equation of state is degenerate. In fact in 3D you have a system of 4 equations (3 NS, 1 for each velocity component and 1 incompressibility relation between the 3 velocity components (div u vanishes). If there were no pressure then we would have 4 equations to determine 3 dependent variables (the velocity components) and the system would be over specified. The consequence of this is that a part of the pressure is kinematically determined and not really arbitrary as such. Generally speaking the pressure is traditionally defined as -1/3 of the trace of the total stress tensor so that it coincides with the inviscid definition where positive pressures are compressive and not tensile as the normal sign convention requires. Of course you can add an arbitrary offset to your pressure variable and this will not affect the velocity field (unless you have a free surface) such that using gauge or absolute pressures makes no difference. I guess in that sense you could say that it is arbitrary to some extent, although perhaps it would be clearer to say that the pressure field is fully determined but for and unknown constant. To solve the incompressible NS equations you should therefore specify at least a single point pressure value in the same way as you would when integrating any other 1st order (wrt pressure) differential equation. Hope this helps to clarify a bit Cheers Adam |