| Author | Message | ||
     Ken-Ming Yin (kenming_yin) New member Username: kenming_yin Post Number: 1 Registered: 03-2005 |
I look at the fluid mechanics application of flexpde: LOWCISC.PDE. This problem transforms the original N-S equations by adding the derivative of U-equation to x and derivative of V-equation to y so that a new equation about pressure is arrived. I wonder how to apply to the cylindrical coordinate system? I have a problem of a cylindrical via (hole). fluid may flow along the outer-surface (90 degree to the axial of the via), or , it may flow perpendicular to the outer surface and right through the via. | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 403 Registered: 06-2003 |
There are many ways to construct an equation for the pressure. The one you mention was used in the Lowvisc.pde sample in version 3. In version 4 and 5, the formulation has been changed to the "Penalty Pressure" approach. The notes to the example "Swirl.pde" explain this formulation as follows: ------------------------------------ In principle, these equations are supplemented by the equation of incompressible mass conservation: div(U) = 0 but this equation contains no reference to the pressure, which is nominally the remaining variable to be determined. In practice, we choose to solve a "slightly compressible" system by defining a hypothetical equation of state p(dens) = p0 + L*(dens-dens0) where p0 and dens0 represent a reference density and pressure, and L is a large number representing a strong response of pressure to changes of density. L is chosen large enough to enforce the near-incompressibility of the fluid, yet not so large as to erase the other terms of the equation in the finite precision of the computer arithmetic. The compressible form of the continuity equation is dt(dens) + div(dens*U) = 0 which, together with the equation of state yields dt(p) = -L*dens0*div(U) In steady state, we can replace the dt(p) by -div(grad(p)) [see Help | Tech Notes | Smoothing Operators in PDEs"] resulting in the final pressure equation: div(grad(p)) = L*dens*div(U) --------------------------------------------- This same "Swirl.pde" example shows the equations in cylindrical geometry, and include the azimuthal velocity component as well. You should take a look at that sample. | ||
| Ken-Ming Yin (kenming_yin) New member Username: kenming_yin Post Number: 2 Registered: 03-2005 |
I looked at the Swirl.pde carefully. 1.(i) about boundary condition from (0,0) -> (rad,0) how natural(vr)=0, natural(vt)=0 is given? (ii) from (rad,ht) -> (0,ht), how natural (vr)=0, natural(vt) is given? Also, can you give comment on value(vz)=0 there? | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 405 Registered: 06-2003 |
The only terms contributing to the definition of the Natural BC in this problem are the viscous terms. Therefore, Natural(vr) means the outward normal component of dens*visc*grad(vr), and similarly for the other velocity components. Setting these zero is essentially a slip condition. The swirl problem was formulated for fixed mesh, so the deformation of the top surface of the fluid is not computed. Value(vz)=0 means the fluid can't escape out the top surface. |