hi,

I'm using a fractional step incomp. FVM NS solver to calculate flow past a square cylinder. Grid used is cartesian type.

I'm trying to solve all the cartesian grids but the cells denoting the cylinder are sort of "blank out". The velocity for these cells are zero. However, what about the pressure for these cells? I know dp/dn=0 at wall but is the pressure the same for the inner cells which are not in contact with the flow boundaries?

Thanks

Exclude those cells from calculations. Set appropriate wall boundary conditions (e.g. dp/dn=0) at the near-wall elements.

regards

dml

hmm i see...

initially i tried to solve all the equations and i got a symmetrical solution, abt the horizontal centre line.

however, when i tried to reforumulate a new matrix which excludes those cells, i got a non-symmetrical solution. that should not be happening, right?

Btw, in both cases, my velocities near the cylinder are usually very big, due to the large pressure gradient. Anyone has this experience before?

thanks

Up to a certain Re number you should obtain a symmetrical solution (the flow without vortex shedding). So if your program gave symmetrical solution before, the new implementation should give similar solution. Absorbing wall boundary conditions into a near-wall fluid elements by itself does not introduce any unsymmetry.

You have probably made a mistake while implementing the changes to your code. You can for example check if your implementation of dp/dn on east/west and north/south wall faces are correct (the signs of the surface integrals should be different).

What concerns velocities; you may get higher velocities, especially close to the corners of your square cylinder.

regards

dml

Probably, there will some problem in the implementation of code near the corner point. Since the cylinder surface is no-slip boundary and so zero velocity on the surface, I dont think the velocity can be very large near the cylinder. The velocity through out the computing domain will be equal to or less than the inlet velocity. Pressue gradients may be very large.

Its well documented in the literature that the flow is symmetric upto Re (based on the side of cylinder) of approximately 45.

hi, regarding the corner cell of the sq. cylinder, i have an enquiry....

since dp/dn=0, that means the pressure(p) of the corner cell (top left corner)= p(north cell)= p(west cell). is that so?

I didn't manage to get them to be equal after solving the poisson eqn, although the eqn was modified at these cells.

i also tried to let p(corner)=p(north)=p(west) but seems like there is no convergence...

i'm wondering if what i did is illegal...