vortex shedding

Hi Dear All,
Can any one know about the following things:

  1. If we have a situation like we not get shedding even for Re=60 for square cylinder. I know that some peoples gives some disturbance initially to start the shedding and then they stop the disturbance and apply the normal inlet condition.
    Can any body tell how to do such things.
  2. In some cases we used the small time step like dt=0.1 but for each time step then we need to convert it to integer value like
    dt=dx=0.5 then step=(int)(xc/ dx+0.1) why we need to change into integer i not find any difference in the results of dt=dx=1 and dt=dx=0.5 except little faster computational time.
  3. How to increase the number of grid size in the same computational domain like if we have a square domain of 1cm and if we used Xmax=20 and Ymax=10 so how i can increase the number of grid size in this 1cm square domain like Xmax=40 and Ymax=20.
    Thanks in advance.
    Khan

Hi dear khan
U can start your run with partitioned velocity in inlet. I mean that u can for some small time in start of your run, set the velocity in part of inlet velocity more than other part. for example u can set u=0.15 in part of inlet that located top of centerline, set u=0.5 in bottom of that, set u=0.1 in remain part of the inlet. But just for little time!!!
u can change your grid size, but the nondimensional number like Re,Ra,Pr,… should be fixed!!
u can change ur grid size from 10 to 20 in y direction for example, but u should now change the velocity to get a fixed Re number!
in this simulation the number of grid in x direction is not important! but for simulate better the outflow B.C should be far enough from obstacle!