Page History

The Strouhal number for flow past cylinder is roughly 0.183 as reported by [Williamson|FLUENT - Unsteady Flow Past a Cylinder - Step 7 #ref1] .
\\
In order to capture the shedding correctly, we should have at least 20 to 25 time steps in one shedding cycle. Let's use 25 for our case.
{latex}
\large
$$
{\rm{Sr}} = {fD \over U} = 0.183
$$
{latex}
For our case, *D = 2, U = 1*
\\
Therefore, shedding frequency *f = 0.0915*
\\
Cycle time,
{latex}
\large
$$
{\rm{t}} = {1 \over f} = 10.9 sec
$$
{latex}
Therefore, *Time Step Size = 10.9/25 = 0.436 sec ~ 0.4 sec*