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The Strouhal number for flow past cylinder is roughly 0.183 as reported by [Williamson (link)|FLUENT - Unsteady Flow Past a Cylinder - Step 7 #ref1] .
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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*
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Therefore, shedding frequency *f = 0.0915*
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Cycle time,
{latex}
\large
$$
{\rm{t}} = {1 \over f} = 10.9 sec
$$
{latex}
Therefore, *Time Step Size = 10.9/25 = 0.436 sec*

To accurately calculate the shedding frequency, open the {{cl-history}} file (saved previously in the same location where the original mesh was read) and plot the data using excel for better data representation and graphing option. Take an average of 10 shedding cycles (e.g 10 C{~}L~ peak).
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{latex}
\large
$$
{\rm{Period}} = {{T_2 - T_1} \over {10}}
$$
{latex}\\
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\\
{latex}
\large
$$
{\rm{f}} = {{1} \over {Period}}$$
{latex}\\
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\\
{latex}
\large
$$
{\rm{Sr}} = {fD \over U}
$$
{latex}\\
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An example of Lift Convergence Plot plotted using excel is shown below:
\\  !Excel Lift convergence_sm.jpg!\\
{newwindow:Higher Resolution Image}https://confluence.cornell.edu/download/attachments/107011456/Excel%20Lift%20convergence.jpg{newwindow}