The Strouhal number for flow past cylinder is roughly 0.183 as reported by Williamson (link).
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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.
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\large
$$
{\rm{Sr}} = {fD \over U}
$$
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*Sr = 0.183 = f*D/U*
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$$
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For our case, *D = 2, U = 1*
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Therefore, shedding frequency *f = 0.0915*
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Cycle time, *t = 1/f = 10.9 sec*
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Therefore, *Time Step Size = 10.9/25 = 0.436 sec*
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