Exercise 10: Potential Flow Example 1: Half body ... - KTH Mechanics
Exercise 10: Potential Flow Example 1: Half body ... - KTH Mechanics
Exercise 10: Potential Flow Example 1: Half body ... - KTH Mechanics
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We get the inviscid flow,The vorticity is,u θ (r) =}{{}Ar + B rsolid <strong>body</strong> rotation }{{}irrotational¯ω = ∇ × ū = {A.32} = 1 r∂∂r (ru θ) ē z¯ω = 0 ⇒ ∂ ∂r (ru θ) = 0 ⇒ u θ = C rConclusion:Irrotational ⇒ inviscidInviscid irrotational5