p0510 - u = k ^ 1 r r ( ru ) 1 r u r = k ^ 1 r r w 2 U R r...

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5.10. CHAPTER 5, PROBLEM 10 459 5.10 Chapter 5, Problem 10 The flow is incompressible if the divergence of the velocity vanishes. To check: · u = 1 r r ( ru r )+ 1 r u θ ∂θ = 1 r r w U R r 2 cos θ W + 1 r ∂θ w 2 U R r sin θ W =2 U R cos θ 2 U R cos θ =0 Therefore this velocity vector does correspond to an incompressible flow. The flow is irrotational if the curl of the velocity is zero. To check:
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Unformatted text preview: u = k ^ 1 r r ( ru ) 1 r u r = k ^ 1 r r w 2 U R r 2 sin W 1 r w U R r cos W = k } 4 U R sin + U R sin ] = 3 U R sin k W = Therefore, this velocity field does not correspond to an irrotational flow....
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