p0432

# p0432 - sin 2 θ ±M k = 1 r l 6 U r 2 R 2 cos 2 θ − 2 U...

This preview shows page 1. Sign up to view the full content.

4.32. CHAPTER 4, PROBLEM 32 517 4.32 Chapter 4, Problem 32 Problem: In cylindrical coordinates for two-dimensional flow, the vorticity is given by ω = 1 r ^ r ( ru θ ) u r ∂θ ± k Compute the vorticity for the following velocity vectors, where U and R are constants. (a) u = U ( r/R ) 2 sin 2 θ e r + U ² R θ /r +2( r/R ) 2 cos 2 θ = e θ (b) u = UR/r ( e r +3 e θ ) (c) u = U ( r/R ) 2 e r 2 U ( r/R ) 2 θ e θ Solution: (a) The vorticity for this velocity field is ω = 1 r l r X U ^ R θ +2 r 3 R 2 cos 2 θ ±~ ∂θ ^ U r 2 R 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: sin 2 θ ±M k = 1 r l 6 U r 2 R 2 cos 2 θ − 2 U r 2 R 2 cos 2 θ M k = 4 Ur R 2 cos 2 θ k (b) The vorticity for this velocity field is ω = 1 r l ∂ ∂ r (3 UR ) − ∂ ∂θ ( UR/r ) M k = (c) The vorticity for this velocity field is ω = 1 r ^ ∂ ∂ r X − 2 U r 3 R 2 θ ~ − ∂ ∂θ X U r 2 R 2 ~± k = − 6 Ur θ R 2 k...
View Full Document

## This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

Ask a homework question - tutors are online