p0436

# p0436 - dx + 8 H − H [ v ( L,y ) − v ( − L,y )] dy...

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4.36. CHAPTER 4, PROBLEM 36 409 4.36 Chapter 4, Problem 36 x y Contour C 2 H 2 L i dx i dx j dy j dy ............... . . . . . . . . . . . . . . . . . . . . . . . . . . ...... ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...................................................................................................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... .... .......................... . . . . . . ............... . . . . . . . . . . . . . . . . . . ... Figure 4.8: Rectangular contour. In general, to evaluate a line integral, we treat the integral as the sum of conventional integrals on each segment of the contour, with the sign determined by the differential distance vector, d s . Hence, referring to Figure 4.8, we see that Γ = 8 L L u ( x, H ) · ( i dx )+ 8 H H u ( L,y ) · ( j dy ) + 8 L L u ( x,H ) · ( i dx )+ 8 H H u ( L,y ) · ( j dy ) = 8 L L [ u ( x, H ) u ( x,H )]
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Unformatted text preview: dx + 8 H − H [ v ( L,y ) − v ( − L,y )] dy Thus, for the given velocity vector, Γ = 8 L − L [ U − U ] dx + 8 H − H ^ 6 U w L H W 2 − 6 U w L H W 2 ± dy = 0 − 0 = 0 To check for consistency, note first that the vorticity is ω = X ∂ v ∂ x − ∂ u ∂ y ~ k = w 2 6 U x H 2 − W k = 2 6 U x H 2 k So, the integral of ω · k is 88 A ω · k dA = 8 H − H 8 L − L 2 6 U x H 2 dxdy = 2 6 U H 2 (2 H ) 8 L − L xdx = 4 6 U H } 1 2 x 2 ] x = L x = − L = 0 This is identical to the result obtained above....
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## This note was uploaded on 10/12/2009 for the course AME 309 at USC.

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