p1010 - 1328 CHAPTER 10. VORTICITY, VISCOSITY, LIFT AND...

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Unformatted text preview: 1328 CHAPTER 10. VORTICITY, VISCOSITY, LIFT AND DRAG 10.10 Chapter 10, Problem 10 Problem: Consider viscous flow close to a solid boundary for which u=U y y +K h h 2 i where y is distance normal to the boundary, U is a constant reference velocity and K is a dimensionless constant. Also, h is the thickness of the viscous layer and is independent of x. (a) Using the closed contour shown, compute the circulation in terms of K , U and ∆x. (b) Is there a value of K for which the circulation is zero? ds = − dx ds = −j dy ds = j dy h y ds = dx .......................................................................................................................................................................................................................................................................................................................................................................................................................... . ......................................................................................................................................................................................................................................................................................................................................................................................................................................... .. . . .. ......... . . .. . .... ..... ........ ....... .... . ..... . ..... ...... . ... . ... ..... ..... . ............................................................................................................................................................................................................................................................................................................................................................................................................................. ............. . . . . . ..... .. . .......................... ... ... . ........................................... .................................................................................................................... ....... ........................................................................................... . . . . . ..... . .. . ......... ........... . .. . ................................................................................................................................................................................................................................................................................................................................................................................................................................... ................................................................................................................................................................................... ... ......................................................................................................................................................................................... . . ...... . . . . . ....... .. . ....... ... ... ....... ..... . ... ...... .. ............................ ............ .... . ............................... ..................................................... . ..... .......... ..................... ....... ... ..... . . . .. . . ...... ... . .. ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ . . ......... . .................. ........... ... . ..... .. .................................................... ............................................................................................................................................................................................................................... ................................................................... ...... . . ....... ... . . .. ......... .... ............. .... ........... .... ....... x = xo x x = xo + ∆x Solution: By definition, the circulation is Γ= C u · ds where we use the standard mathematical convention that integration is positive in the counterclockwise direction. (a) 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, ds. Hence, referring to the figure, we see that xo +∆x Γ= xo xo +∆x + xo xo +∆x = xo u(x, 0) · (i dx) + h 0 u(x, h) · (−i dx) + u(xo + ∆x, y ) · (j dy ) h 0 u(xo , y ) · (−j dy ) h [u(x, 0) − u(x, h)] dx + 0 [v (xo + ∆x, y ) − v (xo , y )] dy 10.10. CHAPTER 10, PROBLEM 10 1329 For the given velocity vector, we have u(x, 0) = 0, u(x, h) = (K + 1)U, v (xo + ∆x, y ) = 0, Therefore, the circulation is xo +∆x Γ= xo [−(K + 1)U ] dx = −(K + 1)U ∆x (b) Obviously, the circulation vanishes when K = −1 v (xo , y ) = 0 ...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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