p0458

p0458 - 4.58. CHAPTER 4, PROBLEM 58 551 4.58 Chapter 4,...

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Unformatted text preview: 4.58. CHAPTER 4, PROBLEM 58 551 4.58 Chapter 4, Problem 58 Problem: The velocity vector is u = U (1+ x/h) i − Uy/h j − U k for an incompressible flow, where U and h are constant reference velocity and length scales, respectively. ˙ For a fluid of density ρ, compute the x-momentum flux, Px , across the square planar area of side h shown, which is parallel to the xz plane at y = h. z ................ . . . . . . . . ..... .. .. .. .. .. .. ... .... . . . . . . .. . . . . . .... . ..... . ...... . . .. . . . ........... . . ........... .. . . ............... . . .............. ....... . .............. ................. . . . . . ... . . ............ .. . ................... .................... .. . . .. . ... .. . . .... ............... . . . .. . . . . .... . . . ....................... ................................................... .. . . ........................ . ........ ........... .. ...... .. . . . .. .. . ....... . . . . .... . . .... . .. .. . . ... . ....................... ............................................ . . ... .. . ............... . .. .............................................. ....................... ... .. ..................... . . .................. . .. ........ .. ... . . .......... .. .. . . . . .. .. ... . . . .. .. ............... . .......... .. . ..... . .. . ............ ....... .. . . .. .. ....... .. .. .. . ... . . . . . ..... . .... .. .. .. .. .. ..... .. ..... ... .. .. .. .. ... .. ... .. .. .... .. h n h y h x ˙ Solution: Call the momentum flux Px . Then, ˙ Px = A h ρu(u · n)dA = h = 0 ρuv dA A 0 = −ρ U 2 ρ U (1 + x/h) (−U y/h)|y=h dxdz h h (1 + x/h)dxdz 0 = −ρ U 2 h2 = −ρ U 2 h2 0 h (1 + x/h)d(x/h) 0 1 0 3 = − ρ U 2 h2 2 1 (1 + ξ )dξ = −ρ U 2 h2 ξ + ξ 2 2 ξ =1 ξ =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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