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Unformatted text preview: = dP dy μd2udy2dy dx = d2udy2 1μdPdx = + dudy 1μdPdx y C1 = + + u 12μdPdx y2 C1y C2 C1 and C2 could be obtained from the boundary conditions: 1) No slip at lower wall U(y=0) = 0 C2 = 0 2) No Slip at top wall U (y = l) = 0 C1 = 121μdPdx l Therefore: = + u 12μdPdx y2 12μdPdx l y Maximum velocity occurs in the center at y = l/2 = ==.*  . = . / uy l2 18μdPdxl2 1811 12E 3Pa s 0 06Pa1m1cm21m100cm2 0 00067m s Problem 5 (10 points) Drag on a flat plat could be found from Eq 18.22 = . = . . * * . = . Cf 1 328Re 1 3281 23 1 11 79E 5 0 005 = = . Drag 12Cf ρV2 A 0 015N Since the plate has 2 faces = * = . Net Drag 2 Plate Drag 0 03 N Boundary layer thickness could be calculated using Eq. 18.23 = δ 5xRe Problem 6 (5 points) Following steps in problem 5 yields: = . Cf 0 0014 = . Drag 3 5 N = . Net Drag 7 03N Boundary Layer thickness Decreases but the drag increase....
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 Fall '09
 Force, John Wiley, boundary layer thickness, Solution Manual Problem

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