HW10 Solution - = -dP dy d2udy2dy dx = -d2udy2 1dPdx = -+...

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Homework 10 Solution Manual Problem 1 (6points) SI (kg /m*s) English (lb*s/ft^2) Air 1.79E-5 3.74E-7 Water 1.12E-3 2.34E-5 10W30 1 0.02 Munson, Bruce, Donald Young, and Theodore Okiishi. Fundamentals of fluid mechanics . 4. Appendix B. John Wiley & Sons Inc, 2002. Print. Problem 2a (5 points) The sheer force is the only force in the tangential direction. = = = . - = . Ft τA μdudyA 1 79E 5Nsm21ms1 mm1000mmm 1m2 0 0179 N Problem 2b (5 points) Stagnation pressure creates the normal force on the plate. The stagnation pressure can be calculated using be Bernoulli equation. = = = . = . Fn PA 12ρV2A 121 23kgm31ms21m2 0 615 N
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Problem 3 (5 points) The ratio of normal to tangential force is presented bellow. = = = = FnFt 12ρV2A μVL A 12ρVLμ ρV12Lμ Re12L Problem 4 (10 points) The force balance will yield the equation shown bellow. Note that positive forces were in negative x direction. - = - P2dydz P1dydz τ2 dx dz τ1 dx dz Next note the following: = = τ2 μdudy | y y2 = = = + τ1 μdudy | y y1 y2 dy = + τ1 μ dudy d2udy2dy (P2-P1) =dP Substituting these equations into the force balance and canceling dz will yield
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Unformatted text preview: = -dP dy d2udy2dy dx = -d2udy2 1dPdx = -+ dudy 1dPdx y C1 = -+ + u 12dPdx 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 = 121dPdx l Therefore: = -+ u 12dPdx y2 12dPdx l y Maximum velocity occurs in the center at y = l/2 = =-=-.-* - . = . / uy l2 18dPdxl2 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 V2 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 = 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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HW10 Solution - = -dP dy d2udy2dy dx = -d2udy2 1dPdx = -+...

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