HW5 sol

HW5 sol - Problem 9.40[2 Problem 9.46[3 Given Find Solution...

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Problem 9.40 [2]
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Problem 9.46 [3] Given: Pattern of flat plates Find: Drag on separate and composite plates Solution: Basic equations: c f τ w 1 2 ρ U 2 = c f 0.730 Re x = For separate plates L 7.5 cm = W 7.5 cm = U1 m s = From Table A.8 at 20 o C ν 1.01 10 6 × m 2 s = ρ 998 kg m 3 = First determine the nature of the boundary layer Re L UL ν = Re L 7.43 10 4 × = so definitely laminar The drag (one side) is F D A τ w d = F D 0 L x τ w W d = We also have τ w c f 1 2 ρ U 2 = 1 2 ρ U 2 0.730 Re x = Hence F D 1 2 ρ U 2 W 0 L x 0.730 Ux ν d = 0.730 2 ρ U 3 2 W ν 0 L x x 1 2 d = The integral is 0 L x x 1 2 d2 L 1 2 = so F D 0.730 ρ W ν L U 3 = F D 0.0150N = This is the drag on one plate. The total drag is then F Total 4F D = F Total 0.0602N = For both sides: 2F Total 0.120N = For the composite plate L 4 7.5 × cm = L 0.30m = F Composite 0.730 ρ W ν L U 3 = F
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This note was uploaded on 05/17/2010 for the course MAE 101b taught by Professor Staff during the Spring '08 term at UCSD.

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HW5 sol - Problem 9.40[2 Problem 9.46[3 Given Find Solution...

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