Problem 4.102 - r d + = R 2 U 1 2 2 u max 2 R 2 2 2 R 4 4 R...

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Problem 4.102 [Difficulty: 3] CS x c y 2 h d Given: Data on flow in 2D channel Find: Maximum velocity; Pressure drop Solution: Basic equations: Continuity, and momentum flux in x direction Assumptions: 1) Steady flow 2) Neglect friction Given data R7 5 m m = Q 0.1 m 3 s = ρ 850 kg m 3 = From continuity QU 1 π R 2 = U 1 Q π R 2 = U 1 5.66 m s = Also ρ U 1 A 1 A ρ u 2 d + 0 = U 1 π R 2 0 R r u max 1 r 2 R 2 2 π r d = 2 π u max R 2 2 R 4 4R 2 = 2 π u max R 2 4 = π u max R 2 2 = Hence u max 2U 1 = u max 11.3 m s = For x momentum p 1 A p 2 A V 1 ρ 1 V 1 A () A 2 ρ 2 u 2 u 2 d + = Note that there is no R x (no friction) p 1 p 2 () π R 2 ρ π R 2 U 1 2 0 R r ρ u max 2 1 r 2 R 2 2 2
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Unformatted text preview: r d + = R 2 U 1 2 2 u max 2 R 2 2 2 R 4 4 R 2 R 6 6 R 4 + + = p p 1 p 2 = U 1 2 1 3 u max 2 + = U 1 2 1 3 2 U 1 ( ) 2 + = U 1 1 3 2 ( ) 2 1 = 1 3 U 1 2 = Hence p 1 3 850 kg m 3 5.66 m s 2 N s 2 kg m = p 9.08 kPa =...
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