Problem 4.96 - − kN ⋅ = Applying y momentum R y L x V x...

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Problem 4.96 [Difficulty: 3] Given: Data on flow out of pipe device Find: Velocities at 1 and 2; force on coupling Solution: Basic equations (continuity and x and y mom.): The given data is ρ 999 kg m 3 = D2 0 c m = L1 m = t2 0 m m = p 3g 50 kPa = Q 0.3 m 3 s = From continuity QA V ave = due to linear velocity distribution V ave 1 2 V 1 V 2 + () = Note that at the exit Vx () V 1 V 2 V 1 () L x + = Hence Q 1 2 V 1 V 2 + () L t = 1 2 V 1 2V 1 + () L t = V 1 2Q 3L t = V 1 10 m s = V 2 2V 1 = V 2 20 m s = At the inlet (location 3) V 3 Q π 4 D 2 = V 3 9.549 m s = Applying x momentum R x p 3g π 4 D 2 + V 3 ρ Q = R x p 3g π 4 D 2 V 3 ρ Q =
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Unformatted text preview: − kN ⋅ = Applying y momentum R y L x V x ( ) ρ ⋅ V x ( ) ⋅ t ⋅ ⌠ ⎮ ⌡ d − = ρ − t ⋅ L x V 1 V 2 V 1 − ( ) L x ⋅ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d ⋅ = Expanding and integrating R y ρ − t ⋅ V 1 2 L ⋅ 2 V 1 ⋅ V 2 V 1 − L ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ L 2 2 ⋅ + V 2 V 1 − L ⎛ ⎜ ⎝ ⎞ ⎠ 2 L 3 3 ⋅ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = R y 4.66 − kN ⋅ =...
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