Problem 5.47 - Problem 5.47 [Difficulty: 4] 6 ft y 4 in 1...

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Problem 5.47 [Difficulty: 4] 4 in 1 in 6 ft x y Given: Flow in a pipe with variable diameter Find: Expression for particle acceleration; Plot of velocity and acceleration along centerline Solution: Basic equations: Assumptions: 1) Incompressible flow 2) Uniform flow Continuity reduces to and for the flow rate QV A V π D 2 4 But DD i D o D i  L x where D i and D o are the inlet and exit diameters, and x is distance along the pipe of length L: D(0) = D i , D(L) = D o . Hence V i π D i 2 4 V π D i D o D i L x 2 4 VV i D i 2 D i D o D i L x 2 V i 1 D o D i 1 L x 2 Vx () V i 1 D o D i 1 L x 2 Some representative values are V0f t ()3 ft s V L 2 7.68 ft s VL () 4 8 ft s The acceleration is given by
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For this flow a x V x V a x V i 1 D o D i 1 L x 2 x V i 1 D o D i 1 L x
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This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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Problem 5.47 - Problem 5.47 [Difficulty: 4] 6 ft y 4 in 1...

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