Problem 5.47

# Problem 5.47 - Problem 5.47[Difficulty 4 6 ft y 4 in 1 in x...

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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 Q V A V π D 2 4 But D D 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 V V i D i 2 D i D o D i L x 2 V i 1 D o D i 1 L x 2 V x ( ) V i 1 D o D i 1 L x 2 Some representative values are V 0 ft ( ) 3 ft s V L 2 7.68 ft s V L ( ) 48 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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