Explanation Given information: Lengths, AB = 2000 ft. BD = 1500 ft. Such that, AC = BD AB = CD Friction factor, f = 0.022 Pressure at A, p A = 90 l b f i n 2 g a g e . Diameters of all pipes, D = 6” Formula used: h f = 8 f L Q 2 π 2 g d 5 p B = p A − ρ g K A B Q A B 2 For water at 20°C, take ρ = 1.94 s l u g f t 3 a n d μ = 2.09 E − 5 s l u g f t . s Each pipe has head loss which is known except for the square of flow rate: Pipe AC: h f = 8 f L Q 2 π 2 g d 5 = 8 ( 0.022 ) ( 1500 ) Q A C 2 π 2 ( 32.2 ) ( 6 12 ) 5 = K A C Q A C 2 p B = p A − ρ g K A B Q A B 2 Where, K A C ≈ 26.58 Similarly, K A B = K C D = 35.44 K B D = 26.58 , and K B C = 44.30 The basic flow relations are: L o o p A B C : 35.44 Q A B 2 + 44.3 Q B C 2 − 26.58 Q A C 2 = 0 L o o p B C D : 44.3 Q B C 2 + 35.44 Q C D 2 − 26.58 Q B D 2 = 0 J u n c t i o n s A , B , C : Q A B + Q A C = 2

8th Edition

ISBN: 9789385965494

Author: Frank White

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 6, Problem 6.130P

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The flow rate in all pipes and pressures at B, C, and D for water at 20°C.

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Chapter 6, Problem 6.129P

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Chapter 6 Solutions

Fluid Mechanics, 8 Ed

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