Question

• The diameter of a pipe carrying water changes gradually from 6 in. at A to 18 in. at B. A is 15 ft. lower than B. What will be the difference in pressure, in psi, between A and B, when 6.2 cfs is flowing, loss of energy being neglected.

Answer 1

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de - 2 2 Given 6 in 2 l = 0.5ft do 18 in = 1.5ft, Q = 6.2 efs hp-hA 15 ft P = Gola 1b/ft² 62-4 g= 32.2 ft/s² (PA - P₂) - ? Using continuity equation, AK - Ag = Q AAVA ABVB * (dA v A (d) va dA VA da VB 1672 (0.5)² VA = (1.5) ve 9 vg 4Q Also, = VB = = 2 - 4x6.2 A B VB A d 3.14 x 1.5) Using Bernoulli's = 3-5 teftis principle, PA + + PVA + Pg ha O P + 1 / P ve + f g ho Po-Pe pg (ho hA) + + P (ve? ) PA - P₂ pg (ho-hA) + (- 400 V ) 62.4* 32.2) (15) - 40 x 62.4 x (3:51) PA - PR =-611.8 Iblft PA - 1 - 611.8 PR B - psi ~ - 4.25 psi

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