Question

Problem 1 (60 points; pipes in parallel & EGL/HGL) A large-size pipeline carrying a water flow (Q) of 0.2 m/s branches into two relatively smaller pipelines (1 and 2). The two branches have the following characteristics: • Pipeline 1: length - 500 m; diameter -0.2 m; friction factor -0.01; two different bends. Pipeline 2: length - 800 m; diameter - 0.3 m; friction factor -0.02; two different bends. The plan view of these pipelines is shown in Figure 8.1. The minor loss coefficients of the bends are annotated in the figure. The minor head losses at the entrance and exit of the junction are negligible. (a) Please calculate the total head loss from Section A to Section B (marked by red dash lines). (b) Assuming the total head at the junction point is 100 m. Please draw the EGL and HGL along the 500 m (-100 m+ 300 m + 100 m) of Pipeline 1. Assume a = 1 throughout all the pipelines. 300 m K= 1.5 K= 1.1 100 m 100 m 2 = 0.2 m/s

Answer 1

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Soln 300m QiA,V 100m P2-AV 30om Pipe Line f001 So0, di0-2 M Rpe Line 2 2800m, dz0.3m, For pipes in parallel, + 0.3 O. 2 = 0-2 ,0.3 = 0.255 0.04 Vt0.09V2 = 0-255 Multiply by loo troughout, = 25.5 4V, t 9V2 25om

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Total bead loss in pipe Total bead Aoss in pipe(2). Total head lois in pipe Minor loss at bends Major loss +Kiv 2gd 29 29 2 .01X 500x V+ (t1+)-5)v 2X9.8 O 2 x 21 6 V,2 2x9.8 1407v, Total bead loss in pipe 2 Major loss Minov loss at bends 23 dr 24 o.02x800xV + (+1s) v 0.3 2x9.8) x55.933 2 2x9.81 2.851 V2 above tso head losses Equating tha I- 407 2.851v,2

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V, 2.026 V2 =) I423 4 V2 V Substitute in 4x 14234V2+ 9V225 S 1-7354 m/s From 24702 m/s Tbere-fore, total head loss from section A to section B in To tal bead Joss O of Total bead loss 407 =1407x (2.4702) 8-5853 m b) Eoergy Gradient Line (E-GL) Total head with respect to any arbitrany dartum is -the Sum of potential bead, pressuve and veloci by head. z+ V Total bea d = 29

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Whn the tiuid Hlows aloog the pipe there is loss of head Cenergy and the tota energy decreases in the directon of fiow. If -the total energy Cor head) at various points along the axis of the pipe is plotted and joined by a line Ene rgy Gradient Line -the Oine so obtaioed is called Hydraulic Gradient Line (HGL) The sun of poten tial bead and pressure bead at any point is called the Piezome tric head. f a Qine is drawn joining the piezome trie levels at vanious Points in a pipe he line so obta ined is call ed Hydra- ulie Gradient Line only Considering pipe line

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HA To tal head at junction A 10om AB Head lo due to friction = flv 2gd 2 0.01X100x(2-410) 2x 9-81x 0.2 =555m Total bead at B To tal bead at A-Head los due to fniction in AB HA-hfAB - (100-1-65)m 98.445m At B, there is head loss due to bend. 2 KVL Mx2.4702) X O.3421 m 2x9.81 .:, Total bead at B, HeB= 98.445-o 342) 98-103 m Head loss due to friction n Bc, o01x 300x (24302) 2gd =4-66Sm 2X 9.81x 0. 2

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Total bead at C.- He- hfec = 48.103 4.665 93.44m At c, there is bead loss due to be nd 2 Ky 5x (2-4+02) 29 0.4665m 2 x9.81 Total head at C, H=93.44-O 4665 92.9735m Head loss due to fniction in cD, 2 flv .lx 100x (2. 4702) 11 1-55m 2gd 2x9.81x0 2 Total bead at D, Ho He-hfep 42.113 5 SSS1 = 91.42m Joining the total head at points A,B, C and Dwe get the Energy Gradiest Line 24702) lo get H.G- L, subtract v 29 0.311 m from 2x9.81 Hh total head at A, B, C and D. (92.9735-0 311) 92-66m He 99 689 m H=too-o-311 41-42-0-31) Ha=98-103-0-311 9.8m 91-11 m

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A BCD = EGL GL EFGH 02 99-689E 8-103 B 98 97-8 F E.G.L 96 H-G.L 94 92-93 92-66 92 42 91-11 Н 90