Question

Water at 20° flows in pipeline system shown below (Fig. 1) at rate of 32400 Vhr. Pipe 1 is 15m long and 0.044 m outer diameter and pipe 2 is 100m long and 0.034m outer diameter. The thickness of both pipes is 2mm. Pipe I is a ductile iron uncoated pipe and pipe 2 is ductile iron coated pipe. There are totally 4 elbows in the system (two elbows in each pipe) and the elbows are 90° street elbows. All pipes are located on the same elevation. Consider major loss and minor loss for both pipes. 1. Calculate the flow rate and head loss in each pipe (10 marks) 2. Calculate the pressure drop between point A and point Blocated on either sides of the main pipeline (5 marks) Note: Attach your calculation on moody diagram to the solution. This forms part of total mark. Branch 1 Main Pipe B Main Pipe Branch 2 Fig. 1

Answer 1

ANSWER :
Deschooge in System : Q = 32400 dlhv = 9415. Assume discharge in pipe 1 and pipe 2. QI 6lls. O-315. clockwise = + 2 ecw=- 0.061 0.06) 44-2 42 EL Relative Roughness of pipet, (&= 0.061 mm. for ductile irons, un coated) 0.00145 0.12 = 0.00375 Relative Roughness of pipe 2, E2 Dz 34-2 m/s. (€ 80-12 min for ductil Tron, coated pppe) kinematic vascosity of wates at 20°C,9=10004x10-6 pppe alus) f2v2 Assumed velocity Elp Rea D friction he=s (ml) x16 (x105) factor of 270 10811 0023 be a ms/ G Branch 1 4.33 | 145 1.8496 1308267 Braunch -3. -7.46875 2.38 10.028 -203.87 67956.67 E 196:02 69264.94 corrected discharge oQ=- She - (-196-02) 2qbla 2x 69264.94 pipe 0Q = 0.001415 MB/s 51.415 419. corrected discharge, a= a to bt1.415 7.415 ils - 3+1415 +1.585 Als-- 9

and Pterations: (Kelbow = 0.5) aceis (m) slo Re = VD of pipe (x105] (x105) h = fly2 2 Kelbow' 294 + 29 9 1 70415 5.35 145 0-02 111 879 2.24 0.628 11602.02 2 -1.585 -1.971 395 0.026 -16.2858 10.275 E -4.4068 1612 • 295 corrected Discharge, a.a = -Ehl QEhlla (-404068) 2x 1612. 295 SQ = 0.001367 m3/3 = 143666 lls. pipe corrected discharge, 8 = 0+08 7.415+1-3666 8.7816 dls. - 2 31.585+1.3666 = -0,2184 bls. 3rd steration: EID Re-Volo flr2 f he pipe Q (4/s)/(m/s) 12 +2K elbow X265) x(105) ha 29 8.7816 6.34 29d 0.022 18.146 145 2.65 2066.367 2 -0-2184 0.21 315 0:086 0·032 -0.378 1730.77 ² 170768 3797.13 corrected discharge DQ = - che - 17.768 QEhlla 2x 3797, 13 0Q==0.002339 mols 3-233966 llo

pipe corrected discharge 1 Q=Q+0Q. 8.7816 ~2:33966 = 6.44194 s. 2. -0.2184 -2,33966 2.558 Ils 2 +2 Kelbow Revo V (1) Discharge in Branch , Q = 6.442 l/s. Discharge in Branch 2, Q2 = 2.558 l/s. Head loss in pppe. 1, H₂, = f, 2, 4² egd, 29 Q, = 6.442 els = 0.006 442 ms. V = (0.006442)/7/4 10.042)2 =406498 tlg. 4.6498x 8.042 1.945x105 1:004x 106 ELD 145X105 resping moody friction factor, f-00024 Head loss, HLE 0.024 x 15x (406498)2 ax0.5x (4.6498.) 2x9.81 X0.042 2x9:8) H21 = 10.547m = 2.558 ills 0.002568 mals. = 3.1806 m/s. 3.180 6x 0·032 Volo 12013X 1045 pooux 166 diagram Q2 Re

375X1001 4 Elo usping moody diagram, 120.03. Head loss in pipe 2, H lg = 0903x100x (8-18)280-5x (3-1892 2x9.81 2x9.81 X0-032 HL₂ = 48288 m. (2) Applying Bernoullis equation tion blw point A and point B PA ในง t VAR PB rw. 29 + ZA 29 + Vo? + Zp + H₂ ZA = 2B 1 29 29 3/3 PA fw PB ow <H₂ = H₂, + H L₂ = 10.5477 48,88 m. PA PB = Vw (59.427) = 9.807x 59.427 = 582.8 kailma pressure drop between A and 6 = 58208 kalm2.