Question

1,5 2500 4" A B 3500 5" 3000 4 2000' Icts 8 c E D 1.5cts 1500' 6000' USE THE MOODY FRICTION FACTOR OF f-0.02 TO CALCULATE THE FLOW IN EACH PIPE IN THE NETWORK SHOWN ABOVE, + CHOOSE CLOCKWISE AS THE POSITIVE DIRECTION SINCE THE MOODY FRICTION FACTOR IS GIVEN THEN USE THE DARCY FRICTION LUSS EDUATION USE THE IDENTIFICATION SYSTEM SHOWN ON THE NETWORK

Answer 1

f ho has Loop zie BCD (1st itoration). Assumed Q (46) 0.25 -0.75 -0.5 0.02 0-02 IIS 35.076 72.499 3.822 6000 20.5 72 0.02 کے 111.397 AQ= Consider Pipe 4 D (ft) helft) BC 3500 9/12 = 0.41867 8.769 1-54.374 DB 2000 %-0.6667 -1.911 1-47.516 -(-47.5116) 0.4265 111.397 Corrected Discha charge for loof 2 pipe Q = (a + AQ! BC 0. 67 65 Cfs. CD -0.75 +0.4265 z BAB - 0.5 +0.4265 = -0.0735 efs. iteration will be done until sh=0 for both Loops. 0.25 +0.4265 a -0.3235 cfs

in the priße, 18 Sol Assime discharge = 75 06, Que - 754, 2 0.50fs, de 0.25 of Roc 0.25 ofs, Qoe-0.75 fs. 1.5645 A 0.75cfs B 3.25.45 lefs Spo 0.24 5.15cts Osafe Head loss, he = fl12-8fhQ2 Ist iteration. 2go gx²15 Consider Loop 1 i.e. ABDE: Loop teie Piße . Assumed (!(182) LA D (ft) helift) AB 0.02 2500 42= •333179.05 BD 0.02/2000 % = .667 1.9012 0.02 /1500/2.667 -0.358 3000 %= -333-206.55 has) 0.75 0.5 229.4 3.824 1.432 DE - 0.25 EA - 0.75 0.02 275.4 510.056 -32.946 0.064593 fts 2 x 510.056 -(-32.946) Corection for discharge, 4Q -- -- 22 hyQ Corected discharge par loop discharge, 2- Bassured +AQ 0.75+0.064593 = 0.8146 cfs. 0:5 +0.064593 -0.5646 fs. -0.25 +0.064593 = -0.18544 cs. - 0.75 +0.064593=-0.68544 ofs. Pipe / corrected AB BA DE EA