Question

QUESTION 6 What is the approximate% flow in middle pipe (red on figure below)? Just whole number percent, not decimal format and no " symbol. LAB 1.050 D -24" • L-1,750 • • EL - 1.675 L,- 1.200 D,- 12" • D, 10" • ,- 1.550 D. 14" . Le 1.100 D 300 EL 1.200 • • * 0.0273 1-0.024 -0.022 -0.02

Answer 1

Solution is attached in the image posted below.

Hope it helps.

Minor head losses has been considered zero.

Head loss due to friction is considered only.

1670ft 3 Is Rooft fmain=0.0275 FE: 0.024 f2= 0.022 FB: 0.02 LAB= 1050 L-1750) 12= 1200 13: 1550 LCO= 1100 DED: 30" % fion in michelle Pipe: ie pipe 2 DAB+2U" D=12" D2=10" 03-14 Total Head LOSS HA-HD =1675-1200 Luas ft

Assume Discharge in AB= QAB Discharge in pipe 1,283 Que Q18 Q2693 Respectively | QAB-Qzt Q.+Qz] aschange in pipe co-QAR [wnole flow will pass through the co] - - - - - heAB= FLAR QAB? + me feo Lco QAB 12.1 DABS 12.1 DCD As pipes are in parallel hence the head loss will be constant along 1,28 3. hfi: hfzhfz. विड CAठ व Oमठ F3 L3 Q32 faci QR-F2 L2023 12.1 05 12.1025 12.1035 converting Diameter from inches too feet DE 10 feet Dan Sfeet 032 feet 0024 x1350 Q2.0.022x1200 * Q & 12.1 x CSS 12.1 x 5 -0.02 x 1550 XQ32 12.1 x

3.471 Qi? = $.429 Q22 - 1.185 032 Now Taking Q2 8 Q3 in terms of Qi 3.471 QR-8.429 Q22 Q-suza Q2 13.471 QI = 1.25 Q2 1125 - Q2] => [0-8 Qi = Q2] 3.471Q,2-1 1.185 Q3" 13.471 Q1= 0.5843 Q3 T17114 Qı= Q3] QAB-Qit Q2+Q3 => Q1 +0.8215 1714 Qi TQAB -> 3.514 Q.7 TQAB-Q11 3.51 % Discharge in Micicue pipe pipe? Pipe 2 x180 Pipe lt Pipe 2+ Pipe 3

X100 0.8Q1 Q1+0.8 Qit laulua, => 22.787 Hence percentage discharge 231. Approximate Discharge in middle Pipe is 23%.