Question

Q2. Find the flows in the loop given the inflows and outflows. The pipes are all 30 cm cast iron (e=0.26 mm), given f as 0.02. 0.28 m3/s T 0.42 mº/s 90 m 10.50 mº/s -0.50 m/s 0.64 m/s 170 m

Answer 1

we have to use. solution:- Given", - A р 9.42 m sec x0.28 m/see Com (sec. Qo9 0.5m/sec ->0.6ungsec * 170m @=0.26mm f=0.02. diameles of each pipe= 30 cm 0.3 m = 0.3m .. Radius= R = 0.3 = 0.15 m. We have one loop i.e. to solve this .. ABDC А. We have to assume we have QAB OBD-Q Co - QAC and then to do modification (i-e.AQ) - Erar {/2nal hedge thes formula 91 fL In 12.15 because; head loss fL Q" 12.125.

For pepe AB and co 913 = 115.63 and for pope BD and he 7:120%se 0-3-0.28 = 0.02 molsce 0.02x1708 12.140.35 -QBD r = 0.02x90 = 61.218 12.180.35 let us first assume for trail 1. as below QAB= 0.3 m/sec = 0.12m 0.42m/A 0.3mile B 0.28 m²/sec QAC 0.42-03 28.64 morse Trail 1- Qepto.o2 -0-64 Qcb=0.64-0.02 - 0.62m² sec Q-03 r=615.63 rz 61.218 D = 61.218 Q=0.02 Q=0.12 r=115.63 Q=0.62 nor Hong Pipe AB BD DC Izral 10.4067 62.378 0.02448 2.44872 -44.4482 143.3812 -0.8815 14.6923 Eram=-34.898/229.2002 СА

Eror _-(-34-898) E1200l 229.9002 -OQ = 0.1518 molec Asley 1st Trail. After ist trail 0.3 +0.1518=0.4518 10.02+0.1518 -0.178 0. 12-0. 1518 -0.0318 0.62 I -0.1518 = 0.4682 because flow is anticlockwise ie opposile to our loop consider ration . After ist trail ; 0.4578 0.0318 0.1718 Prail 2 0.4682 Pipe &Q? Izral 104.4832 23-6027 1.8068 21.934 2759 AB BD DC CA - 25.3474 0.0619 {=0.124 108.2759 3.8934 {2237.6865

-0.124 - ErQw 2]zval 237.6865 AQ=-0.00052169 'AQ ř [AQ Not Therefore no further modification is required QAB=0.45 18 m/sec. Flowing from À to o = 0.1718 m/sec Floroing from 8 toid! O co = 0.4682 m/sec Flowing from 'c' to do CA = 0.0318 myse Flowing from * to é. QBO & Answer. 0.42 rise 0.4518 misec 0.28 m² (see 10.0318 male se 0.1718 m/sec. 0.5 miled a 챰 -> 0.64 m/sec. 0.4682 m/sec