Question

Instruction: 1. Solve the first (1) iteration for the given pipe network. For uniformity of checking, use the following value for flowrate; Pipe AB = 300 Pipe BC = 100 Pipe CD = 200 Pipe DE = 100 Pipe AE = 100 2. You may use excel for computation 200 Rab = 6 Rbc = 6 400A 300 Rae = 8 300 Ko JE Rde=7 Rcd=5 300 200

Answer 1

Question AB = 300 BC = TOO CD - 200 DE = 100 DE = 100 values of Qotos different pipes.

200. hapa 6 B You Roar 300 considering clockwise =+ve Rust 300 200 Pipe QoR 2rQol AB BC CD 300 100 200 loo wow it + 540000 + 3600 + 6oooo 1200 - 200000 2000 +7000 o 1400 1-80000 L 1600 & 390000 9800 - DE AE 1 100 1 8 New discharge from each pipe Q- Qo + DQ from Hardy cross method AQ = - ErQ² {128 Qo)

Page - 39.79 AQ - - 390000 9800 New flow rates After I iteration Qap = 300 - 39.79 = 260:20 Q RC = 100 – 39.79 = 60.21 Q co - 200 -(-39.79) = 239.79. Qar = 100 - 39.29 = 60.21 1 N 11 Q AE = 100 +39.79 = 1.39.79.