PIPE NETWORK ANALYSISPresented byAzaz Ahmed. CIB-09-015. Department of Civil Engineering School of Engineering, Tezpur University, Napaam 784028, Tezpur, Assam, India : Pipe Network D Flo ou w t .Municipal water distribution systems in cities.g. E. Fl o w in F A H G B E C Fig.INTRODUCTION Pipe Network An interconnected system of pipes forming several loops or circuits.. .Pipe network Necessary conditions for any network of pipes The flow into each junction must be equal to the flow out of the junction. The algebraic sum of head losses round each loop must be zero. This is due to the continuity equation. The head loss in each pipe is expressed as hf = kQn. n = 2. For Turbulent flow. As such HARDY CROSS METHOD which uses successive approximations is used.Pipe network problems are difficult to solve analytically. . the head loss in each pipe is calculated according the following equation• hf = kQ2 • Wher e. • With the assumed values of Q. 4f x K L 2g x (∏∕4) = D5 x considering 2 Head loss around each loop is calculated the head loss to be positive in CW-flow and negative in CCW-flow. .Hardy cross method A trial distribution of discharges is made arbitrary in such a way that continuity equation is satisfied at each junction. The correction factor is obtained by- -∑ ΔQ (kQ02) ∑ = (2kQ0 . then the assumed values of Q are correct. But if the net head loss due to assumed values of Q is not zero.If the net head loss due to assumed values of Q round the loop is zero. then the assumed values of Q are corrected by introducing a correction delta Q for the flows. till circuit is balanced. a second trial calculation is made for all loops. .If the Correction factor comes out to be positive. The procedure is repeated till Delta Q becomes negligible. After the corrections have been applied to each pipe in a loop and to all loops. then it should be added to the flows in the CW direction and subtracted from the flows in the CCW direction. D K= 4 2 0 C K= 2 K= 1 4 0 K= 1 A K= 2 9 0 B 3 0 We have to calculate discharge in each pipe of the network. .Let us consider a problem. First Trial D K= 4 A 2 0 3 0 C K= 2 4 0 2 0 1 0 6 0 K= 1 2 0 K= 2 9 0 Discharges are assumed as in the above figure K= 1 B 3 0 . 4 .Loop ADB Pipe k Q Hf= kQ2 2kQ AD DB AB Total 4 1 2 30 10 60 3600 -100 -7200 -3700 240 20 240 500 Q1 = 7. 6 .Loop DCB Pipe k Q Hf= kQ2 2kQ DC CB BD Total 2 1 1 20 20 10 800 -400 100 500 80 40 20 140 Q2 = -3. Corrected flow for second trial. Pipe Correction Flow Direction AD 30 + 7.4 CW AB 60 – 7.6 CCW DC 20 – 3.6 CCW BD 10 – 7.6 16.4 37.4 CW BC 20 + 3.6 CCW BD 2.6 – 3.6 -1 CW .4 2.4 52.6 23. 6 K= 2 9 0 Discharges for the second trial K= 1 B 3 0 .Second Trial D K= 4 A 2 0 C K= 2 4 0 16. 4 1 52. 6 K= 1 23. 4 37. 2 1 2 -5533.1 511.Loop ADB Pipe k Q AD DB AB 4 1 2 37.5 210.6 .4 62.6 Total Hf= kQ2 2kQ 5595 299.4 1 52.54 Q1 = -0. 2 2 114.8 Q2 = 0.2 .Loop DCB Pipe k Q Hf= kQ2 2kQ DC CB BD Total 2 1 1 16.6 1 537.9 -556.6 47.9 -1 -20 65.4 23. . the correction is applied and furthur trials are discontinued.Since Q1 and Q2 are very small. 9 – 0.2 23.2 0.4 + 0. Pipe Correction Flow Direction AD 37.1 37.6 CW BC 23.6 – 0.7 CCW BD 1 – 0.4 – 0.7 CCW .9 CW DC 16.3 CW AB 52.1 0.4 CCW BD 0.1 52.Corrected flow for second trial.2 16.6 + 0. 7 K= 1 23. 4 K= 2 9 0 Final Distribution of discharges.Final Discharge D K= 4 A 2 0 C K= 2 4 0 16. 6 37. 7 52. K= 1 B 3 0 . 3 0. . Thank you for your time and patience.Hereby we conclude Hardy Cross method.