Example Problem - 3 ReservoirsGiven:A three reservoir (branching pipe system)Elev A (ft) =240L1 (ft) =4000Elev C (ft) =170L2 (ft) =3000Dia 1 (in.) =30L3 (ft) =1500Dia 2 (in.) =24Q2 (CFS) =7Dia 3 (in.) =18Required:1. Pressure at Junction2. Elevation of Reservoir BSolution:Derivation of Velocity for 3 Reservoir Problem:Hazen Williams FrictionQ (GPM)L (ft)ID (in)hf (ft)Rearranged for V (velocity)V (fps)D (ft)hf (ft)L (ft)Setting up a table for the pipe characteristics:Pipe123L (ft)400030001500D (ft)2.521.5C130130130Area4.9093.1421.767Setting up a table for Elev P iterations:Elev. Phf1= Elev. A - Elev. Phf3= Elev. P - Elev. Cv1v3Q1Q3Q1-Q2-Q3Move Pftftftft/sft/sCFSCFSCFSUp/Down200403010.4557
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This note was uploaded on 01/13/2011 for the course BRAE 312 taught by Professor Styles during the Fall '10 term at Cal Poly.