*CTC 450 ReviewFriction Loss Over a pipe lengthDarcy-Weisbach (Moodys diagram)Connections/fittings, etc.

*ObjectivesKnow how to set up a spreadsheet to solve a simple water distribution system using the Hardy-Cross method

*Pipe SystemsWater municipality systems consist of many junctions or nodes; many sources, and many outlets (loads)Object for designing a system is to deliver flow at some design pressure for the lowest costSoftware makes the design of these systems easier than in the past; however, its important to understand what the software is doing

*Two parallel pipesIf a pipe splits into two pipes how much flow will go into each pipe? Each pipe has a length, friction factor and diameterHead loss going through each pipe has to be equal

*Two parallel pipesf1*(L1/D1)*(V12/2g)= f2*(L2/D2)*(V22/2g)

Rearrange to:V1/V2=[(f2/f1)(L2/L1)(D1/D2)] .5

This is one equation that relates v1 and v2; what is the other?

*Hardy-Cross MethodQs into a junction=Qs out of a junctionHead loss between any 2 junctions must be the same no matter what path is taken (head loss around a loop must be zero)

*StepsChoose a positive direction (CW=+)# all pipes or identify all nodesDivide network into independent loops such that each branch is included in at least one loop

*4. Calculate K for each pipeCalc. K for each pipeK=(0.0252)fL/D5For simplicity f is usually assumed to be the same (typical value is .02) in all parts of the network

*5. Assume flow rates and directionsRequires assumptions the first time aroundMust make sure that Qin=Qout at each node

*6. Calculate Qt-Qa for each independent loopQt-Qa =-KQan/n |Qan-1| n=2 (if Darcy-Weisbach is used)Qt-Qa =-KQa2/2 |Qan-1| Qt is true flowQa is assumed flowOnce the difference is zero, the problem is completed

*7. Apply Qt-Qa to each pipeUse sign convention of step oneQt-Qa (which can be + or -) is added to CW flows and subtracted from CCW flowsIf a pipe is common to two loops, two Qt-Qa corrections are added to the pipe

*8. Return to step 6

Iterate until Qt-Qa = 0

*Example Problem

2 loops; 6 pipes

By hand; 1 iterationBy spreadsheet

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example

Hardy-CrossSolution to example problem

2-loops (6 pipes)

Lng (ft)Dia. (ft)

ABK'=306AB25000.33

BDK'=7.7BC35000.42

DEK'=5.7DC60000.50

EAK'=368BD20000.67

BCK'=140ED15000.67f=0.02

CDK'=97AE30000.33

DBK'=7.7

Loop 1Loop 2Loop 1Loop 2Corrected Loop 1Corrected Loop 2

IterationQa-bQb-dQd-eQe-aQb-cQc-dQdbcorrectioncorrectionQa-bQb-dQd-eQe-aQb-cQc-dQdb

10.700.400.300.800.300.700.400.080.160.780.320.220.720.460.540.32

20.780.320.220.720.460.540.320.00-0.010.780.330.220.720.450.550.33

30.780.330.220.720.450.550.33-0.00-0.000.780.330.220.720.450.550.33

40.780.330.220.720.450.550.33-0.00-0.000.780.330.220.720.450.550.33

*Next Lecture Equivalent PipesPump Performance CurvesSystem Curves

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