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- 1. Pipe Networks Pipeline systems Transmission lines You are here Pipe networks Measurements Manifolds and diffusers Pumps Transients School of Civil and Monroe L. Weber-Shirk Environmental Engineering
- 2. Pipeline systems: Pipe networks Water distribution systems for municipalities Multiple sources and multiple sinks connected with an interconnected network of pipes. Computer solutions! KYpipes WaterCAD CyberNET EPANET http://www.epa.gov/ORD/NRMRL/wswrd/epanet.html
- 3. Water Distribution System Assumption Each point in the a system can only have one pressure _______ 1 2 The pressure change from 1 to 2 by path a must equal the b pressure change from 1 to 2 by path bp1 V12 p2 V22 + + z1 = + + z 2 + hLγ 2g γ 2g 2 p 2 p1 V1 V22 − = + z1 − a a − z2 − hL Same for path b! γ γ 2g 2g a
- 4. Water Distribution System Assumption V12 V22 V12 V22 a + z1 − a − z2 − hL = b + z1 − b − z 2 − hL 2g 2g a 2g 2g b a 1 2Pressure change by path a b hL = hL a b zero Or sum of head loss around loop is _____. (Need a sign convention) Pipe diameters are constant or K.E. is small Model withdrawals as occurring at nodes so V is constant between nodes
- 5. Pipes in Parallel Find discharge given pressure at A and B Q1 ______& ____ equation energy S-J add flows Qtotal A Q2 B Find head loss given the total flow assume a discharge Q1’ through pipe 1 solve for head loss using the assumed discharge using the calculated head loss to find Q2’ assume that the actual flow is divided in the same proportion _________ as the assumed flow
- 6. Networks of PipesMass __________ at all nodes ____ conservation 0.32 m /s 3 A 0.28 m3/s The relationship between head ? loss and discharge must be maintained for each pipe Darcy-Weisbach equation Swamee-Jain _____________ Exponential friction formula a _____________ Hazen-Williams 1 2 b
- 7. Network Analysis Find the flows in the loop given the inflows and outflows. The pipes are all 25 cm cast iron (ε=0.26 mm). 0.32 m3/s A B 0.28 m3/s100 m 0.10 m3/s C D 0.14 m3/s 200 m
- 8. Network Analysis Assign a flow to each pipe link Flow into each junction must equal flow out of the junction arbitrary0.32 m3/s A B 0.28 m3/s 0.32 0.00 0.040.10 m3/s C D 0.14 m3/s 0.10
- 9. Network Analysis h f = 34.7m Calculate the head loss in each pipe 1 h f = 0.222m 8 fL 2 2 hf = 5 2 Q f=0.02 for Re>200000 h f = −3.39m gD π 3h f = kQ Q Sign convention +CW h f = −0.00m 4 4 8(0.02)(200) s 2 k1,k3=339k1 = = 339 (9.8)(0.25) 5 π 2 m5 k2,k4=169 ∑h fi = 31.53m i=1 0.32 m3/s A 1 B 0.28 m3/s 4 2 0.10 m3/s C 3 D 0.14 m3/s
- 10. Network Analysis The head loss around the loop isn’t zero Need to change the flow around the loop clockwise the ___________ flow is too great (head loss is positive) reduce the clockwise flow to reduce the head loss Solution techniques optimizes correction Hardy Cross loop-balancing (___________ _________) Use a numeric solver (Solver in Excel) to find a change in flow that will give zero head loss around the loop Use Network Analysis software (EPANET)
- 11. Numeric Solver Set up a spreadsheet as shown below. the numbers in bold were entered, the other cells are calculations initially ∆Q is 0 use “solver” to set the sum of the head loss to 0 by changing ∆Q the column Q0+ ∆Q contains the correct flows ∆Q 0.000 pipe f L D k Q0 Q0+∆Q hf P1 0.02 200 0.25 339 0.32 0.320 34.69 P2 0.02 100 0.25 169 0.04 0.040 0.27 P3 0.02 200 0.25 339 -0.1 -0.100 -3.39 P4 0.02 100 0.25 169 0 0.000 0.00 Sum Head Loss 31.575
- 12. Solution to Loop Problem Q0+ ∆Q 0.218 −0.062 −0.202 −0.102 0.32 m3/s A 1 B 0.28 m3/s 4 0.218 2 0.102 0.062 0.202 0.10 m /s3 C 3 D 0.14 m3/sBetter solution is software with a GUI showing the pipe network.
- 13. Network Elements Controls Check valve (CV) Pressure relief valve Pressure reducing valve (PRV) Pressure sustaining valve (PSV) Flow control valve (FCV) Pumps: need a relationship between flow and head Reservoirs: infinite source, elevation is not affected by demand Tanks: specific geometry, mass conservation applies
- 14. Check Valve Valve only allows flow in one direction The valve automatically closes when flow begins to reverse open closed
- 15. Pressure Relief Valve closed open pipeline relief flow Low pipeline pressure High pipeline pressure Valve will begin to open when pressure in exceeds the pipeline ________ a set pressure (determined by force on the spring).Where high pressure could cause an explosion (boilers, water heaters, …)
- 16. Pressure Regulating Valve sets maximum pressure downstream closed openHigh downstream pressure Low downstream pressure Valve will begin to open when the pressure ___________ less downstream is _________ than the setpoint pressure (determined by the force of the spring).Similar function to pressure break tank
- 17. Pressure Sustaining Valve sets minimum pressure upstream closed openLow upstream pressure High upstream pressureValve will begin to open when the pressureupstream greater________ is _________ than the setpoint pressure(determined by the force of the spring).Similar to pressure relief valve
- 18. Flow control valve (FCV) Limits the flow rate ____ ___ through the valve to a specified value, in a specified direction Commonly used to limit the maximum flow to a value that will not adversely affect the provider’s system
- 19. Pressure Break Tanks In the developing world small water supplies in mountainous regions can develop too much pressure for the PVC pipe. They don’t want to use PRVs because they are too expensive and are prone to failure. Pressure break tanks have an inlet, an outlet, and an overflow. Is there a better solution?
- 20. Network Analysis Extended The previous approach works for a simple loop, but it doesn’t easily extend to a whole network of loops Need a matrix method Initial guess for flows Adjust all flows to reduce the error in pressures __________________________ Simultaneous equations _______________________________ Appendix D of EPANET manual
- 21. Pressure Network Analysis Software: EPANETreservoir pipe junction 0.32 m3/s A 1 B 0.28 m3/s 4 0.218 2 0.102 0.062 0.202 0.10 m /s 3 C 3 D 0.14 m3/s
- 22. EPANET network solution 8 fL 2 H i − H j = hij = rQij + mQij n 2hf = 5 2 Q gD π ∑Qj ij − Di = 0 8 fL r = gD 5π 2 ÷ AH = F Aii = ∑ pij n=2 j Aij = − pij 1pij = 1 8 fL pij = n −1 2 Q 5 2 ÷ ij nr Qij + 2m Qij gD π
- 23. Fi = ∑ Qij − Di ÷+ ∑ yij + ∑ pif H f j j f ( nyij = pij r Qij + m Qij 2 ) sgn ( Q ) ij Qij = Qij − yij − pij ( H i − H j )

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