1 / 25

Pipe Networks

Fall 2011. CWR 4540 C. Pipe Networks. Sept 8, 2011. Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University. Magdeburg Water Bridge. Wasserstraßenkreuz opened in October 2003

stacey-joyner
Download Presentation

Pipe Networks

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Fall 2011 CWR 4540 C Pipe Networks Sept 8, 2011 Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University

  2. Magdeburg Water Bridge • Wasserstraßenkreuz • opened in October 2003 • connects the Elbe-Havel Kanal to the Mittellandkanal, crossing over the Elbe River. • longest navigable aqueduct in the world, with a total length of 918 meters (3,012 ft) • previously 12-kilometre (7.5 mi) detour, offloading due to water levels in Elbe • 1350 tons vs 800 tons • 500 million Euros – 6 years • 24,000 tons of steel and 68,000 cubic meters of concrete

  3. Magdeburg Water Bridge

  4. Magdeburg Water Bridge

  5. Magdeburg Water Bridge

  6. Magdeburg Water Bridge

  7. Magdeburg Water Bridge

  8. Magdeburg Water Bridge

  9. Magdeburg Water Bridge

  10. Introduction – Pipe Networks • The purpose of a water distribution network is to supply the system’s users with the amount of water demanded and to supply this water with adequate pressure under various loading conditions. • Water distribution system have three major components; pumping station, distribution storage and distribution piping. • Performance criteria are generally minimum flow rates and pressure (why?)

  11. Today’s Learning Objectives • Apply continuity and energy equations to pipe system networks • Analyze pipe flow and pressure in a network • Illustrate the iterative solution of the loop equations using the Hardy Cross Method • Develop a spreadsheet to solve a simple water distribution system using the Hardy-Cross method

  12. Nodal Method: Friction loss local losses • The energy equation is written for each pipeline in the network as h1: head at the upstream end of a pipe (m) h2: head at the downstream end of a pipe (m) hp : head added by pumps in the pipeline (m) A : cross-sectional area of a pipe (m2) • Positive flow direction is from node 1 to node 2 • Application is limited to relatively simple networks

  13. Example 2.9 (Chin, pg 41-43) The high-pressure ductile-iron pipeline shown in Figure 2.10 becomes divided at point B and rejoins at point C. The pipeline characteristics are given in the following tables. If the flowrate in Pipe 1 is 2 m3/s and the pressure at point A is 900 kPa, calculate the pressure at point D. Assume that the flows are fully turbulent in all pipes. ks Flowrate(Q) Pressure(P)

  14. Example 2.9 (Chin, pg 41-43) – cont. Step 1: Calculate ks/D and f for each pipe Step 2: Calculate the total energy head at location A, hA Step 3: Write down the energy equations for each pipe Step 4: Use continuity equations at two pipe junctions (Qin = Qout1 + Qout2) Step 5: Determine Q of each pipe and hD Step 6: Determine PD using the calculated hD

  15. In-class activity (Example problem) The pipe system shown in the figure below connects reservoirs that have an elevation difference of 20 m. This pipe system consists of 200 m of 50-cm concrete pipe (pipe A), that branches into 400 m of 20-cm pipe (pipe B) and 400 m of 40-cm pipe (pipe C) in parallel. Pipes B and C join into a single 50-cm pipe that is 500 m long (pipe D). For f = 0.030 in all the pipes, what is the flow rate in each pipe of the system?

  16. Hardy Cross Method (Cross, 1936) • The Hardy Cross method is a simple technique for hand solution of the loop system of equations in pipe networks •  The flowrate in each pipe is adjusted iteratively until all equations are satisfied.  • The method is based on two primary physical laws: • The sum of pipe flows into and out of a node equals the flow entering or leaving the system through the node. • the algebraic sum of the head losses within each loop is equal to zero. hL,j: head loss in pipe j of a loop hp,j: head added by any pumps in pipe j n : the number of pipes in a loop

  17. For analysis, basic criteria should be followed: • The total flow entering each point (node) must equal to the total flow leaving that joint (node). ΣQin = ΣQout • The flow in a pipe must follow the pipe’s friction law for the pipe. hf = kQn n is coefficient depend on equation used • The algebra sum of the head loss in the pipe network must be zero. Σ hf = 0

  18. For analysis, basic criteria should be followed: Q : flow rate (m3/s), D : pipe diameter (m) L : length of pipe (m) f : friction factor CH : H-W coefficient • General equation for losses, hf = rQn(SI units) • Darcy-Weisbach • Hazen Williams

  19. For analysis, basic criteria should be followed: Q : flow rate (ft3/s), D : pipe diameter (ft) L : length of pipe (ft) f : friction factor CH : H-W coefficient • General equation for losses, hf = rQn(US units) • Darcy-Weisbach • Hazen William

  20. Hardy-Cross Method (Derivation) For Closed Loop: Qa: assumed flowrate ΔQ : error in the assumption Q : actual flowrate If ΔQ is small, higher order terms in ΔQ can be neglected • n=2.0, Darcy-Weisbach • n=1.85, Hazen-Williams

  21. Steps for Q distribution using Hardy Cross method: Step 1: Build up system configuration and assume a reasonable distribution of flows the pipe network. but Q=0 at each node. Step 2: Calculate head loss of each pipe section with D-W or H-W equ. Step 3: Calculate  (rQ|Q|n-1) and  (n r |Q|n-1) in each loop of the pipe network. Step 4: Get ∆Q value for correction Step 5: Calculate the new flow rate Qnew= Q + ∆Q Step 6: Repeat this step (3 ~ 5) until ∆Q = 0 Step 7: Check the mass and energy balance. Step 8: Compute the pressure distribution in the network and check on the pressure requirement.

  22. Example 2.10 (Chin, pg 46-48) Compute the distribution of flows in the pipe network shown in Figure 2.11 (a), where the head loss in each pipe is given by , and the relative values of r are shown in Figure 2.11 (a). The flows are taken as dimensionless for the sake of illustration. Given condition Assumed Q and direction

  23. In-class activity (Pipe networks)- cont. 25 cfs 100 cfs r = 1 r = 4 Loop I Loop II r = 4 r = 2 r = 3 r = 5 25 cfs 50cfs

  24. In-class activity (Pipe networks)- cont. 25 cfs 61.7 cfs 100 cfs 61.7 cfs r = 1 r = 4 Loop I Loop II r = 4 r = 2 12.1 cfs 14.8 cfs r = 3 38.3 cfs r = 5 25 cfs 35.2 cfs 50cfs

  25. In-class activity (Pipe networks) 1.5 cfs r = 8 r = 15 r = 11 r = 8 0.5 cfs r = 6 r = 25 Determine the flow rate in each pipe of the system with direction, with the data shown! 1 cfs

More Related