1. Pipe Networks Dr. Kristoph-Dietrich Kinzli Fall 2011 CWR 4540 C
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:
The energy equation is written for each pipeline in the network as
Friction loss
local losses
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 = 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:
General equation for losses, hf = rQn (SI units)
Darcy-Weisbach
Hazen Williams
Q : flow rate (m3/s),
D : pipe diameter (m)
L : length of pipe (m)
f : friction factor
CH : H-W coefficient

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

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

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