Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida...
-
Upload
rebecca-gardner -
Category
Documents
-
view
215 -
download
2
Transcript of Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida...
![Page 1: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/1.jpg)
Pipe NetworksPipe Networks
Dr. Kristoph-Dietrich KinzliDepartment of Environmental and Civil Engineering
Florida Gulf Coast University
CWR 4540 C Fall 2011
Sept 8, 2011
![Page 2: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/2.jpg)
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
![Page 3: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/3.jpg)
Magdeburg Water Bridge
![Page 4: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/4.jpg)
Magdeburg Water Bridge
![Page 5: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/5.jpg)
Magdeburg Water Bridge
![Page 6: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/6.jpg)
Magdeburg Water Bridge
![Page 7: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/7.jpg)
Magdeburg Water Bridge
![Page 8: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/8.jpg)
Magdeburg Water Bridge
![Page 9: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/9.jpg)
Magdeburg Water Bridge
![Page 10: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/10.jpg)
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?)
![Page 11: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/11.jpg)
1. Apply continuity and energy equations to pipe system networks
2. Analyze pipe flow and pressure in a network
3. Illustrate the iterative solution of the loop equations using the Hardy Cross Method
4. Develop a spreadsheet to solve a simple water distribution system using the Hardy-Cross method
Today’s Learning Objectives
![Page 12: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/12.jpg)
The energy equation is written for each pipeline in the network as Nodal Method:
pm hQ
Q
gA
QQK
D
fLhh
212
2
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)
Friction loss
local losses
Positive flow direction is from node 1 to node 2 Application is limited to relatively simple networks
![Page 13: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/13.jpg)
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.
D
k
f
S
7.3log2
1ks
Pressure(P)
Flowrate(Q)
![Page 14: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/14.jpg)
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
Example 2.9 (Chin, pg 41-43) – cont.
![Page 15: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/15.jpg)
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?
![Page 16: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/16.jpg)
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.
0)( ,
1
,
jp
n
j
jL hhhL,j : head loss in pipe j of a loop hp,j : head added by any pumps in pipe jn : the number of pipes in a loop
![Page 17: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/17.jpg)
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
For analysis, basic criteria should be followed:
![Page 18: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/18.jpg)
General equation for losses, hf = rQn (SI units) Darcy-Weisbach
Hazen Williams
For analysis, basic criteria should be followed:
25
2
122Q
D
Lf
g
V
D
Lfh f
85.185.187.4
85.1
17.1
67.1082.6 Q
CD
L
C
V
D
Lh
HHf
Q : flow rate (m3/s),
D : pipe diameter (m)
L : length of pipe (m)
f : friction factorCH : H-W
coefficient
![Page 19: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/19.jpg)
General equation for losses, hf = rQn (US units) Darcy-Weisbach
Hazen William
For analysis, basic criteria should be followed:
25
2
7.392Q
D
Lf
g
V
D
Lfh f
85.185.187.4
85.1
17.1
73.482.6 Q
CD
L
C
V
D
Lh
HHf
Q : flow rate (ft3/s), D : pipe diameter
(ft)L : length of pipe
(ft)f : friction factorCH : H-W
coefficient
![Page 20: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/20.jpg)
Hardy-Cross Method (Derivation)
na
nf
QQr
rQh
n
j
n
jaj
n
j
n
jajaj
Qnr
QQr
Q
1
1
,
1
1
,,
For Closed Loop:
])()(
!2
1[ 221 nn
ana
na QQQ
nnQnQQr
n=2.0, Darcy-Weisbach n=1.85, Hazen-Williams
QrnQQr na
na 1
If ΔQ is small, higher order terms in ΔQ can be neglected
QQQ a Qa : assumed flowrate ΔQ : error in the
assumption Q : actual flowrate
QQrnQQrn
a
n
a 11
0, jLh
![Page 21: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/21.jpg)
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.
Steps for Q distribution using Hardy Cross method:
n
j
n
jaj
n
j
n
jajaj
Qnr
QQr
Q
1
1
,
1
1
,,
![Page 22: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/22.jpg)
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.
2QrhL
Given condition
Assumed Q and direction
![Page 23: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/23.jpg)
In-class activity (Pipe networks)- cont.
100 cfs
25 cfs 50cfs
25 cfs
r = 3 r =
4
r = 1 r = 4
r = 5
r =
2
Loop ILoop
II
![Page 24: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/24.jpg)
In-class activity (Pipe networks)- cont.
100 cfs
25 cfs 50cfs
25 cfs
r = 3 r =
4
r = 1 r = 4
r = 5
r =
2
61.7 cfs
12
.1
cfs
38.3 cfs
35.2 cfs
61.7 cfs
14
.8
cfs
Loop ILoop
II
![Page 25: Pipe Networks Dr. Kristoph-Dietrich Kinzli Department of Environmental and Civil Engineering Florida Gulf Coast University CWR 4540 C Fall 2011 Sept 8,](https://reader031.fdocuments.us/reader031/viewer/2022032707/56649e385503460f94b29764/html5/thumbnails/25.jpg)
In-class activity (Pipe networks) 2QrhL
1.5 cfs
1 cfs
r = 15r
= 1
1r = 8
r = 6 r = 25
r =
8Determine the flow rate in each pipe of the system with direction, with the data shown!
0.5 cfs