Transcript Pipe networks - Forum Haiti Org

CWR 4540 C
Pipe Networks
Sept 8, 2011
Dr. Kristoph-Dietrich Kinzli
Department of Environmental and Civil Engineering
Florida Gulf Coast University
Fall 2011
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
Magdeburg Water Bridge
Magdeburg Water Bridge
Magdeburg Water Bridge
Magdeburg Water Bridge
Magdeburg Water Bridge
Magdeburg Water Bridge
Magdeburg Water Bridge
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?)
Today’s Learning Objectives
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 HardyCross method
Nodal Method:
 The energy equation is written for each pipeline in
the network as
Friction loss
 fL
h2  h 1  

 D
h1
h2
hp
A
local losses

Q
 QQ
K m 

hp
2
Q
 2 gA
: head at the upstream end of a pipe (m)
: head at the downstream end of a pipe (m)
: head added by pumps in the pipeline (m)
: cross-sectional area of a pipe (m2)
 Positive flow direction is from node 1 to node 2
 Application is limited to relatively simple networks
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)
 kS 
 2 log

 3.7 D 
f
1
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
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 50cm 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?
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.

n
j 1
( hL , j  h p , j )  0
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
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
For analysis, basic criteria should be followed:
 General equation for losses, hf = rQn (SI units)
 Darcy-Weisbach
Q : flow rate (m3/s),
D
L
f
CH
2
L V
L
2
hf  f
f
Q
5
D 2g
12 D
: pipe diameter (m)
: length of pipe (m)
: friction factor
: H-W coefficient
 Hazen Williams
h f  6.82
L
1.17
D
 V

C
 H
1.85




10.67 L

D
4.87
CH
1.85
Q1.85
For analysis, basic criteria should be followed:
 General equation for losses, hf = rQn (US units)
 Darcy-Weisbach
Q : flow rate (ft3/s),
D
L
f
CH
2
L V
L
2
hf  f
f
Q
5
D 2g
39.7 D
: pipe diameter (ft)
: length of pipe (ft)
: friction factor
: H-W coefficient
 Hazen William
h f  6.82
L
1.17
D
 V

C
 H
1.85




4.73 L

D
4.87
CH
1.85
Q1.85
Hardy-Cross Method (Derivation)
For Closed Loop:
Q Qa  Q
h f  rQ n
n


 r Qa  Q
 r [Qan

r Qan
 nQan1Q 

rnQ an 1Q

Q  
r j Qa, j Qa, j
j 1
n

j 1
r j n Qa, j
nn  1 n2
Qa (Q) 2   (Q) n ]
2!
 r Qa
n
Qa : assumed flowrate
ΔQ : error in the assumption
Q : actual flowrate
n1
n1
n1
Q  rn Qa
n1
Q
If ΔQ is
small, higher
hL, j  0
order terms
in ΔQ can be
neglected
 n=2.0, Darcy-Weisbach
 n=1.85, Hazen-Williams

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
n
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
Q  

r j Qa, j Qa, j
j 1
n

r j n Qa, j
n1
n1
j 1
Step 7: Check the mass and energy balance.
Step 8: Compute the pressure distribution in the network and check on the
pressure requirement.
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 hL  r Q 2 , 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
In-class activity (Pipe networks)- cont.
r=1
r=4
25 cfs
Loop II
r=2
Loop I
r=4
100 cfs
25 cfs
r=5
50cfs
In-class activity (Pipe networks)- cont.
61.7 cfs
r=2
Loop II
14.8 cfs
Loop I
25 cfs
r=4
12.1 cfs
r=1
r=4
100 cfs
61.7 cfs
r=5
25 cfs 35.2 cfs
50cfs
2
h

r
Q
In-class activity (Pipe networks) L
r=8
0.5 cfs
r=8
r = 11
1.5 cfs
r=6
Determine the flow rate in each pipe
of the system with direction, with the
data shown!
r = 25
1 cfs