3. Component of Water Supply System
3
(1). Source (2).Treatment plant
(2). StorageTanks/Reservoirs (3).WaterTransmission/distribution
4. The various natural sources of water can be classified into
two categories:
Surface sources, such as
Ponds and lakes;
Streams and rivers;
Storage reservoirs.
Sub-surface sources or underground sources, such as
Springs; and
Fresh groundwater
Natural Raw Water Sources
4
5. Other non-traditional water sources include:
Ocean water
Desalination of sea water
Brackish underground sources
Desalination
Wastewater reuse
Treatment and re-use of wastewater
Rainwater harvesting (i.e., houses and domes roof, storm water)
Water trading !!
import/export of fresh water
Virtual water trade
Other Water Sources
Brackish water or briny water is water that has more salinity than fresh
water, but not as much as seawater
5
7. Water Quality and Treatment
7
Water to be used in a public water supply is required to be fit for drinking.
This implies that it poses no danger to health, and it should be colourless,
clear, odourless, sparkling and pleasant to taste.
The raw or treated water is analyzed by testing their physical, chemical and
bacteriological characteristics:
Physical Characteristics:
Turbidity; Color;Taste and odor; andTemperature
Chemical Characteristics:
pH;Acidity;Alkalinity; Hardness; Chlorides; Sulphates; Iron;
Nitrate, and Dissolved solids.
Bacteriological Characteristics:
Bacterial examination i.e., pathogenic bacteria or non pathogenic bacteria
such as E.Coli,
8. Distribution Reservoirs/Tanks
Reservoirs in water distribution systems plays an
important role to:
Provide service storage to meet widely fluctuating demands imposed
on water supply distribution systems
Accommodate fire-fighting and emergency requirements
Equalize operating pressures
Fire water tankElevated water tank
Jumaira-UAE
Surface water tanks water tanks in Kuwait
9. Water Transmission
9
Water transmission refers to the transportation of the water from the
source to the treatment plant and to the area of distribution.
It can be realized through
free-flow conduits, (gravity flow)
pressurized pipelines (pumping system) or
a combination of the two (combination of gravity flow and pump).
For small community water supplies through pressurized pipelined
are most common, since they are not very limited by the
topography of the area to be traversed.
Free-flow conduits (canals, aqueducts and tunnels) are preferred in
hilly areas or in areas where the required slope of the conduit more
or less coincides with the slope of the terrain.
10. Methods of Water Transmission/Distribution
10
(pumping system)
(1. Gravity flow)
(2. Pumping system)
(3. Gravity flow and pumping system)
11. Water Distribution System
11
A water distribution system consists of complex interconnected
pipes, service reservoirs and/or pumps, which deliver water from
the treatment plant to the consumer.
Water demand is highly variable, whereas supply is normally
constant. Thus, the distribution system must include storage
elements, and must be capable of flexible operation.
13. Design of Water Supply Networks
General Consideration:
A municipal water distribution system includes a network of mains with storage
reservoirs, booster pumping stations (if needed),fire hydrants, and service lines.Arterial
mains, or feeders, are pipelines of larger size that are connected to the transmission lines
that supply the water for distribution.A major water demand areas in a city should be
served by a feeder loop; where possible the arterial mains should be laid in duplicate.
Parallel feeder mains are cross connected at intervals of one to two kilometers, with
valves to permit isolation of sections in case of a main break. Distribution lines tie to each
arterial loop, forming a complete gridiron “close loops” system that domestic, commercial
consumers and services fire hydrants.The gridiron system illustrated in the Figure 1 is the
best arrangement for distributing water.All of the arterials and secondary mains are
looped and interconnected, eliminating dead ends and permitting water circulation such
that a heavy discharge from one main allows drawing water from other pipes.
The dead-end system shown in Figure 2 is avoided in new construction and can often be
corrected in existing systems by proper looping.Trunk lines placed in the main streets
supply sub-mains, which are extended at right angles to serve individual streets without
interconnections. Consequently, if a pipe break occurs substantial portion of the
community may be without water. Under a some conditions, the water in dead-end lines
develops tastes and odors from stagnation.To prevent this, dead ends may require frequent
flushing where houses are widely separated.
14. Pipe Network Analysis
Pipe network analysis involves the determination of the pipe flow
rates and pressure heads at the outflows points of the network. The
flow rates and pressure heads must satisfy the continuity and energy
equations.
ANALYSIS METHODS
(1). Hardy-Cross Method (Looped Method)
(2). Nodal Method
(3). Newton-Raphson Method
15. (1). The Hardy Cross Method
15
The earliest systematic method of network analysis (Hardy-Cross
Method) is known as the head balance or closed loop method.
This method is applicable to system in which pipes form closed
loops. The outflows from the system are generally assumed to occur
at the nodes junction.
For a given pipe system with known outflows, the Hardy-Cross
method is an iterative procedure based on initially iterated flows in
the pipes.
At each junction these flows must satisfy the continuity criterion, i.e.
the algebraic sum of the flow rates in the pipe meeting at a junction,
together with any external flows is zero.
16. (1).The Hardy Cross Method
The method is based on
Continuity Equation:
Inflow = outflow at nodes
Energy Equation:
Summation of head loss in closed loop is zero
cba QQQ +=
( ) ( )∑ ∑ =∆+⇒= 00
n
l QQKlooph
A B
C D
Qa
Qb
Qc
17. (1).The Hardy Cross Method
17
The relationship between head loss and discharge must be maintained
for each pipe
Darcy-Weisbach Equation
Exponential friction formula Hazen-Williams
52
8
2)(
Dg
fL
KnKQpipeh n
l
π
===
87.485.1
67.10
85.1)(
dC
KnKQpipeh n
l ===
18. (1). Hardy Cross Method (Derivation)
18
QQQ a ∆+=
( ) 0=∆+∑
n
a QQK
( ) ( ) 0...
2
1 221
=+∆
−
+∆+∑ ∑∑ −− n
a
n
a
n
a QQnK
n
QQnKKQ
( )
∑
∑∑ ∑ −
−
−=∆=∆+
1
1
0
n
a
n
an
a
n
a
nKQ
KQ
QQQnKKQ
19. (1). Hardy Cross Method
19
Problem Description
Network of pipes forming one or more closed loops
Given
Demands @ network nodes (junctions)
d, L, pipe material,Temp, and P @ one node
Find
Discharge & flow direction for all pipes in network
Pressure @ all nodes & HGL
20. (1). Hardy-Cross Method (Procedure)
20
1. Divide network into number of closed
loops.
2. For each loop:
a) Assume discharge Qa and direction for
each pipe.Apply Continuity at each node,
Total inflow =Total Outflow. Clockwise
positive.
b) Calculate equivalent resistance K for each
pipe given L, d, pipe material and water
temperature.
c) Calculate hf=K Qa
n for each pipe. Retain
sign from step (a) and compute sum for
loop Σ hf.
d) Calculate hf / Qa for each pipe and sum
for loop Σhf/ Qa.
e) Calculate correction
∆Q = −Σ hf /(nΣhf/Qa).
NOTE: For common members between 2
loops both corrections have to be made.
As loop 1 member, ∆Q = ∆Q 1 − ∆Q 2. As
loop 2 member, ∆ Q = ∆ Q 2 − ∆ Q 1.
f) Apply correction to Qa, Qnew=Qa + ∆ Q.
g) Repeat steps (c) to (f) until ∆Q becomes
very small and Σhf=0 in step (c).
h) Solve for pressure at each node using
energy conservation.
21. Example: 5-1
21
Neglecting minor losses in the pipe,
determine the flows in the pipes and
the pressure heads at the nodes
(kinematic viscosity= 1.13x10-6m2/s)
22. Example: 5-1
22
Solution:
Identify loops
Loop 1 and Loop 2
Allocate estimated flows in
each pipe
Compute head loss
coefficient of each pipe
Compute head loss in each
pipe
Discharge Sign Convention
CW=+ve
CCW=-ve
52
8
Dg
fL
K
π
=
2
KQhhl ==
f
Moody’s Diagram
or Colebrook Eq.
+−=
f
De
f Re
51.2
7.3
/
log2
1
27. Example:
Find the flows in the loop given the inflows and outflows.
The pipes are all 25 cm cast iron (e=0.26 mm).
A B
C D0.10 m3/s
0.32 m3/s 0.28 m3/s
0.14 m3/s
200 m
100 m
28. Example:
Identify Loop
Assign a flow to each pipe link
Flow into each junction must equal flow out of the
junction
A B
C D0.10 m3/s
0.32 m3/s 0.28 m3/s
0.14 m3/s
0.32
0.00
0.10
0.04
assumed
29. Example:
Calculate the head loss in each pipe
f=0.02 for Re>200000
=
=
25
2
8
πgD
fL
K
KQhf
339
)25.0)(8.9(
)200)(02.0(8
25
1 =
=
π
k
k1,k3=339
k2,k4=169
A B
C D0.10 m3/s
0.32 m3/s 0.28 m3/s
0.14 m3/s
1
4 2
3
mh
mh
mh
mh
mh
i
f
f
f
f
f
i
53.31
00.0
39.3
222.0
7.34
4
1
4
3
2
1
=
−=
−=
=
=
∑=
Sign convention
+CW
2
5
s
m
30. Example:
The head loss around the loop isn’t zero
Need to change the flow around the loop
the ___________ flow is too great (head loss is positive)
reduce the clockwise flow to reduce the head loss
Solution techniques
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, WaterCad, etc.)
clockwise
optimizes correction
31. Example:
Numerical 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
31.575Sum Head Loss
32. Example
A B
C D0.10 m3/s
0.32 m3/s 0.28 m3/s
0.14 m3/s
0.218
0.102
0.202
0.062
1
4 2
3
Q0+ ∆Q
0.218
−0.062
−0.202
−0.102
You must be able to make a hand calculation before opting for a
better solution using software with a GUI.
Solution to Loop Problem
33. (2). Nodal Method
33
Fig shows a branched pipe system delivering water from impounding
reservoir A to the service reservoirs B, C and D. F is known direct out flow
from J.
Eq. (1)
34. (2). Nodal Method
34
Eq. (1) can be written as
Eq. (2)
Eq. (3)
Iteration of ZJ can be performed such that QIJ from Eq. (3)
satisfies Eq. (2).
If (∑QIJ-F)≠0 then a correction, ∆ZJ, is applied to ZJ such that
38. Newton Raphson Method
38
Newton–Raphson method is a powerful numerical method for solving
systems of nonlinear equations.
Suppose that there are three nonlinear equations F1(Q1, Q2, Q3) = 0,
F2(Q1, Q2, Q3) = 0, and F3(Q1, Q2, Q3) = 0 to be solved for Q1, Q2,
and Q3.
Adopt a starting solution (Q1, Q2, Q3).
Also consider that (Q1 + ∆Q1, Q2,+ ∆Q2, Q3 + ∆Q3) is the solution
of the set of equations.That is
Eq. (1)
41. Procedure
41
The overall procedure for looped network analysis by the Newton–
Raphson method can be summarized in the following steps:
Step 1: Number all the nodes, pipe links, and loops.
Step 2:Write nodal discharge equations as
where Qjn is the discharge in nth pipe at node j, qj is nodal
withdrawal, and jn is the total number of pipes at node j.
Step 3:Write loop head-loss equations as
where kn is total pipes in kth loop.
42. Procedure
42
Step 4:Assume initial pipe discharges Q1, Q2, and Q3., . . . satisfying
continuity equations.
in all pipe links and compute,ifDetermine friction factors,:5Step
usingiKcorresponding
Step 6: Find values of partial derivatives ∂Fn / ∂Qi and functions Fn,
using the initial pipe discharges Qi and Ki.
Step 7: Find ∆Qi.The equations generated are of the form Ax = b,
which can be solved for ∆Qi.
Step 8: Using the obtained ∆Qi values, the pipe discharges are
modified and the process is repeated again until the calculated ∆Qi
values are very small.
43. Example
43
The pipe network of two loops as shown in Fig. has to be analyzed by the
Newton Raphson method for pipe flows for given pipe lengths L and pipe
diameters D.
The nodal inflow at node 1 and nodal outflow at node 3 are shown in the
figure. Assume a constant friction factor f = 0.02.
Single looped network
44. Example
44
Solution: Write nodal and loop equations
Node1: Q1+Q4-0.6=0
Node 2: Q1+Q2=0
Node 3: Q2+Q3-0.6=0
Node 4: !!!
and loop Eq.
K1Q1
2+ K2Q2
2+ K3Q3
2+ K4Q4
2=0
Q1
Q2
Q4
Q3
52
8
2
)(
Dg
fL
Kn
KQpipeh n
L
π
=
=
45. Example
45
Assume initial pipe discharges based on continuity euqtion:
Q1 = 0.5 m3/s
Apply signs to flowrate in eqns
Nodal discharge functions, F, are
Q1+Q4-0.6=F1
- Q1+Q2=F2
Q2+Q3-0.6=F3
Loop head-loss function
K1Q1
2+ K2Q2
2 - K3Q3
2 - K4Q4
2=F4
In = +
Out = -ve
= 0.5
0.5
0.1
0.1
Apply signs (+ or -)
46. Example
46
The final nodal discharge functions, F, become
and loop head-loss function is
F4=6528Q1
2+ 4352Q2
2 - 6528Q3
2 - 4352Q4
2=0
Take derivatives of equations:
48. Example
48
Now, calculate values of F1 to F4 using previously assumed flows i.e.,
Q1 = 0.5 m3/s
F4=6528Q1
2+ 4352Q2
2 - 6528Q3
2 - 4352Q4
2=2611.2
49. Example
49
Substitute values of F1 to F4 and Q1 to Q4 in the matrix form :
Solve the matrix, we get following flowrate corrections:
50. Example
50
Using discharge corrections, the revised pipe flowrates become:
The process is repeated with the new pipe flowrates. Revised values
of F and derivative ∂F/∂Q values are obtained. Substituting the
revised values, the following new solution matrix is generated:
51. Example
51
As the right-hand side is operated upon null vector, all the
discharge corrections ∆Q = 0.
Thus, the final flowrates are
