design and analysis of water distribution System

Lec # 06 :
Design & Analysis of Water Distribution Systems
Faculty of Engineering,
Lahore Leads University

Content
Design of Water Distribution Systems
Pipe Network Analysis
Water Distribution Systems & networks
2

Part A
Design of Water
Distribution Systems
3

Design of Water Distribution
Systems
Main requirements :
• Satisfied quality and quantity standards
Additional requirements :
• To enable reliable operation during irregular situations (power
failure, fires..)
• To be economically and financially viable, ensuring income for
operation, maintenance and extension.
• To be flexible with respect to the future extensions.
A properly designed water distribution system should fulfill the following requirements:
Design of Water Distribution Systems
4

The design of water distribution systems must
undergo through different studies and steps:
Design Phases
Hydraulic Analysis
Preliminary Studies
Network Layout
5

Preliminary Studies:
4.3.A.1 Topographical Studies:
Must be performed before starting the actual design:
1. Contour lines (or controlling elevations).
2. Digital maps showing present (and future) houses, streets,
lots, and so on..
3. Location of water sources so to help locating distribution
reservoirs.
6

Water Demand Studies:
Water consumption is ordinarily divided into the
following categories:
 Domestic demand.
 Industrial and Commercial demand.
 Agricultural demand.
 Fire demand.
 Leakage and Losses.
7
Details of which are available in previous lecture

Design Criteria
Are the design limitations required to get the most
efficient and economical water-distribution network
Velocity
Pressure
Average Water Consumption
8

Velocity
• Not be lower than 0.6 m/s to prevent
sedimentation
• Not be more than 3 m/s to prevent erosion and
high head losses.
• Commonly used values are 1 - 1.5 m/sec.
9

Pressure
Min. Pressure at peak flow(not less than 150 kPa to avoid
infiltration, proper flow to other buildings)
Max. pressure during low flows
Residential areas (3 stories)-150-300 kPa (15-30m)
Residential areas (firefighting ) -400 kPa (40m)
Commercial areas- 500 KPa (50m)
10

Pressure
• Also, for fire hydrants the pressure should not be less than
150 kPa (15 m of water)
• In general for any node in the network the pressure should
not be less than 25 m of water.
• Moreover, the maximum pressure should be limited to 70 m
of water
11

Pipe sizes
• Lines which provide only domestic flow may be as small as 100 mm (4
in) but should not exceed 400 m in length (if dead-ended) or 600 m if
connected to the system at both ends.
• Lines as small as 50-75 mm (2-3 in) are sometimes used in small
communities with length not to exceed 100 m (if dead-ended) or 200 m
if connected at both ends.
• The size of the small distribution mains is seldom less than 150 mm (6
in) with cross mains located at intervals not more than 180 m.
• In high-value districts the minimum size is 200 mm (8 in) with cross-
mains at the same maximum spacing. Major streets are provided with
lines not less than 305 mm (12 in) in diameter.
12

Head Losses
• Optimum range is 1-4 m/km.
• Maximum head loss should not exceed 10
m/km.
13

Hazen-Williams equation for
pipe flow
Head loss in pipes(water supply network)
Empirical Formula
Named after Allen Hazen and Gardner Stewart Williams.
H= head loss(m)
Q= flow rate(m3/sec)
L= length of pipe(m)
d= diameter(m)
C= Hazen William’s coefficient

Hazen-Williams Equation for
Pipe Flow
Hazen Williamgreatly depends upon Roughness of pipe.
Basic Hazen William Eq is
Where,
V= velocity ,m/s
C=Hazen William co-efficient
gradient (slope) =HL/L

H-head loss in meters
Q=flow in cu meter per sec
D= diameter in mm
L= length of pipe in meters
For safety factor C=100

Hazen-Williams equation for
pipe flow
Advantages
Coefficient C is rough measure of relative roughness
Effect of Reynolds number is included in formula
Effect of roughness on velocity are given directly
Disadvantages
Empirical
Does not differentiate completely between laminar and
turbulent flow
Extremely high and low temp. 20% error in water pipes can not be applied
to all fluids in all conditions

Design Period for Water supply Components
• The economic design period of the components of a
distribution system depends on
• Their life.
• First cost.
• And the ease of expandability.
19

Network Layout
• Next step is to estimate pipe sizes on the basis
of water demand and local code requirements.
• The pipes are then drawn on a digital map (using
AutoCAD, for example) starting from the water
source.
• All the components (pipes, valves, fire hydrants)
of the water network should be shown on the
lines.
20

Part B
Pipe Network Analysis
21

Pipe Network
Analysis
22

Pipe Networks
• A hydraulic model is useful for examining the
impact of design and operation decisions.
• Simple systems, such as those discussed in last
chapters can be solved using a hand calculator.
• However, more complex systems require more
effort even for steady state conditions, but, as in
simple systems, the flow and pressure-head
distribution through a water distribution system
must satisfy the laws of conservation of mass and energy.
23

The equations to solve Pipe network must
satisfy the following condition:
• The net flow into any junction must be zero
• The net head loss a round any closed loop must be
zero. The HGL at each junction must have one
and only one elevation
• All head losses must satisfy the Moody and minor-
loss friction correlation
Pipe Networks
0Q
24

Node, Loop, and Pipes Node
Pipe
Loop
25

After completing all preliminary studies and
layout drawing of the network, one of the
methods of hydraulic analysis is used to
• Size the pipes and
• Assign the pressures and velocities
required.
Hydraulic Analysis
26

Hydraulic Analysis of Water Networks
• The solution to the problem is based on the same
basic hydraulic principles that govern simple and
compound pipes that were discussed previously.
• The following are the most common methods used to
analyze the Grid-system networks:
1. Hardy Cross method.
2. Sections method.
3. Circle method.
4. Computer programs (Epanet,Loop, watercad...)
27

Hardy Cross Method
This method based on:
00
Loop
  f
Junction
hQ
1- A distribution of flows in each pipe is estimated such that
the total inflow must be equal to the outflow at each junction
throughout the network system
The interflow in the network has +ve sign
The outflow from the network has -ve sign
28

2- Neglect Minor loss
3- In each loop
4- If the direction of flow is clockwise it take +ve sign,
otherwise it take –ve sign
5- If the flow is correct other wise, the assumed
flow must be corrected as the flowing:
0
Loop
 fh
0
Loop
 fh
29

WilliamHazenn
ManningDarcyn
kQh n
F
85.1
,2
)1(


 )2( oQQ
   








 
....
2
1
2&1
221
f
n
o
n
o
n
o
n
o
n
Q
nn
nQQkQkkQh
from
  1
f
n
o
n
o
n
nQQkkQh
 
0
0
1




nn
o
n
loop
n
loop
F
nkQkQkQ
kQh
Neglect terms contains term then,
2

For each loop
30

 



 


 
o
F
F
n
o
n
o
Q
h
n
h
nkQ
kQ
1
6-After calculation correct Qo and check 0
Loop
 fh
31

Assumptions / Steps of this method:
1. Assume that the water is withdrawn from nodes only; not
directly from pipes.
2. The discharge, Q , entering the system will have (+) value, and
the discharge, Q , leaving the system will have (-) value.
3. Usually neglect minor losses since these will be small with
respect to those in long pipes, i.e.; Or could be included as
equivalent lengths in each pipe.
4. Assume flows for each individual pipe in the network.
5. At any junction (node), as done for pipes in parallel,
  outin QQ Q  0or
32

6. Around any loop in the grid, the sum of head losses must equal
to zero:
– Conventionally, clockwise flows in a loop are considered (+) and
produce positive head losses; counterclockwise flows are then (-) and
produce negative head losses.
– This fact is called the head balance of each loop, and this can be valid
only if the assumed Q for each pipe, within the loop, is correct.
• The probability of initially guessing all flow rates correctly is
virtually null.
• Therefore, to balance the head around each loop, a flow rate
correction ( ) for each loop in the network should be
computed, and hence some iteration scheme is needed.
hf
loop
 0

33

7. After finding the discharge correction, (one for each loop) ,
the assumed discharges Q0 are adjusted and another iteration is
carried out until all corrections (values of ) become zero or
negligible. At this point the condition of :
is satisfied.
Notes:
• The flows in pipes common to two loops are positive in one
loop and negative in the other.
• When calculated corrections are applied, with careful attention
to sign, pipes common to two loops receive both corrections.


hf
loop
 00.
34

• Note that if Hazen Williams (which is generally used in this method) is
used to find the head losses, then
h k Qf  185.
(n = 1.85) , then  
 

h
h
Q
f
f
185.
• If Darcy-Wiesbach is used to find the head losses, then
h k Qf  2
 


h
h
Q
f
f
2
(n = 2) , then
 



 


 
o
F
F
n
o
n
o
Q
h
n
h
nkQ
kQ
1
35

Example 1
DLpipe
150mm305m1
150mm305m2
200mm610m3
150mm457m4
200mm153m5
1
23
4
5
37.8 L/s
25.2 L/s
63 L/s
24
11.4
Solve the following pipe network using Hazen William Method CHW =100
36

 
24.0
64.085.1
28.0
1 






o
F
F
Q
h
n
h
 
1002.6
1000
0.71
L/s,100
0.71
852.1
852.1
87.4
9
852.1
87.4852.1
852.1
87.4852.1
QKh
Q
D
L
h
Q
DC
L
h
inQCQ
DC
L
h
f
f
HW
f
HW
HW
f

















 
57.0
43.085.1
45.0
2 






o
F
F
Q
h
n
h
1
23 4
5
12
21
2loopin2pipefor
1loopin2pipefor


37

 
15.0
64.085.1
18.0
1 






o
F
F
Q
h
n
h  
 
09.0
42.085.1
07.0
1 






o
F
F
Q
h
n
h
1
23 4
5
12
21
2loopin2pipefor
1loopin2pipefor


38

• Assigning clockwise flows and their associated head
losses are positive, the procedure is as follows:
 Assume values of Q to satisfy Q = 0.
 Calculate HL from Q using hf = K1Q2 .
 If hf = 0, then the solution is correct.
 If hf  0, then apply a correction factor, Q, to all
Q and repeat from step (2).
 For practical purposes, the calculation is usually
terminated when hf < 0.01 m or Q < 1 L/s.
 A reasonably efficient value of Q for rapid
convergence is given by;


Q
H
2
H
Q
L
L
Summary


Q
H
2
H
Q
L
L
39

Example
• The following example contains nodes with different
elevations and pressure heads.
• Neglecting minor loses in the pipes, determine:
• The flows in the pipes.
• The pressure heads at the nodes.
40

Assume flows magnitude and direction
42

First Iteration
• Loop (1)
Pipe
L
(m)
D
(m)
Q
(m3/s)
f
hf
(m)
hf/Q
(m/m3/s)
AB 600 0.25 0.12 0.0157 11.48 95.64
BE 200 0.10 0.01 0.0205 3.38 338.06
EF 600 0.15 -0.06 0.0171 -40.25 670.77
FA 200 0.20 -0.10 0.0162 -8.34 83.42
S -33.73 1187.89
L/s20.14/sm01419.0
)89.1187(2
73.33 3



43

First Iteration
• Loop (2)
Pipe
L
(m)
D
(m)
Q
(m3/s)
f
hf
(m)
hf/Q
(m/m3/s)
BC 600 0.15 0.05 0.0173 28.29 565.81
CD 200 0.10 0.01 0.0205 3.38 338.05
DE 600 0.15 -0.02 0.0189 -4.94 246.78
EB 200 0.10 -0.01 0.0205 -3.38 338.05
S 23.35 1488.7
L/s842.7/sm00784.0
)7.1488(2
35.23 3

44

Second Iteration
• Loop (1)
Pipe
L
(m)
D
(m)
Q
(m3/s)
f
hf
(m)
hf/Q
(m/m3/s)
AB 600 0.25 0.1342 0.0156 14.27 106.08
BE 200 0.10 0.03204 0.0186 31.48 982.60
EF 600 0.15 -0.0458 0.0174 -23.89 521.61
FA 200 0.20 -0.0858 0.0163 -6.21 72.33
S 15.65 1682.62
L/s65.4/sm00465.0
)62.1682(2
65.15 3

14.20
14.20
14.20 7.8414.20
45

Second Iteration
• Loop (2)
Pipe
L
(m)
D
(m)
Q
(m3/s)
f
hf
(m)
hf/Q
(m/m3/s)
BC 600 0.15 0.04216 0.0176 20.37 483.24
CD 200 0.10 0.00216 0.0261 0.20 93.23
DE 600 0.15 -0.02784 0.0182 -9.22 331.23
EB 200 0.10 -0.03204 0.0186 -31.48 982.60
S -20.13 1890.60
L/s32.5/sm00532.0
)3.1890(2
13.20 3



14.20 7.84
7.84
7.84
7.84
46

Third Iteration
• Loop (1)
Pipe
L
(m)
D
(m)
Q
(m3/s)
f
hf
(m)
hf/Q
(m/m3/s)
AB 600 0.25 0.1296 0.0156 13.30 102.67
BE 200 0.10 0.02207 0.0190 15.30 693.08
EF 600 0.15 -0.05045 0.0173 -28.78 570.54
FA 200 0.20 -0.09045 0.0163 -6.87 75.97
S -7.05 1442.26
L/s44.2/sm00244.0
)26.1442(2
05.7 3



47

Third Iteration
• Loop (2)
Pipe
L
(m)
D
(m)
Q
(m3/s)
f
hf
(m)
hf/Q
(m/m3/s)
BC 600 0.15 0.04748 0.0174 25.61 539.30
CD 200 0.10 0.00748 0.0212 1.96 262.11
DE 600 0.15 -0.02252 0.0186 -6.17 274.07
EB 200 0.10 -0.02207 0.0190 -15.30 693.08
S 6.1 1768.56
L/s72.1/sm00172.0
)56.1768(2
1.6 3

48

After applying Third correction
49

Velocity and Pressure Heads:
pipe
Q
(l/s)
V
(m/s)
hf
(m)
AB 131.99 2.689 13.79
BE 26.23 3.340 21.35
FE 48.01 2.717 26.16
AF 88.01 2.801 6.52
BC 45.76 2.589 23.85
CD 5.76 0.733 1.21
ED 24.24 1.372 7.09
1.2121.3
5
13.79 23.8
5
6.52
26.1
6
7.09
50

Velocity and Pressure Heads:
Node
p/g+Z
(m)
Z
(m)
P/g
(m)
A 70 30 40
B 56.21 25 31.21
C 32.36 20 12.36
D 31.15 20 11.15
E 37.32 22 15.32
F 63.48 25 38.48
1.2121.3
5
13.79 23.8
5
6.52
26.1
6
7.09
51

Example
For the square loop shown, find the discharge in all the pipes.
All pipes are 1 km long and 300 mm in diameter, with a friction
factor of 0.0163. Assume that minor losses can be neglected.
52

• Solution:
 Assume values of Q to satisfy continuity equations all at nodes.
 The head loss is calculated using; HL = K1Q2
 HL = hf + hLm
 But minor losses can be neglected:  hLm = 0
 Thus HL = hf
 Head loss can be calculated using the Darcy-Weisbach
equation
g2
V
D
L
h
2
f 
53

First trial
Since HL > 0.01 m, then correction has to be applied.
554'K
Q'KH
Q554H
3.0x
4
Q
x77.2
A
Q
77.2H
81.9x2
V
x
3.0
1000
x0163.0H
g2
V
D
L
hH
2
L
2
L
2
2
2
2
2
L
2
L
2
fL








 



Pipe Q (L/s) HL (m) HL/Q
AB 60 2.0 0.033
BC 40 0.886 0.0222
CD 0 0 0
AD -40 -0.886 0.0222
 2.00 0.0774
54

Second trial
Since HL ≈ 0.01 m, then it is OK.
Thus, the discharge in each pipe is as follows (to the nearest integer).
s/L92.12
0774.0x2
2
Q
H
2
H
Q
L
L 



Pipe Q (L/s) HL (m) HL/Q
AB 47.08 1.23 0.0261
BC 27.08 0.407 0.015
CD -12.92 -0.092 0.007
AD -52.92 -1.555 0.0294
 -0.0107 0.07775
Pipe Discharge
(L/s)
AB 47
BC 27
CD -13
AD -53
55

Assignment # 02
56
Hardy’s Cross Method
(Example)

Assignment # 02
57
Hardy’s Cross Method
(Example)

design and analysis of water distribution System

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