Water Supply and

treatment
(WSEE-3151)

for Ceng 4181 years civil engineering

students.
2024
 Chep 10

Water distribution and supply

3
 Pipelines and appurtenances
pipeline, line of pipe equipped with pumps and valves and other control devices
for moving liquids, gases, and slurries (fine particles suspended in liquid). Pipeline
sizes vary from the 2-inch- (5-centimetre-) diameter lines used in oil-well
gathering systems to lines 30 feet (9 meters) across in high-volume water and
sewage networks. Pipelines usually consist of sections of pipe made of metal (e.g.,
steel, cast iron, and aluminum), though some are constructed of concrete, clay
products, and occasionally plastics. The sections are welded together and, in most
cases, laid underground

4
The selection of pipe materials is based on

• carrying capacity

• strength

• ease of transportation and handling

• availability

• quality of water

• cost (initial and maintenance )

 Pipes
Pipes are used to convey water and wastewater
coming into and out of the fixtures.
So when selecting pipe material for water supply pipes ,we consider
the expected
 water pressure,
water temperature,
durability, ease of installation and cost
Common types of pipe in WS and Waste Water depends on faction.
drainage:

Polypropylene pipes(PP-R) for water supplies

specially designed to provide your buildings with an easy flow of hygienic hot and
cold drinking water

 Polyvinylchloride pipes(PVC) (H2C–CHCl)n, for waste water drainage

PVC pipe offers excellent resistance to conventional corrosion and abrasion. PVC
pipe not threatened by environmental stress cracking due to under loading or
localized tensile stress, and provides soil/water tight joints
Polyvinyl Chloride(PVC) piping is the most widely used plastic piping material. PVC piping
systems are
Environmentally sound
Provide long service life
Easy to install and handle
Resistant to Corrosion and
Cost effective

Chemical formula: (C2H3Cl)n

The following are the various materials,which are used for sewers.
i. Asbestos cement sewers
ii. Cast-Iron sewers
iii. Cement concrete sewers
iv. Stoneware sewers
v. Brick sewers
vi. Corrugated iron sewers
vii. Plastic sewers
viii. Steel sewers
 Laying of Sewers

The construction of sewer consists of the following works:

a. Marking center lines of sewers
b. Excavation of trenches
c. Checking the gradient
d. Preparation of bedding
e. Laying of sewers
f. Jointing
g. Back filling
 Pipes and conduits for water
In water-supply systems the term is usually reserved for covered or closed sections of
aqueduct, especially those that transport water under pressure.

Large conduits may be fabricated of steel sections joined in the field or of

reinforced concrete, often precast and prestressed, or they may be driven at depth
through solid rock, for example, under mountainous terrain or under rivers.

Smaller pressure conduits are usually made of cast iron, steel, or asbestos cement.
Pressure conduits are usually kept below the hydraulic grade line—that is, the line
representing the height to which the water would rise if free—to avoid possible
entrapment of air.
appurtenances
Appurtenance is a general term used to describe things such as valves, fire
hydrants, meters, among other things. There are various names which refer to a
company which distributes water to customers. Some examples include, water
retailer, water utility, water district, water agency, water purveyor, and water
supplier.
Pipe appurtenances are fixtures or attachments that help in isolating and
draining a pipe to perform inspections, tests, cleaning, and repairs.

What are the different types of pipe appurtenances?

Gates,

valves,

manholes,

isolation joints, and

anchorages are pipe appurtenances.

Types of fittings
1. Collar
2. Elbow
3. Gasket
4. Union
5. Reducer
6. Tee
7. Nipple
8. Trap
 Common sanitary fittings and fixtures
Bathtub
• Washbasin
• Water closet
• Faucet
• Sink
• Flushing
• Cistern
• Geyser
 Distribution systems
Depending upon the level of the source of water and the city, topography
of the area, and other local considerations, Gravitational system,

Pumping without storage, and

Pumping with storage

16
Pipeline layout options

17
DISTRIBUTION RESERVOIRS

18
 Purpose of Distribution reservoirs
• Equalizing supply and demand
• Providing water during source or pump failure
• Providing water to meet fire demands
• Increase detention times
• Increasing operating convenience
• Leveling out pumping requirements
• Maintaining pressure levels within acceptable ranges
• Blending water sources 19
Equalizing demand and supply

20
Elevated tank location

21
Distribution reservoir types
1.Ground reservoir

2.Elevated reservoir

3.Stand pipes reservoir

Can be made of
• Concrete or masonry
• Steel tank

22
23
24
25
Capacity Reservoir=Equalizing storage + Emergency reserve (about 25%) + Fire
storage.

• Aesthetics

• Ventilation

• Security and safety

26
Reservoir capacity
Size (m3) Depth of water
(m)
Up to 3500 2.5 – 3.5
3500 – 15 000 3.5 – 5.0
Over 15 000 5.0 – 7.0
For rectangular concrete tank

27
Example=1

1. Design a service reservoir if Qday-max is 2400 m3. Two pumps are working at
constant rate of 150 m3/hr. Determine for how many hours pumping should be done.

• Solution

Step 1. Determine pumping hours Pumping hour = Qdmax/(2 x pumping capacity)

= 2400/(2 x 150)

= 8 hrs for each pump

28
Step 2. Determine reservoir capacity Balancing requirement Pumping is done
for a total of 16 hrs = 2400 m3

But demand for 16 hrs is 2400 x 16/24 = 1600 m3

Excess that needs to be stored = 2400-1600 = 800 m3

29
Emergency requirement = 0.25 x 800 = 200 m3

Ignore fire demand = 0

Total reservoir volume = 1000 m3

Provide 2 reservoirs of each 500 m3

Take depth of 3.0 m

X-sectional area = 500/3.0 = 166.67 m2

Taking circular tank diameter = 14.57 m

30
 Example-2
2.A Town with a Population of 100,000 is to be supplied with water daily at 200 liters per head.
The Variation in demand is as follows
• 6 am – 9am = 40% of total

• 9am - 12 noon = 10% of total

• 12noon – 3Pm = 10% of total

• 3Pm – 6PM = 15% of total

• 6Pm – 9Pm = 25% of total

• Determine the capacity of the service reservoir assuming pumping to be at uniform rate & the period of pumping to be from 6am –
4Pm. Neglect fire demand.

31
Solution: -

Dd= V day = 100,000*200 L/head = 20Ml for ( 6am-9pm)

SS= Q day = 100,000*200 L/head = 20Ml for ( 6am-4pm)10 hrs = 20ML/10hrs = 2 ML/hrs

Period Hours Rate of demand Dd in ML comm dd. SS ML/h Comm ss Excess dd Excess ss
(1) (2) (3) (4) volume (5) (6) Q (7)ML/hr col (5)–(7)=+ (8) col (7)–(5)= + (9)

6am – 9am 3 40% of 20ML 8 8 6 6 2 _

9am-12noon 3 10% of 20ML 2 10 6 12 _ 2

12noon-3Pm 3 10% of 20ML 2 12 6 18 _ 6

3Pm-6Pm 3 15% of 20ML 3 15 2 20 _ 5

6Pm-9Pm 3 25% of 20ML 5 20 Null 20 _ _

Total storage required = Max Excess dd + Max. Excess ss

= 2ML + 6ML = 8ML
Total Volume required = storage + 25% storage( Emergency) + Fire demand
= 8ML + 25% 8ML + 0= 10ML Answer
32
 Assignment(1)
1.A water supply system is proposed to be designed for a small town which has a
maximum daily demand of 515 m3/d. Estimate storage requirement if pumping is
done for 12 hrs. only (from 2 to 14). Use the following demand variation data.

Time (hr) 0-4 4-8 8 - 12 12 - 16 16 - 20 20 - 24

Demand as % of total daily 6.7 9.2 20.8 28.3 25 10

demand

33
Layout of distribution systems

Pipe networks :

Primary or arterial mains from the pumping stations and from storage
facilities to the various districts of the city. valved at intervals of not ≤ 1.5 km

Secondary lines or Sub-mains run from one primary main to another located
at spacings of 2-4 blocks

Small distribution mains or branches Supply water to every consumer and to

the fire hydrants 34
The Layout of the Water Distribution System:
In the layout of the water distribution system, the pipes are generally laid
below the road pavements and generally follow the layouts of roads.
There are four types of water distribution system and for a particular
place, these types either singly or in combinations can be used.
Grid Iron System
Ring System
Radial System
Dead End System
Grid Iron System:
For cities with a rectangular layout where the water branches and mains are laid in
rectangles, the gridiron system is suitable and this system also contains main lines,
branch lines, and sub mains In this system by interconnecting all the lines, dead ends
are eliminated and without stagnating, the water flows continuously in this system.
For well-planned cities, the gridiron system is more suitable.
Advantages of Grid Iron System:
There are the following advantages of this system such as:
• Without any dead ends or sediment deposits, the water will flow continuously.
• Because of the interconnection of pipes, the head loss is minimum in this case.
• For firefighting, the discharge will meet the required discharge.
• Just by closing the cutoff valve in that line which does not affect the other users, the
repair works can be easily done.
Disadvantages of Grid Iron System:
There are the following disadvantages of this system such as:
• The pipes used in this system should be of large diameters and longer lengths
because of circulating flow from all directions.
• The design is difficult because in a particular pipe we cannot determine the
accurate discharge, velocity, or pressure.
• In this system, the cutoff valves required should be more.
Ring System:
In this system, the main pipeline is provided around the city or area and the branch
lines are connected with each other and projected perpendicularly and every street
within the distributed area will get sufficient. This system is also called a circular
system and it is more suitable for a town with well-planned streets and roads.
Advantages of the Ring System:

There are the following advantages of the ring system such as;

• In this system, no stagnation of water occurs.

• In the ring or circular system without affecting the larger network repair work can be done.

• For firefighting a large quantity of water is available.

Disadvantages of Ring System:

There are the following disadvantages of the ring system such as;

• Large diameter and longer length pipes are required in this system.

• While laying pipes, skilled workers are necessary.

• The numbers of cutoff valves are more.

Radial System
In the radial system, the whole area is divided into small distribution zones and for
each distribution zone, an individual distribution reservoir is provided. The pipelines
are laid radially to the surrounded streets from this reservoir. With the mainline, all
distribution reservoirs are connected and pass through the center of the city, and for
areas with radially designed roads this system is suitable
Advantages of Radial System:

There are the following advantages of the radial system such as;

• With the high velocity and the high pressure, the water is distributed.

• Because of quick discharge, the head loss is very small.

• Easy to calculate the sizes of pipes.

Disadvantages of Radial System:

There are the following disadvantages of the radial system such as;

• Because of the number of individual distribution reservoirs, the cost of the project is more.

• Stagnation of water occurs in pipes due to many dead ends.

Dead End Water Distribution System:
• In the dead-end system, the water does not flow continuously because it contains dead ends in
the pipe system, and into several sub-networks, the whole pipe network is divided such as main
lines, sub mains, service connections, and branch lines Firstly, one mainline is laid through the
center of the city or area.

• On both sides of the mainline, sub-mains are laid and then sub-mains are divided into branch
lines and service connections are given from this.
Advantages of Dead End System:
There are the following advantages of the dead-end system such as;
I. The pipes can be easily laid in this system.
II. To make design calculations very simple, the pressure and discharge in each
pipe can be determined easily and accurately.
III. Based on the required demand of the population, the diameters of pipes of
main, sub mains and branches can be designed so the cost can be reduced.
IV. Less number of cutoff valves require.
Disadvantages of Dead End System:
There are the following disadvantages of the dead-end system such as;
I. In the remote parts, the pressure is not constant and is very less.
II. The whole portion should be stopped to repair if there is any damage occurs in
the branch line.
III. For firefighting, limited discharge is available in this system
Design of distribution systems
Design flow: Max day demand + Fire demand
Minimum main sizes:
generally:150mm (6 in);
high value districts: 200mm (8 in);
major streets: 305mm (12 in);
domestic flows only: 100mm (4 in);
small communities: 50-75 mm
Velocity:
• Minimum = 0.6 - 1 m/s;
• Maximum = 2.5 m/s 44
Pressure zones

Pressure reducing valves (PRV).

45
Pressure:

• Minimum value is 140 kPa

• Maximum not to exceed 410 kPa

• P= ƍgh

But mainly depends on pressure ratings of the pipes and appurtenances used and
regulatory requirements

46
 Pipe Network Geometry

Pipe network analysis simulates steady flow of liquids or gases under pressure. Pipe
network can simulate city water system pipe networks, long pipeline networks with
different diameter pipes in series, parallel pipes, groundwater flow into a slotted
well screen, soil vapor extraction well design, and more.
Cont.…
iL = jL + kL - iL = no. of pipes
1 jL = no. of nodes
kL = no. of
loops

48
49
Simple DS Design procedures
D
L2

R
A P-4
P-1 P-3 P-5 E
C
P-2
L3
L1
B
 D Q = 0.75 m3/min
El. 1167

75 mm
50 m
El. 1250 Q=2m3 /min
El. 1207 El. 1185
R 150 m A 200 m 250 m
75 mm E
150mm 125 mm C
El. 1177

75 mm
75 m
El. 1192
B Q = 1 m3/min
 Solution: Velocity Calculation
Velocity is calculated as D Q = 0.75 m3/min

Q = 0.75 m3/min
75 mm
El. 1250 v = 1.6 m/s
R Q = 1 m3/min
Q = 2 m3/min
A
125 mm Q = 0.25 m3/minE
150 mm v = 1.4 m/s C 75 mm
v = 1.9 m/s v = 0.9 m/s
Q = 1 m3/min
75 mm
v = 2.1 m/s

Q = 1 m3/min
 Solution: Head loss calculation
Head loss is calculated as
D

Q = 0.75 m3/min
100 mm, 50 m
El. 1250 hl = 2.36 m
Q = 1 m3/min
R A Q = 0.25 m3/min
Q = 2 m3/min 125 mm, 200 m E
150 mm, 150 m hl = 5.43 m C 75 mm, 250 m
hl = 6.04 m hl = 6.27 m
Q = 1 m3/min
100 mm, 75 m
hl = 6.03 m

B
 Solution: Residual pressure calculation
Residual pressure is calculated as D El. 1167
hres = Elv. 1 – Elv 2 - hl hres = 69.17 m

hl = 2.36 m
El. 1250 El. 1207
hres = 36.96 m
R hl = 6.04 m A hl = 6.27 m E
hl = 5.43 m C
El. 1177
hres, A = 1250 – 1207 – 6.04
hres = 61.53 m
hres, A = 36.96 m
hl = 6.03 m
hres, B = 1250 – 1192 – 6.04 – 6.03
= 45.93 m
B El. 1192
or
hres = 45.93 mhres, B = 1207 + 36.96 – 6.03 – 1192
= 45.93 m
**Simple DS Design procedures **

Assign the required demand at each node

i. Estimate the discharge flowing through the pipes

ii. Assume possible pipe diameters

iii. Calculate the head loss through the pipes

iv. Find the residual pressure at the end of the pipe.

v. Compare this terminal pressure with the desired minimum and maximum pressures.

vi. If the required condition is not satisfied,

vii. then repeat steps (ii) through (vi) until the required conditions are met.
55
 Design methods for networks
i. Trial and error method
ii. Hardy Cross Method (in flow=out flow)

56
 Example.
Q1. A dead end type of a distribution pipe network has been adopted for
supplying water to a certain zone of the city having different blocks and their
populations, as shown in Fig.1 below. The rate of supply is 200 litres per head per
day .The R.L. of the bottom of the elevated storage tank is 225 R.L. of points A, B, C,
D, etc are 210, 200 and 195m respectively. If the minimum pressure head of water
is to be 12m, Design suitable size of pipes. AB and BC lengths are 500 and 400m
respectively. Assume maximum rate of demand to be 2.7 times the average
demand

57
below.

p=300(1) p=500(2) P=250 p=450

p=800(3) (8) (9) F

p=600(4) P=300 (10)

B
A C

P=300(5) p=500 (11)

P=500
(6) P=700(7) p=800 p=900
(12) (13) G
E 58
 At point D, total population P=Block 8+ Block 2+Block3=250+500+800=1550

At point E, population P= Block 12+ Block 7=800+700=1500

At point F, population P=Block 9=450

At point G, population P=Block 13=900

1) Design for AB

No of population served by AB= 6100=(2+3+4+5+7+8+9+10+11+12+13)

Max discharge=2.7x200x6100 l/c/d=0.038m3/s

2.112 x10 3 xLxQ 1.85

hf  Head loss by Hazen-William equation.
D 4.87

Assuming D=180mm, hf=10.60m

Head available at B= Head available at A-hf through pipe AB=225-10.6=214.40

Elevation at point B=200m

59
Net head (pressure head) at B=214.4-200=14.4m>12.0m , hence it is OK!
1) Design of BC

No of population served=2150= (9+10+11+13)

Q=0.0134m3/s, assuming D=140mm

hf=4.20m

head available at C= Head available at B-hf through pipe BC

=214.4-4.20

=210.20

Elevation at point C=195m

60
Net head (pressure head) at C=210.20-195=15.20>12, hence Ok!
Example
ii. Hard-Cross Method

Determine the discharge in each of the pipes using Hard-Cross Method

In flow = Out flow

61
 Complex pipe Networks
Water supply system in municipal districts are constructed
of large number of pipes interconnected from loops and
branches.
Hardy Cross Method can be used
Assign the required demand at each node
Assume the best distribution of flow that satisfies
continuity by careful examination of the network.
 Cont….
• In any network of pipes, the following conditions
must be satisfied:
Flow into each junction (node) must equal flow out
of the junction.
The sum of head losses around each loop must be
zero.
Cont…
 Design procedures
 Assign the required demand at each node.
 Assume the best distribution of flow that satisfies continuity by careful
examination of the network.
The flow entering a node must be equal to the flow leaving the same
node.
 Calculate the head loss, hf, in each pipe.
The algebraic sum of the heads around a closed loop must be zero.
For a loop, take head loss in the clockwise flows as positive and in
the anti-clockwise flows as negative
 Calculate the correction factor for each loop by
 Example
 A single looped network as shown in Fig has to be
analyzed by the Hardy Cross method for given inflow
and outflow discharges. The pipe diameters D and
lengths L are shown in the figure. Use Darcy–Weisbach
head loss–discharge relationship assuming a constant
friction factor f= 0.02.
Iteration 1
Iteration 2
2.Fig. below shows one of the circuits of a distribution system, A city has a population of 100,000
is to be served. Length AB,BC,CD,DE and AE=600,500,650,1196 and 200m respectively.

The pressure at the start point A is 35m head of water and the min. pressure desired at the point D is 20m
head of water. Design the various pipes of this circuit by assuming the per capita demand to be 190
l/p/day and the peak rate of demand for design for design of distribution system as 1.8 times the avg.
assume v=1.3m/s.
70
Solution
Q day-max=1.8x190l/d=342l/d=0.004l/s

At point B=15%x100,000x0.004l/s=60l/s

At point C=10%x100,000x0.004l/s=40l/s

At point D along CD=10%x100,000x0.004l/s=40l/s

At point D along ED=40%x100,000x0.004l/s=160l/s

At point E=25%x100,000x0.004l/s=100l/s

Total=400l/s

Total discharge inflow required at A= 400l/s

i.e ABCD=140 and AED=260l/s 71

Calculate the diameter/size of pipe by using Hazen-William formula or Nomogram
graph for Cost iron pipes.

hf=2.112x10 -3 x L x Q 1.85

D 4.87

72
 Q1. A dead end type of a distribution pipe network has been adopted for supplying water to a certain zone of
the city having different blocks and their populations, as shown in Fig.1 below. The rate of supply is 180
litres per head per day .The R.L. of the bottom of the elevated storage tank is 150 R.L. of points A, B, C, D,
etc are 130, 129, 131 and 128, respectively. If the minimum pressure head of water is to be 10m, Design
suitable size of pipes. AB, BC, and CD lengths are 300, 400 and 500m respectively. Assume maximum rate
of demand to be 2.7 times the average demand.
27
28
800
1000
9 10
1 2 200 300 16
300 400
E F 15 500 G 26
700
11 17 400
3
700 600 600
4 25
600
600
A
B C D 24
5 6
900
12 18
750 850 700 800 20
19 1000
13 14 500
7 23
600 600 22
600 21 500
700 200
H I J
8
900

Fig. 1 73
2.Fig. below shows one of the circuits of a distribution system, A city has a population of 400,000
is to be served. Length AB,BC,CD,DE and AE=40,80,90,100 and 60m respectively.

The pressure at the start point A is 26m head of water and the min. pressure desired at the point D is 12m head of
water. Design the various pipes of this circuit by assuming the per capita demand to be 60 l/p/day and the peak rate
of demand for design for design of distribution system as 1.67 times the avg. assume v=1.3m/s.

74
3.A study is conducted in order to meet the demand of a rural town. The selected site for service
reservoir is located at elevation of 2031m and planned to distribute the water to a town as
shown below in Figure 2. The elevations of each node are A = 1990m, B=1980m and elevation
of C = 1971m.Calculate the size of pipe from Reservoir to A, AB and BC by considering the
minimum free head at Junctions must be greater than 10m and less than 80mand also check the
velocity.

75
Use domestic demand 70 l/cap/day, For water point 30 l/cap/day, For industrial area of 500hectar; 150 l/ha/day,
for school of 500 students 60l/student/day, for a hospital having 60 beds 75l/bed/day, and for commercial area of
700ha; 120 l/ha/day.
Consider Peak Hour Factor =2.0, Minimum Velocity =0.6m/sec and maximum Velocity = 2.5m/sec.

76
77