1.
Course Objectives
 Course Objectives:
◦ Learning objectives can be formulated as:
 The objective of this course is to give students a broad
understanding of population forecasting, water
demand, water source, collection and distribution of
water.
 The general idea about water supply and water
treatment techniques

2.
Some facts
 88% of diarrhoeal disease is attributed to unsafe water
supply, inadequate sanitation and hygiene.
 1.8 million people die every year from diarrhoeal
diseases (including cholera); 90% are children under 5,
mostly in developing countries.
 Every 8 seconds a child dies from drinking contaminated
water (that is 10,000 a day).
 500 million people are at risk from trachoma, 146 million
are threatened by blindness and 6 million people are
visually impaired from this disease.
3

3.
1.QUANTITY OF WATER
 Introduction
 In the design of any water work projects, it is
necessary to estimate the amount of water that is
required. This involves:
– Determination of people who will be served
– per capita water consumption
– Analysis of the factors

4.
1.2 Types of demands
 Various Types of Demands
 Domestic water demand
 Industrial and commercial water demand
 Demand for public use
 Fire demand
 Water required to compensate losses in wastes

5.
Water Demand
Domestic water demand
• This includes the water required in private buildings for
drinking, cooking, bathing, gardening, sanitary purposes, etc.
• The total domestic consumption generally amounts to 55 to 60%
of the total water consumption.
Industrial & commercial water demand
• This includes the quantity of water required to be supplied to
offices, factories, different industries, big hotels, hospitals, etc.
20 to 25% of the total water consumption.

6.
Water Demand
 Demand for public use
• This includes the quantity of water required for public
utility purposes, such as watering of public parks,
gardening, washing and sprinkling on the roads, use in
public fountains and etc.
 Fire demand
• The quantity of water required to fight fire. There are
formulas to estimate Fire demand ,

7.
a. National board of fire underwriter formulas.
Q = 64 √ P (1 – 0.01√P)
Where Q = rate of flow of water in l/sec
P = Population in thousand
b. Freeman formula.
Q = 1135.5 ((P⁄ 10) + 10)
Where Q is in lit/min
c. Kuichling formula.
Q = 3182 √P
Where Q is in lit / min
P is in thousands

8.
Water Demand
 Water required to compensate losses in
wastes
– this includes the water lost in leakage due to
bad plumbing or damage meters, stolen water
due to unauthorized water connections and
other losses.
– Generally ,allowance of 15 – 20% of total quantity of water is
made to compensate for the losses.

9.
Factors Affecting Water Demand
o Size of the city
o Climatic condition
o Characteristics of population
o Quality of water supplies
o Pressure in distribution system
o Cost of water
o System of supply
o Policy of metering and method of charging

10.
Factors Affecting water demand
 Size of City
The per capita water demand of the town, will
increase with the size of the town, because more
water will be required in street washing, running of
sewers, maintenance of parks and gardens
 Climatic condition
The quantity of water required in hotter and dry
places is more due to use sprinkling of water in
garden, washing of clothes and bathing.

11.
Factors Affecting Water Demand
 Living standard of the people
 The per capita demand of the town increase with the
standard of living of the people. The people will start using
room coolers, use of flush latrines, use of automatic dish
washing and others.
 pressure in the distribution system
 the rate of water consumption increases with the increase in
the pressure of the distribution system. The water reaches the
upper storey of the building only if it is released with the
required pressure.
If the pressure  - more water loss due to leakage, wastage

12.
Variation in rate of consumption
 Per capita demand: means the annual consumption of water. It was
therefore, defined as the annual average daily consumption per
person.
per capita Demand = Q / P*365
Where Q: – is the total quantity of water required by a town per
year in liter.
P: - The population of the town
 seasonal variation:
Large amount of water uses in summer season, and lesser user in winter.

13.
Design Period and Population
Forecasting
 Design period
 The number of years for which the designs of water
works have been done is known as design period.
These period should neither be too short or too long.
 Mostly water works are designed for design period of
20 – 30 years.

14.
Factor, which should be kept in view while fixing
the design period:
 Fund
 The life of the material used in project (pipes,
structural materials )
 Anticipated expansion of the town
 The rate of interest on the loan taken

15.
Population Forecasting
 When the design period is fixed the next step is to
determine the population in various periods, because
the population of the towns generally goes on
increasing.
 There are three major factors which affect change in
population:
– Birth
– Death
– Migration

16.
Population forecasting
 Methods of population forecasting
 Arithmetical increase method
 Geometrical increase method
 Incremental increase method
 Decreasing rate method
 Logistic Curve method
 Simple graphical method
 Master plan method
 CSA method

17.
Population forecasting

18.
Population forecasting
◦ By integrating both sides the above equation:
Pn – Po = C (tn – t0 )
Such that;
Pn = P o+ C n
Where; Pn -is population after n decade
Po -is present population
n -is number of years in decade
C - constant
This method is generally applicable to a large and old city.

19.
Population forecasting
o Geometrical increase method
 In this method it is assumed that the percentage increase
in population from decade to decade remains constant.
 If the present population is Po and average percentage growth is k, the
population at the end of n decade will be:
P1 = Po + KPo = Po (1 + k)
P2 = p1 + kp1 = p1 (1 + k) = po (1 + k) (1 + k) = po (1+k)2
P3 = p2+ kp2 = p2 (1 + k) = po (1 + k) (1 + k) (1 + k) = po (1+k)3
Pn = po (1 + k)n where po = initial population ,Pn = popn at n decades ,n = Decades
&
k = Percentage (geometric) increase
 This method is mostly applicable for growing towns and cities having vast scope
of expansion.

20.
 Incremental increase method
This method is the combination of the above two methods
and, therefore gives the advantages of both arithmetic
and geometric increase methods and gives satisfactory
results.
Decreasing rate method
 In this method, the average decrease in the percentage
increase is worked out and is then subtracted from the
latest percentage increase for each successive decades.
 This method is applicable to average size cities growing
under normal condition

21.
Logistic Curve method
When the population of a town is plotted
with respect time the curve so obtained
under normal condition is s – shaped
curve and is known as logistic curve.

22.
Water Sources
 Sources of Water Supply
– The origin of all water is rainfall
– Water can be collected
 as it falls
 as rain before it reaches the ground
 as surface water when it flows over the ground
 as ground water when it percolates in to the ground

23.
Water Sources
 All the sources of water can be broadly
divided into:
– Surfaces sources and
– Sub surface sources
 Surfaces Sources
The surface sources further divided into
– Streams and rivers
– Ponds and Lakes
– Impounding reservoirs etc.

24.
Water Sources
 Streams and Rivers
– Rivers and streams are the main source of surface
source of water
– river and stream water require special treatments
– perennial river
 Natural Ponds and Lakes
– natural basins are formed with impervious bed by
springs and streams are known as “lakes”

25.
Water Sources
 Impounding Reservoirs
– In some rivers the flow becomes very small and
cannot meet the requirements of hot weather
– the water can be stored by constructing weir or a
dam across the river

26.
Water Sources
 Subsurface Sources
 These are further divided into
– Infiltration galleries
– Infiltration wells
– Springs
– Well

27.
Water Sources
 Infiltration Galleries
– A horizontal or nearly horizontal tunnel which is
constructed through water bearing strata for
tapping underground water near rivers, lakes or
streams are called “Infiltration galleries”

28.
Infiltration galleries

29.
Water Sources
 Subsurface Sources
– In order to obtain large quantity of water, the
infiltration wells are sunk in series in the blanks of
river
– The infiltration wells in turn are connected by
porous pipes to collecting sump called jack well
and there water is pumped to purification plant for
treatment

30.
Infiltration well

31.
Water Sources
 Springs
– Sometimes ground water reappears at the ground
surface in the form of springs
– Springs generally supply small quantity of water
and hence suitable for the hill towns
– Types of springs:
 Gravity Springs
 Surface Spring
 Artesian Spring

32.
Water Sources
 Gravity springs
– the surface of the earth drops sharply the water
bearing stratum is exposed to atmosphere and
gravity springs are formed

33.
Water Sources
 Surface Spring
– This is formed when an impervious stratum which
is supporting the ground water reservoir becomes
out crops

34.
Water Sources
 Artesian Spring
– When the ground water rises through a fissure in
the upper impervious stratum

35.
Water Sources
 Wells
– A well is defined as an artificial hole or pit made in
the ground for the purpose of tapping water
– types of wells
 Shallow wells
 Deep wells
 Tube wells
 Artesian wells

36.
Intakes for Collecting Surface Water
 The main function of the intakes works is to
collect water from the surface source
 Intakes are structures which essentially
consist of opening, grating or strainer
through which the raw water from river, canal
or reservoir enters

37.
Intakes for Collecting Surface Water
 Types of Intake structures
– Depending upon the source of water the intake
works are classified as following
 Lake Intake
 Reservoir Intake
 River Intake
 Canal Intake
–

38.
Lake Intake
 These intakes are constructed in the bed of
the lake below the water level; so as to draw
water in dry season also

39.
River Intake
 Water from the rivers is always drawn from
the upstream side, because it is free from the
contamination caused by the disposal of
sewage in it

40.
Reservoir Intake
 It consists of an intake well, which is placed
near the dam and connected to the top of
dam by Foot Bridge

41.
Canal Intake
 An intake chamber is constructed in the
canal section

42.
PUMPING
 The operation of lifting water or any fluid is
called pumping
 Pump, a mechanical machine, is used for
lifting water or any fluid to a higher elevations
or at higher pressures.

43.
The purpose of pumping
 To increase the water pressure at certain points in the
distribution system.
 To lift treated water to elevated storage tanks flow
automatically under gravity into distribution system.
 To lift raw river water to carry it to treatment plant.
 To lift water available from wells to an elevated storage
tank in stages.
 To pump water directly into the distribution system.
 To take out water from basins, sumps, tanks etc

44.
 Classification based on mechanical principle of operation
(i) Displacement pumps
(ii) Centrifugal pumps
(iii) Air lift pumps
(iv) Miscellaneous pumps
 Classification based on type of power required
(i) Steam engine pumps
(ii) Diesel engine pumps
(iii) Electrically driven pumps
 Classification based on the type of service called for
(i) Low lift pumps
(ii) High lift pumps ,(iii) Deep well pumps , (iv) Booster pumps
Types of pumps

45.
 Capacity of pump
 Number of pump units required
 Suction conditions
 Lift (total head)
 Discharge conditions and variations in load
 Floor space requirement
 Flexibility of operation
 Starting and priming characteristics
 Type of drive required
 Initial costs and running costs.
Selection of pump

46.
Water Sources Selection Criteria
 Location: The sources of water should be as near as to
the town as possible.
 Quantity of water: the source of water should have
sufficient quantity of water to meet up all the water
demand through out the design period.
 Quality of water: The quality of water should be good
which can be easily and cheaply treated.
 Cost: The cost of the units of the water supply schemes
should be minimum.

47.
Water quality and pollution
 Absolutely pure water is never found in
nature and contains number of impurities in
varying amounts
 The rainwater which is originally pure also
absorbs various gases, dust and other
impurities while falling
 The water supplied to the public should be
strictly according to the standards laid down
from time to time

48.
Water Quality Characteristics
 impurities present in water may be divided
into the following three categories
– Physical Characteristics
– Chemical Characteristics
– Biological Characteristics

49.
Water Quality Characteristics
 Physical characteristics
– Turbidity
– Color
– Taste and odor
– Temperature

50.
Water Quality Characteristics
 Chemical characteristics
– Total solids
– Alkalinity
– pH
– Dissolved oxygen
– BOD, COD
– Nitrogen
– Hardness (calcium & Magnesium)

51.
Physical water characteristics
 Turbidity
– Turbidity is caused due to presence of suspended
and colloidal solids
– silt, clay rock fragments and metal oxides from soil
– Turbidity is a measure of resistance of water to the
passage of light through it
– Turbidity is expressed as NTU (Nephelometric
Turbidity Units)
– Drinking water should not have turbidity more than
10 NTU.

52.
Physical water characteristics
 Color
– Color is caused by materials in solution or
colloidal conditions
– Colored water is not only undesirable
because of consumer objections to its
appearance but also it may discolor
clothing and adversely affect industrial
processes.

53.
Physical water characteristics
 Temperature
– Temperature increase may affect the portability of
water, and temperature above 150c is unpleasant
to drinking water
– Temperature has an effect on most chemical
reactions that occur in natural water systems
– effect on the solubility of gases in water.

54.
Physical water characteristics
 Tastes and Odor
– minerals, metals, and salts from the soil, and
products from biological reactions, and
constituents of wastewater can produce taste and
odor to water
– Alkaline material imports a bitter taste to water,
while metallic salts may give salty or bitter taste.

55.
chemical water characteristics
 Total Solids
 Total solids include the solids in suspension
colloidal and in dissolved form
 The quantity of suspended solids is determined
by filtering the sample of water through fine filter,
drying and weighing
 The quantity of dissolved and colloidal solids is
determined by evaporating the filtered water
obtained from the suspended solid test and
weighing the residue

56.
chemical water characteristics
 Alkalinity
 defined as the quantity of ions in water that will react
to neutralize hydrogen ions or acids
 pH
– pH is a measure of the concentration of free hydrogen
ion in water
 Dissolved Oxygen (DO)
– The presence of oxygen in the water in dissolved form keeps it
fresh and sparkling.
– But more quantity of oxygen causes corrosion to the pipes
material.

57.
chemical water characteristics
 Oxygen Demand
 Indicators used for estimation of the oxygen
demanding substance in water
– Biological Oxygen Demand (BOD) & Chemical
Oxygen Demand (COD)

58.
chemical water characteristics
 Hardness
– Hardness is caused by the sum of calcium and
magnesium elements present in water.
– Temporary hardness (carbonate hardness)
 Calcium bicarbonate (Ca (HCO3) 2)
 Magnesium bicarbonate (Mg (HCO3) 2)
– Permanent hardness’ (non- carbonate hardness)
 Calcium sulfate (CaSO4)
 Magnesium chloride (MgSO4)
 Calcium chloride (CaCl2)
 Magnesium chloride (Mg Cl2)

59.
Biological Characteristics
 Bacterium
– typhoid fever, cholera, and bacterial dysentery
 Viruses
– hepatitis and poliomyelitis
 Algae
– cause turbidity in water and an apparent color
 Protozoa
– giardiasis and amebic dysentery

60.
Water quality standards
Parameter WHO guideline Recommended for
Ethiopia
pH
Total solids, mg/L
Total hardness, mg/L
Chloride, mg/L
Sulphate, mg/L
Fluoride, mg/L
Iron, mg/L
E. Coli, MPN/100 ml
Nitrate , mg/L
6.5-8.5
1000
500
250
400
1.5
0.3
10
10
5.0-9.5
2000
600
800
600
4
3
30
40

61.
Sources of Water Pollution
 Following are the main sources of water
pollution.
– Domestic sewage
– Industrial wastes
– Agricultural areas
– Distribution system

62.
WATER TREATMENT
 The process of removing the impurities from
water is called water treatment and the
treated water is called wholesome water
 The surface sources generally contains large
amount of impurities therefore they requires
sedimentation, filtration and chlorination as
treatment
 Groundwater which is usually clear may
require only disinfection

63.
Water Treatment
 Methods of Water Treatment
– Aeration
– Screening and grit removal
– Plain sedimentation
– Coagulation and flocculation
– Secondary sedimentation
– Filtration
– Adsorption
– Softening
– Disinfections

64.
Aeration
 It is the process of bringing water in intimate
contact with air, while doing so water
absorbs oxygen from the air
 Aeration may be used to remove undesirable
gases dissolved in water i.e. Co2, H2S,
 Different types of aerators are available
– Gravity Aerator
– Spray aerator
– Air diffuser
– Mechanical Aerator

65.
Aerators
 Gravity aerators
A. Cascade towers
Inlet
chamber
Collection
Chamber

66.
Aerators
B. Inclined apron possibly shaded with plates
Collection
Chamber
Inlet chamber

67.
Aerators
C. Tray aerator
– In tray aerator water falls through a series of trays
perforated with small holes, 5 - 12mm diameter
and 25 - 75mm spacing center to center

68.
Aerators
 Spray aerators
– spray droplets of water into the air from stationary or
moving orifices or nozzles
– Water is pumped through pressure nozzles to spray in
the open air as in fountain to a height of about 2.5m

69.
Aerators
 Air diffuser
– Compressed air is forced into this system through
the diffusers
– This air bubbles up through the water, mixing
water and air and introducing oxygen into the
water

70.
Aerators
 Mechanical Aerator
– These aerators work by vigorously agitating the
water with mechanical mixers

71.
Screening
 Screening usually involves a simple
screening or straining operation to remove
large solids and floating matter as leaves,
dead animals, fish etc.
– Bar screens - with openings of about 75mm
– Mesh screens - with opening of 5 - 20mm

72.
Plain Sedimentation
 Sedimentation is the removal of particles (silt,
sand, clay, etc.) through gravity settling in
basins
 No chemicals is added to enhance the
sedimentation process
 Principle of plain sedimentation - Discrete
particles

73.
Terminal settling velocity

74.
Terminal settling velocity
 i

75.
Terminal settling velocity
 i

76.
Example
 Estimate the terminal settling velocity in water at
a temperature of 15ºc (µ = 0.00113Ns/m2) of
spherical silicon particles with specific gravity
2.40 and average diameter of (a) 0.05mm and
(b) 1.0mm

77.
Ideal horizontal flow sedimentation
basin
 Ideal sedimentation basin has four distinct zones
 Uniform dispersion of water and suspended particles
in the inlet zone
 No particle settling in the inlet and outlet zones
 Continuous flow at constant rate
 Once a particle enters the sludge zone, it remains
there (No resuspension of settled particles)
 The flow-through period is equal to the detention
time
 Particles move forward with the same velocity as the
liquid

78.
Ideal horizontal flow sedimentation basin
Sludge zone
Outlet zone
Inlet zone
Settling zone
Width, B
HO
vc
vf v
v
L
Q Q
a
b
c
d
e
f

79.
sedimentation
vc: critical velocity, m/h
Q: flow rate, m3/h
Td: detention time, h
A: area of top of basin settling zone, m2
SOR: surface overflow rate, m3/m2.h

80.
 The velocity of flow can be reduced by increasing the
length of travel and by detaining the particle for a
longer time in the sedimentation basin
 The size and the shape of the particles can be
altered by the addition of certain chemicals in water
(coagulants)
 Sedimentation Tanks are generally made of
reinforced concrete and may be rectangular or
circular in plan
 Long narrow rectangular tanks with horizontal flow
are generally preferred to the circular tanks with
radial or spiral flow
Sedimentation Tank

81.
Sedimentation Tank

82.
Sedimentation Tank

83.
 For the efficient removal of sediment in the
sedimentation tanks, it is necessary that the flow
is uniformly distributed throughout the cross-
section of the tank.
 If currents, on the other hand, permit a
substantial portion of the water to pass directly
through the tank without being detained for the
intended time, the flow is said to be short
circuited.
Short Circuiting in the Sedimentation Tanks

84.
Inlet zone
 to distribute the water inside the tank
 to control the water's velocity as it enters the
basin
 A most suitable type of an inlet for a rectangular
settling tank is in the form of a channel extending
to full width of the tank with a submerged weir
type baffle wall

85.
Inlet zone
 E

86.
Out let zone
 Outlet arrangement consists of
– weir, notches or orifices
– effluent trough or launder
– outlet pipe

87.
Design parameters of sedimentation tank
1. Detention period ….. 3 to 4 hours for plain settling
2 to 2.5 hours for coagulant settling
1 to 1.5 hours for vertical flow type
2. Overflow rate ……… 15 - 30 m3/m2/day for plain settling
30 - 40m3/m2/day for horizontal flow
40 - 50m3/m2/day for vertical flow
3. Velocity of flow…….. 0.5 to 1.0 cm/sec
4. Weir loading………... 300m3/m/day
5. L:W ………………….. 3:1 to 5:1
Breadth of tank…….. (10 to 12m) to 30 to 50m
6. Depth of tank………. 2.5 to 5m (with a preferred value of 3m)
7. Diameter of circular tank…. up to 60m
8. Solids removal efficiency….. 50%
9. Turbidity of water after sedimentation – 15 to 20 NTU.
10. Inlet and Outlet zones………. 0.75 to 1.0m
11. Free board…………………… 0.5m
12. Sludge Zone…………………. 0.5m

88.
Example
 A water treatment plant has four clarifiers
treating 0.175 m3/s of water. Each clarifier is
4.88m wide, 24.4m long and 4.57m deep.
Determine: (a) the detention time, (b) overflow
rate, (c) horizontal velocity, and (d) weir loading
rate assuming the weir length is 2.5 times the
basin width.

89.
Coagulation (Coagulation Aided with
Sedimentation)
 To remove very fine particles from the water
 Adding of this chemical process is called
coagulation and the chemical used in the
process is called coagulant
 Objective, to form a flocs

90.
Typical coagulants
 Aluminum sulfate: Al2(SO4)3.14 H2O
 Iron salt- Ferric sulfate: Fe2(SO4)
 Iron salt- Ferric chloride: Fe2Cl3
 Polyaluminum chloride (PAC): Al2(OH)3Cl3

91.
What is Coagulation?
 Coagulation is the destabilization of colloids by
addition of chemicals that neutralize the negative
charges
 The chemicals are known as coagulants, usually
higher valence cationic salts (Al3+, Fe3+ etc.)
 Coagulation is essentially a chemical process

92.
10
2
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
- -
-
-
-
-
-
-

93.
Coagulation aim
10
4

94.
Colloidal interaction
10
5

95.
Charge reduction
10
6

96.
Colloid Destabilization
 Colloids can be destabilized by charge
neutralization
 Positively charges ions (Na+, Mg2+, Al3+,
Fe3+ etc.) neutralize the colloidal negative
charges and thus destabilize them.
 With destabilization, colloids aggregate in
size and start to settle
10
7

97.
Jar Tests
 The jar test – a laboratory procedure to
determine the optimum pH and the optimum
coagulant dose
 A jar test simulates the coagulation and
flocculation processes

98.
Jar Tests
 Determination of optimum pH
 Fill the jars with raw water sample
(500 or 1000 mL) – usually 6 jars
 Adjust pH of the jars while mixing
using H2SO4 or NaOH/lime
(pH: 5.0; 5.5; 6.0; 6.5; 7.0; 7.5)
 Add same dose of the selected
coagulant (alum or iron) to each jar
(Coagulant dose: 5 or 10 mg/L)
Jar Test

99.
Determination of optimum pH
 Rapid mix each jar at 100 to 150 rpm for 1 minute. The rapid mix
helps to disperse the coagulant throughout each container
 Reduce the stirring speed to 25 to 30 rpm
and continue mixing for 15 to 20 mins.
This slower mixing speed helps
promote floc formation by
enhancing particle collisions,
which lead to larger flocs
 Turn off the mixers and allow
flocs to settle for 30 to 45 mins
 Measure the final residual
turbidity in each jar
 Plot residual turbidity against pH

100.
Jar Tests – optimum pH
Optimum pH: 6.3

101.
Optimum coagulant dose
 Repeat all the previous steps
 This time adjust pH of all jars at
optimum (6.3 found from first test)
while mixing using H2SO4 or
NaOH/lime
 Add different doses of the selected
coagulant (alum or iron) to each jar
(Coagulant dose: 5; 7; 10; 12; 15; 20 mg/L)
 Rapid mix each jar at 100 to 150 rpm for 1 minute. The rapid
mix helps to disperse the coagulant throughout each container
 Reduce the stirring speed to 25 to 30 rpm for 15 to 20 mins

102.
Optimum coagulant dose
 Turn off the mixers and allow flocs to settle for 30 to 45 mins
 Then measure the final residual turbidity in each jar
 Plot residual turbidity
against coagulant dose
The coagulant dose with
the lowest residual
turbidity will be the
optimum coagulant dose
Coagulant Dose mg/L
Optimum coagulant dose: 12.5 mg/L

103.
Coagulant addition: Rapid Mix

104.
Design of rapid mix
 Dimensions of the tank:
– Determine the tank volume for given detention
time
– Assume a depth.
– Calculate the tank diameter/width
 Power requirements
– Calculate water horsepower.
– Calculate electric horsepower.
– Estimate power costs

105.
Geometry of rapid-mix basin &
Detention time
 Should provide uniform mixing
 Should minimize dead areas and short-circuiting
 Usually square basins for mechanical mixers
 Depth-to width ratio: about 2
 Detention time should provide sufficient time for
homogenization of the chemicals with water
 Typical detention time: 10 s to 5 min
Average detention time
td = V/Q
td: detention time, min; V: volume of tank, m3; Q: flow rate, m3/min

106.
Agitation requirements
2
1









V
P
dy
dv
G
Where,
G: velocity gradient, s-1 (100 to 1000s-1)
P: power imparted to the water, watt
V : tank volume, m3 ; : absolute viscosity, Pa-s
The motor power can be calculated if the motor efficiency
is known (motor efficiency = 80-90%).

107.
Filtration
 Removal of colloidal (usually destabilized)
and suspended material from water by
passage through layers of porous media.

108.
Types of Granular Filters
 Based on filter media
– Slow sand filtration
– Rapid filtration
– High-rate filters
 Based on driving force
– Gravity filters
– Pressure filters
 Based on flow direction
– Down flow filters
– Up flow filters

109.
Particle Removal Mechanisms in
Filters

110.
Filter media size parameters
– Effective size (d10): the size of standard sieve
opening that will pass 10% by weight of the media
– Uniformity coefficient (UC): the ratio of the
standard sieve opening that will pass 60% by
weight of the media (d60) to its effective size.
10
60
d
d
UC 

111.
 Slow sand filter consists of concrete/brick work
rectangular basin containing carefully selected graded
sand supported on gravel and stones.
Slow Sand Filter

112.
Mechanisms of impurities removal in
SSF
 Physical:Mechanical straining/sedimentation
 Chemical: Oxidation of organic matter by
aerobic bacteria
 Biological: Occurs through Schmutzdecke
or “Vital layer”. Schmutzdecke is a layer of
dirt, debris, and microorganisms build up on
the top of the sand

113.
Slow Sand Filter Cleaning
 Periodic raking and cleaning of
the filter by removing the top
two inches of sand. After a
few cleanings, new sand must
be added to replace the
removed sand.
 After a cleaning the filter must
be operated for two weeks,
with the filtered water sent to
waste, to allow the
schmutzdecke layer to rebuild.
 Two slow sand filters should
be provided for continuous
operation.

114.
Advantages and Disadvantages of SSF
 Advantages:
– Simple to construct and operate
– Cost of construction cheaper than rapid sand filter
– Do not usually require coagulation/flocculation before
filtration
– Bacterial count reduction is 99.9% to 99.99% and E.coli
reduction is 99% to 99.9%
 Disadvantages:
– Initial cost is low but maintenance cost is much more than
rapid sand filter
– These filters need a lot of space

115.
Design criteria for slow sand filters
Parameter Recommended level
Design life
Period of operation
Filtration rate
Filter bed area
Height of filter bed
Initial
Minimum
Effective size
Uniformity coefficient
Height of under drains including
gravel layer
Height of supernatant water
10-15 year
24 h/day
0.1 – 0.2 m/h
5-200 m2/filter
0.8-0.9 m
0.5-0.6 m
0.15-0.3 mm
< 3
0.3-0.5 m
1 m

116.
Rapid Filter
 Filtration rate is greater
 Backwashing is incorporated
 Straining is not important removal
mechanism
 Particles adhere to media grains and are
removed
 Each grain is a collector
 Water must be pre-treated to destabilize
negatively charged particles

117.
Components of Rapid Filters

118.
Parts of a rapid filter
 Filter tank: contains all system equipment except the
control equipment. Usually rectangular and made of
concrete and constructed side by side on either side
of a control pipe gallery to minimize piping
 Filter sand: specially manufactured or selected sand
and other filter media
 Gravel support bed: prevents sand from being
disturbed

119.
Parts of a rapid filter
 Under drain system: carries filtered water &
distributes the backwash water uniformly
 Wash water trough: collects back wash water
 Filter bed agitator: agitates sand layer for
cleansing (air-scour or surface wash
systems)
 Control equipment: maintains fairly constant
flow.

120.
Rapid Filter Media
 Filter media control
– the solids holding capacity of the filter bed
– the hydraulic loading rate of the filters
– The finished water quality
 Types
– Single medium (usually sand)
– Dual media (usually sand and anthracite)
– Multi-media (usually sand, anthracite, garnet)

121.
Single-medium filters
 Utilize single medium- usually well
graded sand
 Reverse gradation after
backwashing is the major
drawback of these filters
 Only the top 4-5 cm of the filter
bed are used for filtration
 Filter runs are short due to smaller
solids-holding capacity
 Solution: use large diameter
medium with a uniformity
coefficient close to unity and deep
medium bed

122.
Dual-media Filters
 Avoids the problem of
reverse gradation
 Top: larger and lighter
filter media (e.g.
anthracite)
 Bottom: Smaller and
denser material (e.g.
sand)
 Greater depth of the
filter media is utilized

123.
Multi-media Filters
Common materials
used:
– Anthracite (Sg = 1.55)
– Sand (Sg = 2.65 and )
– Garnet (Sg = 4.05)
– Increased filter run
time
– Improved water quality
– Costly
•Size decreases
•Density increases

124.
Backwashing
 Involves passing water upward through the filter media at a velocity
sufficient to expand (fluidize) the bed and wash out the
accumulated solids
 Done when:
– The head loss through the filter exceeds the design value
– Turbidity breakthrough causes the effluent quality to be less than a
minimum acceptable level
– A pre-selected maximum filter run time has passed since it was last
cleaned
 Filtered water is used which consumes 1-5% of the product water
 Bed expansions are achieved in backwashing

125.
Filter media during filtration and
backwashing

126.
Disinfection
 Treatment used for destruction or removal of
pathogens
 Widely used disinfectants
– Oxidizing agents (halogens, halogen compounds, ozone)
– Physical agents (Ultraviolet radiation)
 Factors that affect efficiency of disinfection
– Type and concentration of microorganisms
– Type and concentration of disinfectant
– Contact time provided
– Character and temperature of the water

127.
Characteristics of an Ideal Disinfectant
 Kill or disable pathogens
 Nontoxic to consumers
 Cheap and easy to use
 Fast acting and long lasting
 Improve water aesthetics

128.
Disinfectants in common use
 Chlorine (Cl2)
 Chloramines (NH2Cl, NHCl2)
 Chlorine dioxide (ClO2)
 Ozone (O3)
 Ultraviolet (UV) radiation.

129.
Chlorination
 Widely used disinfectant
 Advantages: readily available as gas, liquid
or solid; cheap; easy to apply; toxic to most
microorganisms; provides protection in the
distribution system
 Disadvantages: Fatal at high
concentrations; Toxic; Irritant; Corrosive;
Formation of harmful disinfection by-products
(DBPs) like Trihalomethanes (THMs)

130.
Chlorination
 Reactions:
Cl2 + H2O  HOCl + H+ + Cl-
HOCl ↔ H+ + OCl-
 Hypochlorous acid (HOCl) and Hypochlorite ion
(OCl-) are known as “free chlorines” and are the
most effective forms of chlorine.
 HOCl is 80-200 times as strong as OCl-.

131.
Chlorination
 Other substances with which chlorine reacts
 Reducing agents such as S2-, Fe2+, Mn2+, NO2-
H2S + 4Cl2 + 4H2O → H2SO4 + 8HCl
 Organic matter
 Ammonia: HOCl reacts with ammonia to produce
chloramines which are known as “combined chlorine
residuals”
NH3 + HOCl  NH2Cl + H2O (Monochloramine)
NH2Cl + HOCl  NHCl2 + H2O (Dichloramine)
NHCl2 + HOCl  NCl3 + H2O (Trichloramine )

132.
Chlorination
Free residual chlorine = Cl2(l) + HOCl + OCl
Combined residual chlorine = NH2Cl + NHCl2 + NCl3
Total residual chlorine = Free residual chlorine + Combined
residual chlorine

133.
Chloramines
 Advantages
– Less corrosive
– Less toxicity and chemical hazards
– Relatively tolerable to inorganic and organic loads
– No known formation of DBP
– Relatively long-lasting residuals
 Disadvantages
– Not so effective against viruses, protozoan cysts,
and bacterial spores

134.
Chlorine gas cylinder

135.
Hypochlorites
 Less pure and dangerous than chlorine gas
 Strength decreases with time while in storage
 Safer to handle but expensive
 Types:
– Sodium hypochlorite (NaOCl) solution
– Calcium hypochlorite (CaO(Cl)2)
 Reactions:
Ca(OCl)2 + 2 H2O  2 HOCl + Ca(OH)2
NaOCl + H2O  HOCl + NaOH

136.
Breakpoint Chlorination

137.
Example
 Referring to the breakpoint-chlorination curve shown
below, calculate the daily chlorine requirement in kg
to attain a 0.4 mg/L free chlorine residual. The
design discharge is 7500 m3/d.
0
0.2
0.4
0.6
0.8
1
0 0.5 1 1.5 2
Applied chlorine, mg/L
Residual
chlorine,
mg/L

138.
UV Disinfection
 Physical method of inactivating pathogens.
 Mechanism of UV Disinfection:
– Radiation with an wavelength of around 260 nm
penetrates the cell wall and cell membrane of
microorganisms and is absorbed by cell material
such as DNA and RNA and promotes changes
that prevents replication to occur

139.
Miscellaneous Water Treatment Processes
 Removal of dissolved gases
 Removal of Iron and manganese
 Removal of silica
 Fluoridation
 Removal of oil
 Softening (refer the removal of hardness in
chapter three)

140.
WATER DISTRIBUTION SYSTEM
 After treatment, water is to be stored
temporarily and supplied to the consumers
through the network of pipelines called
distribution system.
 The distribution system also includes
– pumps,
– reservoirs,
– pipe fittings,
– instruments for measurement of pressures, flow leak
detectors etc

141.
Requirement of Distribution System
 The system should convey the treated water up to
consumers with the same degree of purity
 The system should be economical and easy to
maintain and operate
 It should safe against any future pollution. As per as
possible should not be laid below sewer lines.
 Water should be supplied without interruption even
when repairs are undertaken
 The system should be so designed that the supply
should meet maximum hourly demand

142.
System of Distribution
 Depending upon the methods of distribution,
the distribution system is classified as the
follows
– Gravity system
– Pumping system
– Dual system or combined gravity and pumping
system

143.
Gravity System
 When some ground sufficiently high above
the city area is available

144.
Pumping System
 Constant pressure
can be maintained
in the system by
direct pumping into
mains
 Supply can be
affected during
power failure and
breakdown of
pumps

145.
Combined Pumping and Gravity
System

146.
Methods of Supply of Water
 Continuous System
– This is the best system and water is supplied for
all 24 hours. This system is possible when there
is adequate quantity of water for supply.
 Intermittent System
– If plenty of water is not available, the supply of
water is divided into zones and each zone is
supplied with water for fixed hours in a day or on
alternate days

147.
 The system has following disadvantages:
– Consumers have to store water for non-supply hours.
– Bigger sized pipes are to be laid, because full day’s
supply is to be provided within few hours of supply.
– Pipelines are likely to rust faster due to alternate
wetting and drying. This increases the maintenance
cost.
– There is also pollution of water by ingress of polluted
water through leaks during non-flow periods.

148.
Layouts of Distribution System
 Generally in practice there are four different
systems of distribution which are used. They
are:
– Dead End or Tree system
– Grid Iron system
– Circular or Ring system
– Radial system

149.
Dead End or Tree System
 This system is suitable for irregular developed
towns or cities.

150.
Dead End or Tree System
 Advantages:
– Discharge and pressure at any point in the
distribution system is calculated easily
– The valves required in this system of layout are
comparatively less in number.
– The diameter of pipes used are smaller and
hence the system is cheap and economical
– The laying of water pipes is used are simple.

151.
Dead End or Tree System
 Disadvantages:
– There is stagnant water at dead ends of pipes
causing contamination.
– During repairs of pipes or valves at any point the
entire downstream end are deprived of supply
– The water available for firefighting will be limited
in quantity

152.
Grid Iron System
 From the mains water enters the branches at
all junctions in either direction into sub-mains
of equal diameters.

153.
Grid Iron System
 Advantages
– As water is supplied from both the sides at any
point, very small distribution area will be affected
during repair.
– Every point receives supply from two directions
and with higher pressure
– In case of fire, more quantity of water can be
diverted towards the affected area, by closing the
valves of nearby localities.
– There is free circulation of water and hence it is
not liable for pollution due to stagnation.

154.
Grid Iron System
 Disadvantages:
– More length of pipes and number of valves are
needed and hence there is increased cost of
construction
– Calculation of sizes of pipes and working out
pressures at various points in the distribution
system is laborious, complicated and difficult.

155.
Circular or Ring System
 Supply to the inner pipes is from the mains
around the boundary. It has the same
advantages as the grid-Iron system. Smaller
diameter pipes are needed. The advantages
and disadvantages are same as that of grid-
Iron system.

156.
Circular or Ring System

157.
Radial System
 Water is pumped to the distribution
reservoirs and from the reservoirs it flows by
gravity to the tree system of pipes.
 The pressure calculations are easy in this
system. Layout of roads needs to be radial to
eliminate loss of head in bends. This is most
economical system also if combined
pumping and gravity flow is adopted.

158.
Radial System

159.
Pressure in the Distribution System
 When the water enters in the distribution
main, the water head continuously is lost due
to friction in pipes, at the entrance of
reducers, due to valves, bends, meters etc till
it reaches the consumers tap
 The effective head available at the service
connection to a building is very important,
because the height up to which the water can
rise in the building will depend on this
available head only

160.
The pressure in the distribution system
depends upon the factors as listed below:
– The height of highest building where water should
reach with adequate pressure, without boosting
– Pressure required for fire hydrant
– The distance of the locality from the distribution
reservoir.

161.
The following pressures are considered
satisfactory in the case of multi-storied
buildings.
 One storey only 7m head
 Two storey building 12m head
 Three storey building 17m head
 3 to 6 storey heights 2.1 to 4.2kg/cm2
(21 to 42m)
 6 to 10 storey heights 4.2 to 5.2kg/cm2
 Above 10 storey 5.27 to 7kg/cm2

162.
Service/Distribution Reservoirs
 A service reservoir has four main functions:
– To balance the fluctuating demand from the
distribution system, permitting the source to give
steady or differently phased output.
– Provide a supply during a failure or shutdown of
treatment plant, pumps or trunk main leading to
the reservoir.
– To give a suitable pressure for the distribution
system and reduce pressure fluctuations therein.
– To provide a reserve of water to meet fire and
other emergency demands.

163.
Types of Service Reservoirs
 Generally, there are two types of service reservoirs:
– Surface reservoir (Ground Reservoir or Non-elevated)
– Elevated reservoir ( Over head Tank)

164.
Accessories of Service Reservoirs
 Inlet Pipe : For the entry of water
 Ladder : To reach the top of the reservoir
and then to the bottom of the reservoir, for
inspection and cleaning
 Lightening Conductor : In case of elevated
reservoirs for the passage of lightening
 Manholes : For providing entry to the inside
of reservoir for inspection and cleaning
 Outlet pipe: For the exit of water

165.
Cont’d
 Outflow Pipe : For the exit of water above
full supply level
 Vent pipes : For free circulation of air
 Washout pipe : For removing water after
cleaning of the reservoir
 Water level indicator: To know the level of
water inside the tank from outside.

166.
Accessories of service reservoir
LIGHTENING CONDUCTOR
OUTLET PIPE
INFLOW PIPE
WATER LEVEL INDICATOR
WASH OUT PIPE
DITCH
OVER FLOW PIPE
MANHOLE
LADDER

167.
Design Capacity of Service Reservoirs
 Analytically by finding out maximum
cumulative surplus during the stage when
pumping rate is higher than water
consumption rate and adding to this
maximum cumulative deficit which occurs
during the period when the pumping rate is
lower than the demand rate of water

168.
 By drawing mass curves (graphical
method)
– A mass diagram is the plot of accumulated inflow
(i.e. supply) or outflow (i.e. demand) versus time.
The mass curve of supply (i.e. supply line) is,
therefore, first drawn and is superimposed by the
demand curve.

169.
Example 1:
 A small town with a design population of 1600 is
to be supplied water at 150liters per capita per
day. The demand of water during different
periods is given in the following table:
Time
(hr)
0 - 3 3 - 6 6 - 9 9 - 12 12 - 15 15 - 18 18 - 21 21- 24
Demand
(1000lit
ers)
20 25 30 50 35 30 25 25
• Determine the capacity of a balancing reservoir if pumping is done 24 hours
at constant rate.

170.
Solution:
 Per capita water consumption = 150l/c/d
 Total water demand = demand * population =
150*1600 = 240,000liters
 Rate of pumping = 240,000/24 = 10,000lit/hr
= 30,000lit/3hr

171.
Analytical Method
Time Pumping Demand Cum. Supply Cum. Demand Surplus Deficit
0 - 3 30,000 20,000 30,000 20,000 10,000
3 - 6 30,000 25,000 60,000 45,000 15,000
6 - 9 30,000 30,000 90,000 75,000 15,000
9 - 12 30,000 50,000 120,000 125,000 5,000
12 - 15 30,000 35,000 150,000 160,000 10,000
15 - 18 30,000 30,000 180,000 190,000 10,000
18 - 21 30,000 25,000 210,000 215,000 5,000
21 - 24 30,000 25,000 240,000 240,000 0 0

172.
Cont’d
 Maximum cumulative surplus = 15,000 liters
 Maximum cumulative deficit = 10,000 liters
 Balancing storage = 15000 + 10000 =
25,000lit = 25m3
If the reservoir is circular with depth, h = 3.0 m,
m
d 4
.
3
3
4
*
25




173.
Mass curve method

174.
Example 2
Consider example 1, if the pumping is done for:
– Eight hours from 8 hrs to 16 hrs
– Eight hrs from 4 hrs to 8 hrs and again 16 hours
to 20 hrs.
Calculate the capacity of the balancing reserve.
Solution:
 Total water demand = 240,000lit/hr
 Rate of pumping = 240,000/8 = 30,000l/h
= 90,000lit/3hrs

175.
Eight hours from 8 hrs to 16 hrs
A) Analytical Method
Time Pumping Demand
Cum.
Supply
Cum.
Demand
Surplus Deficit
0 - 3 0 20000 0 20000 20000
3 - 6 0 25000 0 45000 45000
6 - 8 0 20000 0 65000 65000
8 - 9 30000 10000 30000 75000 45000
9 - 12 90000 50000 120000 125000 5000
12 - 15 90000 35000 210000 160000 50000
15 - 16 30000 10000 240000 170000 70000
16 - 18 0 20000 240000 190000 50000
18 - 21 0 25000 240000 215000 25000
21 - 24 0 25000 240000 240000 0 0

176.
Maximum cumulative surplus = 70,000
Maximum cumulative deficit = 65,000
Balancing storage, S = 135,000lit =
135m3

177.
B) Graphical Method

178.
Depth and Shape of Service Reservoirs
 Depth
Size (m3) Depth of water (m)
Up to 3500 2.5 to 3.5
3500 to 15,000 3.5 to 5.0
Over 15,000 5.0 to 7.0

179.
Factors influencing depth for a given
storage are:
 Depth at which suitable foundation conditions are
encountered
 Depth at which the out let main must be laid
 Slope of ground, nature and type of back fill
 The need to make the quantity of excavated material
approximately equal to the amount required for backing,
so as to reduce unnecessary carting of surplus material
to tip.
 The shape and size of land available

180.
Shape
 Circular reservoir is geometrically the most
economical shape, giving the least amount of walling
for a given volume and depth
 A rectangular reservoir with a length to width ratio
1.2 to 1.5

181.
Pipes Used in the Water Distribution
System
Pipe Materials
 Pipe materials used in transmission and
distribution systems must have the following
characteristics
– Adequate tensile strength and bending strength to withstand
external loads.
– High bursting strength to withstand internal water pressure
– Ability to resist impact loads to water flow suitable for
handling and joining facilities
– Resistance to both internal and external corrosion

182.
The types of pipes used for distributing
water include:
 Cast iron pipe
 Steel pipe
 Concrete pipe
 Plastic pipe
 Asbestos cement pipe
 Copper pipe
 Lead pipe

183.
A pipe material is selected based on
various conditions:
 Cost
 Type of water to be conveyed
 Carrying capacity of the pipe
 Maintenance cost
 Durability, etc.

184.
Cast iron pipes
 Advantages:
– The cost is moderate
– The pipes are easily joined
– The pipes aren’t subjected to corrosion
– The pipes are strong and durable
– Service connections can be made easily
– Disadvantage:
– The breakage of this pipe is large
– Carrying capacity decreases with increase in life
– The pipes become heavy and uneconomical when their
sizes increase (especially beyond 1200mm)
Advantages:

185.
Galvanized Iron Pipes
 Advantages:
– The pipes are cheap
– Light in weight and easy to handle and transport
– Easy to join
 Disadvantage:
– These pipes are liable to incrustation (due to
deposition of some materials inside part of pipe)
– Can be easily affected by acidic or alkaline water
– Short useful life

186.
Plastic Pipes
 Advantages:
– The pipes are cheap
– The pipes are flexible and possess low hydraulic
resistance (less friction)
– They are free from corrosion
– The pipes are light in weight and it is easy to
bend, join and install them
– The pipes up to certain sizes are available in coils
and therefore it becomes easy to transport

187.
 Disadvantage:
– The coefficient of expansion for plastics is high,
the pipes are less resistant to heat
– Some types of plastics may impart taste to the
water

188.
Appurtenances in the Distribution
System
 The various devices fixed along the water
distribution system are known as
appurtenances.
 The following are the some of the fixtures used
in the distribution system.
– Valves
– Fire hydrants and
– Water meter

189.
Types of Valves
 The following are the various types of valves
named to suit their function
– Sluice valves
– Check valves or reflex valves
– Air valves
– Drain valves or Blow off valves
– Scour valve

190.
Valves
Sluice valves
– These are also known as gate-valves or stop
valves.
– These valves control the flow of water through
pipes.
Check Valve or Reflux Valve
– These valves are also known as non-return
valves. A reflux valve is an automatic device
which allows water to go in one direction only

191.
Sluice valve

192.
Air Valves
 These are automatic valves and are of two
types namely
– Air inlet valves
– Air relief valves
 Air Inlet Valves
– These valves open automatically and allow air to
enter into the pipeline so that the development of
negative pressure can be avoided in the
pipelines.

193.
– The vacuum pressure created in the down
streamside in pipelines due to sudden closure of
sluice valves
– This situation can be avoided by using the air inlet
valves.
 Air Relief Valves
– Sometimes air is accumulated at the summit of
pipelines and blocks the flow of water due to air
lock. In such cases the accumulated air has to be
removed from the pipe lines.

194.
 Drain Valves or Blow off Valves
– These are also called wash out valves they are
provided at all dead ends and depression of
pipelines to drain out the waste water.

195.
Water Meter
 These are the devices which are installed on
the pipes to measure the quantity of water
flowing at a particular point along the pipe.
 The readings obtained from the meters help
in working out the quantity of water supplied
and thus the consumers can be charged
accordingly.

196.
Fire Hydrants
 A hydrant is an outlet provided in water pipe
for tapping water mainly in case of fire.
 They are located at 100 to 150 m a part
along the roads and also at junction roads.

197.
Determination of Pipe Sizes
 Permissible velocities for best results for
different pipe sized pipes are within the
range of 0.3 to 2m/s.
 Once the velocity of flow is established loss
of head due to friction, bends and other
reasons can be computed
 The head required to develop a particular
velocity in a particular sized pipe is then
calculated.

198.
 The size of the pipe used in the water
distribution system or the velocity of flow
through the pipe can be determined by one
of the following formulas:
 Darcy –Weisbach formula:
 Hazen-Williams formula:
 Manning’s Formula:
Determination of Pipe Sizes
gD
fLV
hf
2
2

L
h
S
S
CD
Q f

 ,
278
.
0 54
.
0
63
.
2
n
S
AR
Q
2
/
1
3
/
2


199.
Determination of Pipe Sizes
 The most common pipe flow formula used in
design and evaluation of a water distribution
system is the Hazen-Williams’ formula.
L
h
S
S
CD
Q f

 ,
278
.
0 54
.
0
63
.
2

200.
Water supply pipes sizes commercially available are
given in the following table:
Metric
sizes (mm)
10 20 25 30 40 50 60 80 100 150 200 250 300
English
(In)
½ 3/4 1 11/4 11/4 2 21/2 3 4 6 8 10 12
Metric
sizes
(mm)
350 375 400 450 500 525 600 675 750 900 950 1050
English
(In)
14 15 16 18 20 21 24 27 30 36 38 42

201.
Example 1:
 Given
– Total population of a town = 80,000
– Average daily consumption of water =
150liters/capita/day
– If the flow velocity of an outlet pipe from intake =
1.5 m/s, determine the diameter of the outlet pipe.

202.
Exp. Cont’d..
 Solution
 Total flow, Q = Demand* Population =
150*80,000 = 12x106 lit/day
 Required pipe area,
 But the pipe size available on the market is
300mm & 350mm, then take D = 350mm
mm
V
Q
D
V
Q
D
V
Q
A 343
*
5
.
1
4
*
1389
.
0
4
4
4
2










203.
Example 2:
 A town has a population of 100,000 persons.
It is to be supplied with water from a reservoir
situated at a distance of 6.44km. It is
stipulated that one-half of the daily supply of
140lit/capita should be delivered in 6 hours. If
the loss of head is estimated to be 15m,
calculate the size of pipe. Assume f = 0.04.

204.
Exp. Cont’d..
Solution
 Total daily supply =
 Since half of this quantity is required in 6
hours
 Maximum flow =
 According to the Darcy-Weisbach formula:
Where, hf = 15m, f = 0.04, L = 6440m
3
3
000
,
14
10
000
,
100
*
140
m

s
m /
324
.
0
)
60
*
60
*
6
*
2
(
000
,
14 3

5
2
1
.
12 d
fLQ
hf 

205.
 But available pipe sizes 675mm & 750mm,
take 750mm diameter pipe
mm
m
d
d
683
683
.
0
10
.
12
*
15
)
324
.
0
(
*
6440
*
04
.
0
*
1
.
12
)
0324
(
*
6440
*
04
.
0
15
2
5
2






206.
Energy Losses in Pipes
 The major energy loss (head loss) in pipes can
be found by one of the three formulas:
 Darcy-Weisbach
Where, hf = head loss (m)
f = friction factor (which is related to the relative
roughness of the pipe material & the fluid flow
characteristics)
L = length of pipe (m)
gD
fLV
hf
2
2


207.
 Hazen-Williams formula
Where, C = Coefficient that depends on the material and
age of the pipe
S = Hydraulic gradient (m/m)
L
h
S
S
CD
Q
f

 ,
278
.
0 54
.
0
63
.
2

208.
 Table - Values of C for the Hazen-Williams
formula
Pipe Material C
Asbestos Cement 140
Cast Iron
 Cement lined
 New, unlined
 5years-old, unlined
 20 years old, unlined
130 – 150
130
120
100
Concrete 130
Copper 130 - 140
Plastic 140 - 150
New welded Steel 120
New riveted Steel 100

209.
 Nomographs solve the equation for C = 100.
Given any two of the parameters (Q, D, hf or
V) the remaining can be determined from the
intersections along a straight line drawn
across the nomograph.
 Manning’s Formula
, R = D/4, S = hf/L
n
S
AR
Q
2
/
1
3
/
2


210.
 Where, n = Coefficient of roughness
depending on pipe material, usually
n = 0.013 GI pipes
n = 0.009  Plastic pipes
n = 0.015  Clay concrete pipes

211.
Exercise :
 For Q = 30l/s, D = 200mm, n = 0.013, L = 1500
 From Nomograph, hf/L = 0.00825
 = 0.00825*1500 = 12.38m
 
550
.
12
1500
*
1000
/
200
)
100
/
30
(
)
013
.
0
(
936
.
10
*
936
.
10
936
.
10
3
/
16
2
2
3
/
16
2
2
3
/
16
2
2





 L
D
Q
n
hf
D
Q
n
L
hf
S

212.
Procedure of Analyzing Pipe Size
and Pressure
 Assume any internally consistent distribution of flow. The sum
of the flows entering any junction must equal the sum of the
flows leaving
 Compute the head losses in each pipe by means of an equation
or diagram. Conventionally, clockwise flows are positive and
produce positive head losses.
 With due attention to sign, compute the total head loss around
each circuit: hL = KQn.
 Compute, without regard to sign, for the same circuit, the sum
of: KnQn-1.
 Apply the corrections obtained from equation to the flow in each
line. Lines common to two loops receive both corrections with
due attention to sign.

213.
Example.
 The pipe network of two loops as shown in Fig.
below has to be analyzed by the Hardy Cross
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.

214.
 To obtain initial pipe discharges applying
nodal continuity equation, the arbitrary pipe
discharges equal to the total number of loops
are assumed. The total number of loops in a
network can be obtained from the following
geometric relationship:
Total number of loops = Total number of pipes
- Total number of nodes+1

215.
 Moreover, in this example there are five pipes and
four nodes. One can apply nodal continuity equation
at three nodes (total number of nodes - 1) only as,
on the outcome of the other nodal continuity
equations, the nodal continuity at the fourth node
(last node) automatically gets satisfied. In this
example there are five unknown pipe discharges,
and to obtain pipe discharges there are three known
nodal continuity equations and two loop head-loss
equations.

216.
 To apply continuity equation for initial pipe
discharges, the discharges in pipes 1 and 5
equal to 0.1 m3/s are assumed. The obtained
discharges are
Q1= 0.1 m3/s (flow from node 1 to node 2)
Q2= 0.1 m3/s (flow from node 2 to node 3)
Q3= 0.4 m3/s (flow from node 4 to node 3)
Q4= 0.4 m3/s (flow from node 1 to node 4)
Q5= 0.1 m3/s (flow from node 1 to node 3)

217.
 The discharge correction ∆Q is applied in
one loop at a time until the ∆Q is very small
in all the loops. ∆Q in Loop 1 (loop pipes 3,
4, and 5) and corrected pipe discharges are
given in the following table:

218.
 Thus the discharge correction ∆Q in loop 1 is 0.15
m3/s. The discharges in loop pipes are corrected as
shown in the above table. Applying the same
methodology for calculating ∆Q for Loop 2:

219.
 The process of discharge correction is in
repeated until the ∆Q value is very small as
shown in the following tables:

220.
 The discharge corrections in the loops are
very small after five iterations, thus the final
pipe discharges in the looped pipe network in
Fig. above are
– Q1= 0.223 m3/s
– Q2= 0.223 m3/s
– Q3= 0.192 m3/s
– Q4= 0.192 m3/s
– Q5= 0.182 m3/s