This document provides an overview of water supply engineering and sanitary engineering. It discusses environmental engineering and its focus on managing natural resources and protecting the environment. Sanitary engineering is defined as supplying communities with potable water and treating wastewater. The document then covers various water sources, their characteristics, water treatment processes, and the components and design of water intake and collection works.

1. SANITARY AND
ENVIRONMENTAL
ENGINEERING (1)
LECTURES NOTES
Dr. Ahmed Seaf
Dr. Emad Hamdy
WATER SUPPLY ENGINEERING
3RD
YEAR CIVIL ENG.
ENVIRONMENTAL ENGINEERING DEP.

2. INTRODUCTIO
N
Environmental engineering is
essential for development of facilities
for protection of the environment and
for the proper management of natural
resources.
The environmental engineer places
special attention on the biological,
chemical, and physical reactions in
the air, land, and water environments
and on improved technology for
integrated management systems,
including reuse, recycling, and
recovery measures.

3. SANITARY
ENGINEERING
•It is the branch of engineering
responsible for supplying the
communities with potable water
and getting rid of the generated
waste water.
•Sanitary engineering including
these four categories
water treatment systems
water distribution network
waste water collection system
waste water treatment systems
3rd
year
4th
year

4. Rain
1
2
3 4
Water sources in nature

5. CHARACTERISTICS OF
WATER SOURCES
1- Rain water
•The most pure water source
•Rich with dissolved oxygen
(corrosion) and may cause acidic
rains over industrial zones
•Small suspended solids content
(dust or sand) due to land washing
•Could be stored and used after
filtration

6. CHARACTERISTICS OF
WATER SOURCES CONT.
2- Ground water
•High dissolved solids content
•Different properties according
to confining soil
•Due to nature filtration almost
no Suspended solids content
•Could be used from depths
more than 40 m

7. 3- Surface water
•Low dissolved solids content
with high suspended solids and
bacterial content
•Highly polluted due to misuse
•Relatively large quantities
•Could be used after treatment
(purification)
CHARACTERISTICS OF
WATER SOURCES cont.

8. 4- Sea water
•Very high dissolved solids
content more than 35000 (p.p.m)
(part per million- mg/liter – g/m3
)
•Could be used after treatment
(Desalination) cost must be
considered
CHARACTERISTICS OF
WATER SOURCES cont.

9. FLOW LINE DIAGRAM OF SURFACE
WATER TREATMENT
Water
source
Plain sedimentation, chemical
coagulation, plate - tube settler
(Lamella), Pulsator
Collection
works
Sedimentation Filtration
Storage
Heat, Chemicals,
Light / Radiation Disinfection
Distribution network
Ground reservoir,
Elevated tanks
Slow and Rapid sand
filters, Dual and
multimedia filters

10. DATA REQUIRED TO SUPPLY
A CITY BY WATER
1- Design period
2-Design population (current
and forecasting)
3-Design flow
4- Master plan (water source,
city development plans.....etc)

11. DESIGN
PERIOD
Factors affecting on design period
1- Useful life of different water
system components
Concrete structures 40 – 50 years
Pipes 40 – 50 years
Mechanical parts 20 – 25 years
Electrical parts 15 – 20 years

12. Factors affecting on design
period cont.
2- Rate of population growth
High rate → Decrease design
period
Low rate → Increase design
period
3- Easy of extension
Easy extension → Decrease design
period
Hard extension → Increase design
period
4- Rate of Interest
5- Primary performance of system
units

13. DESIGN POPULATION

14. POPULATION
FORECASTING-There are various methods to
estimate the future population:
1- Arithmetic method: this method
represents stage 2 of population growth
diagram the increase in population is
assumed to be constant for every constant
duration
Pn = P0 + ka . (t n – t 0)
Where: Pn = future population at time n
P0 = present population
ka = rate of change of populationt
P
∆
∆
=

15. POPULATION
FORECASTING2- Geometric method: this method
represents stage 1 of population growth
diagram the rate of increase in
population is assumed to be constant for
every constant duration
ln Pn = ln P0 + kg . (t n – t 0)
Where: Pn = future population at time n
P0 = present population
kg = rate of change of
population
t
P
kg
∆
∆
=
ln

16. POPULATION
FORECASTING
3- Decreasing Rate of increase "saturation"
method : this method represents stage 3 of
population growth diagram as the rate of
increase in population is decreasing as the
population approaching saturation level
(S)
P n = S – (S - P 0) * e –K d *(t n – t 0)
Where: Pn = future population at time n
P0 = present population
kd = rate of change of population
t
PS
kd
∆
−∆−
=
)ln(

17. POPULATION
FORECASTING4- Rate of growth method :
P n = P 0 (1 + r) n
Where: Pn = future population at time n
P0 = present population
r = annual rate of growth
5- Population density method :
a- P = area * population density
b- P = No. of residential units *
population density per unit

18. POPULATION
FORECASTING
Graphical methods
6- Graphical extension method
7- Graphical comparison method

19. WATER CONSUMPTION
•Amount of water consumed (liter /
capita / day)
•Types of water consumption
according to uses
1- Domestic (50%)
2- Industrial (15%)
3- Commercial (15%)
4- Public (20%)
Losses and fire demand (F.D.)
F.D. (litre/sec) = 20 * population /
10000
F.D. (m3
/d) = 120 * population / 10000

20. Factors affecting on water
consumption
1- Size of city
2- Standard of living
3- Climate
4- Pressure and quality of water
5- Sewage facilities
6- Cost

21. CONSUMPTION

23. •Types of water consumption
according to design
Q ave = W C ave * Population
comparison between cities
Q max month = [1.2-1.6] * Q ave
Design of water collection and
treatment plant
Q max day = [1.6-1.8] * Q ave
Design of storage and main
distribution lines
Q max hour = 2.5 * Q ave
Design of minor distribution lines

24. WATER CONSUMPTION
FORECASTING1- Rate of increase method :
W.C n = W.C 0 (1 + r*) n
Where: W.Cn = future W.C. at time n
W.C0 = present population
r* = rate of increase in water
consumption ≈ 10% rate of population
growth
2- % increase method
a) % increase in WC = (( P n / P 0) 0.125
– 1(
b) % increase in WC = (( P n / P 0) 0.11
– 1(
W.C n = W.C 0 (1 + % increase)

25. Proposes of studying water quality:
1.Determine the degree of pollution.
2.Determine of design steps for water
treatment process, (drinking water –
industrial water – swimming ponds).
3.Assessment of treatment units.
4.Check the effluent of WTP with
environmental.
WATER QUALITY

26. CHARACTERISTICS OF
WATER
1. Physical characteristics.
1.1 Temperature.
1.2 Color : Colorless.
1.3 Odor : Odorless.
1.4 Turbidity : Turbidity measurements
are made by turbidity-meters in terms
of (NTU), (FTU), and (JTU). There is no
direct relationship between NTU or FTU
readings and JTU readings. The NTU is
the standard measure, requiring use of a
nephelometer, which measures the
amount of light scattered, usually at 90o
from the light direction, by suspended
particles in the water test sample.

27. TURBIDITY
METERS

28. CHARACTERISTICS OF
WATER
1. Physical characteristics.
1.5 Suspended solids : Those solids which
are retained by a glass fiber filter and
dried to constant weight at 103-105o
C.
Method: A well-mixed sample is filtered
through a standard GF/F glass fiber
filter, and the residue retained on the
filter is dried to constant weight at
103-105o
C.

29. CHARACTERISTICS OF
WATER
1. Physical characteristics.
1.6 Dissolved solids :
Method: A well-mixed sample is filtered
through a standard glass fiber filter.
The filtrate is evaporated to dryness in
a weigh dish and dried to constant
weight at 180°C. The increase in dish
weight represents the total dissolved
solids.
Note:
Suspended and dissolved solids could be
measured using Suspended and
dissolved solids-meters.

30. SOLIDS
METERS

31. CHARACTERISTICS OF
WATER
2. Chemical characteristics.
A.Organic tests
Ammonia, Nitrite and Nitrate
B. Inorganic tests
B.1 pH : measured by pH-Meters. pH is
the measurement of the hydrogen ion
concentration, [H+].All human beings
and animals rely on internal
mechanisms to maintain the pH level
of their blood. The blood flowing
through our veins must have a pH
between 7.35 and 7.45.

33. CHARACTERISTICS OF
WATER
B.2 Electrical conductivity: is a
measurement of the dissolved material in
an aqueous solution, which relates to the
ability of the material to conduct
electrical current through it.

34. CHARACTERISTICS OF
WATERB.3 Alkalinity, acidity and salinity.
B.4. Hardness.
B.5. Chlorides.
B.6. Minerals.(Fe, Mn, Mg, Ca,………etc)
B.7. Gases (O2, CO2, H2S, ………..ets)

35. 3. Biological characteristics.
• Source, (Micro-organisms, bacteria, virus,
protozoa…etc)
• Pathogens = (Harmful bacteria)
• Indicator = Used to indicate the present of
pathogens.
• Properties of an ideal indicator:
1. Applicable for all types of water.
2. Always present when pathogens are
present.
3. Non-pathogen for the lab. Personal.
4. Have a longer survival time outside the
CHARACTERISTICS OF
WATER

36. Impurities in water, their causes and effects
Impurities Causes Effects
Suspended
solids Bacteria
Some causes
disease
Silt and clay Turbidity
Algae and
protozoa
Odor, color and
turbidity

40. FLOW LINE DIAGRAMS OF GROUND
WATER TREATMENT (HIGH D.S.)
GW
source
Sand – Ceramic - Cartridge
filters
Collection
Wells
Filtration
Nano filters /
RO
Storage
Disinfection
Distribution network
Ground reservoir,
Elevated tanks

41. WATER TREATMENT (IRON AND
MANGANESE)
GW
source
Cascade or diffused air
Collection
Wells
Aeration
Sedimentation
and filtration
Storage
Disinfection
Distribution network
Ground reservoir,
Elevated tanks

42. WATER TREATMENT (HARDNESS
REMOVAL)
GW
source
Heat – Lime – Soda – Ion exchange
Collection
Wells
softness
Sedimentation
and filtration
Storage
Disinfection
Distribution network
Ground reservoir,
Elevated tanks

44. COLLECTION WORKS
COMPONENTS
1. Intake structure.
2. Intake conduits.
3. Raw water lift pump and
sump.
4. Transmission lines (Force
main).
COLLECTION
WORKS
The work that is performed on the source of
water for the purpose of the transfer of sufficient
quantities of raw water to the treatment plant

45. TYPES OF
INTAKES
1. Pipe intake
2. Shore intake
3. Submerged (Tower) intake
4. Temporary intake
The primary functions of an intakes is to
- To supply highest quantity of water from the
sources
- To protect piping and pumps from damage or
clogging as a result of floating and submerged
debris.

46. FACTORS AFFECTING THE
CHOICE OF INTAKE STRUCTURE
TYPE
1.Width of water source.
2.Fluctuation in water level
3.Depth of water & character of
the source bottom.
4.Navigation requirements.
5.Effect of currents, floods and
storms upon the structure.
6.Shore pollution condition.

47. FACTORS AFFECTING THE
CHOICE OF INTAKE
STRUCTURE LOCATION
1. Upstream the served city to prevent the
direct pollution.
2. On straight part of the water source to
prevent settling and scoring.
3. Restricted area taken around the intake
structure (150 m upstream and 50 m
downstream).

48. TYPES OF
INTAKES
1. Pipe intake
(Wide cannels W ≥ 50 m)
2. Shore intake
(Narrow cannels W < 50 m, non-polluted
shore)
3. Submerged (Tower) intake
(Narrow cannels W < 50 m, polluted
shore)
4. Temporary intake

49. Pipe intake
conduit
Main header
Raw P.S.

50. SLUICE VALVES
GATE VALVE < 300
MM
BUTTERFLY VALVE >
300 MM

51. NON RETURN
VALVES
FOOT
VALVE
CHECK
VALVE

52. TYPES OF PIPE
1) (Ductile iron)
2) (Cast iron)
3) (SS)
4) (GRP)
5) (PVC-u PVC)
6) PE (HDPE and LDPE)
7) (PP)
8) )
9) .

53. PUMP
S
PUMPS CHARACTERISTIC
CURVES

54. H,Q CURVE
SYSTEM HEAD
CURVE
PUMPS DUTY POINT
T.D.H = H st.
+ H l
+ H m
+ H v

57. Shore intake

58. Submerged intake

59. DESIGN OF INTAKE CONDUITS
• Number (n) ≥ 2
• Diameter (Φmm) = 200– 250 – 300 - … – 500 – 600 - … –
1000 mm.(up to 3200)
• Design flow = Qd = 1.10 * 1.5 * Qav(or) 1.10 * P.F m* Qav
• Ordinary velocity = 0.6 – 1.5 m/s
• Maximum velocity at one pipe is broken ≤ 2.5 m/s
DESIGN FOR PRESENT AND FUTURE
1. Number (n) ≥ 2 in the future.
2. Assume velocity at future is 1.4 to 1.5 m/s,
3. Get the present velocity
V= (V.future) *[(Qdpresent)/(Qdfuture)] ≥ 0.6 m/s.
If unsafe close some pipes in the present

60. Head losses calculation
Calculated for the maximum velocity condition (present or
future)
V act = 0.355 * C * D 0.63
* S 0.54
Such that
V act =Maximum velocity (present or future) (m/s)
C = Fraction coefficient (80-150) take 120
D = Intake conduit diameter (m)
S = Hydraulic gradient line slope (m/m)
get (S)
H L = L * S
L = Intake conduit Length (m)

61. •Total width of each screen (L) = intake conduit Φ + 0.40 m
•The bar width (b) = 1.0 – 2.0 cm.
•The spacing between two bars (S) = 2.0 – 5.0 cm.
•The inclined angle of the bar screen (θ) = 30 – 60o
.
•The minimum screen depth (d) = (LWL – BL) - 0.5 m
•L = (n+1)*b + n*S
•Assume b & S get n
•H L = 1.4 ( v2
2
– v1
2
) / ( 2 g ) ≤ 10 cm
•v1 = Q 1 screen / ( L * d)
•v2 = Q 1 screen / (n * s * d)
DESIGN OF SHORE INTAKE SCREEN

62. Design of Main header (Pipe intake)
• Qd = 1.1 *Q mm Future
• R.T. = 1 min
• Volume = Qd * R.T.
• Length (depended on No. of pumps)
• Get Φ
Design of Sump
• R.T. = 2 min at maximum flow(1.50 *
Qav.f)
• R.T. = 5 min at minimum flow(0.80 * Qav.f)
• Volume = Qd * R.T.
• Length (depended on No. of pumps)
• d = HWL – BL – ∑ HL m
• W ≥ 1.50 m (for maintenance purpose)

63. 1. Improve the physical characteristics of
water, by removing turbidity, color and
taste.
2. Destroy any contained bacteria,
special pathogenic bacteria.
3. Removal of hardness, iron and
manganese salts and excessive amount
of gases and salts soluble in water.
PURPOSES OF WATER
PURIFICATION WORKS

64. 1.The slow sand filter plant –
which consists of plain sedimentation
followed by slow sand filtration and
disinfection.
2. The rapid sand filter plant –
which consists of chemical coagulation
followed by rapid sand filtration and
disinfection.
In most surface water, two systems of
water purification are in common use:

65. 1.Settling of discrete (non flocculent) particles.
Theory of sedimentation
( ) 2
18
d
g
V l
s
µ
ρρ −
=Stoke’s LawStoke’s Law

66. 2. Settling of flocculent particles.
Theory of sedimentation
3. Zone settling.
4.Compression settling.

67. Factors affect the sedimentation
efficiency:
1.Retention time.
2.Horizontal velocity.
3.State of flow.
4.Shape ,size and Specific gravity of
solids.
5.Relationship between tank dimensions.
6.Surface loading rate.
7.Hydraulic load on out let weir.
8.Inlet and outlet arrangement.
9.Suspended solids concentration in water
to be treated.
10.Temperature of water to be treated.
(Specific gravity, Viscosity)

68. 1.The flow is laminar flow.
2.Impurities particles are evenly distributed
on the whole area of the tank​​
3.the case of entrance and exit does not affect
the sedimentation efficiency
4.The settled particles does not resuspended
Assumptions of ideal sedimentation tanks
Action zones of typical sedimentation tank

69. 1 - The walls of the tank to be completely smooth
and vertical.
2 - Tank body be impermeable to water.
3 - Weirs installed on the entrance and exit for the
distribution of water in the horizontal plane.
Requirements of typical sedimentation tank

71. 4 - Baffles and barriers for the distribution of
water in the vertical plane.
Requirements of typical sedimentation tank

72. 5 – Provide a slope in the bottom of the tank to
assemble sludge.
6 – Sludge should be removed periodically.
Requirements of typical sedimentation tank

73. According to operation technique:
1.Fill and draw (Batch System)
In this type, the raw water stays a sedimentation
period inside a sedimentation basin.
2.Continues flow
The flow inter the sedimentation basin from inlet
arrangement, and in the same time exit from outlet
arrangement, the retention time in the basin is the
required sedimentation time.
TYPES OF SEDIMENTAION TANKS
According to shape:
1.Rectangular.
2.Circular

77. Design of Plain Sedimentation Tanks
•Qd = 1.1* Qmm= 1.10 * P.F m* Qav(m3
/d),
•Get Qd (m3
/hr) = Qd (m3
/d)/working period (hr/d)
•Retention Time = 2– 5 hrs
•SLR = 25 – 40 m3
/m2
/d = Qd / S.A
•WLR = 150 – 300 m3
/m/d = Qd / Lw
W L R ( rectangular weir) ≤ 150 (m3 / m / d)
W L R ( V-notch weir) ≤ 300 (m3 / m / d)
•For rectangular tanks only Vh ≤ 0.3 m/min
•Velocity in inlet and outlet pipes = 0.60 – 1.50 (m/s)

78. Rectangular Sed.
Tanks
Circular Sed. Tanks
(Clarifiers)
d = 3 – 5 m d = 3 – 5 m
B = 2 – 4 d Ø ≤ 35 m
L = (4 – 5 B) ≤ 40 m n ≥ 2
n ≥ 2 Volume = n (/4) Ø2
*d
Volume = nLBd S.A = n /4 Ø2
S.A = nLB Lw
= nØ
Cross.A = nBd
Lw
= nB

79. VS = ( Q * S.S. * R.R. ) / ( ∂ * 106
* (1-WC
) * N * n )
Such that
VS = Sludge volume
Q = Q d
( m3
/ day )
S.S. = Suspended solids = 80 P.P.M
R.R. = Removal ratio = 90 – 95 %
∂ = sludge density = 1.02 (t/m3
)
WC
= sludge water content = 95 %
n = No of sedimentation tanks
N = no of withdrawals per day
Velocity in sludge pipe = 1.00 – 2.00 (m/s)
Sludge withdrawal time = 10 – 20 min.
Minimum sludge pipe diameter = 200 mm
Sludge removal

80. COAGULATION PROCESS
•Purpose
Removal of most quantity of solids present in the raw
water by chemical action.
•Theory of Coagulation.
Impurity particles are of small size and carries a negative
electric charge, which means the occurrence repulsion
between each other and the stability of these impurities in
place which prevents the deposition.
This theory is based on breaking the state of stability that
exist between particles (Destabilization) as well as a
compilation of the work of these molecules (Aggregation)
There are two theories that are used to explain this theory

81. • Types of Coagulants
1. Alum or [Aluminum sulphate (AL2 (SO4)3 +
18H2O)].
2. Ferric and Ferrous sulphate.
3. Ferric chloride (spicily for colored water)
•Chemical theory
Addition of a chemical matter (coagulant) to raw water
that reacts with water alkalinity and produce a gelatinous
forming (flocs.).
•Physical theory
The flocs. carries a positive charge at its surface, in the
other side, suspended solids carry a negative charge at
their surface. Attraction force appears between them, the
suspended solids attaches to the flocs surface that causes
increasing of flocs weight. Faster settling appears,
sedimentation efficiency will increase.

82. •Factors affect the coagulation efficiency
1.Coagulant dose.
2.pH of raw water.
3.Mixing eff.
4.turbidity .
Coagulant Optimum pH
Alum 4 – 7
Fe. compounds ≥ 8.5

83. •Methods of alum feeding
1.Dry feeding
Use the alum as a powder in case of insoluble materials.
2. Wet feeding
Use the alum in liquid form (solution), better than dry
feeding, need concentrated alum solution tank to prepare
the alum solution.

84. Jar test
Jar test is used to determine the daily coagulant
dose.
Steps of the test
1. 5 vessels each 1 liter put in them different coagulant doses.
2. Flash mixing for 30 sec. (100 - 300 rpm)
3. Gentle mixing for 10 min. (10 - 30 rpm)
4. Sedimentation for 30 min.
Get removal efficiency for each vessel.

85. Different mixing methods
1- Mechanical mixing (Impellers)
2- Hydraulic mixing
3- Inline mixing
Chemical
coagulation
process
components
1- Alum solution tanks
2-Flash mixing tank(s)
3- Clari-flocculator / rectangular
flocculation and sedimentation tanks

86. Design of Alum Solution Tanks
V = (Q * S) / (∂ * 106
* C)
Such that
V = Alum solution volume
Q = Q d = 1.1* Qmm(m3
/ day)
S. = Dosage = 30 → 80 P.P.M
C = Concentration = 5 → 20 %
∂ = Alum solution density = 1.02 ( t / m3
)
V1 = V / no. of tanks (2 → 3)
d= 1 → 3 m
A = W2
= V1/d
Alum weight = (Q * S * no. of days) / 106
calculated for 30 days

87. Design of Flash Mixing Tank(s)
R T = 5 – 60 sec { take 30 sec{
Q d =1.1 * Q mm
V = Q d * R T
Depth of chamber d = 3 m
Area of chamber A = V / d
A =  * Ø 2
/ 4
Ø = chamber diameter

88. Design of impellers
G = √ (P / µ * V) = 300 – 700 sec -1
P = Theoretical power
µ = dynamic viscosity of water = 1.14 * 10 -3
V = Volume of flash tank
Assume G = 500 sec -1
Get P
P = K * S * n3
* D5
K = impeller coefficient = 1
S = water density = 1000 ( kg / m3
(
n = no of rotation per sec (1 → 2) r.p.s.
D = impeller diameter (m(
Assume n get D
Such that D = (0.33 → 0.50) chamber diameter

89. Clari-flocculator

90. Q d
= 1.1 * Q max month
Assume SLR = 25 – 40 ( m3
/ m2
/ d )
SA = Q d
/ SLR [the working hours is important]
Area of tanks S A
S A = [n *  / 4] * (Ø s
2
- Ø f
2
)
n ≥ 2
Ø s = total tank diameter ≤ 35 m
Ø f = Flocculation zone diameter = (0.33 – 0.50) * Ø s
Assume Ø f = 0.40 * Ø s & Assume n get Ø s
Flocculation zone
Assume d f
= 2 → 5 m [3 m[
Vol = d f
* [n *  / 4] * Ø f
2
RT = Vol /Q d
R T = 15 → 40 minute
Design of Clari-flocculator tanks

91. Design of paddles
G = √ (P / µ * V) = 20 – 70 sec -1
P = Theoretical power
µ = dynamic viscosity of water = 1.14 * 10 -3
V = Volume of one flocculation tank
Assume G = 50 sec -1
Get P
P = 0.50 * C d * A * ρ * v r
3
C d = impeller coefficient = 1
ρ = water density = 1000 (kg / m3
(
v r = relative velocity of paddles [ 0.45 → 0.70
(m/s( [
Get area of paddles

92. Assume d s
= d f
+ (0.5 → 1.00) m
Vol = d s
* [n * π / 4] * (Ø s
2
- Ø f
2
)
RT = Vol /Q d
R T = 2 → 3 hr
Checks
1) W L R = Q d
/ n *  * Ø s
W L R (rectangular weir) ≤ 150 (m3
/ m / d)
W L R (V notch weir) ≥ 300 (m3
/ m / d)
Sedimentation zone (Clari-zone)

93. Rectangular flocculation and sedimentation

94. Q d
= 1.1 * Q max month
Assume SLR = 25 – 40 ( m3
/ m2
/ d )
SA = Q d
/ SLR [the working hours is important]
Area of tanks S A
S A = n * B * Ls
n ≥ 2
L s = Sedimentation tank length ≈ 0.80 L t
B = width of tank = (6 → 12) m
Assume L s = 0.80 L t & Assume n get L t
L t ≤ 50 m, L f = 0.2 L t
Flocculation zone
Assume d f
= 2 → 5 m [3 m[
Vol = d f
* n * B * L f
RT = Vol /Q d
R T = 15 → 40 minute
Design of Rectangular flocculation and
sedimentation tanks

95. Design of paddles
G = √ (P / µ * V) = 20 – 70 sec -1
P = Theoretical power
µ = dynamic viscosity of water = 1.14 * 10 -3
V = Volume of one flocculation tank
Assume G = 50 sec -1
Get P
P = 0.50 * C d * A * ρ * v r
3
C d = impeller coefficient = 1
ρ = water density = 1000 (kg / m3
(
v r = relative velocity of paddles [ 0.45 – 0.70
(m/s( [
Get area of paddles

96. Assume d s
= d f
+ (0.5 → 1.00) m
Vol = d s
* n * B * L s
RT = Vol /Q d
R T = 2 → 3 hr
Checks
1) V h = [Q d / n * B * d s ] ≤ 0.30 ( m / min )
2) W L R = Q d
/ n * B
W L R (rectangular weir) ≤ 150 (m3
/ m / d)
W L R (V notch weir) ≥ 300 (m3
/ m / d)
Sedimentation zone
Velocity in inlet pipe = 0.60 – 1.50 (m/s)
Velocity in outlet pipe = 0.30 – 0.60 (m/s)

97. Other chemical sedimentation units
Plate Settler or Tube Settler

98. FILTRATION
Filtration can be defined as a physical-chemical process for
separating suspended and colloidal impurities from water by
passing it through a bed of granular material. Water fills the
pores of the filter media, and impurities are absorbed on the
surface of the grain or trapped in the openings.
The purposes of filtration in water purification are:
•Removal of the remaining suspended solids .
•Removal of turbidity.
•Removal of iron and manganese salts.
•Removal of taste, odor and color.
•Removal of at least 90% of bacteria.
•Removal of algae.

99. 1. Straining mechanism
Impurities solids bigger size than voids
between filter bed particles are arrested on
it and removed from water. The major
removal takes place in the upper few
centimeters of the filter bed. The impurities
which deposited on the filter bed surface
help in straining the small particles also.
Theory of filtration
Filtration theory depends on passing water through a
porous material that removes the undesirable impurities
from it. The theory of filtration could be divided into two
main mechanisms the straining mechanism and the
transportation (non screening ) mechanism.

100. 2. Transportation (non straining) mechanism
In this mechanism several types of removal could take place
such as Sedimentation action, Adsorption Action,
Electrolytic Action and biological action.
Sedimentation Action
Removal of suspended particles between the filter bed
particles whose act as sedimentation basins. The suspended
particles settle on the sides of filter bed particles.
Adsorption Action
Adsorb the colloidal matters on the filter
bed particles as a result to coat it by a
gelatinous layer from bacteria and
microorganisms.

101. Electrolytic Action
The filter bed particles are electrically charged by negative
charge opposite to the charged of impurities present in water
to be filtrate. Due to that the filter bed particles attract the
impurities. When their charges get neutralized, the washing
of filter bed renews the charges.
Biological Action
The organic impurities in water like algae,
plankton…etc deposit on the filter bed
capturing different microorganisms into them.
The microorganisms find the source of food on
the water particles, this leads to some
important biological and chemical change in
water quality.

102. CLASSIFICATION OF FILTERS
1. A.T. Type of filter media
1. Sand (the most popular filtration media type)
2. Carbon (to remove odor)
3. Volcanic (in case of colored water)
2. A.T. No. of filter media
1. Single media.
2. Dual media
3. Multi media
3. A.T. rate of filtration
1. Slow filters (3 – 10 (m3/m2/d))
2. Rapid filters (120 – 200 (m3/m2/d))

103. CLASSIFICATION OF FILTERS
4. A.T. Direction of flow
1. Down flow.
2. Up flow
5. A.T. Characteristic of flow
1. Gravity.
2. Pressure.

104. Dual Filter

105. Slow sand filters
1.Filtration rate is 3-10 m3
/m2
/d
2.Minimum operation and maintenance
requirements.
3.Usually does not require chemical pretreatment.
4.Large land area required to construct.
5.Filter is cleaned by removing the top 10 cm of
sand.
6.Operated with Gravity force.
7.Depends mainly on the straining mechanism and
Biological action.
8.Most likely used in small systems.

106. Slow sand filters

107. Design Criteria of Slow Sand Filter
Effective size of the sand = 0.25 – 0.35 mm
Dirty skin layer = 3 – 8 cm
Washing time (removal time) = 1 – 15 days ( 1 day if
mechanical & 15 day if manual)
Ripening the filter is taking = 7 – 15 days
The whole cleaning process is taking = 8 – 30 days
The operation time (between two washes) is = 2 – 6
month.
Rate of filtration (ROF) = 3 – 10 m3
/m2
/d
Area of filter = 1000 – 2500 m2
The filter is Rectangular (L*B)
L & B ≤ 50 m
n ≥ 2
L/B = 1 – 1.25

108. Example
Design the SSF for a WTP working 16 hr/d, if the
design flow is 32,000 m3
/d
Solution
Qd = 32000 m3
/d = 32000/16 = 2000 (m3
/h)
Assume that ROF = 6 m3
/m2
/d = 6/24
(m3
/m2
/hr)
SA = 2000/(6/24) = 8000 m2
Assume L = 50 m, B = L/1.25 = 40
SA = 50 * 40 = 2000 m2
n = 8000/2000 = 4 filters (ok)
Take total No. of filters = 4 + 1 = 5 filters

109. •The Rapid Sand Filter (RSF) differs from the Slow Sand
Filter in a variety of ways, the most important of which
are:
1- Higher filtration rate
2- Ability to clean automatically using backwashing.
3- Follows the pre-disinfection and coagulation process.
4-Depends mainly on the transportation (non straining)
mechanism for the removal of S.S.
In RSF the complete filtration cycle (filtration and back
washing) occurs successively.
Rapid sand filters

112. Design Criteria of Rapid Sand Filter
Effective size of the sand = 0.6 – 1.5 mm
Sand uniformity coefficient = 1.35 – 1.5
Sand specific gravity = 2.55 – 2.65
Wash water speed = 2.5 – 3.5 m/s
Cleaning period = 25 – 35 min
Ripening the filter is taking = 15 – 20 min
Washing by compressed air = 2 -5 min
Washed by pressured water = 10 min & 15 -20 min if
no air
The operation time (between two washes) is = 12 – 36
hrs
Rate of filtration (ROF) = 100 – 200 m3
/m2
/d
Area of filter = 40 – 64 m2

113. Empirical equation to determine minimum number of filters in
the WTP = 0.044 * [Qmm (m3
/d)]0.5
The filters numbers:
If nw ≤ 5 take nT = any no. + 1 for wash
If nw > 5 take nT = even no. + 2 for wash
If nw ≥ 30 take nT = no. divisible by 4 + 4 for wash
Amount of wash water (m3/d) = no. of washing by day * time of
washing (10 min) * nT * ROW (m3
/m2
/d)/(24*60 min/d) * SA
(m2
)
The washing: 1. every 12 hrs. 2. every 24 hrs. 3. every 36 hrs.
The filter is Rectangular in surface area (L*B)
L & B ≤ 8 m
B : L = 1 : 1.25 up to 1 : 2
Rate of washing (ROW) = 5 – 6 ROF

114. Example
Design the RSF for a WTP working 16 hr/d, if the design flow
is 32000 m3
/d
Solution
Qd = 32000 m3
/d = 32000/16 = 2000 (m3
/h)
Assume ROF = 200 m3
/m2
/d = 200/24 m3
/m2
/hr = 5 (m3
/m2
/hr)
SA = 2000/5 = 400 m2
Assume L = 8 m, B = L/1.25 = 6.25
SA1 = 8*6.25 = 50 m2
nw = 400/50 = 8 filters (ok)
nT = 8 + 2 = 10 filters
Assume that ROW = 5 ROF = 25 m3
/m2
/hr
Amount of wash water (m3
/d) = no. of washing by day (1) * time
of washing (10 min)* nT (10) * ROW (25) (m3
/m2
/hr)/(60
min/hr) * SA (50) (m2) = 2083 m3
/d
% WW = (2083/32000) *100 = 6.5 %

115. DISINFECTION
•PURPOSE
The main purpose of disinfection is to
reduce the potential health risk
associated of drinking water by
inactivating pathogens. This prevents
the possible spread of water-born
diseases.

116. 1. Contact time and dosage
The longer contact time and dosage the greater the kill is.
2. Temperature
As temperature increase the rate of kill increase.
3. Characteristics of water
Suspended solids may shield bacteria from the
action of the disinfectant.
Some compounds may adsorb the disinfectant.
Viruses, cysts and ova obstruct the disinfection
process as they are more resistant to
disinfectants than are bacteria.
•FACTORS AFFECTING DISINFECTION

117. •Requirements of good disinfectant
1. Effective in destroying all kinds of pathogenic bacteria.
2. Do its task within a reasonable contact time at normal
temperature.
3. Economical and easily available.
4. Give residual concentration to safe guard against re-
contamination in water supply system.
5. Not toxic and objectionable to user after the water
treatment.
6. Adaptability of practical, quick and accurate assay
‘techniques for determining disinfection concentration
for operation control and as a measure of disinfecting
efficiency.

118. METHODS OF DISINFECTION

119. (BOILING)
Process:
The water should be allowed to boil for at
least 20 minutes.
•Advantages
–Simple and effective method of purification
–Will kill many waterborne bacteria through
the intense heat
–Uses local available materials
•Disadvantages
–Can sometimes be difficult, time consuming,
and cost inefficient because of the high
volume of fuel used
–Will not remove suspended or dissolved
compounds

120. EXPOSURE
Advantages:
Kills harmful bacteria and pathogens
Simple, convenient and inexpensive
If used correctly, the water is as clean as
boiled water
Will not change the taste of water
Disadvantages:
Requirement of huge surface area.
A 6-12 hour waiting period

121. ET
Ultra violet rays → Wave length of about, 1000-
4000mµ

Produced by passing electric current through
mercury enclosed in quarts bulb

The bulb is then immersed in water 10cm or
below

122. ET
Advantages

Pure odour free, colourless water with
turbidity of below 15mg/lit

Kills all type of bacteria

Normally used for sterilizations at
hospitals
Disadvantages

Very costly

Possible interruption by electricity

No residual for networks disinfection
When UV radiation penetrates the cell wall
of an organism, it damages genetic material,
and prevents the cell from reproducing.

123. Bromine and Iodine treatment

8mg/lit for 5 mint. Contact period

Available in the form of pills also
OZONE
•OZONE is Strongest oxidant/disinfectant
available.
•More effective against microbes than
chlorination.
•But, costly and difficult to monitor, control
under different condition and leaves no
residual.
•Mostly being used as pre-disinfecting for
water bodies containing organics.

124. The advantages and disadvantages of
chlorination
•Advantages
1. Cheap
2. Residual for network
3. Available
4. Easy to store and use.
5. Simple equipment required.
•Disadvantages
1. High chlorine dose may cause change in the
water colour and taste due to damage of
pipes or it self.
2. Chlorine reacts with organic compound that
appears in water and the results are cancer
compounds (Trihalomethane –THM-).

125. H2O + CL2 → HOCL + HCL
HOCL →H+
+ OCL-
HCL →H+
+ CL-
Chlorination of tab water

126. Break point chlorination
Ι - Destruction of CL2 by reducing agents.
Π - Formation of chloro-organic of chloramines
Ш - Destruction of components

128. STORAGE WORKS
Types of storage
1.Ground Storage.
(Appears in water treatment plant after
disinfection stage and before high lift
pump station)
2. Elevated Storage.
(Appears in different position according
to its function)

129. GROUND STORAGE
Purpose
1. Produce contact time for disinfection = (0.5) hr
C1 (m3
) = (0.5)hr * Qmm (m3
/hr), Qmm (m3
/hr) = Qmm
(m3
/d)/wp
2. Balancing difference between maximum daily and
maximum monthly flow through one day
C2 (m3
) = [Qmd (m3
/d) – Qmm (m3
/d)] * 1 day
3. Saves Emergency Storage = (15 % - 40 %) of daily
production
C3 (m3
) = (0.15 – 0.4) * Qmm (m3
/d) or (4 – 10 hr) * Qmm
(m3
/hr)
4. Saves 80% of fire Storage
C4 (m3
) = 0.8 * Fire requirements

130. Design Capacity of Ground
Reservoir
C (m3
) = take bigger of [C1 or C2 or
C3] + C4
L ≤ 50 m,
L = 1.0 – 2.00 B
d = 3 – 5 m
n ≥ 2 tanks
The ground reservoirs are built of
reinforced concrete and they have
be coated by isolating materials to
prevent any percolations.

132. • The reservoir placed beneath the surface
of the earth and the water level of the
reservoir is equal to the level of the
surrounding land.
• The tank has wall baffles to prolong the
path of the water to ensure that there is
enough contact with chlorine and to
support the tank roof.
• There is the upper ceiling vents (Air
vents) in order to refresh the air in the
tank and if the tank is fill out the air get
out of the tank through these openings.
• There is a tendency in the bottom of the
tank at the end and increasing the depth at
the exit pipe so as to ensure the discharge
of water in the reservoir is fully.

133. Example
It's required to design the ground storage of a WTP
serves 300,000 capita with average summer water
consumption of 420 l/c/d. if the summer peak factor is
1.40
Solution
Calculations of flows
qmm = 420 l/c/d
qav = 420/1.4 l/c/d = 300 l/c/d
Qav = 300 * 300,000 = 90,000,000 l/d = 90,000 m3
/d = 3,750
m3
/hr
Qmm = 1.4* Qav = 5,250 m3
/hr = 126,000 m3
/d
Qmd = 1.8* Qav = 6,750 m3
/hr = 162,000 m3
/d

134. Design Capacity
C1 = 0.5 hr * 5250 m3
/hr = 2625 m3
C2 = 162,000 – 126,000 = 36,000 m3
C3 = 6 hrs * 5250 = 31,000 m3
C4 = 0.8 * [ (300,000/10,000) * 120] = 2,880 m3
Cd = 36,000 + 2,880 = 38,880 m3
Take [4 tanks each (50m*40m*5m)]
That design volume will be 40,000 m3
that saves about 7 hrs emergency (ok)

135. ELEVATED STORAGE
Purpose
First: with respect to quantity
1.Cover the fluctuation in water
consumption through day.
2.Cover the difference between the
maximum consumption and
maximum production through one
day (maximum day) = Qmh - Qmd
3.Save 20 % of fire demand.

136. Second: with respect to pressure
The locations of elevated tank:
1.Just after high lift pump to:
•Fix the head on pumps, then the pumps work at
maximum efficiency.
•Prevent the effect of water hammer action on the
high lift pumps.
•And at this case the elevated tank is called (Surge
Tank).
2. At middle of city (at higher points) to:
•Improve water pressure in the network.
3. At extreme points to:
•Improve the water pressure in the network near to
the city boundaries.
•Give ability to city extended in the future.

137. Types of elevated tanks according to
its function
1.Balance elevated tank (only one pipe for
filling and drawing & pipe to waste during
empty to be washed).
2.Storage or feeding elevated tank (Pipe to fill
and pipe to draw & pipe to waste during
empty to be wash).

139. Float valve Pressure release valve

140. Design
C1
= [Q max hour
- Q max day
] * 3 / 24=
0.70*Q ave
* 3 / 24
C2
= from total mass curve (S curve(
C = max of (C1
, C2
) + 0.20 * F D
F D = 120 * pop / 10000
C = n * d * π * Ø 2
/ 4
d =(1/3 to 2/3) Ø
Ø = 10 to 20 m
get n

141. W T P works 24 hr with constant rate
W
C
curve
Pum
ping
C2
= [ a - b] * pop * 1.8

142. W T P works <24 hr with constant rate
W
C
curve
Pum
ping
C2
= [ a + b] * pop * 1.8

143. W T P works 24 hr with variable rate
W
C
curve
Pum
ping
C2
= [ a + b] * pop * 1.8

144. Distribution networks
REQUIREMENT OF A DISTRIBUTION
SYSTEM:
1. The system should convey the treated
water up-to consumers with the same degree
of purity
2. The system should be economical and easy
to maintain and operate
3. The diameter of pipes should be designed
to meet the fire demand
4. It should safe against any future pollution.
As per as possible should not be laid below
sewer lines.
5. Water should be supplied without
interruption even when repairs are
undertaken

145. Types of networks systems
1-Tree (Dead end) system
This system is suitable for
irregular developed towns or
cities. In this system water
flows in one direction only
into sub-mains and branches.
The diameter of pipe
decreases at every tree
branch.
1-Tree (Dead end) system
2- Loop (Ring) system
3- Radial system
4- Grid iron system

146. ADVANTAGES
1. Discharge and pressure at any point in the distribution
system is calculated easily
2. The valves required in this system of layout are
comparatively less in number.
3. The diameter of pipes used are smaller and hence the
system is cheap and economical
4. The laying of water pipes is used are simple.
DISADVANTAGES
1. There is stagment water at dead ends of pipes causing
contamination.
2. During repairs of pipes or valves at any point the entire
down stream end are deprived of supply
3. The water available for fire fighting will be limited in
quantity

147. 2-Loop ( Ring ) system
Supply to the inner pipes is from the mains around the
boundary. It decreases the effect of damage of pipes.
Smaller diameter pipes are needed.

148. 3-Radial system
This is a zoned 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 need to be radial to eliminate
loss of head in bends.

149. 4-Grid iron system
From the mains water enters the branches at all
Junctions in either directions into sub-mains of equal
diameters. At any point in the line the pressure is
balanced from two directions because of interconnected
network of pipes.

150. ADVANTAGES
1. In the case of repairs a very small portion of
distribution are a will be affected
2. Every point receives supply from two directions and
with higher pressure
3. Additional water from the other branches are available
for fire fighting
4. There is free circulation of water and hence it is not
liable for pollution due to stagnation.
DISADVANTAGES
1. More length of pipes and number of valves are needed
and hence there is increased cost of construction
2. Calculation of sizes of pipes and working out pressures
at various points in the distribution system is laborious ,
complicated and difficult.

151. Distribution system design
Losses are calculated based on Hazen Williams Equ.
V act = 0.355 * C * D 0.63
* S 0.54
Such that
V act =Velocity (m/s)
C = Fraction coefficient (80-150) take 120
D = Pipe diameter (m)
S = Hydraulic gradient line slope (m/m)
get (S)
H L = L * S
L = Pipe Length (m)

152. Hardy-Cross Method
This method used to determine
1- Discharge & flow direction for all
pipes in network
2-Pressure @ all nodes & HGL
Network of pipes forming one or more
closed loops
Limitation of the method
1- ∑Qin = ∑Qout
2- the pipes forms closed loops
Given
Demands @ network nodes (junctions)
d, L, pipe material, Temp, P @ one node

153. ∑
∑
∑
∑
∑
∑
−=
−=
−= −
af
n
a
a
n
a
n
a
n
a
n
a
Qhn
KQ
QKQn
KQ
KQn
KQ
δ
δ
δ 1
Hardy-Cross Method (Derivation)
( )
( ) 0
0
0
0
1
=+
=+
=
=
∑
∑
∑
∑
−
δ
δ
n
a
n
a
n
a
n
f
nQQK
QK
KQ
h
For Closed Loop:
( ) ( ) 221
!2
1
δδδ −− −
++=+ n
a
n
a
n
a
n
a Q
nn
nQQQ
n=2.0, Darcy-
Weisbach
n=1.85, Hazen-
Williams

154. Hardy-Cross Method (Procedure)
1. Divide network into number of closed loops.
2. For each loop:
a) Assume discharge Qa and direction for each pipe.
Apply Continuity at each node, Total inflow = Total
Outflow. Clockwise positive.
b) Calculate hydraulic gradient slope (S) for each pipe
given Qa, d, pipe material.
c) Calculate hf=S*L for each pipe. Retain sign from step
(a) and compute sum for loop ∑ hf.
d)Calculate hf / Qa for each pipe and sum for loop ∑ hf/
Qa.
e) Calculate correction δ =- ∑ h /(n ∑ h /Q ).

155. Hardy-Cross Method (Procedure)
NOTE: For common members between 2 loops both
corrections have to be made.
As loop 1 member, δ = δ 1 - δ 2.
As loop 2 member, δ = δ 2 - δ 1.
f) Apply correction to Qa, Qnew= Qa + δ.
g) Repeat steps (c) to (f) until δ becomes very small
and ∑ hf=0 in step (c).
h) Solve for pressure at each node using energy
conservation.