WATER SUPPLY

WATER SUPPLY

Bakun Dam – to be completed in 2011

Water supply scheme • Before a water supply scheme is developed and implemented, there are several major activities that need to be carried out. • Among the major activities involved: – Establish the need for investment and demonstrate that development is justified – Water resources study – Water demand forecasting – Water quality and treatment requirements – Formulation of alternative schemes – Financial / economical analysis

WATER RESOURCES • Surface water – streams, rivers, lakes, ponds, reservoir • Ground water

WATER DEMAND / CONSUMPTION • Domestic demand • Industry and commercial demand • Public demand • Fire-fighting demand • Unaccounted-for-Water (UFW) –wastage, leakage, illegal tapings, damage meters etc.

WATER DEMAND • WATER DEMAND FOR URBAN SUPPLIES - Based on population served, per capita consumption, service factor, industrial & other special demands.

Basic formula for water demand estimation The basic formula: Wdn = (P n x C x F ) + Dn where Wdn = water demand at the end of year “n” P n = projected population at the end of year “n” C = per capita consumption at the end of year “n” F = service factor at the end of year “n” Dn = additional demand at the end of year “n” e.g. to cater for new industrial estate, new town etc. As a guideline: light industries : 22000 L/ha/day heavy industries: 45000 L/ha/day

e.g. Estimate the water required in year 2010 for town A. The relevant data is given below. P = 20,000 people C = 230 L/person/day F = 1.0 Da = A 10 hectare heavy industrial area will be developed. Water demand is assumed 45000 L/ha/day Wd n = (P n x C x F ) + D a = (20000 person x 230 L/person/day x 1.0) + (45000 L/ha/day x 10 ha) = 4.6 x 106 + 0.45 x 106 = 5.05 x 106 L/day

SERVICE FACTOR • Reflects the potential percentage of population served State

Urban

Rural

1985

1993

2000

1985

2000

2008

Johor Kedah Kelantan

0.92 0.95 0.65

0.99 0.99 0.82

0.99 0.99 0.99

0.61 0.58 0.30

0.78 0.92 0.50

0.85 0.95 0.70

Melaka N.Sembilan Pahang

0.99 0.89 0.95

0.99 0.99 0.99

0.99 0.99 0.99

0.82 0.75 0.65

0.85 0.86 0.80

0.95 0.95 0.90

Perak Perlis P.Pinang

0.98 0.93 0.98

0.99 0.99 0.99

0.99 0.99 0.99

0.75 0.50 0.85

0.80 0.93 0.87

0.90 0.95 0.95

Sabah Sarawak Selangor Terengganu

0.99 0.95 0.95 0.85

0.99 0.99 0.99 0.92

0.99 0.99 0.99 0.99

0.38 0.33 0.73 0.40

0.55 0.40 0.80 0.82

0.75 0.70 0.90 0.95

Malaysia

0.93

0.99

0.99

0.58

0.74

0.86

POPULATION PROJECTION • Methods - Graph

POPULATION PROJECTION - Zone (for new township) Housing e.g. 500 apartments. Average number per house: 5 people

Housing e.g. 500 terrace houses. Average number per house: 5 people

Industrial

Commercial

- Geometric progression P n = Po (1 + r) n P n = projected population at the end of year “n” Po = population at the beginning of the year zero r = assumed population growth rate n = number of years

e.g From population census in year 2000, mukim A has a population of 53000 people. If the population growth rate is 1.27 %, estimate its population in 2008.

P 2008 = P2000 (1 + 1.27 ) 2008 – 2000 100 = 53000 (1.0127) 8 = 58630 people

PER CAPITA CONSUMPTION Per capita demand (consumption) include normal commercial & industrial use, domestic use and accounted–for-water losses. As a guideline: Urban – 230 to 320 liter/head/day Semi urban – 180 to 230 liter/head/day Rural – 135 to 180 liter/head/day Average daily per capita demand = quantity required daily population served

e.g. Calculate the average daily per capita demand of a town if the daily water demand is 10 million liter and the population is 35000. Average daily per capita demand = quantity required daily population = 10 x 10 6 L/day 35000 = 285.7 L/person/day

FACTORS AFFECTING PER CAPITA DEMAND • • • • • • •

Standard of living Types of industries Climate Water quality Pressure in distribution system Cost of water and metering Sewerage system

WATER TREATMENT

Aerial View of Sg. Johor Water Treatment Plant

WATER TREATMENT Objectives of water treatment To bring raw water quality up to drinking water quality standard – chemically & bacteriologically safe for human consumption; aesthetically acceptable, free from apparent turbidity, objectionable taste and other. Collection of surface water Water intake - a structure constructed at a surface water source to facilitate the conveying of water to the treatment plant.

Water intake

Cross section of a weir intake

Water intake

Direct pumped extraction intake

Collection well intake

Water intake

gate

Water intake at a reservoir

Water Treatment Processes

WATER TREATMENT Screening Placed at water intake Coarse screen Remove floating / large objects such as twigs, leaves that can foul or damage equipment. Fine screen Remove aquatic plants / small plants.

Grit removal • Grit - Inorganic particles e.g. sand, silt > 0.2 mm diameter; specific gravity of 2.65. • Characteristics: Abrasive. Cause problems to machinery e.g. pumps. • Removed in grit chamber (a channel with reduced velocity to allow settlement of particles.)

Grit chamber

AERATION Purpose: ØIncrease oxygen content ØTo liberate dissolved gases e.g. CO2 and H2S; reduce corrosiveness & odour Ø Provide oxygen for the oxidization of dissolved iron and manganese to their insoluble form 4 Fe 2+ + O2 + 10 H2O --------► 4 Fe (OH)3 (s) + 8 H +

Types of aerators • Water- into- air aerators - Cascade aerators

Cascade aerators at Rasa Water Treatment Plant (Sungai Selangor Water Supply scheme)

Multiple platform aerators

Spray aerators

Air-into-water aerators

Venturi aerator

Draft tube aerators

Coagulation & flocculation The primary purpose of the coagulation / flocculation process is the removal of turbidity from the water (by agglomerating the colloidal particles into larger and settleable floc)

Colloidal particles - negatively charged, stable, repel each other. Causes turbidity.

Coagulation • Chemicals are added to destabilize the

colloidal particles so that the tendency to repel each other is reduced and the particles can coagulate forming flocs. • Chemicals

are called coagulants. Coagulants have positive charges. • Common coagulants:

- Aluminium sulphate (Al2SO4) - Ferric sulfate (Fe2(SO4)3 - Ferric Chloride (FeCl3)

Coagulation Negatively charged particles repel each other due to electricity. Positively charged coagulants attract to negatively charged particles due to electricity. Neutrally charged particles attract due to van der Waal's forces.

Particles and coagulants join together into floc.

Coagulation reaction If alkalinity is present in water, A12(SO4)3.14H2O + 6 HCO 3 2 Al(OH)3 (s)

+ 6 CO2 + 14 H2O + 3 SO4 2-

(Aluminum hydroxide particles: jelly- like & positive charge, neutralize negative charge particles collide & stick together forming large floc) Without alkalinity, Al2 (SO4)3.14H2O 2 Al(OH)3 (s) + H2SO4 + 8 H2O H2SO4 is neutralize with hydrated lime, Ca(OH)2

Coagulant aids e.g. polymers, polyelectrolyte are added to improve coagulation process. Lime is added to increase alkalinity

Selection and optimum dosages of coagulants are determined experimentally by jar test. Procedure 1. Six jars is filled with water (turbidity, ph and alkalinity have been predetermined) 2. The jars are dosed with different amounts of coagulant. 3. Water is mixed rapidly for about 1 min to ensure complete dispersion of chemical. (Coagulation takes place) 4. Then mixed slowly for 15 to 20 minutes to aid the formation of flocs. (Flocculation)

Jar test

5. The water is allowed to settle ~30 minutes (Sedimentation) 6. Portions of the settled water are tested to determine the remaining turbidity.

With the addition of coagulants, small particles starts to attach to each other, forming floc.

Larger particles start to settle. The top portion is beginning to clear up

Most of the particles have completely settled out.

Example From jar test, the optimum dosage of alum is 35 mg/L. Determine the alum requirement if the plant processes 3000 m3 water daily. Solution Alum dosage = 35 mg/L Treatment plant flow rate = 3000 m3/day Alum required = 3000 m3 / day x 35 mg/L = 105 kg/day

Q5. The results from jar test shows that 3 mL of an alum solution (10 g/L concentration) and 1.6 mL lime solution (5 g/L concentration) result in optimum floc formation. Determine the daily amount of alum and lime to coagulate 350 L/sec of water. 3 mL of alum solution

1L raw water in 3 mL contains alum solution (10 g/L concentration in 1 liter contain 10 g alum)

(3 ml /1000 mL) x 10 g = 0.03 g Alum dosage = 0.03 g in 1 Liter of raw water = 0.03 g/L

Rapid mixing is required to mix the chemicals uniformly. Special tanks are built for this purpose. Methods 1. Hydraulic mixing: makes use of the turbulence created due to loss of head across obstruction to flow. Simple, reliable. Normally for capacity of plant < 45 MLD. coagulant

water

coagulant

Baffled plate

Overflow weir

coagulant

Hydraulic jump is formed creating turbulence

water

Mixing flume

Mechanical mixing - flash mixer

Mechanical mixer – uses pump to mix coagulants

Rapid mixing tank. Mechanical mixer is used to mix coagulants /coagulant aids

Design criteria for mixing Mixing is directly proportional to the velocity gradients (G) established in the water by stirring action. Velocity Gradient, G G=ÖP/m"

(unit: s-1)

where P = Power input, W (N.m/s or kg m2/s3) m = dynamic viscosity (kg/ m.s) " = volume of tank (m3) Coagulation (rapid mixing) : G = 700 – 3000s-1 Flocculation (slow mixing) : G = 20 – 80 s-1

Q6. Determine the velocity gradient of a rapid mixing tank if the input power is 5.7 kW, the temperature of the water is 30 0C and the volume of the tank is 8 m3.

Q7. If the G value is 2000 s-1, what is the input required?

Detention time, t Theoretical time the water is detained in a basin / tank t="/Q where Q = flow rate , m3/s " = volume of tank (m3) l Q in

Q out h w

e.g. Q = 4 m3/s " = 15 m x 30 m x 4 m t = (15 x 30 x 4) / 4 = 450 sec. = 7.5 min

Flocculation Slow / gentle mixing to promote particle contact leading to formation of large floc. Flocculation takes place in a flocculation tank. G = 20 – 80 s –1 Camp Number (G t) = 12000 - 270000 Methods of mixing • Hydraulic flocculator - Flow is directed over & around baffles in a tank • Mechanical flocculator Using slowly rotating paddles inside a tank

Hydraulic flocculator Flow is directed around baffles

Design criteria for baffled tank (Round – the end type) Detention time, t = 20 - 30 minutes Velocity of water flowing in channel, v = 0.09 - 0.24 m/sec Channel

Cross section area of channel, A

Water inlet

h

Distance between baffle y > 0.2 m

Baffle, thickness: 5 – 8 cm

b x Distance between baffle and end of tank, x = 1.5 y

b = channel length

Determination of tank depth, h Volume of water, " = flow rate, Q x detention time, t Distance of water traveled = velocity of water x detention time Cross sectional area of channel, A = Channel depth (h)

Volume of water Distance of water traveled

= Cross sectional area of channel Distance between baffles = A > 1.0 m y

Determination of tank length, L Channel effective length = channel length - x No. of channel

= Distance of water traveled Channel effective length

Length of tank = (No. of channel x Distance between baffles) + (No. of baffles x thickness of baffle)

Determination of tank width, w w = channel length x no. of set of channel

Power input for hydraulic tank P=rgQh where r = density of fluid, kg / m3 g = gravitational constant, m/s2 h = head loss, m G

=ÖP/m" =ÖrgQh/m" =Örg h/mt

Mechanical flocculation

Mechanical flocculation

Rapid mixing tank Coagulants Flash mixer

Paddle flocculator

Flocculation tank

Mechanical flocculator Design criteria t = 20 -30 minutes h water < 5 m w, number of revolutions of paddle per minute = 2 -15 rev/min v p, velocity of paddle tip (relative velocity between paddles and fluid) = 0.3- 0.7 m/s = 75 % from actual velocity of paddle, va where va = 2prw r = distance from shaft to center of paddle plate

L vp

r

Paddle

w

Power input, P P = FD v p where FD= drag force on paddles, N Therefore, where

FD = 1CD A p r v p2 2

(i) (ii)

CD = dimensionless coefficient of drag, 1.8 for flat plates A = area of paddles, m2 = no. of plates x length of plate x width of plate Substituting (ii) into (i) P =1/2 CD Ap r vp3 G = Ö P/ m " = Ö 1 CD AP r vp3 2m"

Sedimentation

• Separation of suspended particles by gravitational settling. • The process takes place in a tank or basin called sedimentation tank. • The principle involved is the reduction of flow so that particles settle during the detention time. • Particles which accumulate at the bottom of a sedimentation tank are called sludge.

Shape of tank • Rectangular • Circular

Type of flow

Horizontal flow

Up flow

Radial flow

A rectangular tank can be divided into four different functional zones: • Inlet zone to disperse influent flow and suspended matter uniformly across the tank. • A settling zone in which settling takes place. • An outlet zone in which clarified water is collected uniformly over the cross section of the tank and directed to the outlet • A sludge zone at the bottom (which the settled solids accumulate and from which they are withdrawn for disposal. baffle Outlet zone

Inlet zone

Settling zone

Sludge zone Scraper

Sludge removal

Inlet and outlet details Baffle

Baffle

Multiple Openings

Tank Bottom

Overflow Inlet Weir

Multiple Openings

Overflow Weir

TYPES OF INLETS FOR SEDIMENTATION TANKS

Outlet Pipe

OUTLET DETAILS OF SEDIMENTATION TANKS

Outlet Pipe

The design of a sedimentation tank depends on • Type of particle / type of sedimentation - Type 1 sedimentation (sedimentation of discrete particle) Particles which have little or no tendency to flocculate e.g. sand. No changes in size, shape and density during settling.

- Type II sedimentation (sedimentation of flocculent particle) Particles which agglomerate as they settle in the tank. Size, velocity and shape changed.

• Detention time, t t="/Q • Settling velocity of particles to be removed, vs For discrete particle, Stokes Law is widely used to determine the settling velocity of discrete particles. Assuming particle is spherical and flow is laminar, settling velocity is v s = g d 2 (r p - r w) 18 m where d = diameter of particle r p = density of particle r w = density of water

e.g. Determine the settling velocity of a spherical particle with a diameter 0.5 mm and s.g. 2.65 settling through water at 20 0C. Assume flow is laminar. v s = g d 2 (r p - r w)

18 m = 9.81 m/s (2650 – 998.2) kg/m3 x (5.0 x 10 -4)2 m2 18 x 1.002 x 10 -3 N.s/m2 = 0.22 m/s

For flocculent particle, the settling properties are determined by carrying out settling column analysis.

• Surface overflow rate, vo vo

surface area, As vs

v1 sludge out Up flow tank Q vo = Q /As

(m3/day) m2

= ("/t) / As = (As h/t) / As vo = h/t where As = surface area of tank h = depth of tank For an upflow tank, in order for a particle to be removed, vs > vo

Ideal sedimentation tank.

settled solids

Rectangular sedimentation tank w

L A s= w L h vh Q Ac = w H

An ideal rectangular sedimentation tank illustrating the settling of discrete particle A vh h

vs

B In order for particle A to settle in the tank, it must have a settling velocity, vs great enough so that it reaches point B during the detention time, t. In other words, settling velocity, vs must equal to the depth of tank divide by the detention time. But t = " / Q

vs = h/t vs = h / (" / Q) = h Q/ " = h Q / (w x L x h) = Q/wxL = Q / As = vo

Therefore, in order for a particle to be removed from a sedimentation tank, vs must be equal to or greater than vo : vs ³ vo

Removal of particles in an ideal sedimentation tank

A vh v s = vo

h hs

1.

2.

3.

vs

B

Particle which enters the settling zone at point A and has a settling velocity, vs equals to vo will settle at point B. Its settling time, ts is equivalent to t, the detention time. This particle will be removed from the tank. v s = vo ts = t Particles with v s > vo will also be removed regardless of the height at which it enters the tank. Its settling time, ts < t , detention time. Particle with v s < vo will be removed if it enters the tank at a distance from the bottom not greater than h s where h s = v s t The proportion of particles with v s < vo which will be removed is given by P = v s (100) vo

Type 1 sedimentation - Discrete particle (sand, grit material.) - Application: e.g. grit chamber design. - Determine the settling velocity, vs of the particle to be removed in the grit chamber and set the overflow rate, vo lower than vs. e.g. vo = 0.33 to 0.7 x vs and

vo = Q / A s = h/t

where vh= 2 – 7 cm/s (to determine Ac ) = Q/ Ac h = 3.0 – 4.0 m

• Type II sedimentation (sedimentation of flocculent particles) Stokes equation cannot be used since the particle size and velocity will continually change. To determine the settling properties of flocculent particle, a settling column test / analysis has to be carried out. Two graphs are obtained from settling column analysis. From these graphs, vo and t are obtained for design purposes.

Percentage removal of SS versus vo

Total percentage removal, RT

80 % 70 % 60 % 50 % 40 % 30 % 50

100

150

200

vo (m/day)

For 65 % removal of SS in sedimentation tank, v o = 52 m/day.

Percentage removal of SS versus t

Total percentage removal, RT

80 % 70 % 60 % 50 % 40 % 30 % 0.5

1.0

1.5

2.0

Detention time, h

For 65 % removal of SS in sedimentation tank, t = 1.25 hour

From graph, vo = 52 m/day and t = 1.25 hr Apply scale-up factor for design purposes vo design

= vo (obtained from graph) x 0.65 = 52 m/day x 0.65 = 33.8 m/day = 1.41m / hr

t design

= t x 1.75 = 1.25 x 1.75 = 2.2 hours

Next, the size of the tank (length, width, height) is obtained through vo = Q / A s = h/t

Settling column test sampling port / outlet H1 H2 H3 H4 H5

Settling column Height of 2.5 m, external diameter of 0.12 m Sampling outlets / ports, at different levels. Samples are collected, at defined time steps, into small containers installed in a rotating supporting device. The concentration of SS of each sample is evaluated.

Procedure • • •

Initial concentration of suspended solid, Co is determined. A settling column is filled with suspension to be analyzed and the suspension is allowed to settle. Samples are withdrawn from the sample ports at selected time intervals. Depth, m Sampling time (min) t1 t2 t3 t4 H1 H2 H3 H4

•

C11 C12 C13 C14

C21 C22 C23 C24

C31 C32 C33 C34

C41 C42 C43 C44

The concentration of suspended solids is determined for each sample and the percent removal is calculated. R% =1 -

Ct (100) Co where R % = percent removal at one depth and time Ct = concentration at time t and given depth, mg/L Co = initial concentration, mg/L

Depth, m t1 H1 H2 H3 H4

Sampling time (min) t2 t3

R%11 R% 12 R% 13 R% 14

R% 21 R% 22 R% 23 R% 24

R% 31 R% 32 R% 33 R% 34

t4 R% 41 R% 42 R% 43 R% 44

•

Plot percent removal versus depth. Interpolations are made between these plotted points to construct curves of equal concentration at reasonable percentages, i.e. 5 – 10 %.

•

Each intersection point of an isoconcentration line and the bottom of the column defines an overflow rate. vo = H/ti where H = height of column, m ti= time defined by intersection of isoconcentration line and the bottom of column (x – axis) where the subscript, i, refers to the first, second, third etc. interception points

•

A vertical line is drawn from ti to intersect all the isoconcentration lines crossing the ti time. The midpoints between isoconcentration lines define heights H1, H2, H3, etc. used to calculate the fraction of solids removed. For each time, ti defined by the intersection of the isoconcentration line and the bottom of the column (x-axis), you can construct a vertical line and calculate the fraction of solids removed: RTa = Ra + H1 (Rb – Ra) + H2 (Rc – Rb) +… H H where Rta = total fraction removed for settling time, ta Ra,Rb,, Rc = isoconcentration fractions a, b, c ,etc.

•

The series of overflow rates and removal fractions are used to plot curves of suspended solids removal versus detention time and suspended solids removal versus overflow rate that can be used to size the settling tank.

•

Eckenfelder recommends that scale-up factors of 0.65 for overflow rate and 1.75 for detention time be used to design the tank.

e.g. Determine the surface overflow rate and detention time that is able to remove 65 % of suspended solids. Use scaleup factor of 0.65 for vo and 1.75 for detention time. The initial concentration of SS is 40 mg/L. Data Concentration of SS at specified time and depth Sampling time (min) Depth (m)

10

20

30

40

60

90

120

0.5

24

20

18

14

12

11

10

1.0

32

26

22

16

14

12

11

2.0

34

28

24

18

16

15

12

Solution

1. Calculate percentage removal (Percentage passing each sampling port). R % = Co - Ct j)

(100)

Co e.g. at depth of 0.5 m and sampling time 10 minutes, R % = (40 – 24 ) x 100 40 = 40 %

Percentage removal of SS at specified time and depth Sampling time (min) Depth (m)

10

20

30

40

60

90

120

0.5

40%

50%

60%

65%

70%

73%

75%

1.0

20%

35%

45%

60%

65%

70%

73%

2.0

15%

30%

40%

55%

60%

63%

70%

2. Plot the isoconcentration percentage removal.

lines versus based on

0 0.5 1.0

40% 50% 60% 65%

70%

73%

75%

20% 35% 45% 60%

65%

70%

73%

15% 30% 40% 55 %

60%

63%

70%

1.5 2.0

10

20

30 40 50 60 70 80 Sampling time (min)

90 100 120

0 0.5 1.0

40% 50% 60% 65%

70%

73%

75%

20% 35% 45% 60%

65%

70%

73%

15% 30% 40% 55 %

60%

63%

70%

1.5 2.0

R a = 30 % R b = 40 % 10

20

30

ta

tb

40

50

60

70 80

90 100 120

Sampling time (min)

RTa = Ra + H1 (Rb – Ra) + H2 (Rc – Rb) +… H

H

0 0.5 1.0

40% 50% 60% 65%

70%

73%

75%

20% 35% 45% 60%

65%

70%

73%

15% 30% 40% 55 %

60%

63%

70%

1.5 2.0 30%

Sampling10 time (min) Point a

b

c

d

20 ta

40%

50%

30 40 tb t c

50

60%

60 td

70 80

65%

70%

90 100 120 te

tf

R T (Total SS removal)

t (hr)

R T = 30 % + 1.3 (40% – 30%) + 0.6 (50% – 40%) + 2.0 2.0 0.4 (60% – 50%)+0.25 (65% – 60%)+0.2 (70%–65%) 2.0 2.0 2.0 + 0.15 (75%–70%) = 43 % 2.0 R T = 40 % + 1.4 (50% – 40%) + 0.65 (60% – 50%) + 2.0 2.0 0.4 (65% – 60%) + 0.30 (70% – 65%) + 2.0 2.0 0.20 (75% – 70%) = 52.5 % 2.0 R T = 50 % + 1.4 (60% – 50%) + 0.6 (65% – 60%) + 2.0 2.0 0.4 (70% - 65%) + 0.25 (75%- 70%) = 57.3 % 2.0 2.0

20 min = 0.33 hr

v0 = h/t (m/day) 2.0 m / 20 min = 144 m/day

30 min = 0.5 hr

2.0 m / 30 min = 96 m/day

38 min = 0.63 hr 60 min = 1 hr

2.0 m / 38 min = 76 m/day 2.0 m / 60 min = 48 m/day

R T = 60 % + …………….. = 68 %

Point e

R T (Total SS removal)

t (hr) 98 min = 1.63 hr

R T = 65 % + …………….. = 71 %

v0 = h/t (m/day) 2.0 m / 98 min = 29.4 m/day

Percentage removal of SS versus v o and versus t Overflow rate, vo (m/day) 50

100

150

200

Total percentage removal, RT

80 % 70 % 65 % 60 %

65 % removal = v o = 50 m/day t = 1 hour

50 % 40 % 30 % 0.5

1.0

1.5

detention time, t (hour)

2.0

Apply scale-up factor for design purposes vo design

= vo (obtained from graph) x 0.65 = 50 m/day x 0.65 = 33.8 m/day = 1.41m / hr

t design

= t x 1.75 = 1.0 x 1.75 = 1.75 hours

• Design criteria for rectangular sedimentation tank vo t L:w h

= 0.85 – 1.5 m3/m2.hour = 2 – 4 hours (typical) = 3:1 to 5:1 (typical 4:1) = 3.0 to 5.0 m

FILTRATION A process for separating suspended matter from water by passing it through porous medium.

Isometric view of a sand gravity filter

Classification of filter • Type of filter media - One media

sand

(Sand or anthracite)

- Dual media

Anthracite

(Sand and anthracite)

- Multi media (Sand, anthracite and garnet)

Garnet

600 – 1000 mm

Diameter : 0.25 – 0.35 mm

100 – 450 mm

800 – 1000 mm 350 – 500 mm

Fine sand

Gravel

Dual Media filter

450 mm 250 mm 125 mm

Gravel

100 – 450 mm

100 – 450 mm

450 mm 200 mm

Fine sand s.g. = 2.6

Diameter : 0.55 – 0.95 mm

Gravel

Rapid sand filter

Slow sand filter

Coarse anthracite s.g. = 1.5

Medium Sand

Coarse anthracite s.g. = 1.5

Medium Sand s.g. = 2.6 Fine Garnet s.g. = 4.0 Gravel

Multi Media filter

Q As = wL

w h L

• Filtration rate, v = Q/As As = surface area of filter, m 2 - Slow rate (0.15 – 0.2 m3/m2. hour) - Rapid rate (5 – 10 m3/m2. hour) - High rate (10 – 15 m3/m2. hour) • Type of operation - Gravity filter - Pressure filter

Rapid sand pressure filter

Dual media Pressure Filter

Types of Sand Filter • Slow sand filter • Rapid sand filter

Schmutzdecke (Biological layer) Electron photomicrograph of the complex biological matrix found in the schumtzdecke

Slow sand filter

Slow Sand Filter Schumtzdecke, or biological layer

Slow sand filter Effective size of sand : 0.25 – 0.35 mm Depth of filter sand : 800 – 1000 mm Depth of gravel layer : 350 – 500 mm Tank depth : 2.5 – 3.0 m Filtration rate : 0.15 – 0.2 m3/m2.hour Method of cleaning : Scraping off surface layer of sand. Replace with new sand. Length of run between cleaning : 20 to 360 days Size of unit : large, 50 to 2000 m2 Length: width = 10:9 – 10:6 For low turbidities; not exceeding 30 NTU.

Rapid sand filter

Rapid sand filter Effective size of sand : 0.55 – 0.95 mm Depth of filter sand

: 600 – 1000 mm

Depth of gravel layer

: 100 – 450 mm

Tank depth

: 3.0 – 4.0 m

Filtration rate

: 5 – 10 m3/m2.hour

Method of cleaning : Backwash (sending air, air-water or water upwards through the filter bed by reverse flow) Length of run between cleaning : 24 to 72 hours Size of unit : small, 150 m2 Length: width = 10:9 – 10:6

Back wash process

Typical section of a Rapid Gravity filter

Dual media filter • e.g. anthracite sand. • More efficient than single media. • During backwash, the sand (smaller but heavier) will settle first followed by the coarser but lighter anthracite.

Anthracite

Sand

Disinfection A process of killing microorganisms which still remains in water after filtration. Methods of disinfection • Chemical methods such as chlorine, chlorine dioxide. (Most common) • Ozone gas (O3) – powerful disinfectant • Ultra-violet radiation • Heat

UV lamp

UV water disinfection unit

Ozone gas generator

Chlorine gas

Chlorine Widely used as disinfectant because: • It is cheap, readily available (gas, liquid, powder) • It is easy to apply (high solubility) • It leaves residual in solution which provides protection to pollution in the distribution system. • It is very toxic to most micro-organisms. Reaction of chlorine with water Cl2 + H2O

HCl + HOCl

HOCl

H + + OCl –

Chlorine existing in water as hypochlorous acid and hypochlorite ion is defined as Free chlorine.

When chlorine is added to water containing ammonia, the ammonia reacts with HOCl to form various chloramines. HOCl + NH3

H2O + NH2Cl (monochloramine)

HOCl + NH2Cl

H2O + NHCl2 (dichloramine)

HOCl + NHCl2

H2O + NCl3 (trichloramine)

Chlorine existing in water in chemical combination with ammonia or organic nitrogen is defined as combined chlorine. Combined chlorine is less powerful than free chlorine.

Vacuum Chlorinator

Breakpoint chlorination • The principle of free-residual chlorination is to add sufficient chlorine to oxidise ammonia, organic matter, iron, manganese and other substances in the water. • At break point, all ammonia has reacted with chlorine and further addition of chlorine produces free chlorine. Breakpoint chlorination Mole Ratio (C1 2:NH3-N) Chlorine demand = amount of chlorine reduced in chemical reactions = Chlorine dosage – chlorine residual

Chlorine Dosage (mg/L)

Normal practice in Malaysia, • Free residual chlorine - minimum : 0.1 mg/L • Contact time -

minimum: 10 – 15 minutes normal > 30 minutes

Recommended Minimum Bactericidal Chlorine Residuals for Disinfection

pH Value

Minimum Free Available Chlorine Residual after 10-Min Contact (mg/1)

Minimum Combined Available Chlorine Residual after 60-Min Contact (mg/1)

6.0

0.2

1.0

7.0

0.2

1.5

8.0

0.4

1.8

9.0

0.8

> 3.0

10.0

0.8

> 3.0

Other treatment Fluoridation – adding fluoride compounds to water the purpose of reducing tooth decay. Softening – to reduce hardness in water. CaO or Na2CO3 is added to water Adsorption process using activated carbon to reduce taste and odour.

Application of activated carbon to reduce taste and odour

Water distribution system The objective of a water distribution system is to supply potable water, at sufficient pressure and quantity, to the consumers. The water distribution system consists of pipelines, reservoir/storage tank, pumps, fire hydrant, control valves and meters.

Types of water distribution system - reliable, economical.

Gravity system - costly Direct pumped system

- most common Gravity and pumped combination

Daily water consumption versus time

Storage tank /reservoir • Flow equalization. (Balance the fluctuations in demand) • Provide storage for fire protection. • Meet emergency demands e.g. pump fails, maintenance. • Improve water quality due to storage.

Types of pipe network

Dead end / tree network

Grid or looped network Ring or circular network