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/dayheavy 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/dayF = 1.0Da = 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

State Urban Rural1985 1993 2000 1985 2000 2008

JohorKedahKelantan

MelakaN.SembilanPahang

Perak Perlis P.Pinang

Sabah Sarawak Selangor Terengganu

0.92 0.99 0.99 0.95 0.99 0.99 0.65 0.82 0.99

0.99 0.99 0.99 0.89 0.99 0.99 0.95 0.99 0.99

0.98 0.99 0.99 0.93 0.99 0.99 0.98 0.99 0.99

0.99 0.99 0.99 0.95 0.99 0.99 0.95 0.99 0.99 0.85 0.92 0.99

0.61 0.78 0.85 0.58 0.92 0.95 0.30 0.50 0.70

0.82 0.85 0.95 0.75 0.86 0.95 0.65 0.80 0.90

0.75 0.80 0.90 0.50 0.93 0.95 0.85 0.87 0.95

0.38 0.55 0.75 0.33 0.40 0.70 0.73 0.80 0.90 0.40 0.82 0.95

Malaysia 0.93 0.99 0.99 0.58 0.74 0.86

• Reflects the potential percentage of population served

SERVICE FACTOR

• Methods- Graph

POPULATION PROJECTION

- Zone (for new township)

POPULATION PROJECTION

Industrial

Commercial

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

Housinge.g. 500 apartments. Average number per house: 5 people

- 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.gFrom 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 demand (consumption) include normal commercial & industrial use, domestic use and accounted–for-water losses.

As a guideline:Urban – 230 to 320 liter/head/daySemi urban – 180 to 230 liter/head/dayRural – 135 to 180 liter/head/day

Average daily per capita demand = quantity required daily

population served

PER CAPITA CONSUMPTION

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

Aerial View of Sg. Johor Water Treatment Plant

WATER TREATMENT

WATER TREATMENT

Objectives of water treatmentTo 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 waterWater intake - a structure constructed at a surface water source to facilitate the conveying of water to the treatment plant.

Cross section of a weir intake

Water intake

Collection well intake

Direct pumped extraction intake

Water intake

Water intake

Water intake at a reservoir

gate

Water Treatment Processes

ScreeningPlaced at water intake

Coarse screenRemove floating / large objects such as twigs, leaves that can foul or damage equipment.

Fine screenRemove aquatic plants / small plants.

WATER TREATMENT

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

AERATIONPurpose:ØIncrease oxygen contentØTo liberate dissolved gases e.g.

CO2 and H2S; reduce corrosiveness & odourØ 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)

Spray aerators

Multiple platform aerators

Venturi aerator

Draft tube aerators

Air-into-water 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)

Negatively charged particles repel each other due to electricity.

Particles and coagulants jointogether into floc.

Positively charged coagulants attract to negatively charged particles due to electricity.

Neutrally charged particles attract due to van der Waal's forces.

Coagulation

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 SO42-

(Aluminum hydroxide particles:jelly- like & positive charge, neutralize negative charge particles collide & stick together forminglarge 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 aidse.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.

Jar test

Procedure1. Six jars is filled with water (turbidity, ph and alkalinity have been predetermined)

3. Water is mixed rapidlyfor 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)

5. The water is allowed to settle ~30 minutes (Sedimentation)

6. Portions of the settled water are tested to determine the remaining turbidity.

2. The jars are dosed with different amounts of coagulant.

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

Most of the particles have completely settled out.

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

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/LTreatment plant flow rate = 3000 m3/dayAlum 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.

alum solution (10 g/L

concentrationin 1 liter

contain 10 g alum)

3 mL of alum solution

raw water1 L

in 3 mL contains

(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

Overflow weirBaffled plate

coagulant

water

Mixing flume

coagulant

water

Hydraulic jump is formed creating turbulence

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 = √ P / µ ∀ (unit: s-1)

where

P = Power input, W (N.m/s or kg m2/s3)µ = 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

whereQ = flow rate , m3/s∀ = volume of tank (m3)

Q in Q out

h

l

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 flocculatorFlow is directed around baffles

ChannelWater inletCross

sectionarea of channel, A

Distance between baffley > 0.2 m

x

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

b

b = channel length

Baffle, thickness: 5 – 8 cm

h

Detention time, t = 20 - 30 minutes

Velocity of water flowing in channel, v = 0.09 - 0.24 m/sec

Design criteria for baffled tank (Round – the end type)

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 = Volume of waterDistance of water traveled

Channel depth (h) = Cross sectional area of channelDistance between baffles

= A > 1.0 my

Determination of tank length, L

Channel effective length = channel length - x

No. of channel = Distance of water traveledChannel effective length

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

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

Power input for hydraulic tank

P = ρ g Q h

where ρ = density of fluid, kg / m3

g = gravitational constant, m/s2

h = head loss, m

G = √ P / µ ∀

= √ ρ g Q h / µ ∀

= √ ρ g h / µ t

Mechanical flocculation

Mechanical flocculation

Paddle flocculator

Flash mixerCoagulants

Rapid mixing tank

Flocculation tank

Mechanical flocculatorDesign criteria

t = 20 -30 minutesh water < 5 mω, number of revolutions of paddle per minute = 2 -15 rev/minv p, velocity of paddle tip(relative velocity between paddles and fluid)

= 0.3- 0.7 m/s= 75 % from actual velocity of paddle, vawhere

v a = 2 π r ωr = distance from shaft to center of paddle

plate

Paddle

L w

r

v p

Power input, P

P = FD v p

where FD= drag force on paddles, N (i)

Therefore, FD = 1CD A p ρ v p2 (ii)2

where

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 ρ vp3

G = P/ µ ∀

= 1 CD AP ρ vp3

2 µ ∀√

√

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 bottomof 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.

Inlet zoneSettling zone

Sludge zone

baffle

Outlet

zone

Scraper

Sludge removal

Inlet and outlet detailsBaffle Baffle

Tank Bottom Multiple Openings

Overflow Inlet Weir Overflow Weir

TYPES OF INLETS FOR SEDIMENTATION TANKS

OUTLET DETAILS OF SEDIMENTATION TANKS

Multiple Openings

Outlet Pipe 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 (ρ p - ρ w)18 µ

where d = diameter of particleρ p = density of particleρ 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 (ρ p - ρ w)18 µ

= 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 area, As

vs

v1

sludge out

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

vo

Up flow tank

For an upflow tank, in order for a particle to be removed, vs > vo

• Surface overflow rate, vo

L

h

w

Q

A s = w L

Ac = w H

Rectangular sedimentation tank

Ideal sedimentation tank.

vh

settled solids

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.

Av h

v sh

B

An ideal rectangular sedimentation tank illustrating the settling of discrete particle

In other words, settling velocity, vs must equal to the depth of tank divide by the detention time.

vs = h/tBut t = ∀ / Q

vs = h / (∀ / Q)

= h Q/ ∀

= h Q / (w x L x h)

= Q / w x L

= 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 ≥ vo

Removal of particles in an ideal sedimentation tank

vh

h

h s

1. 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 = t2. 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.

3. Particle with v s < vo will be removed if it enters the tank at a distance from the bottom not greater than h swhere h s = v s t

The proportion of particles with v s < vo which will beremoved is given by

P = v s (100)vo

v s = vo

v s

A

B

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 / As = 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

30 %

70 %

60 %

50 %

40 %

80 %

Tota

l per

cent

age

rem

oval

, RT

10050 150 200

vo (m/day)

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

Percentage removal of SS versus t

1.00.5 1.5 2.0

30 %

70 %

60 %

50 %

40 %

80 %

Tota

l per

cent

age

rem

oval

, RT

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 / As = h / t

H 1

H 2

H3

H 4

H 5

sampling port / outlet

Settling column test

Settling columnHeight 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 C11 C21 C31 C41

H2 C12 C22 C32 C42H3 C13 C23 C33 C43

H4 C14 C24 C34 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 timeCt = concentration at time t and given depth,

mg/LCo = initial concentration, mg/L

Depth, m Sampling time (min)t1 t2 t3 t4

H1 R%11 R% 21 R% 31 R% 41

H2 R% 12 R% 22 R% 32 R% 42

H3 R% 13 R% 23 R% 33 R% 43

H4 R% 14 R% 24 R% 34 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, mti= 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, tidefined by the intersection of the isoconcentrationline 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, taRa,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 timebe used to design the tank.

DataConcentration of SS at specified time and depth

Sampling time (min)

j)

e.g. Determine the surface overflow rate and detention time that is able to remove 65 % of suspended solids. Use scale-up factor of 0.65 for vo and 1.75 for detention time. The initial concentration of SS is 40 mg/L.

120906040 302010Depth (m)

121516182428342.0

111214162226321.0

101112141820240.5

1. Calculate percentage removal (Percentage passing each sampling port).

R % = Co - Ct (100)

Co

e.g. at depth of 0.5 m and sampling time 10 minutes,

R % = (40 – 24 ) x 100

40

= 40 %

Solution

120906040 302010Depth (m)

70%63%60%55%40%30%15%2.0

73%70%65%60%45%35%20%1.0

75%73%70%65%60%50%40%0.5

Percentage removal of SS at specified time and depthSampling time (min)

2. Plot the isoconcentration lines versus based on percentage removal.

0

0.5

4030

1.0

2010

1.5

2.0

50 70 12060 80 90 100Sampling time (min)

40% 50% 60%

65% 70% 73%

70%63%60%55 %40%30%15%

65%

45%35%20%

70% 73%

60%

75%

0

0.5

4030

1.0

2010

1.5

2.0

50 70 12060 80 90 100

Sampling time (min)

40% 50% 60%

65% 70% 73%

70%63%60%55 %40%30%15%

65%

45%35%20%

70% 73%

60%

75%

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

H H

t a t b

R a = 30 % R b = 40 %

0

40302010 50 70 12060 80 90 100

0.5

1.0

1.5

2.0

Sampling time (min)

40% 50% 60%

65% 70% 73%

70%63%60%55 %40%30%15%

65%

45%35%20%

70% 73%

60%

75%

t a t c t d t etb

30% 50% 60% 70%40% 65%

2.0 m / 60 min= 48 m/day

60 min = 1 hr

R T = 60 % + …………….. = 68 %d

2.0 m / 38 min= 76 m/day

38 min= 0.63 hr

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

c

2.0 m / 30 min= 96 m/day

30 min= 0.5 hr

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

b

2.0 m / 20 min= 144 m/day

20 min= 0.33 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

a

v0 = h/t(m/day)

t (hr)

R T (Total SS removal)Pointt f

Percentage removal of SS versus vo and versus t

1.00.5 1.5 2.0

Overflow rate, vo (m/day)

detention time, t (hour)

10050 150 200

30 %

70 %

60 %

50 %

40 %

80 %

Tota

l per

cent

age

rem

oval

, RT

2.0 m / 98 min

= 29.4 m/day

98 min= 1.63 hr

R T = 65 % + …………….. = 71 %e

v0 = h/t(m/day)

t (hr)R T (Total SS removal)Point

65 %

65 % removal =v o = 50 m/dayt = 1 hour

• Design criteria for rectangular sedimentation tank

v o = 0.85 – 1.5 m3/m2.hourt = 2 – 4 hours (typical)

L : w = 3:1 to 5:1 (typical 4:1)h = 3.0 to 5.0 m

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

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

FILTRATION

Isometric view of a sand gravity filter

Classification of filter

• Type of filter media

- One media

(Sand or anthracite)

- Dual media

(Sand and anthracite)

- Multi media

(Sand, anthracite and garnet)

Anthracite

Garnet

sand

Medium Sand

Diameter : 0.55 – 0.95

mm

Finesand

Diameter : 0.25 –0.35 mm80

0 –

1000

mm

600

–10

00 m

m

350

–50

0 m

m

100

–45

0 m

mGravel Gravel

Finesands.g. = 2.620

0 m

m10

0 –

450

mm

450

mm Coarse

anthracites.g. = 1.5

Medium Sands.g. = 2.6

125

mm

100

–45

0 m

m

Fine Garnets.g. = 4.0

Coarseanthracites.g. = 1.545

0 m

m25

0 m

m

Gravel Gravel

Dual Media filter Multi Media filter

Slow sand filter Rapid sand filter

• 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

QAs = wL w

L

h

Rapid sand pressure filter

Dual media Pressure Filter

Types of Sand Filter

• Slow sand filter • Rapid sand filter

Slow sand filter

Schmutzdecke(Biological layer)

Electron photomicrograph of the complex biological matrix found in the schumtzdecke

Slow Sand FilterSchumtzdecke, or biological layer

Slow sand filter

Effective size of sand : 0.25 – 0.35 mmDepth of filter sand : 800 – 1000 mmDepth of gravel layer : 350 – 500 mmTank depth : 2.5 – 3.0 mFiltration 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 daysSize of unit : large, 50 to 2000 m2

Length: width = 10:9 – 10:6For low turbidities; not exceeding 30 NTU.

Rapid sand filter

Rapid sand filterEffective 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

Ozone gas generator

UV water disinfection unit

UV lamp

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 hypochlorousacid 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 oxidiseammonia, 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.

Chlorine Dosage (mg/L)

Breakpoint chlorination

Mole Ratio (C12:NH3-N)

Chlorine demand = amount of chlorine reduced in chemical reactions

= Chlorine dosage – chlorine residual

Normal practice in Malaysia,

• Free residual chlorine - minimum : 0.1 mg/L

• Contact time - minimum: 10 – 15 minutes

normal > 30 minutes

> 3.00.810.0

> 3.00.89.0

1.80.48.0

1.50.27.0

1.00.26.0

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

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

pH Value

Recommended Minimum Bactericidal Chlorine Residuals for Disinfection

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

Gravity system

Direct pumped system

Gravity and pumped combination

- reliable, economical.

- costly

- most common

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