Sanitary engineering-1-Water treatment and water supply

SANITARY AND ENVIRONMENTAL ENGINEERING (1)

LECTURES NOTES

Dr. Ahmed SeafDr. Emad Hamdy

WATER SUPPLY ENGINEERING

3RD YEAR CIVIL ENG.

ENVIRONMENTAL ENGINEERING DEP.

INTRODUCTION

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.

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

Rain

1

2

3 4

Water sources in nature

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

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

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.

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.

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

DATA REQUIRED TO SUPPLY A CITY BY WATER

1- Design period2-Design population (current and forecasting)3-Design flow 4- Master plan (water source, city development plans.....etc)

DESIGN PERIOD

Factors affecting on design period 1- Useful life of different water system components

Concrete structures 40 – 50 yearsPipes 40 – 50 yearsMechanical parts 20 – 25 yearsElectrical parts 15 – 20 years

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 period4- Rate of Interest5- Primary performance of system units

DESIGN POPULATION

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 population tP

∆∆=

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)



kg = rate of change of

population t

Pkg

∆∆= ln

POPULATION FORECASTING3- 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)



kd = rate of change of population

t

PSkd

∆−∆−= )ln(

POPULATION FORECASTING4- Rate of growth method :

P n = P 0 (1 + r) n



r = annual rate of growth5- Population density method :

a- P = area * population density

b- P = No. of residential units *

population density per unit

POPULATION FORECASTING

Graphical methods 6- Graphical extension method

7- Graphical comparison method

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

Factors affecting on water consumption1- Size of city2- Standard of living3- Climate 4- Pressure and quality of water5- Sewage facilities6- Cost

DAILY WATER CONSUMPTION

•Types of water consumption according to designQ 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

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)

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

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.

TURBIDITY METERS


1. Physical characteristics.1.5 Suspended solids : Those solids which

are retained by a glass fiber filter and dried to constant weight at 103-105oC.

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 103-105oC.


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.

SOLIDS METERS


2. Chemical characteristics.

A.Organic testsAmmonia, Nitrite and Nitrate

B. Inorganic testsB.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.

CHARACTERISTICS OF WATERB.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.

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)

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

human body (24 hrs)


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

FLOW LINE DIAGRAMS OF GROUND WATER TREATMENT (HIGH D.S.)

GW source

Sand – Ceramic - Cartridge filters

Collection Wells

Filtration Nano filters / RO

Storage

Disinfection



FLOW LINE DIAGRAMS OF GROUND WATER TREATMENT (IRON AND MANGANESE)

GW source

Cascade or diffused air

Collection Wells

Aeration Sedimentation and filtration

Storage

Disinfection



FLOW LINE DIAGRAMS OF GROUND WATER TREATMENT (HARDNESS REMOVAL)

GW source

Heat – Lime – Soda – Ion exchange

Collection Wells

softness Sedimentation and filtration

Storage

Disinfection



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

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.

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.

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).

TYPES OF INTAKES1. 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

Pipe intake

conduit

Main header

Raw P.S.

SLUICE VALVES

GATE VALVE < 300 MM

BUTTERFLY VALVE > 300 MM

NON RETURN VALVES

FOOT VALVE

CHECK VALVE

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) .

PUMPS

PUMPS CHARACTERISTIC CURVES

H,Q CURVE

SYSTEM HEAD CURVE

PUMPS DUTY POINT

T.D.H = H st. + H l + H m + H v

Shore intake

Submerged intake

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

Head losses calculationCalculated 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 * SL = Intake conduit Length (m)

•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 ( v22 – v1

2 ) / ( 2 g ) ≤ 10 cm•v1 = Q 1 screen / ( L * d)•v2 = Q 1 screen / (n * s * d)

DESIGN OF SHORE INTAKE SCREEN

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 ΦDesign 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)

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

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:

1.Settling of discrete (non flocculent) particles.

Theory of sedimentation

( ) 2

18d

gV ls µ

ρρ −=Stoke’s LawStoke’s Law

2. Settling of flocculent particles.

Theory of sedimentation

3. Zone settling.4.Compression settling.

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)

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

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

4 - Baffles and barriers for the distribution of water in the vertical plane.


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


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 flowThe 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

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)

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 = nØ

Cross.A = nBd

Lw = nB

VS = ( Q * S.S. * R.R. ) / ( ∂ * 106 * (1-WC) * N * n )Such that

VS = Sludge volumeQ = Q d( m

3 / day )S.S. = Suspended solids = 80 P.P.MR.R. = Removal ratio = 90 – 95 %∂ = sludge density = 1.02 (t/m3)WC = sludge water content = 95 %n = No of sedimentation tanksN = 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

COAGULATION PROCESS•PurposeRemoval 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

• Types of Coagulants1. Alum or [Aluminum sulphate (AL2 (SO4)3 +

18H2O)].2. Ferric and Ferrous sulphate.3. Ferric chloride (spicily for colored water)

•Chemical theoryAddition of a chemical matter (coagulant) to raw water that reacts with water alkalinity and produce a gelatinous forming (flocs.).

•Physical theoryThe 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.

•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

•Methods of alum feeding

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

Jar testJar test is used to determine the daily coagulant dose.Steps of the test1. 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.

Different mixing methods

1- Mechanical mixing (Impellers)2- Hydraulic mixing 3- Inline mixing

Chemical coagulation process components

1- Alum solution tanks2-Flash mixing tank(s)3- Clari-flocculator / rectangular flocculation and sedimentation tanks

Design of Alum Solution Tanks

V = (Q * S) / (∂ * 106 * C)Such that

V = Alum solution volumeQ = Q d = 1.1* Qmm(m3 / day)S. = Dosage = 30 → 80 P.P.MC = Concentration = 5 → 20 %∂ = Alum solution density = 1.02 ( t / m3 )

V1 = V / no. of tanks (2 → 3)d= 1 → 3 mA = W2 = V1/d

Alum weight = (Q * S * no. of days) / 106

calculated for 30 days

Design of Flash Mixing Tank(s)

R T = 5 – 60 sec { take 30 sec {Q d =1.1 * Q mm

V = Q d * R TDepth of chamber d = 3 m

Area of chamber A = V / dA = * Ø 2 / 4Ø = chamber diameter

Design of impellers

G = √ (P / µ * V) = 300 – 700 sec -1

P = Theoretical powerµ = dynamic viscosity of water = 1.14 * 10 -3

V = Volume of flash tankAssume G = 500 sec -1

Get P P = K * S * n3 * D5

K = impeller coefficient = 1S = 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

Clari-flocculator

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 ≥ 2Ø s = total tank diameter ≤ 35 mØ f = Flocculation zone diameter = (0.33 – 0.50) * Ø s

Assume Ø f = 0.40 * Ø s & Assume n get Ø s

Flocculation zoneAssume 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

Design of paddlesG = √ (P / µ * V) = 20 – 70 sec -1


V = Volume of one flocculation tankAssume G = 50 sec -1

Get P P = 0.50 * C d * A * ρ * v r

3 C d = impeller coefficient = 1ρ = water density = 1000 (kg / m3(

v r = relative velocity of paddles [ 0.45 → 0.70 (m/s ( [

Get area of paddles

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 hrChecks1) 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)

Rectangular flocculation and sedimentation

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 ≥ 2L s = Sedimentation tank length ≈ 0.80 L t

B = width of tank = (6 → 12) mAssume L s = 0.80 L t & Assume n get L t

L t ≤ 50 m, L f = 0.2 L tFlocculation zoneAssume 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

Design of paddlesG = √ (P / µ * V) = 20 – 70 sec -1


V = Volume of one flocculation tankAssume G = 50 sec -1

Get P P = 0.50 * C d * A * ρ * v r

3 C d = impeller coefficient = 1ρ = water density = 1000 (kg / m3(

v r = relative velocity of paddles [ 0.45 – 0.70 (m/s ( [

Get area of paddles

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 hrChecks 1) V h = [Q d / n * B * d s ] ≤ 0.30 ( m / min ) 2) W L R = Q d / n * BW 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)

Other chemical sedimentation units

Plate Settler or Tube Settler

FILTRATIONFiltration 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.

1. Straining mechanismImpurities 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 filtrationFiltration 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.

2. Transportation (non straining) mechanismIn this mechanism several types of removal could take place such as Sedimentation action, Adsorption Action, Electrolytic Action and biological action.

Sedimentation ActionRemoval 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 ActionAdsorb the colloidal matters on the filter bed particles as a result to coat it by a gelatinous layer from bacteria and microorganisms.

Electrolytic ActionThe 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 ActionThe 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.

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 media3. Multi media

3. A.T. rate of filtration1. Slow filters (3 – 10 (m3/m2/d))2. Rapid filters (120 – 200 (m3/m2/d))

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.

Dual Filter

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.

Slow sand filters

Design Criteria of Slow Sand FilterEffective size of the sand = 0.25 – 0.35 mmDirty skin layer = 3 – 8 cmWashing time (removal time) = 1 – 15 days ( 1 day if mechanical & 15 day if manual)Ripening the filter is taking = 7 – 15 daysThe whole cleaning process is taking = 8 – 30 daysThe operation time (between two washes) is = 2 – 6 month.

Rate of filtration (ROF) = 3 – 10 m3/m2/dArea of filter = 1000 – 2500 m2

The filter is Rectangular (L*B)L & B ≤ 50 mn ≥ 2L/B = 1 – 1.25

ExampleDesign the SSF for a WTP working 16 hr/d, if the design flow is 32,000 m3/d

SolutionQd = 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

•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

http://www.thewatertreatments.com/

Design Criteria of Rapid Sand Filter

Effective size of the sand = 0.6 – 1.5 mmSand uniformity coefficient = 1.35 – 1.5Sand specific gravity = 2.55 – 2.65Wash water speed = 2.5 – 3.5 m/sCleaning period = 25 – 35 minRipening the filter is taking = 15 – 20 minWashing by compressed air = 2 -5 minWashed by pressured water = 10 min & 15 -20 min if no airThe operation time (between two washes) is = 12 – 36 hrsRate of filtration (ROF) = 100 – 200 m3/m2/dArea of filter = 40 – 64 m2

Empirical equation to determine minimum number of filters in the WTP = 0.044 * [Qmm (m3/d)]0.5The filters numbers:If nw ≤ 5 take nT = any no. + 1 for washIf nw > 5 take nT = even no. + 2 for washIf nw ≥ 30 take nT = no. divisible by 4 + 4 for washAmount 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 mB : L = 1 : 1.25 up to 1 : 2Rate of washing (ROW) = 5 – 6 ROF

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 = 6.25SA1 = 8*6.25 = 50 m2

nw = 400/50 = 8 filters (ok)nT = 8 + 2 = 10 filtersAssume that ROW = 5 ROF = 25 m3/m2/hrAmount 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 %

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.

1. Contact time and dosageThe longer contact time and dosage the greater the kill is.

2. TemperatureAs 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

•Requirements of good disinfectant1. 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.

METHODS OF DISINFECTION

HEATING (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

SUNLIGHT EXPOSURE

Advantages:Kills harmful bacteria and pathogensSimple, convenient and inexpensiveIf used correctly, the water is as clean as boiled waterWill not change the taste of water

Disadvantages:Requirement of huge surface area. A 6-12 hour waiting period

ULTRAVIOLET

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

ULTRAVIOLET

Advantages Pure odour free, colourless water with turbidity of below 15mg/lit Kills all type of bacteria Normally used for sterilizations at hospitalsDisadvantages Very costly Possible interruption by electricityNo residual for networks disinfection

When UV radiation penetrates the cell wall of an organism, it damages genetic material, and prevents the cell from reproducing.

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.

The advantages and disadvantages of chlorination

•Advantages1. Cheap2. Residual for network3. Available4. Easy to store and use.5. Simple equipment required.

•Disadvantages1. 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-).

H2O + CL2 → HOCL + HCL

HOCL →H+ + OCL-

HCL →H+ + CL-

Chlorination of tab water

Break point chlorination

Ι - Destruction of CL2 by reducing agents.Π - Formation of chloro-organic of chloraminesШ - Destruction of components

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)

GROUND STORAGEPurpose1. Produce contact time for disinfection = (0.5) hrC1 (m3) = (0.5)hr * Qmm (m3/hr), Qmm (m3/hr) = Qmm

(m3/d)/wp2. Balancing difference between maximum daily and

maximum monthly flow through one dayC2 (m3) = [Qmd (m3/d) – Qmm (m3/d)] * 1 day3. Saves Emergency Storage = (15 % - 40 %) of daily

productionC3 (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

Design Capacity of Ground ReservoirC (m3) = take bigger of [C1 or C2 or C3] + C4

L ≤ 50 m, L = 1.0 – 2.00 Bd = 3 – 5 mn ≥ 2 tanks

The ground reservoirs are built of reinforced concrete and they have be coated by isolating materials to prevent any percolations.

• 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.

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 flowsqmm = 420 l/c/dqav = 420/1.4 l/c/d = 300 l/c/dQav = 300 * 300,000 = 90,000,000 l/d = 90,000 m3/d = 3,750 m3/hrQmm = 1.4* Qav = 5,250 m3/hr = 126,000 m3/dQmd = 1.8* Qav = 6,750 m3/hr = 162,000 m3/d

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)

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.

Second: with respect to pressureThe 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.

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).

Float valve Pressure release valve

DesignC1 = [Q max hour - Q max day] * 3 / 24 =

0.70 * Q ave * 3 / 24C2 = from total mass curve (S curve(

C = max of (C1 , C2 ) + 0.20 * F D

F D = 120 * pop / 10000C = n * d * π * Ø 2 / 4d =(1/3 to 2/3) ØØ = 10 to 20 m

get n

W T P works 24 hr with constant rate

WC curve

Pumping

C2 = [ a - b] * pop * 1.8

W T P works <24 hr with constant rate

WC curve

Pumping

C2 = [ a + b] * pop * 1.8

W T P works 24 hr with variable rate

WC curve

Pumping

C2 = [ a + b] * pop * 1.8

Distribution networksREQUIREMENT OF A DISTRIBUTION SYSTEM: 1. The system should convey the treated water up-to consumers with the same degree of purity2. The system should be economical and easy to maintain and operate3. The diameter of pipes should be designed to meet the fire demand4. 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

Types of networks systems

1-Tree (Dead end) systemThis 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) system2- Loop (Ring) system3- Radial system4- Grid iron system

ADVANTAGES1. Discharge and pressure at any point in the distribution system is calculated easily2. 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 economical4. The laying of water pipes is used are simple. DISADVANTAGES1. 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

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.

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.

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.

ADVANTAGES 1. In the case of repairs a very small portion of distribution are a will be affected2. Every point receives supply from two directions and with higher pressure3. Additional water from the other branches are available for fire fighting4. There is free circulation of water and hence it is not liable for pollution due to stagnation.

DISADVANTAGES1. 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.

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 * SL = Pipe Length (m)

Hardy-Cross Method

This method used to determine 1- Discharge & flow direction for all pipes in network2-Pressure @ all nodes & HGL Network of pipes forming one or more closed loopsLimitation of the method1- ∑Qin = ∑Qout 2- the pipes forms closed loopsGivenDemands @ network nodes (junctions)d, L, pipe material, Temp, P @ one node

∑∑∑∑

∑∑

−=

−=

−= −

af

na

ana

na

na

na

Qhn

KQ

QKQn

KQ

KQn

KQ

δ

δ

δ1

Hardy-Cross Method (Derivation)

( )( ) 0

0

0

0

1 =+

=+

=

=

∑∑∑∑

− δ

δna

na

na

n

f

nQQK

QK

KQ

h

For Closed Loop:

( ) ( ) 221

!2

1 δδδ −− −++=+ na

na

na

na Q

nnnQQQ

n=2.0, Darcy-Weisbachn=1.85, Hazen-Williams

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 =- ∑ hf /(n ∑ hf/Qa).

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.

Date post:	23-Jan-2018
Category:	Environment
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