chapter 3- water supply

WATER SUPPLYEnvironmental Engineering Sem1 2012/2013*

Environmental Engineering Sem 2 2011/2012

INTRODUCTIONEnvironmental Engineering Sem 1 2012/2013

*Malaysia was reported to receive rainfall of 324billion m3/year in year 20001. It distributed to surface runoff and evapo-transpiration about 152 billion m3/year respectively and only 20 billion m3/year becomes groundwater.

Malaysias water status are diverse, increase and competing needs. However the future forecast ( done by SMHB) indicated that water supply demand cannot be met in many locations due to uneven rainfall distribution, temporally or spatialy degradation in the raw water supply catchments, because of Un-control opening of forested areas (logging, roads, urbanizations) resulting in massive sedimentation flow into rivers Pollution in rivers as it passes urban areas, reaches between raw water source catchment/storages & downstream water supply intake

1 -Ir Salmah Zakaria (2008) , Water Resources and Global Warming: Focus, Water Supply , Meeting of Water Supply Senior Officer, Equatorial Hotel, KL.


INTRODUCTIONEnvironmental Engineering Sem 1 2012/2013

*Humans are defending on water supply for the daily basis routine. . Therefore we should not keep throwing everything into the river. Consequently, the cost of water treatment will be higher and not affordable any longer. Freshwater is finite and water should be treated as a commodity.

We should learn from other on what they have been doing to have a sustainable water supply. A Singapore Success StoryOn 27 February 1977, at the opening of the Upper Pierce Reservoir, the prime Minister, Lee KuanYew said It should be a way of life to keep the waters clean .. In ten years let us have fishing in the Singapore River and in the Kallang River.In October 1977, The Ministry of the Environment , submitted an Action Plan to the Prime MinisterSingapore River and in Kallang Basin How did they do it?5,000 hawkers were relocated into food centres26,000 squatter families were resettled to HDB flatsMoved factories to JTC, then onlyRemoved more than 200 tonnesof rubbishUpgrade the riverside walkway along the riverUpgrade parking lots along riversDeveloped recreational areas along riversSand was brought in to form beaches for recreation


INTRODUCTIONEnvironmental Engineering Sem 2 2011/2012*In Malaysia, the worst water crisis happen in year 1998 for Klang Valley residents. The main reasons for the water crisis were due:

development of the resource and treatment facilities could not meet the rapid pace of urbanisation and industrialisation; a prolonged drought causing the reduction of flows in the rivers and the subsequent decrease in the reservoir levels; the low flows in the rivers were insufficient to dilute the pollutant loads resulting frequent closures of water treatment plants and High water losses due to leakages in the distribution systems and pilferages


WATER QUANTITY REQUIREMENTEnvironmental Engineering Sem 2 2011/2012*Water supply is required for domestic and industrial, agriculture, environment and Bio-Diversity, hydropower, transportation and recreation.

In Malaysia at year 20072, about 14,278 MLD water treated from 462 numbers of water treatment plants. The water demand is 12,330 MLD, most serves for urban area compared to rural .The water being used by 27 Million people.2 -Ir Noor Azahari Zainal (2008) , Operational Aspects of WaterDemand Management- For Domestic & Industrial Use, Meeting of Water Supply Senior Officer, Equatorial Hotel, KL.


WATER QUANTITY REQUIREMENTEnvironmental Engineering Sem 2 2011/2012*


WATER QUANTITY REQUIREMENTEnvironmental Engineering Sem 2 2011/2012*Most of treated water was used for domestic and industrial besides agricultural. At year 2010 (Table 3. 1), the water consumption is slightly higher for irrigation, However by the year 2020 to 2050, more treated water will be used for domestic and industrial activity.

Table 3.1: Water Demand for Malaysia from 2000 to 2050

200020102020203020402050Domestic & Industrial(mil m3/yr)34835579742389371026811544Irrigation(mil m3/yr)735065176517613261326132Total (mil m3/yr)108331208613940150691640017676


WATER SOURCESEnvironmental Engineering Sem 2 2011/2012*

Seawater

Groundwatershallow wellsdeep wells

Surface waterriverslakesReservoirs

Groundwater Vs Surface water

GroundwaterSurface waterconstant compositionhigh mineral contentlow turbiditylow colorlow or no D.O.CO2, H2S may be presenthigh hardnesshigh Fe, Mn variable compositionlow mineral contenthigh turbiditycoloredD.O. presentlow hardnesstaste and odorClay, silt, NOM


WATER TREATMENT*In practice, all public water supplies undergo some form of treatment, with the degree of the treatment being dependent on the quality of the raw water supply.

The quality of treated water is now almost standardized in the developed world, with treatment facilities having to satisfy many water quality parameters on a frequent monitoring basis.

The aim of water treatment is to produce water;That is palatableesthetically pleasingpresence of chemicals does not pose a threat to human healthincludes chloride, color, corrosivity, iron, manganese, taste and odorThat is potableSafe to drink protective of human healthNot necessarily esthetically pleasingThat meets regulatory criteria/standardsWithout interruptionAt reasonable cost.

Environmental Engineering Sem 2 2011/2012*TREATED WATER QUALITY STANDARD** SAJ Holdings Sdn Bhd


TYPES OF WATER TREATMENT PLANTEnvironmental Engineering Sem 2 2011/2012*Objective: to remove turbidity and color from surface waterPlant : Filtration Plant /Coagulation Plant/Conventional Plant


TYPES OF WATER TREATMENT PLANTEnvironmental Engineering Sem 2 2011/2012*Objective: to remove the hardness of the mineralPlant : Softening Plant


TYPES OF WATER TREATMENT PLANTEnvironmental Engineering Sem 2 2011/2012*Objective: to remove sulfidePlant : Groundwater Plant


Environmental Engineering Sem 2 2011/2012*THE GENERAL PURPOSE OF WATER TREATMENT PROCESS

Pre chlorination-used for disinfection of the raw water only if it does not result in formation of thrihalomethane.Pre sedimentation-To reduce silt and settleable organic matter prior to chemical treatmentCoagulation -alum/ other chemicals are added to water to form tiny sticky particle called floc.Flocculation-is the process of the aggregation of the destabilized particles and precipitation products.Sedimentation-To remove the particles and precipitate by gravitational settling.Filtration-To remove even smaller particles.Flouridation-Fluoride is added into water to prevent incident in dental caries.Disinfection-Chrorine/disinfectant is added to kill any bacteria or microorganism that may be in the water.Post Chlorination-To form and remain the residual chlorine in the distribution system.Activated Carbon-To remove odor and taste producing compound.


Environmental Engineering Sem 2 2011/2012*WATER TREATMENT PROCESSESWater treatment processes are depending on the water sources. There are 4 classes of water treatment as shown in Table 3.2. The treatment processes may be difference from one to another as shown in Table 3.3. The typical water treatment process is shown in Figure 3.2. Table 3.2 Classes of Water Treatment

ClassDescriptionSourceAB

C

DNo treatmentDisinfection only

Standard Water Treatment

Special Water TreatmentSome borehole waterOccasional upland waterSome borehole waterOccasional upland waterLowland rivers and reservoirsSome rural supplies (Fe and Mn), Colour removalTrace element removal, Industrial Water Electronics industry requirement, Algae removalOrganic removal


Environmental Engineering Sem 2 2011/2012*WATER TREATMENT PROCESSESTable 3.3 Flow Chart Outline of Water Treatment Processes

Treatment CategorizationGroundwater Supply Class BStandard Water TreatmentClass CSpecial Water Treatment Class DIntakeIntakeIntakePre treatmentAerationCoarse screening, Fine Screening,Pumping ,Storage,Equalization,Neutralization,AerationChemical pre-treatmentScreening, Pumping,Storage, EqualizationNeutralization, AerationSoftening, Algae removalPrimaryTreatmentCoagulation and mixingFlocculation, SedimentationCoagulation and mixing,Flocculation,SedimentationSecondarytreatmentFiltrationFiltrationDisinfectionDisinfectionDisinfectionDisinfectionAdvance TreatmentAdsorption,Activated carbon,Membrane ProcessHalogenated compound removal, Fe and Mn removalFluoridationFluoridationFluoridationFluoridationDistributionnetworkYesYesYes


Environmental Engineering Sem 2 2011/2012*What are the common raw water sources used in Malaysia?

As an engineer, you are asked to select the appropriate water treatment plant. What will be your selection? Why? List the primary treatment.


What are the common raw water sources used in Malaysia? Answer : River water and lake As an engineer, you are asked to select the appropriate water treatment plant. What will be your selection? Answer : Coagulation plant/ Conventional plant ( Refer Slide 19, the common practice in Malaysia) Why? Answer : We use the river water as an intake List the primary treatment. Answer: Coagulation and mixing Flocculation, SedimentationEnvironmental Engineering Sem 2 2011/2012*


Environmental Engineering Sem 2 2011/2012*Raw water source- - - Our concern


Environmental Engineering Sem 2 2011/2012*Raw water intakes withdraw water from a river, lake, or reservoir ( surface water source) over a predetermined range of pool levels.

Intake site selections depend on :Water qualityWater depthStream or current velocitiesAccessPower availabilityProximity to water treatment plantEnvironmental impactHazard to navigate

Raw water intake structures control withdrawal of raw water from a surface water source. Intake structure contains gates, screens, control valves, pumps, chemical feeders, flow meters, offices and machine shop.Intake design consideration;Intake velocity High velocities head loss, entrain suspended matter, trap fish, and other aquatic animals.Velocity below 8 cm/s allows aquatic animal to escape, and minimize the suspended matterIntake-port location- Water quality in each stratum may vary. To achieve, multiple intake ports set at various levels are generally providedTop intake less than 2 m below normal level. Bottom intake least 1 m above the bottom Gates - Usually sluice gates . Large cast iron gates that slide vertically on a guide track.INTAKE


Environmental Engineering Sem 2 2011/2012*


Environmental Engineering Sem 2 2011/2012*Types of intake structures; floating intake ( Figure 3.3), submerged intake, pier intake, tower intake ( Figure 3.4), exposed or tower intake and shore intake

Figure 3.4: Tower intakeFigure 3.3 Floating intakeINTAKE


Environmental Engineering Sem 2 2011/2012*AERATIONTo create turbulence to provide for the maximum contact between water and air to achieve desired dissolved oxygen content at ambient temperature and pressure .

It can be a simple mechanical process of spraying water into the air and allowing it to fall over a series of cascades (waterfalls) ( Figure 3.5), multiple platform aerator- tray ( Figure 3.6) venturi aerator, draft-tube aerator while absorbing or desorbing oxygen in its journey.

Figure 3.5 Cascade Figure 3.6 Tray


Environmental Engineering Sem 2 2011/2012*To supply of O2 from the atmosphere to water to effect beneficial changes in the water quality.To release excess H2S gas which may cause undesirable tastes and odor.To release excess CO2 which may have corrosive tendencies on concrete materials.To increase the O2 content of water which may have negative taste, color and stain properties due to the presence of Fe and Mn in solution ( Mostly for groundwater) . The addition of oxygen assists the precipitation of Fe and Mn as following:AERATION4Fe2+ + O2 + 10 H20 4 Fe(OH)3 (s) + 8H+2Mn2+ + O2 + 2 H20 2 MnO2 (s) + 4H+


Environmental Engineering Sem 2 2011/2012*AERATIONExample

Parit Raja Water Treatment Plant has a plant capacity of 60 MLD. Calculate the required water surface area of a cascade used for aeration.

Solution:

To calculate exposed water surface area, we used design criteria of 10 m2 of exposed water surface for every 50L/s of design flow.

Design capacity = 10 m2 of exposed water surface 50 L/s of designed flow Water surface area = 10 m2 x 60 MLD 50 L/s = 10 m2 x 60 x 10 6 L/D 50 L x 60 s x 60 min x 24 hr s min hr day = 10 m2 _x 60 x 10 6 L/D 4320000 L/D = 138.8 140 m2


Environmental Engineering Sem 2 2011/2012*SOFTENINGThe removal of ions that cause hardness is called SOFTENING. This process is common for groundwater source.

Hardness in natural waters comes from the dissolution of minerals from geologic formation that contain calcium and magnesium . Two common minerals are calcite and dolomite. The natural process by which water become hard is shown below.


Environmental Engineering Sem 2 2011/2012*SOFTENINGHardness is a term often used to characterize the ability of a water to:cause soap scum Ca2+ + (Soap)- Ca(Soap)2 (s) increase the amount of soap neededcause scaling on pipescause valves to stick due to the formation of calcium carbonate crystals leave stains on plumbing fixtures

Total Hardness (TH)Technically - the sum of all polyvalent cations Practically - the amount of calcium and magnesium ions (the predominant minerals in natural waters) (TH = Ca2+ + Mg2+)It is divided into carbonate (CH) and noncarbonate hardness (NCH), (TH = CH + NCH)

DescriptionHardness range (mg/L as CaCO3)Extremely soft0 - 50Very soft50 100Moderately hard100 150Hard150 - 300Very hard> 300


Environmental Engineering Sem 2 2011/2012*SOFTENINGCarbonate Hardness (CH)associated with HCO3-, CO32- CH = TH or Total alkalinity, whichever is lessOften called "temporary hardness" because heating the water will remove it. When the water is heated, the insoluble carbonates will precipitate and tend to form bottom deposits in water heaters. Ca2+ + 2HCO3- CaCO3(s) + CO2(g) + H2O

Non-Carbonate Hardness NCH = TH - CHIf Alkalinity Total hardness, then NCH = 0Called permanent hardness because it is not removed when the water is heated. It is much more expensive to remove non-carbonate hardness than carbonate hardness.Ca2+, Mg2+ associated with other ions, Cl-, NO3-, SO42-


Environmental Engineering Sem 2 2011/2012*SOFTENINGExample A sample of water having a pH of 7.2 has the following concentrations of ionsCa2+40 mg/LMg2+10 mg/LNa+11.8 mg/LK+7.0 mg/LHCO3-110 mg/LSO42-67.2 mg/LCl-11 mg/LCalculate the TH, CH, NCH, Alkalinity, and construct a bar chart of the constituents Solution *Sample Calculation: Concentration of Ca2+ in mg/L as CaCO3 = (Concentration in meq/L) * (Equivalent Weight of CaCO3) = (1.995 meq/L) X (50 mg/meq) = 99.8 mg/L as CaCO3

IonConc.mg/LM.W.mg/mmolnEq. Wt.mg/meqConc.meq/L (ion conc. / EW)Conc.mg/L asCaCO3Ca2+40.040.1220.051.99599.8*Mg2+10.024.3212.150.82341.2Na+11.823.0123.00.51025.7K+7.039.1139.10.1798.95HCO3-110.061.0161.01.80090.2SO42-67.296.1248.051.40069.9Cl-11.035.5135.50.03115.5


Environmental Engineering Sem 2 2011/2012*SOFTENINGCheck Solution(Cation)s = (Anion)s 175.6 = 175.6 Note: to within 10% mg/L as CaCO3

Total Hardness = of multivalent cations = (Ca2+) + (Mg2+) = 99.8 + 41.2 = 141 mg/L as CaCO3

Alkalinity = (HCO32-) + (CO32-) + (OH-) - (H+)Since pH = 7.2 (neutral pH, OH- & H+ are negligible)Alkalinity (HCO32-) = (1.80 x 10-3) eq/L Alkalinity = (1.80 x 10-3 eq/L)(50 g/eq)(1000 mg/g)= 90.1 mg/L as CaCO3

Carbonate Hardness (the portion of the hardness associated with carbonate or bicarbonate) Alkalinity = 90.1 mg/L as CaCO3 TH = 141 mg/L as CaCO3CH = 90.1 mg/L as CaCO3 (Note: if TH < Alk then CH = TH; and NCH = 0 )

Non-carbonate HardnessNCH = TH - CH = 141 - 90.1 =50.9 mg/L as CaCO3


Environmental Engineering Sem 2 2011/2012*SOFTENING

Softening can be accomplished by the lime soda process, ion exchange, nanofiltration and reverse osmosis. Lime soda softening is discussed in this chapter.

Lime-Soda Softening-it is possible to calculate the chemical doses necessary to remove hardness-hardness precipitation is based on the following two solubility reaction:To supply hydroxyl ions, the economic way is to buy LIME ( CaO) , then mix with water to produce Ca(OH)2 (hydrated lime)


Environmental Engineering Sem 2 2011/2012*SOFTENINGSoftening Reactions

The softening reactions are regulated by controlling the pH.FIRST any free acids are neutralizedTHEN- pH is raised to precipitate the CaCO3; if necessary - pH is raised further to remove Mg(OH)2FINALLY if necessary, CO32- is added to precipitate the noncarbonate hardness

123456Six important softening reactions are shown.

In each case, the chemical that has been added to the water is printed in bold type.

Designation (s) is for solid, hence indicates that the substance has been removed from the water.

In reality they occur simultaneously.


Environmental Engineering Sem 2 2011/2012*SOFTENINGMagnesium is MORE EXPENSIVE to remove than calcium, then we leave as much Mg 2+ as possible in the water.

The removal of noncarbonate hardness is MORE EXPENSIVE , because we must add the CO32- ( in term of SODA) therefore we leave as much noncarbonate hardness as possible in the water.

Traditionally, the final total hardness is set of 75 to 120 mg/L as CaCO3, however due to the economic constraints, many utilities will operate at total hardness of 140 -150 mg/L as CaCO3.

Concurrent removal of other speciesNatural Organic Matter (NOM)TurbidityOther metals LimitationBecause of the solubility of CaCO3 (s) and MgOH2 (s), ideal mixing is prohibited some hardness ions remain in solutioninsufficient time for reactionsMinimum Ca hardness ~ 30 mg/L as CaCO3Minimum Mg hardness ~ 10 mg/L as CaCO3


(1) NEUTRALIZATION OF CARBONIC ACID(H2CO3)Add limeTo neutralize any free acids ( acid carbonic) that may be present in the waterNO hardness is removed in this step CO2 + Ca(OH)2 CaCO3(s) + H2O

Environmental Engineering Sem 2 2011/2012*SOFTENING(2) PRECIPITATION OF CARBONATE HARDNESS DUE TO CALCIUM-add lime-pH must be raised up to 10.3 to percipitate the calcium carbonate-To achieve this pH , convert all of the bicarbonate to carbonate-The carbonate then serves as common ion for percipitation reactionCa2+ + 2HCO3- + Ca(OH)2 2CaCO3(s) + 2H2O


With the aid of chemical equation, show the formation of carbonic acid



Environmental Engineering Sem 2 2011/2012*SOFTENING(3) Precipitation of carbonate hardness due to magnesium

Must add more lime to achieve a pH about 11.The reaction may be considered to occur in two stages.First stage occurs when we convert all of the bicarbonate to carbonateMg2+ + 2HCO3- + Ca(OH)2 MgCO3+ CaCO3(s) + 2H2O

It is SOLUBLE.so the hardness of water did notCHANGE

Second stage- addition of more limeMg2+ + CO32- + Ca(OH)2 MgOH2(s)+ CaCO3(s) 4) Removal Of Noncarbonate Hardness Due To CalciumNO further increase in pH is required.Provide additional carbonate in the form of soda ashCa2+ + Na2CO3 CaCO3(s) + 2Na+


Environmental Engineering Sem 2 2011/2012*SOFTENING5) Removal Of Noncarbonate Hardness Due To Magnesium-First add lime

Mg2+ + Ca(OH)2 MgOH2(s)+ Ca2+

not much changes in hardness . We still HAVE Ca2+

-to eliminate Ca2+ , add soda

Ca2+ + Na2CO3 CaCO3(s) + 2Na+


Environmental Engineering Sem 2 2011/2012*SOFTENINGFlow diagram for solving softening problem ( all addition as mg/L as CaCO3 )


Environmental Engineering Sem 2 2011/2012*SOFTENINGExample

From the water analysis presented below, determine the amount of lime and soda (mg/L as CaCO3) to soften the water to 120 mg/L as CaCO3

.

Solution

1) Plot the bar chart as shown below

149.2 + 65.8185 + 29.82) From the bar chart , we note following

TH = 215 mg/L as CaCO3CH = 185 mg/L as CaCO3 HOW?? if TH > alk CH = AlkNCH = 30 mg/L as CaCO3 HOW?? NCH = TH -CH

Water composition (mg/L as CaCO3)Ca2+ = 149.2CO2 = 29.3HCO3-= 185.0Mg2+ = 65.8SO42-=29.8Na+ = 17.4Cl-= 17.6


Environmental Engineering Sem 2 2011/2012*SOFTENING3) Follow the logic of Figure in Slide 38, calculate the lime dose as follows

4) Calculate NCH left and NCH removed

NCH left = final desired hardness 40 NCH removed = NCH NCH left Therefore, NCH left = 120 40 = 80 mg/L

NCH removed = 30 80 = -50 ( negative is indicating there is no need to remove NCH, no SODA required )

StepDose (mg/L as CaCO3)Lime = CO229.3Lime = HCO3-185.0Lime = Mg2+ - 40 = 65.8 - 4025.8Lime = excess25.8TOTAL = 265.9


WHY COLLOIDS ARE SUSPENDED IN SOLUTION

HOW TO REMOVE COLLOIDS ????



Environmental Engineering Sem 2 2011/2012*COAGULATION and FLOCCULATIONSurface water contains organic and inorganic particles.

Particle such as clay, and colloids remain in suspension without aggregating for long periods of time. Consequently the particle cannot be removed by sedimentation in a reasonable amount of time.

Majority of ions in surface water consist of negatively charged particle/colloids which are stable in nature( stable = existing in ionized form) .

They repel other colloidal particles before they collide with one another. The colloids are continually involved in Brownian movement.


Environmental Engineering Sem 2 2011/2012*COAGULATION and FLOCCULATION

How to destabilize the particles??? Neutralize the charge by addition of an ion opposite to it ( Destabilization)Destabilization (Coagulation) ( refer slide 29)ParticlesFlocculation(Refer slide 30) Al(OH)3 (s) @ Fe(OH)3 (s)Settle down at the bottom of the flocculation tankAl 3+ / Fe3+Al 3+ / Fe3+



Coagulation process utilizes what is known as a chemical coagulant is mixed thoroughly with the water and various species of the positively charged particles adsorb to the negatively charged colloids such as colour, clay, turbidity and other particles through the processes of charge neutralisation to produce microfloc.

Once the charge is neutralized, the small suspended particles are capable of sticking together. The slightly larger particles, formed through this process and called microflocs, are not visible to the naked eye.

The water surrounding the newly formed microflocs should be clear. If it is not, all the particles' charges have not been neutralized, and coagulation has not been carried to completion. More coagulant may need to be added.

Microfloc itself is not yet settleable , then flocculation process takes place.



Flocculation is the process in which the destabilised particles are bound together by hydrogen bonding of Van der Waals forces to form larger particle flocs.

High molecular weight polymers, called coagulant aids, may be added during this step to help bridge, bind, and strengthen the floc, add weight, and increase settling rate. Once the floc has reached it optimum size and strength, the water is ready for the sedimentation process.



COAGULANT: is the substance (chemical) that is added to the water to destabilize particles and accomplish coagulation

PROPERTIES OF COAGULANTTrivalent cations

Nontoxic: obvious for the production of safe waterinexpensiveInsoluble in the neutral pH. The coagulant that is added must percipitate out of solution so that high concentration of the ion are not left in the water. Such precipitation greatly influenced the colloidal removal process


Types of coagulant commonly used


Coagulant typeexamplesInorganic metallic coagulantAluminium sulfate (Al2(SO4)314H2O, sodium aluminate, aluminium chloride, ferric sulfate and ferric chloridePrehydrolyzed metal saltsMade from alum and iron salts and hydroxide under controlled condition; polyaluminium chloride (PAC) Organic polymersCationic polymers, anionic polymers, and nonionic polymersNatural plant-based materialsOpuntia spp. And Moringa Oleifera (used in many parts of the world esp. developing country.


Environmental Engineering Sem 2 2011/2012*COAGULATION and FLOCCULATIONHow does alum works?

In sufficient alkalinity in the water

1Al2(SO4)314H2O + 6HCO3-2Al(OH) 3 3H2O(s) + 6CO2 +8H2O + 3SO42-1 mole of alum added uses 6 moles of alkalinity and produces 6 moles of CO2Alum sludge , settle in the flocculation tank The above reaction shifts the carbonate equilibrium and decreases the pH

However, as long as sufficient alkalinity is present and CO2 (g) is allowed to evolve, the pH is not drastically reduced and is generally not an operational problemHCO3- + H+ = H2CO3(Acid Carbonic)


Environmental Engineering Sem 2 2011/2012*COAGULATION and FLOCCULATIONExampleCalculate the amount of alum sludge produced and alkalinity (HCO3- ) consumed when 1 mg/L alum was used.

SolutionChemical reactionAl2(SO4)314H2O + 6HCO3- 2Al(OH)3(s) + 6CO2 + 8 H2O + 3SO42- + 14H2O

Molecular weight( MW)MWalum = 594 g/moleMWalkalinity= 61 g/moleMWalum sludge= 78 g/mole

Solid removed when 1 mg/L alum was used, 1 mg/L = 1.684 x 10-6 moles/L( 594 g/mole)(1000 mg/g)

Known that 1 mole/L alum yield 2 mole/L of alum sludge, thereforeSolid removed = 2 (1.684 x 10-6 moles/L) ( 78 g/mole)= 2.63 x 10-4 g/L= 0.26 mg/L

Alkalinity consumed when 1 mg/L alum was used,Known that 1 mole/L alum yield 6 mole/L of alkalinity, thereforeAlkalinity removed = 6 (1.684 x 10-6 moles/L) ( 61 g/mole)= 61.6 mg/L HCO3-Expressed in CaCO3= 61.6 mg/L HCO3- x EW CaCO3 EW HCO3- = 61.6 mg/L HCO3- x 50 g/eq 61 g/eq = 50 .5 mg/L HCO3- as CaCO3


Environmental Engineering Sem 2 2011/2012*COAGULATION and FLOCCULATIONExampleA 50 mg/L alum dose is used to coagulate a turbid surface water. Calculate the amount of the floc ( alum sludge) produces ( kg/d) if the flow is 0.04 m3/s.

Solution

= 156 ( 50 g/m3 )(0.04 m3/s) (24 x 3600 s/d) ( 1 kg/1000g) 594= 45.4 kg/d

From chemical equation: Al2(SO4)314H2O + 6HCO3- 2Al(OH)3(s) + 6CO2 + 8 H2O + 3SO42- + 14H2O


Why trivalent cations considered as good coagulant?




Therefore, coagulation and flocculation designed to removeMicroorganisms and colloids that caused turbidity Toxic compounds that are sorbed to particlesNOM (precursor of DBPs)JAR test (Figure 3.6) is a laboratory works to illustrate the coagulation and flocculation concepts associated to nature water. From this experiment the optimal pH, coagulant dose ,and coagulant aid could be determined.


Environmental Engineering Sem 2 2011/2012*COAGULATION and FLOCCULATIONExample

A typical test is conducted by first dosing each jar with the same alum dose and varying the pH in each jar. The result s are shown in below. Find the optimal pH, coagulant dose, and the theoretical amount of alkalinity that would be consumed at the optimal dose.


Environmental Engineering Sem 2 2011/2012*COAGULATION and FLOCCULATIONSolution

Conduct the second jar test with pH 6.0 for six beakers ( Why 6? Refer to the jar which has the lower turbidity in jar test 1). The results are shown belowConstruct the graph turbidity remaining vs alum Dose

From the graph, the optimal alum dosage was estimated to be 12.5 mg/L



theoretical amount of alkalinity that would be consumed at the optimal dose.

TRY YOURSELF !!!! REFER to SLIDE 50-51

Answer: 6.31 mg/L HCO3- as CaCO3



Effectiveness of Coagulation

The crux of efficient coagulation is the efficiency of MIXING the coagulant with the raw water.

Coagulation happens in two mechanisms;Adsorption/destabilization of the soluble hydrolysis species on the colloid and destabilization Sweep coagulation where the colloid is trapped in the hydroxide precipitate



Mixing equipment is need in coagulation. Why?To dispersion of the coagulant into the raw water. -Dispersion of the coagulant into water is called flash mixing or rapid mixing.-Rapid mixing aims to produce the high G.

Common alternatives for mixing when the mechanism of coagulation is adsorption/destabilization are;Diffusion mixing by pressured water jetsIn line mechanical mixingIn line static mixing

Common alternatives for mixing when the mechanism of coagulation is sweep coagulation are;Mechanical mixing in stirred tanksDiffusion by pipe gridHydraulic mixing



Efficiency of MIXING is depending on the 1) velocity gradient and 2) mixing time

Velocity gradient, G G , can be thought as the amount of shear taking place;

For coagulation, G must be higher enough. When chemical be added, the different G should e take into the consideration. Adsorption/destabilization: 3000 s-1 < G < 5000 s-1 , t = 0.5 s Sweep coagulation : 600 s-1 < G < 1000 s-1 , 1 s < t < 10 s

Different chemicals require different velocity gradients

Power of mixture imparted to water could be calculated;

Power = G2 V where, G = velocity gradient, s -1 V = volume in m3 = dynamic viscosity of water, Pa.s



Mixing Time

The time that a fluid remains in the reactor and affects the degree to which the reaction goes to completion.

In the ideal reactor, t = V/Q where , t = time ( in second) V = volume ( m3) Q = flowrate ( m3/s) Adsorption/destabilization: t = 0.5 s Sweep coagulation : 1 s < t < 10 s

note : real reactor do not behave as ideal reactor because of density difference due to temperature or other causes.



Example A rapid mixer is used for the dispersion of the coagulant to achieve the adsorption/destabilization reaction. If the water ( temperature 170 C) flows at 300 MLD, determine the volume of coagulation tank. Calculate the power. Given ; velocity gradient, G = 2000 s-1

SolutionFor the adsorption/destabilization reaction, t should be fixed as 0.5 s.Volume = Q t = ( 300 x 106 L) ( 0.5 s x d x hr x min ) d 24 hr 60 min 60 s = 1740 L x m3 1000L = 1.74 m3Refer to water properties, dynamic viscosity of water @170 C =1.081 x 10-3 Pa.s

Power = G2 V = (2000) 2 (1.74) (1.081 x 10-3) = 7524 WattNote: recalculate the power by assuming the temperature of water is 250 C. Give comment.



Effectiveness of Flocculation

The crux of efficient flocculation is the efficiency of MIXING to bring the particles into contact with one another so that they will collide, stick together and grow to a size that will readily settle. The mixing to to flocculate the coagulated water.

For flocculation, high enough to cause particles contact and keep the floc from settling but low enough to prevent the floc from tearing apart.

Mixing Time

For conventional treatment where settling follows flocculation, the flocculation times ranges from 20 -40 minutes.

Efficiency of MIXING is depending on the 1) velocity gradient and 2)mixing time.

Velocity gradient, G



Flocculation is normally accomplished with

1) paddle flocculator or 2) baffled chamber2) baffled chamberCoagulationFlocculationSedimentation



The Recommended Standards for Water Works Great Lakes Upper Missisippi River Board of State Sanitary Engineers ( GLUMRB) recommended the criteria for flocculation tank.

1. Inlet and outlet design shall prevent short circuiting and destruction of floc

2. Minimum flow-through velocity shall not be less than 2.5 to 7.5 mm/s with the detention time for floc formation of at least 30 min.

3. Agitators shall be drivenby variable speed drives with the peripheral speed of paddles ranging from 0.15 to 0.91 m/s

4. Flocculation and sedimentation basin shall be asa close together as possible


In flocculation tank, flow-through velocity is normally to 2.5 - 75 mm/s. Explain what will be happen if the velocity is less than 2.5 mm/s and more than 7.5 mm/s.

In designing a flocculation tank, which one of the following parameters take more priority: horizontal /flow through veloccity or detention time?

Flocculation and settling tanks should be as close as possible . Why?



Environmental Engineering Sem 2 2011/2012*SEDIMENTATION

Sedimentation basin = clarifier = settling tankWater flows into the settling basins, where the flow is almost devoid of turbulence.The water resides here for time periods ranging from 2 to 8 hours and flocculated particles settle out as sludgeThe sludge is mechanically removed periodicallySedimentation basin are usually rectangular or circular with either a radial or upward water flow pattern.

The key parameters and typical values in the design of settling tank are: - surface over flow rate 20- 35 m3/day/m2 - detention times 2-8 h - weir overflow rate 150 300 m3 /day/mEither rectangular or circular , normally 1 unit sedimentation system comprises of 2 tanks



The design can be divided into four zones: inlet, settling, outlet and sludge storage

Inlet zone - to evenly distribute the flow and suspended particles across the section of the settling zone*

Sludge storage zone- depends upon the method of cleaning, the frequency of cleaning and the quantity of sludge estimated to be produced.

Outlet zone- to remove the settled water from the basin with out carrying away any of the floc particles

Zones of sedimentation: (a) horizontal flow clarifier , (b) upflow clarifier**



Three classes of sedimentation

TypesDescriptionExampleWater treatmentWastewater treatmentI Settling as discrete particles at a constant settling velocityNo flocculation during sedimentation1. Pre-sedimentation 2. in filter bed after backwashing1. grit chamber IIParticles that aggregated or flocculate during sedimentationSedimentation after alum or iron coagulation1. Primary sedimentation2. In settling tanks after trickling filtration3. In upper portions of secondary clarifiers after activated sludge treatment



TypesDescriptionExampleWater treatmentWastewater treatmentIIIParticles settle as a zone or blanketUsually have a clear interface between the settling sludge and the clarified effluentsettling in lime soda ash sedimentation activated sludge sedimentation sludge thickeners


Environmental Engineering Sem 2 2011/2012*SEDIMENTATIONIdeal sedimentation basins (Type 1)- UPFLOW CLARIFIER

settling velocity, vs of the particle to be removed Velocity of water decreases as the water flows upward (overflow rate = vo, hydraulic surface loading)

Velocity of the particle remains unchangedIf vs vo, then 100% of particles remain in tankIf vs < vo, then 0% of particles remain in tank


Environmental Engineering Sem 2 2011/2012*SEDIMENTATIONExample The settling velocity of calcium carbonate floc formed during flocculation is 2.1 mm/s. If the detention time in the settling zone is 1.0 h and upflow rate is 1.75 GPM/sq ft, what is the minimum depth of water required to ensure removal of the floc by gravity settling. Given, I ft = 0.3048 m, 1.0 US gal =3.785L

Solution1.75 GPM/ ft2 = 1.18 mm/sV= 2.1 -1.18 mm/s = 0.92 mm/s

Velocity = depth/ hourDepth = (velocity )( hour) =( 0.92 mm/s) ( 1 h) = 3.3 m


Environmental Engineering Sem 2 2011/2012*SEDIMENTATIONExample Calculate the diameter and depth of a circular clarifier for a design flow of 3800 m3/d and an overflow rate of 0.00024 m/s and a detention time of 3 h. Calculate the weir loading rate by assuming the total effluent weir is 20 m.Solution Volume , V = Qt = (3800 m3/d) ( 3/24)= 475 m3

Q = 3800 m3/d = 0.044 m3/s

Surface overflow rate = Q/A 0.00024 m/s = 0.044 m3/s A m2Area, A = 183.3 m2

Volume, V = AD D=V/A= 475 m3/183.3 m2 = 2.59 mDiameter = 15.3 mWeir loading rate= Q/ Lw = 3800 m3/d 20 m= 190 m3/day.m ( OK!)



Ideal sedimentation basins (Type 1)- REGTANGULAR BASINParticle removal is dependent on the overflow rate, v0

In order for particle to be removed settling velocity , vs must be sufficient so that it reaches the bottom during the time the water resides in the tank (td).If Vs = Vo , then vs > v0 , 100 % particles should be easily removed vs = v0 , 100 % particles removed vs < v0 , some fraction of the particles will be removed P = 100 vs v0


Environmental Engineering Sem 2 2011/2012*SEDIMENTATIONExample A water treatment plant has a horizontal flow sedimentation tank with an overflow rate of 17 m3/d. m2 and wishes to remove particles that have settling velocities of 0.1 mm/s. What percentage of removal should be expected for each particle in an ideal sedimentation tank?

Solution vs = 0.1 mm/s

v0 = 17 m3/d. m2 = ? mm/s, ( do the conversion so, v0= 0.2 mm/s

Note that vs < v0 , P = 100 vs / vo = 100 ( 0.1)/(0.2) = 50%

Recalculate by considering v0 are 0.2 mm/s and 1 mm/s respectively.


Environmental Engineering Sem 2 2011/2012*SEDIMENTATIONExample

Determine the surface area of a sedimentation tank . The design flow is 0.044 m3 /s. Use a design overflow rate of 20 m / day. Find the depth of the sedimentation for the given overflow rate and detention time.

Solution:

Find the surface area.

First change the flow rate to compatible units.(0.044 m3/ s)(86,400 s / day) = 3801.6 m3 /dayThe surface area is = 3801.6 m3 /day 20 m/ day =190.m2

2) Find the length to width dimensionCommon length-to-width , L: W ratios ( 2:1 < L:W < 5:1 , and lengths seldom exceed 100 m. A minimum of two tanks is always provided.

use two tanks, each with a width of 5 m, a total surface area of 190 m2 ,

Length = 190 m2 /(2 tanks)(5 m wide) = 19 mmeet our length-to-width ratio of 3.8 : 1 ( OK!)Tank 1Tank 2


Environmental Engineering Sem 2 2011/2012*SEDIMENTATION3) Find the tank depth.

Rule of thumb that the detention time should be 2-8 h.Assumed the detention time of 120 min

Q = V/tV = Q t V = (1900.8 m3)(120 min)( d ) = 158 m3 d 1440 minDepth= 158 m3 =1.684 m,= 1.7 m 95 m2

The final design would then be two tanks, each having the following dimensions: 5 m wide x 19 m long x 1.7 m deep plus sludge storage depth.


Pilferage to steal (small amount)*Palatable pleasant, tasty, ediblePotable drinkable, clean, filtered****Occasional rare, infrequent, special*********************

chapter 3- water supply

Documents

water resources

water supply demand

sustainable water supply

cost of water treatment

raw water supply catchments

worst water crisis

high water losses

malaysias water status