Water Treatment: Coagulation and Flocculation 1
Surface Water Treatment Removal of turbidity rapid mix tank flocculation tanks settling (sedimentation) tanks 2
Rapid Mixing Used to blend chemicals and water being treated Retention time from 10-30 sec. Mechanical mixing using vertical-shaft impeller in tank with baffles (Source: Water Supply and Pollution Control, 5th ed. W. Viessman, Jr. and M.J. Hammer, Harper Collins College Publ. 1993) 3
Rapid Mixing 4
Goal: To alter the surface charge of the particles that contribute to color and turbidity so that the particles adhere to one another and are capable of settling by gravity 5
Colloids Small particles (0.001 to 1 µm) Usually negatively charged Particles repel so suspension is considered stable 6
Coagulation (process) Colloidal particles (0.001-1 µm) + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + floc (1-100 µm) 7
Coagulant Non-toxic and relatively inexpensive Insoluble in neutral ph range - do not want high concentrations of metals left in treated water Concentration 8
Coagulants Alum: Al 2 (SO 4 ). 3 14H2 O Ferric chloride: FeCl 3 Ferric sulfate: FeSO 4 9 Polyelectrolytes
How does alum work? Al 2 (SO 4 ) 3 14H 2 O 2Al 3+ + 3SO 4 2- + 14H 2 O 2Al 3+ + colloids neutralize surface charge 2Al 3+ + 6HCO 3-2Al(OH) 3 (s) + 6CO 2 If insufficient bicarbonate is available: Al 2 (SO 4 ) 3 14H 2 O 2Al(OH) 3 (s) + 3H 2 SO 4 - + 14H 2 O Optimum ph: 5.5 to 6.5 Operating ph: 5 to 8 10
Flocculation Paddle units rotate slowly, usually <1 rpm Velocity of water: 0.2-0.5 m/s Detention time of at least 20 min (Source: Water Supply and Pollution Control, 5th ed. W. Viessman, Jr. and M.J. Hammer, Harper Collins 11 College Publ. 1993)
Flocculation 12
Why colloids are suspended in solution and can t be removed by sedimentation and filtration??? Answer :The particles in colloid range are too small to settle in a reasonable time period and too small to be trapped in the pores of the filter How to remove colloids? Answer: Coagulation and Flocculation 13
1. Surface water contains organic and inorganic particles. 2. 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. 3. Majority of ions in surface water consist of negatively charged particle/colloids which are stable in nature( stable = existing in ionized form). 4. They repel other colloidal particles before they collide with one another. T he colloids are continually involved in Brownian movement. 14
How to destabilize the particles??? Neutralize the charge by addition of an ion opposite to it (Destabilization) Particles (- ve charge) Destabilization (Coagulation) ( Refer slide 29) Flocculation (Refer slide 30) Al 3+ / Fe 3+ Al 3+ / Fe 3+ Al(OH) 3 (s) @ Fe(OH) 3 (s) Settle down at the bottom of the flocculation tank 15
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 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. 16
Flocculation is the process in which the destabilised particles are bound together by hydrogen bonding of Van der Waal s 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. 17
COAGULANT : is the substance (chemical) that is added to the water to destabilize particles and accomplish coagulation PROPERTIES OF COAGULANT Trivalent cations Nontoxic: obvious for the production of safe water inexpensive Insoluble 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 18
Table 1 : Types of coagulant commonly used Coagulant type Inorganic metallic coagulant Prehydrolyzed metal salts Organic polymers Natural plant-based materials examples Aluminium sulfate (Al 2 (SO 4 ) 3 14H 2 O, sodium aluminate, aluminium chloride, ferric sulfate and ferric chloride Made from alum and iron salts and hydroxide under controlled condition; polyaluminium chloride (PAC) Cationic polymers, anionic polymers, and nonionic polymers Opuntiaspp. And MoringaOleifera (used in many parts of the world esp. developing country. 19
How does alum works? In sufficient alkalinity in the water Alum sludge=alum floc =dry sludge (without water, H 2 O=Al(OH) 3 ), settle in the flocculation tank 1Al 2 (SO 4 ) 3 14H 2 O + 6HCO 3-2Al(OH) 3 (s) + 6CO 2 +14H 2 O + 3SO 4 2-1 mole of alum added uses 6 moles of alkalinity and produces 6 moles of CO2 The above reaction shifts the carbonate equilibrium and decreases the ph However, as long as sufficient alkalinity is present and CO 2 (g) is allowed to evolve, the ph is not drastically reduced and is generally not an operational problem 20
Example Calculate the amount of alum sludge produced and alkalinity (HCO 3- ) consumed when 1 mg/l alum was used. Solution Chemical reaction Al 2 (SO 4 ) 3 14H 2 O + 6HCO 3-2Al(OH) 3 (s) + 6CO 2 + 3SO 4 2- + 14H 2 O Molecular weight( MW) MW alum = 594 g/mole (Al 2 (SO 4 ) 3 14H 2 O) MW alkalinity = 61 g/mole (HCO 3- ) MW alum sludge = 78 g/mole (Al(OH) 3 ) 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) Unit conversion mg/l mole/l Known that 1 mole/l alum yield 2 mole/l of alum sludge, therefore Solid removed = 2 (1.684 x 10-6 moles/l) ( 78 g/mole) = 2.6262 x 10-4 g/l = 0.263 mg/l 21
(Cont.) Alkalinity consumed when 1 mg/l alum was used, Known that 1 mole/l alum yield 6 mole/l of alkalinity, therefore Alkalinity removed = 6 (1.684 x 10-6 moles/l) ( 61 g/mole) Expressed in CaCO 3 = 0.616 mg/l HCO 3 - = 0.616 mg/l HCO 3 - x EW CaCO 3 EW HCO 3 - = 0.616 mg/l HCO 3 - x 50 g/eq 61 g/eq = 0.505 mg/l HCO 3- as CaCO 3 3 Unit conversion mole/l mg/l 22
Example Calculate the amount of alum sludge produced and alkalinity (HCO 3- ) consumed when 1 mg/l alum was used. Solution Chemical reaction Al 2 (SO 4 ) 3 14H 2 O + 6HCO 3-2Al(OH) 3 (s) + 6CO 2 + 8 H 2 O + 3SO 4 2- + 14H 2 O Molecular weight( MW) MW alum MW alkalinity MW alum sludge = 594 g/mole = 61 g/mole = 78 g/mole 23
(Cont..) 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, Therefore; Solids removed = 2 (1.684 x 10-6 moles/l) ( 78 g/mole) = 2.62 x 10-4 g/l = 0.26 mg/l 24
Alkalinity consumed when 1 mg/l alum was used, Known that 1 mole/l alum yield 6 mole/l of alkalinity, therefore Alkalinity removed = 6(1.684 x 10-6 moles/l)(61 g/mole) = 0.616 mg/l HCO 3 - Expressed in CaCO 3 = 0.616 mg/l HCO 3 - x EW CaCO 3 EW HCO 3 - =0. 616 mg/l HCO 3 - x 50 g/eq 61 g/eq = 0.50 mg/l HCO 3- as CaCO 3 25
Example (Assignment) A treatment plant with an average flow of 0.044 m 3 /s treats its water with alum Al 2 (SO 4 ) 3 14H 2 O at dose of 25 mg/l. Alum coagulation is used to remove particulate matter, reduce the concentration of organic matter and reduce the alkalinity of water according to equation; Al 2 (SO 4 ) 3 14H 2 O 2Al 3+ + 3SO 4 2- + 14H 2 O If the organic matter concentration is reduced from 8 mg/l to 3 mg/l, determine the total mass of alkalinity consumed and the total mass of dry solids removed per day. 26
Why trivalent cations considered as good coagulant? Answer: i) Cations such as Fe 3+ and Al 3+ has a higher molecular weight, easy to settle. i) It removes turbidity effectively even a small dose was used. 27
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. Therefore, coagulation and flocculation designed to remove Microorganisms and colloids that caused turbidity Toxic compounds that are sorbed to particles Inorganic materials 28
Example A typical test is conducted by first dosing each jar with the same alum dose and varying the ph in each jar. The results are shown in below. Find the optimal ph, coagulant dose, and the theoretical amount of alkalinity that would be consumed at the optimal dose. 29
Solution 1) 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 below 2) Construct the graph turbidity remaining vs alum dose 3) From the graph, the optimal alum dosage was estimated to be 12.5 mg/l 30
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4) Calculate the theoretical amount of alkalinity that would be consumed at the optimal dose. TRY YOURSELF!!!! Answer: 6.31 mg/l HCO 3- as CaCO 3 32
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