Coagulation and Flocculation in Water Treatment

1. Coagulation and Flocculation in Water Treatment
Lecturer: Dr. Mahmoud Nasr
18th October, 2014

2. Introduction
Coagulation is the destabilization of colloids by addition of chemicals that neutralize the negative charges
The chemicals are known as coagulants, usually higher valence cationic salts (Al3+, Fe3+, Fe2+ etc.)
Flocculation is the agglomeration of destabilized particles into a large size particles known as flocs which can be effectively removed by sedimentation or flotation.

3. Scientific background
Colloid Stability
-Colloids have a net negative surface charge
-Electrostatic force prevents them from agglomeration
-Brownian motion keeps the colloids in suspension
-Difficult to remove colloids by conventional gravity settling
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Repulsion
Colloid
Colloid

4. Colloid Destabilization
-Colloids can be destabilized by charge neutralization
-Positively charges ions (e.g. Al3+and Fe3+) neutralize the colloidal negative charges and thus destabilize them
-After destabilization, colloids aggregate in size and start to settle
Coagulation
Flocculation
Sedimentation
Colloidal particles
Alum

5. Optimum coagulant dose
-Fill the jars with raw water sample 1L – usually 6 jars
-Adjust pH of all jars at 6.8
-Add different doses of the selected coagulant (alum or iron) to each jar (Coagulant dose: 5; 10; 15; 20; 30; 40 mg/L)
-Rapid mix each jar at 100 to 150 rpm for 1 minute. The rapid mix helps to disperse the coagulant throughout solution
-Reduce the stirring speed to 25 to 30 rpm for 15 to 20 mins
-Turn off the mixers and allow flocs to settle for 30 to 45 mins
-Then measure the final residual turbidity in each jar
-Plot residual turbidity against coagulant dose
Jar Test

6. Optimum pH
-Fill the jars with raw water sample 1L – usually 6 jars
-Add coagulant of all jars at 10 mg/L
-Adjust pH of the jars while mixing using H2SO4 or NaOH/lime (pH: 5.0; 5.5; 6.0; 6.5; 7.0; 7.5)
-Rapid mix each jar at 100 to 150 rpm for 1 minute. The rapid mix helps to disperse the coagulant throughout solution
-Reduce the stirring speed to 25 to 30 rpm for 15 to 20 mins
-Turn off the mixers and allow flocs to settle for 30 to 45 mins
-Then measure the final residual turbidity in each jar
-Plot residual turbidity against pH
Jar Test

7. Effect of coagulant dose on turbidity removal
-At low coagulant dosage, the electrophoretic mobility (EM) of the particles is still negative, and hence colloidally stable
-The optimum coagulation dosage corresponds with the condition where the EM is very close to zero
- At EM=0, charge neutralisation is responsible for the destabilisation of the particles.
- At slightly higher alum dosages, the EM becomes positive and the residual turbidity increases, indicating that charge reversal causes restabilisation of the particles
-At an overdose of the coagulants, additional precipitation mechanism might be occurred when the coagulants tend to form fairly thick layers around the particles. At this state, organic substances are removed by incorporation into or sorption onto hydroxide flocs. This phenomenon is known as "sweep-floc coagulation".
Operating conditions

8. Q. 1 In the following figure, illustrate the three zones (EM<0, EM=0, and EM>0), and the corresponding values of coagulant and turbidity

9. Effect of pH on coagulant solubility
-The isoelectric point (IP), is the pH at which a particular molecule carries no net electrical charge
-The IP of aluminium hydroxide is around pH 7-9
-Aluminium hydroxide possesses lowest charge, as well as lowest solubility at the IP.
-When alum is added to the wastewater, a series of soluble hydrolysis species are formed. These hydrolysis products possess a positive charge at pH below the IP of the aluminium hydroxide. Therefore, at pH<IP, charge neutralization is prevailing
-On the contrarily, negatively charged species are predominate at pH above the IP.
-Therefore, at pH>IP, particle removal is achieved mainly through adsorption and/or bridge formation rather than by charge neutralization.
Operating conditions

10. Q. 2 From the following figure, illustrate the two zones responsible for charge neutralization and sweep flocculation
Q. 3 What is the effect of pH on iron coagulant solubility

11. Scientific background
-Electrocoagulation (EC) is equivalent to electrolytic cells, where electrical energy is converted to chemical energy.
-In this technology a source of direct current (DC) is connected to a pair of electrodes (i.e. anode and cathode) immersed in a solution that serves as the electrolyte.
-When power is supplied, coagulating positive ions are generated from dissolution of the anode. Sacrificial metal anodes produce metal ions into solution according to Eq. 1. M(s) → Mn+(aq) + ne- (Eq. 1)
-The generated positive metal ions neutralized the negative surface charges of contaminants leading to larger agglomerates named "flocs".
Electrocoagulation

12. -At the cathode, hydrogen gas and hydroxide ions are generated according to the following equation (Eq. 2): Cathode reaction: 3H2O + 3e− → H2(g) + 3OH−(aq) (Eq. 2)
-The aluminium ion liberated from the anode (Eq. 1) may also react with the hydroxide ions produced at the cathode (Eq. 2) to form insoluble hydroxides (Eq. 3).
-The formed amorphous Al(OH)3(s) characterized by large surface areas, which are positive for a rapid adsorption of soluble organic compounds and trapping of colloidal pollutants. Al3+(aq) + 3H2O ↔ Al(OH)3(s) + 3H+(aq) (Eq. 3)

13. Q. 4 From the following figure, illustrate the two reasons for charge neutralization and sweep flocculation
Q. 5 What is the effect of hydrogen bubbles on turbidity removal

14. 1000
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z F
I t M
w
Q. 6 According to Faraday’s law of electrolysis (Eq. 4):
What is the effect of current intensity on AL+3 ions dissolution
What is the effect of electrolysis time on AL+3 ions dissolution
What is the effect of gap between electrodes on AL+3 ions dissolution
(Eq. 4)
Where:
w=aluminum dissolving (mg-Al);
I=current intensity (A);
t= time (s);
M= molecular weight of Al (=27 g/mol);
z=number of electrons involved in the reaction (equals to 3 for Al3+);
F= Faraday’s constant (96,485 C/mol);

15. V  I R
Q. 7 According to Ohm's law (Eq. 5):
What is the effect of resistance on voltage input
How to decrease the required electric energy (cost saving)
(Eq. 5)
Where:
V=Voltage (V)
I=current intensity (A);
R= resistance (Ohm);