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LESSON PLAN CEE 437
Topic Water treatment –
Clarification process
(Part I)
Learning Objectives :
- Understand the general process of water treatment facilities
- Understand the fundamentals of coagulation and flocculation
- Be able to design coagulation and flocculation tanks
Reading material:
Chapter 2 on coagulation
and flocculation,
Lecture note
Contaminants to be removed
Different treatment configura0ons
Unit processes involved
Coagula3on -‐ par3cle destabiliza3on
Floccula3on -‐ agglomera3on; small par3cle →
bigger par3cles
Sedimenta3on -‐ removal of large par3cles by
gravity
Filtra3on -‐ removal of small par3cles that don’t
seQle
http://il.youtube.com/watch?v=D4_UTjRHee
Physical property of par0cles
Figure 11.
Coagula0on
Coagulants (coagulant aids)
Most Common: hydrolyzed metal coagulants:
- Alum: Al 2 (SO 4 ) 3 *(H 2 O) 14 (5 – 50 mg/L; pH 5.5-‐8)
- Fe(OH) 3
- FeCl 3
- Fe 2 (SO 4 ) 3
- (pH 4-‐9; broader than alum; beQer performance, but more $$)
- Mechanism intermediate between simple ions and ca3onic polymers
- Form +-‐charged hydrolyzed species (e.g., AlOH2+, Al(OH) 2 +)
- Highly charged, soluble hydrolysis products compress charged layer, reducing repulsive forces
- Large mul3nuclear hydrolyzed metal species sorb onto colloids
- Reduce effec3ve surface charge (so less repulsion)
- Create bridges between par3cles (help bring par3cles together)
- Synthe3c ca3onic polymers are also some3me added as coagulants or as coagulant aids - Work by bridging mechanism - + charges adsorb to surface of 2 par3cles, and bring together
How much coagulant/flocculants to
add?
• Experimental
– Jar test
• Theore3cal
– Dissocia3on equilibrium
Coagulant Added No Coagulant
Jar test
1. 100 rpm for 1 minutes, rapid mix
2. 20-70 rpm, 10-30 min,
flocculation
3. 0 rpm, sedimentation
http://il.youtube.com/watch?v=D4_UTjRHee
Example 3-‐
Different pH
Alum = 10 mg/L
NTU
Different Alum conc.
pH = 6.
NTU
pH can be controlled by adding lime (CaO), calcium hydroxide (Ca(OH) 2 ), sodium hydroxide (NaOH), sodium carbonate (NaCO 3 ) or soda ash.
Normalized Form Logarithmic form
[Al3+] = 109.7[H+]^3 log[Al3+] = 9.7 - 3pH
log[AlOH 2 +] = 4.7 - 2pH
log[Al(OH) 2 +] = -0.4 - pH
log[Al 2 (OH) 2 4+]= (13.4 - 4pH)/log
log[Al 6 (OH) 15 3+] = (12-3pH)/log
log[Al8(OH) 20 4+] = (9.5-4pH)/log
log[Al(OH) 4 - ] = -13.3+pH
2 5 3 5 9.^733104. 7 [ ] 2
[ ]
1010 [ ][ ]
[ ]
[ ]^10 [^ ] +
− + +
- −^ + AlOH = (^) H^ Al = HAl^ H = H
10 [ ]
[ ]
[ ( ) ]^10 [^ ]^0.^4
2
- 1 3 (^2) + −^ +
= = H
H
Al OH^ Al
[ Al 2 ( OH ) 24 +^ ] = (^10
− 6. 3 )[ Al 3 + ] 2
[ H +^ ]^2
= 1013.^4 [ H +^ ]^4 / 2
[ Al 6 ( OH ) 135 +^ ] = (^10
− 47 )[ Al 3 + ] 6
[ H +^ ]^15
= 1012.^0 [ H +^ ]^3 / 6
[ Al 8 ( OH ) 240 +^ ] = (^10
− 69 )[ Al 3 + ] 8
[ H +^ ]^20
= 109.^5 [ H +^ ]^4 / 8
[ ]
[ ]
[ ( ) ]^10 [^ ]^13.^3
4 23 3 (^4) + −
− −^ +
H H
Al OH^ Al
Solubility is best described using logarithmic Conc. vs. pH plots
Group 1, 6
Group 2, 7
Group 3,
Group 4,
Group 5,
Axis Title Axis Title Al3+ AlOH2+ Al(OH)2+ Al2(OH)2 4+ Al6(OH)15 3+ Al8(OH)20 4+ Al(OH)4-‐
Minimum solubility of Al
Al ( OH ) 2 +^ = Al ( OH )^ − 4 , 10 −^0.^4 [ H +^ ] = 10 − 13. 3 [ H +^ ]
[ H +^ ]^2 = 10
− 13. 3 10 −^0.^4 = 10 −^12.^9 ,[ H +^ ] = 10 −^6.^45 − − > pH = 6. 45 CT , Al = 2 Al ( OH ) 2 +^ = 2 Al ( OH ) 4 −, CT , Al = 2 x 10 −^0.^4 [ H +^ ] = 2 x 10 −^6.^85 = 10 −^6.^55 = CT^ m ,i A n l = 7. 6 mg / LAl
Addi0on of coagulants based on pC-‐pH diagram (what are the differences to the last slides?) Hi pH due to water softening at Lincoln plant. Figure 11. Clarifica0on Figure 10.
Calcula3on
Effective tank diameter (for square or rectangular)
Types of Floccula0on Tanks
Sample Mechanical Flocculator
