1. Water Treatment Part 3 Groundwater Treatment
Dr. Abdel Fattah Hasan

2. Groundwater (GW) are usually:
Cool and uncontaminated
Has uniform quality
Usually used directly for municipal use (just chlorine is added to avoid post contamination)
Sometimes GW is polluted or contaminated with:
Hardness
Fertilizers WW
Pesticides
Radionuclides
Toxic metals, such as Arsenic

3. GW Treatment Options

4. Hardness Removal - Precipitation softening
Hardness of water is caused by divalent cations, such as Ca & Mg ions
Max. hardness for public supply: mg/l as CaCO3
Moderate hardness for public supply: mg/l as CaCO3
Precipitation softening uses lime CaO and soda ash Na2CO3 to remove Ca and Mg
Lime slurries are usually has the form Ca(OH)2
Lime treatment has the incidental benefits of bacterial actions, removal of iron and aid in clarification of turbid surface water
Carbon dioxide can be applied after lime treatment to lower pH by converting the excess hydroxide ion and carbonate ion to bicarbonate ion

5. Converting Ca and Mg into mg/l CaCO3
Ca Hardness as mg/l CaCO3 = Ca (meq/l) X 50
Mg Hardness as mg/l CaCO3 = Mg (meq/l) X 50

6. Chemical reactions in precipitation softening
1- Lime added to water reacts first with any available CO2
CO2 + Ca(OH)2 = CaCO3 + H2O
Ca(HCO3)2 + Ca(OH)2 = 2 CaCO3 + 2H2O
Mg(HCO3)2 + Ca(OH)2 = CaCO3 + MgCO3+ 2H2O
MgCO3 + Ca(OH)2 = CaCO3 + Mg(OH)2
Mg(HCO3)2 + 2Ca(OH)2 = 2 CaCO3 + Mg(OH)2 + 2H2O
MgSO4 + Ca(OH)2 = CaSO4 + Mg(OH)2
CaSO4 + Na2CO3 = CaCO3+ Na2SO4
1 eq to one eq
2- Then Lime reacts with any calcium bicarbonate present in water
3- Then Lime reacts with magnesium bicarbonate
2eq of lime to one eq Mg(HCO3)2
4- Non-carbonate Ca (sulfate or chloride) require addition of soda ash and non-carbonate Mg (sulfate or chloride) require both lime and soda ash
1 eq to one eq
1 eq to one eq
Note: Ca ion can be effectively removed by lime addition (pH = 10.3), but Mg ion demand higher pH, so lime should be added in excess of about (35 mg/l; 1.25 meq/l)

7. Re-carbonation
Used to stabilize excess lime of treated water by adding CO2:
Ca(OH)2 + CO2 = CaCO3 + H2O
CaCO3 + CO2 + H2O = Ca(HCO3)2

8. Excess lime softening
Calcium can effectively reduced by lime addition, but magnesium removal need excess lime
Lime and Soda ash dosages to be estimated by chemical equations PLUS excess lime for Mg removal
Practical limits (remains after estimation of theoretical dosages from chemical equations) for hardness removal are:
- CaCO3: 30 mg/l as CaCO3 (= 0.6 meq/l Ca++)
Mg(OH)2 :10 mg/l as CaCO3 (= 0.2 meq/l Mg++)
Note: Sodium (Na) concentration is usually increased by the amount added in the Soda ash

9. Excess Lime Softening

10. Selective Calcium Carbonate Removal
Used to soften water with low Mg hardness (less than 40 mg/l as CaCO3)
Enough lime is added to remove Ca without Excess.
Soda ash mayor may not be required, depending on the contents of non-carbonate hardness.
Recarbonation is usually performed to reduce scale formation.

11. Selective Calcium Carbonate Process

12. Split-Treatment Softening
By dividing the raw water into two portions for softening in a two stage system
QP MgR QR
MgR QE
MgE
Split treatment can result in chemical savings
Recarbonation my not be required
Split around 1st stage is determined by the level of Mg desired in treated water
Mg in treated water = (QP X MgR + QE X 10)/QR

13. Example:
Water defined by the following analysis is to be softened by excess lime treatment. Assume that the practical limit of hardness removal for CaCO3 is 30 mg/l and of Mg(OH)2 is 10 mg/l as CaCO3
CO2= 8.8 mg/l
Ca++ = 40mg/l
Mg++ =14.7mg/l
Na+ = 13.7mg/l
Alk (HCO3-) =135 mg/l as CaCO3
SO4= 29mg/l
Cl- = 17.8mg/l
(a) Sketch a meg/l bar graph and list the hypothetical combination of chemical compounds in solution
(b) Calculate the softening chemicals required, expressing lime dosage as CaO and soda ash as Na2CO
(d) Draw a bar graph for softened water before and after recarbonation. Assume that half the alkalinity in the softened water is in the bicarbonate form.

14. Calcium hardness = 2X 50 = 100 mg/l as CaCO3
component
mg/l EW
Meq/l
CO2
8.8 22
0.4
Ca+ 40 20 2
Mg++
14.7l
12.2
1.21
Na+
13.7 23
0.6
Alk (HCO3-)
135 mg/l as CaCO3 50
2.7
SO4= 29 48
Cl-
17.8 35
0.51
Calcium hardness = 2X 50 = 100 mg/l as CaCO3
Magnesium hardness = 1.2X 50= mg/l as CaCO3
component
Meq/l
Lime
Soda Ash
CO2
0.4
Ca(HCO3)2 2
Mg(HCO3)2
0.7
1.4
MgSO4
0.51
4.31
Required lime dosage = 4.31 X = 156
Dosage of Soda ash = 0.51* 53= 27 mg/l Na2CO3

15. (A) (B) (C) Na+ Mg++ Ca++ CO2 C1- SO4= HCO3- 0.4 Ca++ Na + Mg ++ Ca++
3.21
0.0
2.00
3.81
Na+
Mg++
Ca++
CO2
C1-
SO4=
HCO3-
0.4
(A)
0.0
2.70
3.30
3.81
0.0
0.6
0.8
1.91
Ca++
Na +
Mg ++
Ca++
OH-
C1-
SO4
CO3=
OH-
(B)
1.25 of excess lime
0.0
0.2
0.8
1.40
1.91
0.0
0.6
0.8
1.91
Na+
Mg++
Ca++
C1-
SO4=
CO3=
HCO3-
(C)
0.0
0.4
0.8
1.40
1.91

16. Iron and manganese removal
Fe++ and Mn++ soluble in groundwater exposed to air these reduced to insoluble Fe+++ and Mn++++
Rate of oxidation depend on pH, alkalinity, organic content and present of oxidizing agents
Filtration – sedimentation and filtration
Fe++ ( ferrous) + oxygen Fe Ox ( ferric oxidizes)
Manganese can not oxidized as easily as iron need to increase pH
ِaeration –chemical oxidation – sedimentation- filtration
Fe++ + Mn++ + oxygen FeOx + MnO2 ( ferric oxidizes)
Free chlorine residual

17. Iron and manganese removal
Fe (HCO3)2 + KMn O4 Fe (OH) 3 + Mn O2
Mn(HCO3)2 + KMnO4 MnO2
Potassium permanganate
Potassium permanganate

18. Iron and manganese removal
Manganese zeolite process
Figure 7-20

19. Figure 7-20
Aeration optional
Well water
Anthracite medium
Dry KMnO4
Manganese treated greensand
Dissolving tank and solution feeder
Under drain
Pressure filter
Finished water

20. Water Stabilization
Ferrous metal when placed in contact with water results in an electric current caused by the reaction between the metal surfaces and existing chemicals in water
Fe Fe++ + 2electron
2 elec + H2O + ½ O2 OH-
2Fe++ + 5H2O + ½ O2 Fe(OH)3 + 4 H+
To Protect ductile iron pipe against internal corrosion is by lining with thin layer of cement mortar placed during manufacturing

21. Ion- exchange softening and nitrate removal
Ions of a particular species in solution are replaced by ions of a different species attached to an insoluble resin

22. Ion Exchanger

23. Cation exchange softening
Mg ++
CaR
MgR
+ Na+
In Process of Removal
+ Na2R
CaR
MgR
Ca ++
Mg ++
excess
Regeneration
+ NaCl
Na2 R +
NaCl

24. Anion exchange for Nitrate Removal
So =4
NO-3
RSO4
RNO3
RCl +
+ Cl -
Regeneration
with NaCl
Disadvantages : high operating costs and problem
of brine disposal

25. Removal of dissolved salts
Distillation : (desalination of sea water)
heating sea water (35000 mg/l mostly NaCl) to boiling point and converting it into steam to form water vapor that is condensed yielding salt free water

26. Removal of dissolved salts
Reserves osmosis
Forced passage of the natural osmotic pressure to accomplish separation of water and ions

27. Semi permeable Reverse Osmosis (b) (c) (a)
Membrane
P> P0 Po
Saline
water
Fresh
water
Osmosis
Reverse
osmosis
Osmosis
equilibrium
(b)
(c)
(a)

29. Reverse osmosis system
Pretreatment unit
Pump to provide high pressure
Post-treatment
Brine disposal

30. Permeate (product water)
Reverse osmosis models
Alkali
Saline water
Permeate (product water)
Scale inhibitor
Waste brine
Pump
Granular-media filter
Acid
10-30% of saline feed
Cartridge filter

31. Source of wastes in water treatment
Residue from chemical coagulation
Precipitation from softening
Filter back wash
Settled solid from pre-sedimentation
Total Sludge Solids produced in WT (lb/mil gal) =
8.34 x [0.44 x alum dosage (mg/l) x Turbidity (NTU)]

32. Example 7-16
A surface water treatment plant coagulant a raw water having a turbidity of 9 units by applying an alum dosage of 30 mg/l.
Estimate the total sludge solids production in pounds per million gallons of water processed.
Compute the volume of sludge from the settling basin and filter backwash water using 1% solid concentration in the sludge and 500 mg/l of solids in the waste water. Assume 30% of total solids are removed in the filter.

33. Solids in sludge = 0.70 X 166 = 116 lb/ mil gal
Applying Eq. 7-39
Total sludge solids =
8.34 (0.44 X X 9)= 166 lb/ mil gal
Solids in sludge = 0.70 X 166 = 116 lb/ mil gal
Solids in backwash water = 0.30 X 166
= 50 lb/ mil gal
Volume =
Sludge solids (lb)
Solids fraction X 8.34 (lb/ gal)
116
Sludge volume =
= 1390 gal/mil gal
1.0
100
X 8.34
Wash- water volume =
500
1,000,000
X 8.34
= 12,000 gal/mil gal

34. PRECIPITATE PRODUCED COMPONENT IN WATER Formula MEQ/L APPLICABLE
EQUATION
CaCO3
Mg (OH)2
CO2
0.40
7-15
CA(HCO3)2
2.00
7-16
4.00
MG(HCO3)2
0.70
7-17
1.40
MGSO4
0.51
7-18 & 7-19
EXCESS LIME
1.25
7-20
7.56
1.21
Minus the practical limits (solubility)
- 0.60
6.96
- 0.20
1.01
Resident = CaCO3 + Mg(OH)2
= 6.96 X X 29.2 = 378 mg/ l

35. Dewatering and waste disposal of wastes from water treatment plants
Lagoons
Drying beds
Gravity thickening
Centrifugation
Pressure filter

36. Gravity thickening Sludge in flow Inlet baffle Weir Pickets
Supernatant
overflow
Weir
Pickets
Scraper blades
Sludge discharge

37. Filtration pressure

38. Centrifugation

39. Example
Case 1: Groundwater source with small infrequent possible contamination used for domestic use
Solution : chlorination or ozonation or filtration

40. Example 2
Surface water: floating matter, high suspended matter, high turbidity, considerable, biological contamination, clay
Solution: screening, sedimentation, coagulation, flocculation, filtration, chlorination