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Ion Exchange (3rd DC 197; 4th DC 255) Definition
- the reversible interchange of ions between a solid and a liquid phase in which there is no permanent change in the structure of the solid. Types of Ion Exchange Resins
1) Cation Resin – remove cations 2) Anion Resin – remove anions
Type Remove Example Regenerant ----------------------------------------------------------------------------------------------------------------------------------------------------------Cation Resin Strong Acid Salts of strong Ca2+ H2SO4 base, weak base NH4+ HCl - Nonselective ..........................................……………………………............... Weak Acid Salts of strong Ca2+, Mg2+ H2SO4 base Na+ HCl - Selective no NH4+ Type Remove Example Regnerant ----------------------------------------------------------------------------------------------------------------------------------------------------------Anion Resin Strong Base Salts of strong SO42-, ClNaOH acids, weak acid HCO3-, - non selective NO3......……..................................................... Weak Base Salts of strong SO42NaOH, acid Cl- ,NO3- selective ---------------------------------------------------------------------------------------------------------------------------------------------------------
Applications of Ion Exchange Processes
a. Water Softning (Ca 2+, Mg2+) b. P-removal (PO 43-) c. N-removal (NH4+, NO3−) d. Demineralization (Ca2+, Mg2+, Fe3+, Mn2+, Na+) e. Radionuclides removal (Sr-85, Cs-134, I-131)
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Types of ion exchange material
a. b.
Natural Synthetic
Natural ion exchange material: 1) Zeolite – Naturally occurring clays, used for water softening to remove remove Ca2+ and Mg2+. Na2SO4 + Al 2(SO4)3 -----> precipitate ---------> Na 2O Al2O3 heat
Ion exchange capacity of Zeolite: 20 kilograins of hardness as CaCO3 -------------------------------------------------ft3 zeolite
Units: 1 grain / gal = 17.1 mg / L 20 kgr ft3 103 gr --------- ------------ ---------- = 2673.8 gr/gal ft3 7.48 gal kgr 2673.8 gr 17.1 mg/L ----------------------------- ------------------------------- = 45,722 mg/L gal gr /gal 45722 mg kg 103 L -------------- ----------- ---------- = 45.7 kg/m 3 L 106 mg m3
2) Sulfonated coal Synthetic (Commercial) ion exchange resin Characteristics: - cross linked polystyrene - very rigid - porous for good hydraulic characteristics - higher exchange capacity high melting point require less regenerant (e.g., NaCl) more expensive ($)
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Ion Exchange Equilibria
a. Ion exchange reaction: H –R
+ Na+
↔
Na –R + H+
where R = ion exchange resin H –R = Hydrogen resin (Strong acid resin) Na –R = Sodium resin {Na –R} {H+} Keq = ------------------{H –R} {Na+} where Keq = selectivity coefficient; { } = activity b. Regeneration Reaction Na –R
+
H+ Cl−
H –R
+
Na+
Cl−
Note: Equilibrium: aA + bB <===> cC c
+ dD
d
{C} {D} K = ----------------------------a b {A} {B}
Note: - We want the selectivity coefficient to to be large, but not too large. If it's too large, then regeneration will be difficult. Water Softning by Ion Exchange
1. General chemical reactions with zeolites - Zeolites are used in the ion exchange process for water softening to adsorb calcium and magnesium. The calcium and magnesium ions are exchanged for sodium ions.
(2Na)-Z + Ca2+
Ca-Z + 2 Na+
(2Na)-Z + Mg2+
Mg-Z + 2 Na+
Z = ion exchange resin such as zeolites - Other ions may be substituted for calcium and magnesium in the reaction.
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If calcium or magnesium is in the form of bicarbonate or sulfate, the equation may be written in molecular form as: (2Na)-Z + Ca(HCO3)2 (2Na)-Z + Mg(HCO3)2
(2Na)-Z + CaSO4 (2Na)-Z + MgSO4
Ca-Z + 2 Na(HCO3) Mg-Z + 2 Na(HCO 3) Ca-Z + Mg-Z +
Na2 SO4 Na2 SO4
Regeneration After the zeolite bed is exhausted it may be regenerated with sodium chloride according to the equation: Ca-Z + 2NaCl
(2Na)-Z + Ca2+ + 2 Cl−
Mg-Z + 2NaCl
(2Na)-Z + Mg2+ + 2 Cl−
The ion-exchange material can now be used to remove more hardness. • Ca2+ and Mg2+ is a water stream that must be disposed of. Advantages and Disadvantages (Water softening using Ion Exchange)
Advantages 1) Easy to operate - Properly designed systems are relatively easy to operate
2) Near zero hardness - Near zero hardness levels can be achieved in most cases
3) Regeneration with common salt - Regeneration of ion resin with common salt can be readily accomplished
Disadvantages 1) No reduction of total solids or alkalinity - Reduction in alkalinity or total solids cannot be achieved
2) Waste brine disposal problems - Waste regenerant brine may cause disposal problems, if contaminated.
3) Health problem - Relatively high sodium level in treated water may cause health problem.
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General Rules
a. Effect of Charge Charge on Exchange Potential •
The higher the charge, the greater the attraction attraction to to the resin (easier (easier to remove) - but tough to get off removed ions during regeneration.
Higher affinity Lower affinity Easier to remove Harder to remove <-----------------------------------------------------------------> Examples: PO43– SO42– Cl− Th4+ Fe3+ Ca2+ Na+ Th4+ Al3+ Ca2+ Na+ b. Effect of Hydrated Radius on Exchange Potential •
Larger hydrated radius, easier to remove.
- The Extended Debye-Hückel Equation: -
The effective hydrated radius of the ion, a is the radius of the ion and its closely bound water molecules. Large ions and and less highly charged ions bind water less tightly and have smaller hydrated hydrated radii than smaller, more highly charged ions.
c. High ion concentration •
High ion concentration can reduces or reverse the effect of charge and the effective hydrated radius of the ion - This gives clues to the best flow scheme (including regeneration)
d. High temperature - High temperature reduces the effect of atomic number Notes: 1) Exchange potential may be approximated by their activity coefficients 2) Organic ions and m etalic complexes (e.g., MnO 3-) have very high exchange exchange potentials
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Breakthrough curve (continuous flow-through run) rd
th
- see Fig 3-21 (3 DC 199); Fig 4-23 (4 DC 256)
Influent hardness
Water, Cin 3
Exhausted resin
Fresh resin
O C a C s a L / g m , s s e n d r a H
Effluent hardness
Due to saturation of ion exchange material
Time, hr Treated water Cout
Breakthrough Breakthrough time tim e
Breakthrough time • • •
time to breakthrough, time to saturate (exhaust) the exchange resin the length of time between regeneration cycle
Total Capacity of the resin Breakthrough time = ------------------------------------------------------------------------------Mass of ions removed / time In the water softening problems, ions = hardness Example 3-19 (3rd DC 199); Example 4-19 (4th DC 256) A home water softener has 0.1 m 3 of ion-exchange resin with an exchange capacity of 57 kg /m 3 (i.e., 57 kg of hardness as CaCO 3 per m3 of resin volume). The occupants use 2,000 L of water per day. The water contains 280 mg/L of mg/L of hardness as CaCO3 and it is desired to soften it to 85 mg/L as mg/L as CaCO3. In this example, it is assumed all (100%) hardness in the water which passes through the ion exchange column is removed.
(1) How much water should be bypassed? (2) What is the time between regeneration cycle “Breakthrough time”?
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(Solution) C = Concentration Q = flow rate
Q, Cin
Mass loading rate = CQ
(Q-Qb), Cin Qb Cin
(Q-Qb), Ce (Q-Qb), Ce~0 Qb,Cin
Q, Cp
Q, Cp
1) Mass Balance: Accum = Inputs – Outputs + Reactions Because there is no accumulation at the pipe joint and no reactions, Inputs = Outputs (Q - Q b) Ce + Qb Cin = QCp If Ce = 0 Qb Cin = QCp Qb Cp 85 mg/L ----- = ------ = --------------------------- = 0.30 (30% of total flow) Q Cin 280 mg/L Qb = 0.3 Q = (0.3)(2000 L/d) = 600 L/d
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1) Loading Rate = (1- 0.3) 0.3) Q Cin = 0.7 (2000 L/d) (280 mg/L) = 392,000 mg/d 2) Total Capacity of the resin Break through time = ---------------------------------------------------------------------------------Mass of ions removed / time (57 kg/m3)(0.1 m3) = --------------------------------------------------------------------------- = 14.5 days -6 (392,000 mg/d)(10 kg/mg)
Note that it is assumed all (100%) hardness in the water which passes through the ion exchange column is removed.
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Example ) An ion exchange process is to be used to soften water with an analysis as shown in Table 1. A synthetic zeolite resin is to be used with an exchange capacity of 20 Kilograins of hardness as CaCO 3 per ft 3 when regenerated at the rate of 15 lb of salt per ft3. Flow rate is 300 gpm. Assume Assume a final hardness level of 2 mg/L (as CaCO CaCO 3) is required. (1 grain/gal = 17.1 mg/L) Use the following design criteria: a. The maximum loading rate < 5 gpm/ft 3 of resin b. The bed depth = 30 < 72 inches. Assume a 24 hour regeneration cycle for each softner unit. (Actual frequency of regeneration would depend upon operating conditions.) Table 1. Water Analysis =================================== Component Conc. (mg/L as CaCO 3) ----------------------------------------------------------------------------------------------------------------------Calcium 85 Magnesium 26 Sodium 37 Chloride 10 Sulfate 60 Bicarbonate 78 ----------------------------------------------------------------------------------------------------------------------1. Write general chemical reactions that occur i n the ion exchange process. 2. Calculate total hardness (mg/L as CaCO 3) 3. Calculate hardness to be removed (grains/gal) 4. Calculate hardness to be removed (Kilograins/d) 3 5. Calculate total resin required (ft ) 6. Calculate the amount of salt required for regeneration. 7. Calculate the bed depth and shell diameter for the ion exchange equipment. Given: 3 a) use Zeolite (20 kgr/ft resin) 3 b) 15 lb NaCl/ft resin for regeneration/day c) Q = 300 gpm d) final hardness = 2 mg/L as CaCO 3 1. Write general chemical reactions that occur i n the ion exchange process. process. ..... see above above 2. Total Hardness = (Ca 2+ + Mg2+) = 85 + 26 = 111 mg/L as CaCO 3 3. Hardness to be removed = (Initial Hardness) - (Final Hardness) = 111 mg/L - 2 mg/L = 109 mg/L as CaCO 3 1 grain/gal = 109 mg/L (----------------) (----------------) = 6.37 grains/gallon 17.1 mg/L 4. Hardness removed --------------------------------------- = 1 day (24 hrs)
300 6.37 1440 2.752E6 2.752E6 gal grains min grains -------- ----------- ---------- = ----------------------------min gal day day = 2752 Kilograins/day
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5. Total resin required
Hardness removed
2752 Kgr
= ------------------------------------------------------- = --------------------------------- = 138 ft 3 Exchange capacity 20 Kgr/ft3
6.Calculate the lbs of salt required per month for zeolite regeneration. 15 lbs salt / d 3 Salt required = (138 ft )(--------------------) 1 ft3 resin = 2070 lbs salt/day = 62,100 lbs/month 7. Calculate the bed depth and shell diameter for the ion exchange equipment.
For 2 units, Actual design resin volume/unit = 138 ft 3 /2 units = 69 ft3 /unit When one unit is in regeneration (flow rate through each unit at normal flow), Maximum loading rate
300 gpm
------------------------------------------------------------- = --------------- = 4.34 gpm/ft 3 /unit unit 69 ft3 /unit (Criteria: <5 gpm/ft 3) Using a 60 inch (5 ft) diameter shell, Surface area, A =
2 = 3.14 (2.5) 2 ft2 = 19.6 ft 2
Π r
V Bed depth, H = ----------A 69 ft3 Bed depth = ----------------------------- = 3.52 ft = 42 inches 19.6 ft 2 (Criteria: 30 < 72 inches) Flow rate through each unit at normal flow, Normal loading rate 300 gpm / 2 ------------------------------------------------------- = --------------------------------------unit 69 ft3 /unit = 2.2 gpm /ft 3 /unit (Criteria: <5 gpm / ft 3)
Example An ion exchange process is to be used to soften water at the rate of 500 gpm. A synthetic zeolite resin will be packed in shells with diameter of 5 ft. The resin has an exchange capacity of of 20 Kilograins of CaCO 3 per ft3 when regenerated at the rate of 15 lb of salt per ft 3. The raw water has total hardness hardness of 100 mg/L as CaCO 3. Assume this process process can achieve 95% hardness removal efficiency. efficiency. 1 grain/gal grain/gal = 17.1 mg/L. Use the design criteria: a. The maximum loading rate = 5 gpm/ft 3 of resin
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b. The bed depth =
30 < 72 inches. inches.
a. Calculate the bed depth for the ion exchange equipment. Total Hardness = 100 mg/L as CaCO 3 Hardness to be removed = 100 mg/L - 5 mg/L = 95 mg/L as CaCO 3 1 grain/gal = 95 mg/L (----------------) = 5.5556 grains/gallon 17.1 mg/L Hardness removed 500 gal 5.5556 grains 1440 min --------------------------------------------------- = ----------- ----------------------------------------- ------------1 day (24 hrs) min gal day 4,000,000 grains = --------------------------------------------------- = 4,000 Kirograins/day day Total resin required
4000 Kgr = ----------------------------------- = 200 ft 3 20 Kgr/ft 3
For 2 units, Actual design resin volume/unit = 200 ft 3 /2 = 100 ft3 /unit When One unit is in regeneration (flow rate through eeach unit at normal flow), Maximum loading rate 500 gpm -------------------------------------------------------------- = ----------------------------------unit 100 ft3 /unit = 5.0 gpm/ft 3 /unit
OK
Using a 60 inch (5 ft) diameter shell, Surface area = pi r 2 = 3.14 (2.5) 2 ft2 = 19.6 ft 2 100 ft3 Bed depth = --------------- = 5.1 ft = 61 inches 19.6 ft2 (30 < 72 inches)
OK
Flow rate through each unit at normal flow, Normal loading rate
500 gpm/2
------------------------------------------------------- = ------------------------------------- = 2.5 gpm/ft 3 /unit unit 100 ft3 /unit OK 3. Calculate the lbs of salt required per day for zeolite regeneration assuming a 24 hour regeneration cycle for each softener unit. For zeolite regeneration, 15 lbs salt Salt required = (200 ft 3)(---------------) )(---------------) = 3000 lbs salt/day 1 ft3 resin
