ion exchange. presentation outline ion exchange reactions unit operations of ion exchange sodium,...
TRANSCRIPT
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ION EXCHANGE
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Presentation Outline
Ion Exchange Reactions Unit Operations of Ion Exchange Sodium, Hydrogen Cycle and Regeneration Production of Pure Water Active or Exchange Zone Design of Ion Exchangers Quantity of Regenerant Wastewater Production
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Ion Exchange Reactions
Ion Exchange is the displacement of ion by another
Ion Exchange is a reversible chemical reactions wherein an ion from solution is exchanged for a similarly charged ion attached to an immobile solid particles
The displaced ion moves into solution and the displacing ion becomes a part of the insoluble materials (ResinResin)
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Ion Exchange
Two types of ion exchange materials are used The cation exchange material
The anion exchange material
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Ion Exchange
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Ion Exchange
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Ion Exchange
The insoluble part of the exchange materials is called the host
The cation exchange materials may be represented by rr : the number of active sites in the insoluble
material r n/mr n/m: the number of charged exchangeable
particles attached to the host materials -n-n: is the charge of the host +m+m: is the charge of the exchangeable cation
mmrn
mr
n CR /
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Ion Exchange Reactions
cation exchange reaction as well as the anion exchange reaction are as follows
Ion exchange reaction are governed by Equilibrium. For this reason, effluents from ion exchange processes never yield pure water
mqrn
qsr
nqsmrn
mr
n Cm
rnCRC
q
rnCR //
ptro
tsr
otspro
pr
o Ap
roARA
t
roAR //
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Ion Exchange Reactions
Displacement Series for Ion ExchangeDisplacement Series for Ion Exchange
• Displacement series for ion exchange materials is shown in left side table
• when an ion species high in table is in solution, it can displace ion species in the insoluble material below it in the table.
• to remove any cation in solution, the displaceable cation must be the proton, and to remove any anion, the displaceable anion must be the hudroxyl ion.
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ION Exchange Reactions
Examples of exchange materials Zeolites (Natural Material) Synthetic resins
Synthetic resins are insoluble polymers
These polymers are either acidic or basic group, and they are called functional group
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Ion Exchange Reactions
These groups are capable of performing reversible exchange reactions with ions in solution
The total number of these groups determine the exchange capacityexchange capacity of the exchange material
The type of functional group determines ion ion selectivityselectivity
The exchanger may be regenerated by the reverse reactions (upon exhaustion)
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Unit Operation of Ion Exchange
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Unit Operation of Ion Exchange
In both units, the influent is introduced at the top of the vessel
the bed of ion exchanger materials would be inside the vessels
As the to be treated passes through, exchange of ions takes place
This exchange of ions is the chemical reaction of the unit process of ion exchange
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Sodium, Hydrogen cycle, And Regeneration Sodium and Hydrogen are the logical choices for the
exchangeable ions.
The cation exchange resin using sodium to remove the Ca+2 may be represented by the following reactions
NarnCaRCarn
NaR rnrn
rnrn
2/22
2
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Sodium, Hydrogen Cycle, And Regeneration As soon as the resin is exhausted, it may be regenerated
The resin is regenerated by using a concentration of NaCl of a bout 5 to 10%, thus, driving the reaction to the left
Operations where regeneration is done using NaCl, the cycle is called the Sodium CycleSodium Cycle
Operations where regeneration is done using acids (H2SO4), the cycle is called the Hydrogen CycleHydrogen Cycle
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Sodium, Hydrogen Cycle, And Regeneration
The following table shows approximate exchange capacities and regeneration requirements for ion exchangers
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Sodium, Hydrogen Cycle, And RegenerationExchanger, cycleExchanger, cycle Exchange Exchange
CapacityCapacity (geq/m3)
RegenerantRegenerant Regenerant Regenerant RequirementRequirement
)geq/m3(
Cation exchangers:
Natural zeolite, Na 175-350 NaCl 3-6
Synthetic zeolite, Na 350-700 NaCl 2-3
Resin, Na 350-1760 NaCl 1.8-3.6
Resin, H 350-1760 H2SO4 2-4
Anion exchanger:
Resin, OH 700-1050 NaOH 5-8
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Sodium, Hydrogen Cycle, And Regeneration
In order to determine the exchange capacities and regeneration requirements we have to do the following:
Perform an actual experiment
Obtain data form the manufacturer
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Sodium, Hydrogen Cycle, And Regeneration
Table below shows some additional properties of exchangers
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Sodium, Hydrogen Cycle, And Regeneration
The strongly acidic (cation) exchangers readily remove cations from solutions
The weakly acidic exchangers have limited ability to remove certain cations
The strongly basic (anion) exchangers can readily remove all the anions
The weakly basic one remove mainly the anions of strong acid such as SO4
-2 and Cl
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Production of “PURE WATER’’
Theoretically, It would seem possible to produce pure water by combining the cation exchanger and the anion exchanger
The following equation for the hydrogen cycle is
rnHCRCq
rnHR qrn
qsr
nqsrnr
n/
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Production of “PURE WATER’’
Letting the molar concentration ofbe
The corresponding concentration in geq / L is
qsC
gmol/L][ qsC
][][ q
sqs
qs
qs Cq
q
C
CC
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Production of “PURE WATER’’
Therefore, the total concentration in gram equivalents per liter of removable cations in solutions is the sum of all the cations. Thus,
As, [CatT]eq of cations is removed form solution, a corresponding number of equivalent concentrations of anions pair with the H+ ions displaced from the cation bed
mi
i
qsieq
i
iCqCatT
1
][][
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Production of “PURE WATER’’
The total anions and the hydrogen ions displaced is expressed as follows
Practically, we may say that “ pure water” is
produced and expressed as follows
eqeqeq TCatHTTAnion ][][][
The units of ti are equivalents per mole
mi
i
tsieq
i
iAtTAnion
1
][][
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Production of “PURE WATER’’
Example: A wastewater contains the following ions:
mg/L12024 CrO
mg/L 15 Zmg/L, 30 2n
2 Cumg/L 202 Ni
Calculate the total equivalents of cations and anions, assuming the volume of the wastewater is 450 m3.
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Production of “PURE WATER’’
Solution:Solution:
Ions (mg/L) Equiv. Mass Cations (meq/L) Anions (meq/L)
58a __ 2.069b
31.75 0.945 __
32.7 0.469 __
29.35 0.681 __
12024 CrO
302 Cu
152 Zn
202 Ni
a Equiv. mass b120/58 = 2.069 .582/)]16(452[24 CrO
395.2 069.2
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Production of “PURE WATER’’
Solution (Cont’d) Solution (Cont’d)
Total equivalents of cations = 2.395(450) = 1077.75 1077.75 Ans
Total equivalents of cations = 2.069(450) = 931.05931.05 Ans
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Active or Exchange Zone
Active zone is a segment of exchanger bed engaged in exchanging ions
ultps
nnnnobx
M
XA
CCC
2
][][)(][)(2 1
1
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Active or Exchange Zone
Where:
x
b
][ oC
1n
= total volume of water or wastewater treated at complete exhaustion of bed
= = length of active zone
= volume treated at breakthrough
= influent concentration to
= total volume treated at time
1nt
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Active or Exchange Zone
(Cont’d)
n = total volume treated at time nt
= concentration of solute at effluent of at time 1nt][ 1nC
= concentration of solute at effluent of at time nt][ nC
= surficial area of exchanger bed sA
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Active or Exchange Zone
Active zone at various times during adsorption and the breakthrough curve
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Active or Exchange Zone
Example 2: A breakthrough experiment is conducted for a wastewater producing the results below. Determine the length of the active zone. The diameter of the column used is 2.5 cm. and the packed density of the bed is 750 kg/m3. is equal to 2.2 meq/L. And
The experiments results are tabulated on the next slide
][ oCmeq/g5.6)/( ultMX
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Active or Exchange Zone
C, meq/L
0.06 1
0.08 1.2
0.09 1.3
0.1 1.4
0.2 1.48
0.46 1.58
1.3 1.7
1.8 1.85
2.1 2
L,
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Active or Exchange Zone
Solution:
ultps
nnnnobx
M
XA
CCC
2
][][)(][)(2 1
1
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Active or Exchange Zone
0.06 1.0 0.20 0.07 0.014
0.08 1.20 0.10 0.085 0.0085
0.09 1.30 0.10 0.095 0.0095
0.10 1.40 0.08 0.15 0.012
0.20 1.48 0.10 0.33 0.033
0.46 1.58 0.12 0.88 0.1056
1.30 1.70 0.15 1.55 0.2325
1.80 1.85 0.15 1.95 0.2925
2.10 2.00
meq/L,C L, )( 1 nn
2
][][ 1 nn CC )( 1 nn
2
][][ 1 nn CC
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Active or Exchange Zone
22
00049.04
025.0mAs
Therefore,
= 1.2 mm= 1.2 mm
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Design of Ion Exchangers
Designs of ion exchangers should include the following:
Quantity of exchange materials
Quantity of regenerant
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Quantity of Exchange Materials
The amount of exchange bed materials required can be determined by the calculating the amount of displacing ions in solution to be removed
The equivalents of ion displaced from the bed is equal to the equivalents of displacing ion in solution
The mass of bed materials CatTBedMass CatTBedMass in kilograms is
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Quantity of Exchange Materials
24
1000)()(])[(
24
1000)()()]([
int1
int
ult
qs
mi
i i
ult
eq
M
X
tQCq
M
X
tQCatTsCatTBedMas
i
i
Q is the m3/d of flow and tint is the interval of regeneration in hours
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Quantity of Exchange Materials
By analogy, the mass bed materials for the anion exchanger in Kiograms is:
24
1000)()(])[(
24
1000)()()]([
int1
int
ult
ts
mi
i i
ult
eq
M
X
tQAt
M
X
tQAnionassAnionTBedM
i
i
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Quantity of Exchange Materials
The volume in m3 for CatBedVolCatBedVol
24
1000)1()()(])[(
24
1000)1(
)()(])[(
int1
int1
ultp
qs
mi
i i
p
ult
qs
mi
i i
M
X
swelltQCq
swellM
X
tQCq
CatTBedVol
i
i
i
i
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Quantity of Exchange Materials
the volume in m3 For AnionTBedVolAnionTBedVol
24
1000)1()()(])[( int1
ultp
ts
mi
i i
M
X
swelltQAtolAnionTBedV
i
i
The percentage of swell of the exchanger bed is a very important property
It determines the final size of the tank into which the material is to be put
This value can be obtained through
Experiments
from the manufacturer
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Quantity of Exchange Materials
Example: Using a bed exchanger, 75 m3 of water per day is to be treated for hardness removal between regenerations having intervals of 8 h. the raw water contains 400 mg/L of hardness as CaCO3. The exchanger is a resin of exchange capacity of 1412.8 geq/m3. Assume that the packed density of the resin is 720 kg/m3. Calculate the mass of exchanger material to be used and the resulting volume when the exchanger is put into operation.
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Quantity of Exchange Materials
Solution:
Assume cation exchanger:
meq/L 0.004meq/L850
400][
1
mi
i
qsi
i
iCq
24
1000)()(])[( int1
ult
qs
mi
i i
M
X
tQCqsCatTBedMas
i
i
Also, assume that all of the cations are removed
g
meq96.1
g 720(1000)
meq )1000(8.1412
geq8.1412
3
mM
X
ult
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Quantity of Exchange Materials
Therefore,
nsA kg 02.5124
1000
96.1
)8)(75(004.0
sCatTBedMas
24
1000)1()()(])[( int1
ultp
qs
mi
i i
M
X
swelltQCqCatTBedVol
i
i
Assume swell= 0.8
nsA m 13.0)8.01(720
02.51 3CatTBedVol
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Quantity of Regenerant
24
1)()()(][ int
1
tQRCqegenerantCatR i
i
qs
mi
ii
The Kilogram equivalents of regenerant, CatRegenerant, used to regenerate cation exchangers is
The Kilograme equivalents of regenerant, Anion Regenerant, used to regenerate anion exchange is
24
1)()()(][ int
1
tQRAtegenerantAnionR i
i
ts
mi
ii
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Quantity of Regenerant
Example: Using a bed exchanger, 75 m3 of water per day is to be treated for hardness removal between regeneration having intervals of 8 hours. The raw water contains 80 mg/L of Ca+2 and 15 mg/L of Mg2+. The exchanger is a resin of exchange capacity of 1412.8 geq/m3. Assume that the packed density of the resin is 720 kg/m3. Calculate the kilograms of sodium chloride regenerant required assuming R = 2 and that all of the cations were removed
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Quantity of Regenerant
Solution:
24
1)()()(][ int
1
tQRCqegenerantCatR i
i
qs
mi
ii
33
3
1
m
keq)10(22.5
L
geq)10(22.5
)1000)(2/3.24(
15
)1000)(2/1.40(
80][
mi
i
qsi
i
iCq
Therefore,
nsAon regenerati of intervalper kg 0.26
24
1)8)(75)(2)(10(22.5 3
egenerantCatR
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Wastewater Production
In the operation of ion exchangers, wastewater are produced. These come form:
Solvent water (used to dissolve the regenerant)
Backwash and rinse requirements
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Wastewater production
In the sodium cycle, the concentration of NaCl is about 5 to 10% for an average of 7.5%
If the quantity of regenerant required is 0.26 Kg
The volume of wastewater produced from regeneration can be calculated as follows The total mass of regenerant solution is
0.26 / 0.075 = 3.47 Kg
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Wastewater Production
The corresponding volume is 3.47/ 1000 = 0.0035 m3
For an interval of regeneration of 8 h and assuming a rate of flow for the water treated of 75 m3 / d
The volume of water treated is 75/24(8) = 25m3
Thus, the wastewater produced is 0.0035/25 * 100 = 0.014 % by volume
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Wastewater Production
The quantity of backwash and rinse water requirements should be determined by experiment on the actual exchanger bed to be used in the design and is expressed as a function of bed volume
For the cation exchanger the volume of bed was previously derived as
3int1 m24
1000)()(])[(
ultp
qs
mi
i i
M
X
tQCqCatTBedVol
i
i
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Wastewater Production
With the swelling not being considered. Thus,
24
1000)()()(])[( int1
ultp
qs
mi
i i
M
X
nseVBackwashRitQCqhRinseVolCatBackwas
i
i
For the anion exchangers
24
1000)()()(])[( int1
ultp
ts
mi
i i
M
X
nseVBackwashRitQAtlashRinseVoAnionBackw
i
i
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Wastewater Production
Example: using a bed exchangers, 75 m3 of water per day is to be treated for hardness removal between regeneration having 8 hours. The raw water contains 80 mg/l of Ca and 15 mg/l of Mg. the exchanger is a resin of exchange capacity of 1412.8 geq/m3. Assume that the packed density of the resin is 720 kg/m3. calculate the total volume of rinse and backwash requirement if the backwash and rinse per unit volume of bed is 18m3/m3
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Wastewater Production
Solution:
24
1000)()()(])[( int1
ultp
qs
mi
i i
M
X
nseVBackwashRitQCqhRinseVolCatBackwas
i
i
33
1 m
keq )10(22.5[
mi
i
qsi
i
iCq
g
meq 96.1
g 720(1000)
meq )1000(8.1412
m
meq 8.1412
3
ultM
X
nsA m 66.124
1000
)96.1(720
)18)(8)(75)(10(22.5 33
hRinseVolCatBackwas
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