1 Department of Civil Engineering-IIT Delhi CEL 212: Environmental

Department of Civil Engineering-I.I.T. Delhi
CEL 212: Environmental Engineering
Second Semester 2011-2012
Home Work 7 Solution
Q1. For a single-stage softening, following water (see characteristics below) is softened using the
Lime-soda ash process (pH =9). No magnesium removal is required here.
[5+10
=15
points]
Species
Concentrations (milli-equivalents/L)
Carbon dioxide
1.0
Ca2+
4.0
Mg2+
2.0
Na+
3.0
HCO32.5
SO425.0
Answer the following:
(i) Write balanced equations for removing CO2, calcium bicarbonate and calcium sulfate using
the Lime-soda process.
(ii) Calculate amount of CaO and Na2CO3 required? (unit: milli-equivalents/L)? For a flow of
10,000 m3/day, calculate the daily chemical requirement and the mass of solids produced?
Assume that the lime used is 90% pure and the soda ash is 80% pure.
Answer: See solution of Q3.
Q2. In a WWTP, residuals from the sedimentation basin consist of solids carried by the raw water
and the precipitates formed by adding the chemicals. Summary of the chemical dosages and raw
water constituents producing sludge solids are given below.
Parameter
Value
Design maximum day flow
113, 500 m3/day
Design average day flow
57,900 m3/day
Maximum turbidity
17 NTU
Maximum seasonal iron
0.7 mg/L
Maximum seasonal manganese concentration
0.4 mg/L
Optimum coagulant aid, cationic polymer
0.05 mg/L
Optimum coagulant ferric sulfate dosage
25 mg/L
Hydrated lime (Ca(OH)2 for pH adjustment
15 mg/L
Seasonal potassium permanganate (KMnO4) dosage
4 mg/L
Seasonal PAC for taste and odor control
5 mg/L
* Assume 1 mg-total suspended solids = 1 NTU
Under maximum flow conditions, calculate the following:
(i)
Amount of solids from raw water?
(ii)
Amount of solids produced due to precipitation of iron as ferric hydroxide?
(iii)
Assuming 20% of Ca (OH)2 precipitate as CaCO3, amount of lime solids produced
during pH adjustment?
Answer:
Maximum turbidity = 17 NTU =17 mg total suspended solids (this is total solids present in raw
wastewater).
Maximum seasonal iron = .0.7 mg/L =(0.7*0.001)/56=1.25*10-5 moles/L. Given that one mole of iron
produces one mole of ferric hydroxide. So 1.25*10-5 moles/L of iron produces 1.25*10-5 moles/L moles
of ferric hydroxide (i.e., 1.25*10-5 *107g/L=0.00133 g/L).
Hydrated lime (Ca(OH)2 used for pH adjustment = 15 mg/L = (15*0.001)/74) =2.07*10-4 moles/L
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If 20% of Ca (OH)2 precipitate as CaCO3, i.e., 0.2*(2.07*10-4 moles/L)=0.414*10-4 moles/L Ca(OH)2
precipitates as CaCO3 (i.e., 0.414*10-4 moles/L).
Amount of lime solids produced during pH adjustment = (0.414*10-4 moles/L)*(100g/mole)
= 41.4*10-4 g/L
As maximum daily flow =113, 500 m3/day, lime produced =(113, 500 m3/day)*( 41.4*10-4 *1000g/m3)
=469.8 Kg/day
Q3. The AXX well water has bicarbonate hardness equal to 100 mg/L as CaCO3. Calculate
amounts of lime and soda ash required during the lime-soda ash process (Given: Daily flow rate:
10,000 m3/day).
[10 points]
Answer:
Bicarbonate hardness = 100 mg/L as CaCO3 = (100 mg/L)/ [(100g/mole) × (1000mg/g)] =0.001 mole/L
The balanced equation:
Ca(HCO3)2+CaO+H2O 2CaCO3 (s) +2H2O
i.e., 1 mole of calcium bicarbonate consumes 1 mole of CaO and produces 2 moles of calcium carbonate.
As there is no non-carbonate hardness is assumed here, no soda ash is required.
Thus: Lime (CaO) required = 0.001 moles/L (i.e., 0.001moles/L ×56g/mole=0.056 g/L CaO)
If lime is 90% pure, so we need = 0.056g/0.9=0.0632 g/L. So Daily lime requirements = (0.0632g/L
×0.001Kg/g) × (10,000 m3/day ×1000 L/m3) = 632 Kg/day (answer)
Soda-ash required =0 (answer)
CaCO3 produced = 2×0.001 moles/L (i.e., 0.002moles/L ×100g/mole=0.2 g/L CaCO3) (answer)
Another Variation: If all hardness were non-carbonate hardness only and due to only calcium sulfate
only, following calculation steps would have used to determine soda ash requirements.
Calcium sulfate-associated hardness = 100 mg/L as CaCO3
= (100 mg/L)/ [(50g/equivalents) × (1000mg/g)] =0.002 equivalents/L CaCO3
Or 0.002 equivalents/L CaSO4
Or (0.002 equivalents/L) × [(136/2 g/equivalents)]/ (136g/mole) =0.001 moles/L
The balanced equation:
CaSO4+Na2CO3 CaCO3 (s) +Na2SO4
i.e., 1 mole of calcium sulfate consumes 1 mole of soda ash and produces 1 mole of calcium carbonate.
As there is no carbonate hardness is assumed here, no lime is required.
Thus: Soda ash required = 0.001 moles/L (i.e., 0.001moles/L ×106g/mole=0.106 g/L Na2CO3)
As soda ash is 80% pure, so we need = 0.106g/0.8=0.133 g/L. So Daily soda ash requirements =
(0.133g/L ×0.001Kg/g) × (10,000 m3/day ×1000 L/m3) = 1330 Kg/day (answer)
Lime required =0 (answer)
CaCO3 produced = 1×0.001 moles/L (i.e., 0.001moles/L ×100g/mole=0.1 g/L CaCO3) (answer)
Q4. Look at the following water quality data. Say one needs to treat this water using ion-exchange
process (capacity: 90 Kg hardness/m3 material at 0.4m/minute flow rate). For this material,
regeneration is conducted using 10% NaCl solution which requires 100 Kg sodium chloride/m3
resin. Answer the following:
[7+3=10 points]
(i) Calculate mass of resin required and chemical required for regeneration process?
(ii) Can I use this ion-exchanger for removing anions from water? Provide reasons.
Species
Free CO2 Ca2+ Mg2+ Na+ HCO3- SO42Conc. (mequiv/L)
1.0
4.0
1.0
2.0 2.5
4.5
Answer:
Hardness to be removed = removal of hardness due to Ca2+, Mg2+ ions
=> removal required =4+1=5 mequiv/L as CaCO3 = 0.005 equiv/L as CaCO3
= 0.005*50g/L CaCO3 = 0.25 g/L = 0.25 Kg hardness/m3 solution
2
given capacity of resin =90 Kg hardness/m3 material at 0.4m/minute flow rate).
volume of resin required = (0.25/90) m3 resin material/m3 solution=0.0027 m3 resin material/m3 solution
If density of resin is given, mass of resin required can be calculated.
Amount of NaCl required = (0.0027 m3 resin material/m3 solution)*[(100 Kg NaCl/m3 resin)/0.10]
=2.7 Kg NaCl/m3 solution
As this resin is used for removing hardness, i.e., divalent ions, this cannot be used for removing anions.
The given resin does not have capacity for removing anions.
Q5. Calculate the amount of lime-soda ash required for treating 1 MLD of water? How much
solid waste is produced from this process?
Free carbon dioxide
Mg2+
3 mg/L
18 mg/L
Ca2+
Na+
44 mg/L
16 mg/L
Alkalinity (HCO3-)
122 mg/L
Purity in lime
Purity in Soda
85%
100%
Hint:
Here Soda is NaOH. Soda will be used to react with CO2, alkalinity, magnesium and calcium. Sodium
will not contribute to soda requirement.
CO2+ NaOH Na2CO3 + H2O
Ca (HCO3)2 + 2NaOH CaCO3 (s) +Na2CO3 + 2H2O
Mg (HCO3)2 + 4NaOH Mg (OH) 2(s) +2Na2CO3 + 2H2O
Follow steps used in Q3 solution.
Q6. Calculate daily lime-soda ash requirement during softening of water (100 mg/L carbonate
hardness (as CaCO3) and 100 mg/L MgSO4) (Given: Daily water production: 1000 m3/day; lime is
90% pure and soda ash is 80% pure).
Answer:
Part 1: Assume all carbonate hardness is due to calcium bicarbonate (Ca (HCO3)2), now follow the
methodology used in Q3.
Part 2: For removal of 100 mg/L magnesium sulfate, now calculates amount of soda ash required.
Magnesium sulfate-associated hardness = 100 mg/L as CaCO3 = (100 mg/L)/ [(50g/equivalents) ×
(1000mg/g)] =0.002 equivalents/L CaCO3
Or 0.002 equivalents/L MgSO4
Or (0.002 equivalents/L) × [(120/2 g/equivalents)]/ (120g/mole) =0.001 moles/L
The balanced equation:
MgSO4+Na2CO3 MgCO3 (s) +Na2SO4
i.e., 1 mole of magnesium sulfate consumes 1 mole of soda ash and produces 1 mole of magnesium
carbonate. As there is no carbonate hardness, no lime is required.
Thus: Soda ash required = 0.001 moles/L (i.e., 0.001moles/L ×106g/mole=0.106 g/L Na2CO3)
As soda ash is 80% pure, so we need = 0.106g/0.8=0.133 g/L. So Daily soda ash requirements =
(0.133g/L ×0.001Kg/g) × (10,000 m3/day ×1000 L/m3) = 1330 Kg/day (answer)
Lime required =0 (answer)
MgCO3 produced = 1×0.001 moles/L (i.e., 0.001moles/L ×84g/mole=0.084 g/L MgCO3) (answer)
Q7. Draw a typical breakthrough curve for ion-exchanger unit and show different parameters on
this curve. Discuss its importance in deciding regeneration frequency of ion-exchanger unit.
[2+2+6= 10 points]
Hint: See lecture notes on Ion Exchange
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Q8. Excess sodium intake can results in high blood pressure and inner ear problems for some
people. The regulatory body recommends maximum allowable concentration to be 20 mg/L sodium
in drinking water. Now sodium ions are used in ion exchange process. For reducing hardness from
6 to 1.5 meq/L in water, how much sodium ion is produced (mg/L) and if it poses any health risk
based on given maximum concentration guideline.
Solution:
Reduction in hardness value = 6-1.5 meq/L =4.5 meq/L (during the ion exchange process).
Sodium ions produced during hardness reduction = 4.5 meq/L
Concentration of sodium ions produced = 4.5 × (0.001) × (23g/mole/1)=0.1035 g/L= 103.5 mg/L
As regulatory body recommends maximum allowable concentration to be 20 mg/L sodium in drinking
water, concentration of sodium ions produced during the ion exchange process violates this requirement
and thus this process poses health risk.
Hints:
QA3: One mole of CaO is required to react with one mole of CaCO3 to produce calcium bicarbonate.
So, calculate CaO required depending on amount of CaCO3 produced. Generally upto 40 mg/L CaCO3
and 10 mg/L Mg (OH) 2 remain in softened water, resulting in continued release of these ions in water.
QA5: For this, follow Jar test procedure in detail, analyze residual turbidity versus dose used and then
finding desired residual turbidity in water for finding optimal dose (should be lowest and versatile
covering broad pH ranges, and cost-effective).
QA7: Acid will be used to solubilise precipitated CaCO3 and then it is removed using softening process.
This happens due to supersaturation of CaCO3.
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Answer 1:
Non-carbonate hardness (NCH) to be left=80 mg/L-35 mg/L= 45 mg/L
Removal of Non-carbonate hardness=NCH in raw water- NCH in treated water =92-45 mg/L=47 mg/L
For the lime process, lime is required for CO2 reaction, carbonate hardness and magnesium.
Moles of Mg2+ present = 15mg/L/ (24 ×1000 mg/mole) =6.25×10-4 moles/L.
As 1 mole of Mg2+ consumes 2 moles of bicarbonate, so 6.25×10-4 moles/ Mg2+ consumes 12.4×10-4
moles/L bicarbonate (i.e., 75.64 mg/L bicarbonate ions).
Alkalinity present =68 mg/L as CaCO3
CO2
Initial
amount
(mg/L)
3
Mg2+
15 mg/L
Amount of CaO
required (mg/L)
Sludge produced
Reason
=(56/44)×(3)
=3.82
CaCO3 produced
=(100/44)×(3)
=6.82 mg/L
=(56/24)×(15)
=35
Mg(OH)2 produced=
=(58/24)×(15)
=36.25 mg/L
1 mole of CO2 requires 1 mole of
CaO and produces 1 mole of
CaCO3 (i.e., 44 mg/L CO2 reacts
with 56 mg/L CaO)
1 mole of Mg2+ requires 1 mole
of CaO and produces 1 mole of
Mg(OH)2 and 1 mole of CaCO3
(i.e.,24 mg/L Mg reacts with 56
mg/L CaO)
Ca(CO3) produced=
=(100/24)×(15)
=62.5 mg/L
Alkalinity
Total
68 mg/L
as
CaCO3
=(56/100)×(68)
=38.08
Lime=3.82+35+3
8.08 mg/L
Assume all to be carbonate
alkalinity;
1 mole of CaCO3 requires 1 mole
of CaO (i.e., 44 mg/L CO2 reacts
with 100 mg/L CaO)
Sludge as Mg(OH)2=
36.25 mg/L
Sludge as
CaCO3=62.5+6.28=68.78
mg/L
Total lime required = 3.82+35+38.08=76.9 mg/L
As one mole of CaO results in formation of one mole of calcium hydroxide [Ca (OH) 2]
Amount of hydrated lime required = (74 g Ca(OH)2/mole)/(56 g/mole)×(76.9 mg/L)
=101.6 mg/L (answer)
Now calculate amount of lime required
= (101.6 mg/L) × (volume of water treated in liters/day)/ (0.85) = X Kg/day
Cost of using lime everyday
= (X Kg/day) × (cost as Y Rs/Kg lime) = Z Rs/day
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Hint Q3: Hardness removal required = hardness in raw water-hardness desired in treated water = 400-50
= 350 mg/L (or say X meq/L) (Note: this info can be given in terms of water quality ion concentrations;
then you need to find total of hardness either in terms of mass or in terms of milli-equivalents removal
required).
Volume of water treated per day =25000 liters/h × 24h/day=600000liters/day
Mass of Hardness removal required = (350mg/L×10-6 Kg/mg) × (600000liters/day)=210 Kg/day (or X
meq/L×600000liters/day=600000X meq/day)
Ion exchange capacity of zeolite = 10 Kg hardness/m3 of zeolite
Volume of zeolite everyday for desired water quality = (210 Kg/day)/ (10 Kg/m3 zeolite) =21 m3/day
Volume of NaCl required for regeneration = (21 m3/day) × (50Kg/m3 resin)/ (0.1) = 10500 Kg/day
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