Department of Civil Engineering-I.I.T. Delhi CEL 212: Environmental Engineering Second Semester Home Work 7 Solution

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1 Department of Civil Engineering-I.I.T. Delhi CEL 212: Environmental Engineering Second Semester 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 Ca Mg Na HCO SO Answer the following: (i) Write balanced equations for removing CO 2, calcium bicarbonate and calcium sulfate using the Lime-soda process. (ii) Calculate amount of CaO and Na 2 CO 3 required? (unit: milli-equivalents/l)? For a flow of 10,000 m 3 /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. 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 m 3 /day Design average day flow 57,900 m 3 /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) (iii) Amount of solids produced due to precipitation of iron as ferric hydroxide? Assuming 20% of Ca (OH) 2 precipitate as CaCO 3, amount of lime solids produced during ph adjustment? 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= g/l). Hydrated lime (Ca(OH) 2 used for ph adjustment = 15 mg/l = (15*0.001)/74) =2.07*10-4 moles/l 1

2 If 20% of Ca (OH) 2 precipitate as CaCO 3, i.e., 0.2*(2.07*10-4 moles/l)=0.414*10-4 moles/l Ca(OH) 2 precipitates as CaCO 3 (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 m 3 /day, lime produced =(113, 500 m 3 /day)*( 41.4*10-4 *1000g/m 3 ) =469.8 Kg/day Q3. The AXX well water has bicarbonate hardness equal to 100 mg/l as CaCO 3. Calculate amounts of lime and soda ash required during the lime-soda ash process (Given: Daily flow rate: 10,000 m 3 /day). [10 points] Bicarbonate hardness = 100 mg/l as CaCO 3 = (100 mg/l)/ [(100g/mole) (1000mg/g)] =0.001 mole/l The balanced equation: Ca(HCO 3 ) 2 +CaO+H 2 O 2CaCO 3 (s) +2H 2 O 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 = 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= g/l. So Daily lime requirements = (0.0632g/L 0.001Kg/g) (10,000 m 3 /day 1000 L/m 3 ) = 632 Kg/day (answer) Soda-ash required =0 (answer) CaCO 3 produced = moles/l (i.e., 0.002moles/L 100g/mole=0.2 g/l CaCO 3 ) (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 CaCO 3 = (100 mg/l)/ [(50g/equivalents) (1000mg/g)] =0.002 equivalents/l CaCO 3 Or equivalents/l CaSO 4 Or (0.002 equivalents/l) [(136/2 g/equivalents)]/ (136g/mole) =0.001 moles/l The balanced equation: CaSO 4 +Na 2 CO 3 CaCO 3 (s) +Na 2 SO 4 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 = moles/l (i.e., 0.001moles/L 106g/mole=0.106 g/l Na 2 CO 3 ) 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 m 3 /day 1000 L/m 3 ) = 1330 Kg/day (answer) Lime required =0 (answer) CaCO 3 produced = moles/l (i.e., 0.001moles/L 100g/mole=0.1 g/l CaCO 3 ) (answer) Q4. Look at the following water quality data. Say one needs to treat this water using ion-exchange process (capacity: 90 Kg hardness/m 3 material at 0.4m/minute flow rate). For this material, regeneration is conducted using 10% NaCl solution which requires 100 Kg sodium chloride/m 3 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 CO 2 Ca 2+ Mg 2+ Na HCO 3 SO 4 Conc. (mequiv/l) Hardness to be removed = removal of hardness due to Ca2+, Mg2+ ions => removal required =4+1=5 mequiv/l as CaCO 3 = equiv/l as CaCO 3 = 0.005*50g/L CaCO 3 = 0.25 g/l = 0.25 Kg hardness/m 3 solution 2

3 given capacity of resin =90 Kg hardness/m 3 material at 0.4m/minute flow rate). volume of resin required = (0.25/90) m 3 resin material/m 3 solution= m 3 resin material/m 3 solution If density of resin is given, mass of resin required can be calculated. Amount of NaCl required = ( m 3 resin material/m 3 solution)*[(100 Kg NaCl/m 3 resin)/0.10] =2.7 Kg NaCl/m 3 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 3 mg/l Mg mg/l Ca mg/l Na + 16 mg/l Alkalinity (HCO - 3 ) 122 mg/l Purity in lime 85% Purity in Soda 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. CO 2 + NaOH Na 2 CO 3 + H 2 O Ca (HCO 3 ) 2 + 2NaOH CaCO 3 (s) +Na 2 CO 3 + 2H 2 O Mg (HCO 3 ) 2 + 4NaOH Mg (OH) 2(s) +2Na 2 CO 3 + 2H 2 O Follow steps used in Q3 solution. Q6. Calculate daily lime-soda ash requirement during softening of water (100 mg/l carbonate hardness (as CaCO 3 ) and 100 mg/l MgSO 4 ) (Given: Daily water production: 1000 m 3 /day; lime is 90% pure and soda ash is 80% pure). Part 1: Assume all carbonate hardness is due to calcium bicarbonate (Ca (HCO 3 ) 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 CaCO 3 = (100 mg/l)/ [(50g/equivalents) (1000mg/g)] =0.002 equivalents/l CaCO 3 Or equivalents/l MgSO 4 Or (0.002 equivalents/l) [(120/2 g/equivalents)]/ (120g/mole) =0.001 moles/l The balanced equation: MgSO 4 +Na 2 CO 3 MgCO 3 (s) +Na 2 SO 4 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 = moles/l (i.e., 0.001moles/L 106g/mole=0.106 g/l Na 2 CO 3 ) 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 m 3 /day 1000 L/m 3 ) = 1330 Kg/day (answer) Lime required =0 (answer) MgCO 3 produced = moles/l (i.e., 0.001moles/L 84g/mole=0.084 g/l MgCO 3 ) (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 3

4 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 = 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)= g/l= 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. 4

5 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/ ( mg/mole) = moles/l. As 1 mole of Mg2+ consumes 2 moles of bicarbonate, so moles/ Mg2+ consumes moles/l bicarbonate (i.e., mg/l bicarbonate ions). Alkalinity present =68 mg/l as CaCO3 Initial amount (mg/l) Amount of CaO required (mg/l) CO 2 3 =(56/44) (3) =3.82 Mg mg/l =(56/24) (15) =35 Alkalinity Total 68 mg/l as CaCO3 =(56/100) (68) =38.08 Lime= mg/l Sludge produced CaCO3 produced =(100/44) (3) =6.82 mg/l Mg(OH)2 produced= =(58/24) (15) =36.25 mg/l Ca(CO 3 ) produced= =(100/24) (15) =62.5 mg/l Sludge as Mg(OH)2= mg/l Reason 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 Mg 2+ 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) 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 CaCO3= =68.78 mg/l Total lime required = =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 5

6 Hint Q3: Hardness removal required = hardness in raw water-hardness desired in treated water = = 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 liters/day=600000X meq/day) Ion exchange capacity of zeolite = 10 Kg hardness/m 3 of zeolite Volume of zeolite everyday for desired water quality = (210 Kg/day)/ (10 Kg/m 3 zeolite) =21 m3/day Volume of NaCl required for regeneration = (21 m 3 /day) (50Kg/m 3 resin)/ (0.1) = Kg/day 6