# Department of Civil Engineering-I.I.T. Delhi CEL 212: Environmental Engineering

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1 Department of Civil Engineering-I.I.T. Delhi CEL 212: Environmental Engineering Q1. An anaerobic reactor, operated at 35 o C, treats wastewater with a flow of 2000m 3 /d and a biological soluble COD (bscod) concentration of 500g/m 3. At 90% bscod removal and a biomass synthesis yield of 0.04 g Volatile Suspended Solids/ g bscod used, estimate the amount of methane produced in m 3 /d. Step 1: Prepare a steady state mass balance for COD to determine the amount of influent COD converted to methane. COD in = COD eff + COD VSS(biomass) + COD methane COD in = COD concentration in influent x Wastewater Inflow COD eff = COD concentration in effluent x Wastewater Inflow COD VSS = 1.42 g COD/g VSS x Yield coeff g VSS/g COD x Efficiency of system COD methane =? Step 2: Determine the volume of gas occupied by 1 mole of gas at 35 o C. Step 3: The CH 4 equivalent of COD converted under anaerobic conditions = (L/ mole)/ (64 g COD/ mole CH 4 ) Step 4: CH 4 production = COD methane x CH 4 equivalent of COD converted COD in = COD concentration in influent x Wastewater Inflow = ( 500g/m 3 ) *(2000m 3 /d)=1000 Kg/d Biological soluble COD removal = 90% COD eff = 0.10* 1000 Kg/d = 100 Kg/d COD used = 0.90* 1000 Kg/d = 900 Kg/d Biomass synthesis yield = 0.04 g Volatile Suspended Solids/ g bscod used Biomass synthesis yield = (900*1000g/day)*(0.04 g VSS)=36000 g/day COD corresponding to biomass = (36000 g VSS/day)*(1.42 g COD/g VSS)= Kgday COD in = COD eff + COD VSS(biomass) + COD methane COD methane = COD in -{COD eff+ COD VSS(biomass) } COD methane = (1000 Kg/d)- {100 Kg/d } = Kg/d (answer) COD methane = Kg/d (answer) As 64 g COD comes from 1mole of CH 4. So Kg/d COD would result from = (1 mole CH 4 /64)*( Kg/d) = mole CH 4 /day Now we know that for STP conditions (i.e., 273 K and 1 atm pressure, 1 mole gas occupies 22.4 Liter volume). Now we need to calculate volume of methane gas at 35 C (i.e., 308K) and 1 atm pressure: V/T=constant (as P is constant here) [22.4/273] = [V2/308] ==> V2=25.27 Liters/mole (this is true for methane gas also) Now as biological process is producing moles methane gas/day, it indicates that liters/day (or m3/day). 1

2 Q2. The water content of solids slurry (WW sludge) is reduced from 98 to 95 %. What is the percent reduction in volume assuming that solids contain 70% organic matter of specific gravity 1.0 and 30% mineral matters of specific gravity 2.0? What is the specific gravity of 98 and 95% slurry? w1=98% (i.e., 2% solids) w2=95% (i.e., 5% solids) (this solid content is increased due to dewatering process) W solids /(S solids *ρ w ) = W mineral /(S mineral *ρ w )+ W organic matter /(S organic matter * ρ w ) Here S: specific gravity W: weight ρ w = density of water (i.e.,1 Kg/liter) Case 1: Water content =98% (i.e., 2% solids); Organic matter content =70% =0.7 W solids Mineral matter =30% =0.3W solids ρ w = 1 Kg/liter S mineral =specific density of mineral = 2 S organic matter =specific density of organic matter=1 For determining specific gravity of all solids (S solids ): W solids /(S solids ) = 0.3W solids /(2)+0.7W solids /(1) 1/S solids = = 0.85 =>S solids =1/0.85 =1.18 (this is specific density of solids) Now if specific density of water = 1 and water content =98% (i.e., solid content =2%), we can calculate specific gravity of sludge = 1/S sludge = (0.98)/1+(0.02)/1.18 = S sludge = 1/ =1.003 (or Density of sludge = 1.003*1=1.003 Kg/L) Density of sludge = Mass of sludge /Volume of sludge =>Volume of sludge = Mass of sludge/density of sludge For 1 Kg mass of sludge, volume of sludge = 1/(solid content*density of sludge) = 1/(0.02*1.003) = liters (initial volume of sludge per Kg mass of sludge) After solid waste treatment, water content is 95% (i.e., 5% solids). V1=49.85 liters/kg Assuming specific gravity of sludge remains same (i.e., S sludge = 1.003) or Density =1.003 Kg/L New volume of sludge after reduction in water content per Kg of sludge =V2= 1/(0.05*1.003) = liters So volume reduction = [1-(19.94/49.85)] *100 =60% 2

3 Q3. Assume that the data given in Q2 belongs to primary sludge. What will be the volume of digested slurry in case 60% of the volatile solids are destroyed and water content is reduced to 90%? Say 60% volatile solids (or organic matter) is destroyed. Initially 70% solid was organic matter and now remaining organic matter = (70%)*(0.40) = 28% (or mineral content =1-0.28=0.72 (i.e., 72%). Revised water content =90% (and thus solid content =10%). 1/(S sludge ) = 0.72/(2)+0.28/(1) =0.64 S sludge =1/0.64 =1.56 Density of sludge = 1.56*1=1.56 Kg/L Volume of sludge = Mass of sludge/density of sludge For 1 Kg mass of sludge, volume of sludge = 1/(0.10*1.56) =6.41 liters Q4. What are three stages of anaerobic digestion process? Discuss their importance and BOD requirements at different stages. Hint:See Lecture Notes. Q5. Some substances can be toxic to bacterial growth. How can you incorporate this effect in determining specific biomass growth rate constant? Some substance can impart toxicity to bacteria and thus it might affect their growth. For incorporating this effect, microbial decay constant due to toxic substance (i.e., k decay,substance ) need to be determined.then this aspect needs to be incorporated in the overall biomass rate equation. R overall =R lag +R growth +R stationary +R endogenousdecay +R substance decay Q6. Why do we keep SRT greater than HRT and how does it affect plant s performance? Keeping SRT> HRT results in long retention of biological solids (i.e., biomass) in aeration tank where aerobic biological processes occur. This increased retention of biological solids results in higher removal of organic matter compared to SRT=HRT situation (i.e., no recirculation case). Q7.Comment on performance of chlorination unit placed before and after aeration tank. Placing chlorination unit before aeration tank can kill bacteria however; it would not reduce microorganisms and pathogen in wastewater effluent which we might discharge to river. In this case, the discharge stream might have pathogens unsuitable for discharging in river water. 3

4 Q8. Calculate BOD and alkalinity requirements for removing 14 mg NH 4 + -N during nitrification process? Also explain the need for maintaining alkalinity in the nitrification reactor. Q9. A completely mixed activated sludge plant is to treat m 3 /d industrial wastewater (BOD 5 = 1200 mg/l; BOD effluent = 100 mg/l prior to discharge). For 5-day SRT, 5000 mg/l MLSS is required (Y= 0.7 kg/kg and k d =0.03/day). Calculate following: (i) The volume of the reactor (ii) The mass and volume of solids wasted each day (iii)the sludge recirculation ratio. (iv) To reduce BOD effluent to 30 mg/l would increase SRT a good idea? Discuss. Q10. Write overall oxidation and cell synthesis reactions during nitrification process and calculate amount of BOD required? Hint:See solution of Q8. Q11. Calculate total carbonaceous and nitrogen oxygen demand of a water sample that contains 5 mg/l organic compounds having chemical formula C 6 H 6 N 2 O 2. Assume that nitrogen is converted to ammonia and then to nitrate. Hint: See notes. This question has been discussed in class and asked in HW and Exam. Q12. Discuss different stages of anaerobic digestion process in the order they occur. Hint: See notes Q13. Name three disinfection kinetics models generally used to model disinfection process. Can CT concept be applied to all three kinetic models? Why or why not? Hint: For part 1: See notes. For part 2 Ct concept is applied only for first order kinetic model (i.e., for Chick s Law when n=1 in the Watson s Law). Q14. Comment on difference between domestic and hospital solid waste management (SWM) treatment methodologies. domestic SWM hospital SWM more treatment is required for inert more treatment is required for sharps, infectious and biodegradable materials generally, segregration, screening, biological processes are used and then land fill disposal and methan gas capturing is done. microorganisms and unused drugs, etc. generally segregation, followed by incineration is done to kill pathogens; biological process and advanced oxidation processes such as ozonation, UV radiation and hydrogen peroxide usage for properly treating unused drug before doing land filling and methane gas capturing if its feasible. Q15. Look at the following information for a completely mixed biological reactor. 4

5 Influent water information: Parameter Influent water Effluent water Flow rate 50 m 3 /d 50 m 3 /d Biomass 0 mg VSS/L?? Substrate 95 mg BOD 5 /L?? Tank volume 200 m 3 VSS: volatile suspended solids Calculate following: 1. effluent BOD 5 2. biodegradable organic matter removal efficiency of tank 3. biomass leaving the tank Here HRT =θ= 200m3/(50m3/d)= 4 days X in aerobic reactor =? S 0 =95 mg BOD 5 /L S final =? 1/θ =[(µ m S)/(K S +S)]-k d so: determine expression for S. S=[ K S (1+ θ k d )]/[( θµ m -1- θ k d )] Given K S =60 mgbod 5 /L (this is average value) k d =0.06/day (this is average value) µ m =3/day (this is average value) Solve for S now. It is coming out to be: S final = 6.9 mg BOD 5 /L S removal efficiency =(1-6.9/95)100% = 92.74% For determining X coming out of the system: Do a mass balance for substrate first. 0=QS 0 -QS f +Vr S here r S is rate of substrate utilization. sor S = -Q(S0-S)/V And we also know that r S = -kxs/[ K S +S] so {-Q(S0-S)/V} = -kxs/[ K S +S] X = [ (K S +S)(S0-S)(Q/V)]/(kS) here Q/V = 1/HRT solve for X. X=43 mg VSS/L Q16. Discuss removal efficiencies of different units for different contaminants in wastewater treatment plant with following schematic: Influent water Primary settling tank Biological aeration Secondary settling- Effluent water: Parameter Influent water After settling (and influent to aeration tank) 5 Effluent water BOD 200 mg/l 130 mg/l 30 mg/l suspended solids 240 mg/l 120 mg/l 30 mg/l phosphorous 7 mg/l 6 mg/l 5 mg/l nitrogen 35 mg/l 30 mg/l 26 mg/l

6 Removal efficiency table Parameter After settling after aeration tank BOD =(200- =(130- remarks more removal is observed in aeration tank than settling basin 130)/200=35% 30)/130=76.9% suspended solids =50% 75% more removal is observed in aeration tank than settling basin phosphorous 14.29% 16.67% more removal is observed in aeration tank than settling basin nitrogen 14.29% 13.33% comparable removal is observed in settling tank and in aeration tank Now for every contaminant, remaining concentration values need to be compared with receiving body standard (say for river) to determine if further water treatment is required and which parameters need to be removed and then one needs to decides about another unit process in the given treatment train. 6

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