Page 1 of 19 End Semester Examination PES-203: Unit Processes and Operations II (Biological) M.Tech. Environmental Science and Technology Instructor: Dr. Akepati S. Reddy Date: 26-05-2013Time: 3 hours (2-00 PM to 5-00 PM) Max. Marks: 100 Roll No.: Note: Please assume if any requisite data is not given. Attempt all parts of a question together. Q.11.1A 10 MLD STP is treating sewage with the following characteristics: BODu300 mg/L TSS280 mg/L The STP has a primary clarifier and an activated sludge process with secondary clarifier. The ASP is run at 10 day SRT. Values of Y and kd parameters of the ASP are 0.4 and 0.1/day respectively. The primary clarifier is removing suspended solids and BOD by 70% and 40% efficiencies respectively. Assume soluble BODu of the treated effluent as 5 mg/L. The treated effluent is having 40 mg/L of suspended solids. Consistencies of the primary sludge and secondary sludge are 4% and 1% respectively.The STP sludge is stabilized to the level of 10% biodegradable solids fraction in an anaerobic sludge digester (Bheema digester) at SRT/HRT equal to 20 days. Hemispherical dome of the digester has the capacity to store only 50% of the biogas generated. Find the quantities and biodegradable fractions of the primary sludge and the secondary sludge generated?Find the biogas generation potential of the total sludge? If a Bheema digester is used for the sludge stabilization, find the dimensions of the digester? Quantity of primary sludge generated q TSS Q =TSS level in the influent: 280 mg/L Flow rate (Q): 10 MLD and 10000 m3/day Removal efficiency: q: 70% Primary sludge generated: 1960 kg/day TSS of the clarified effluent: TSS(1-q): 84 mg/L Biodegradable primary sludge generated equivalent uOM BOD Q q =BODu = 300 mg/L Removal efficiency: q: 40% Page 2 of 19 Organic Matter equivalent (OMequivalent): 1/1.07 Biodegradable sludge generated: 1122 kg/day Biodegradable fraction of the primary sludge: 1121.5/1960 = 57.22% BODu of the clarified effluent: 300 (1-0.4) = 180 mg/L Volume of the primary sludge generated: 1960/0.04/1000 = 49 m3/day here consistency of the primary sludge is 4% Quantity of secondary sludge generated = Sludge generated in ASP Sludge washed out Sludge generated in ASP = net biomass synthesis + Cell debris generated + Sludge contributed by influent Net biomass synthesis ) ( 1) ( .SRT kS S Q YNBSRde i+=Flow rate of the influent: 10000-49 = 9951 m3/day Yield coefficient (Y): 0.4 Influent BODu (Si): 180 mg/L Treated effluent BODu: 5 mg/L Autooxidation coefficient (kd): 0.1/day Solids retention time: 10 days Net biomass synthesis rate: 248 kg/day Cell debris generation rate SRT kSRT k S S Q Yf CDGRdd e id. 1. ). ( .+= Flow rate of the influent: 10000-49 = 9951 m3/day Cell debris fraction of the autooxidized biomass (fd): assumed as 15% Cell debris generation rate: 52 kg/day Sludge contributed by the influent Flow rate of the influent: 10000-49 = 9951 m3/day TSS of the influent: 84 mg/L Biodegradable fraction of the influent TSS: 57.2% Sludge contributed by the influent: (10000-49) x 0.084 x (1-0.5722) =358 kg/day Sludge washed out in the secondary clarifier Flow rate of the influent: 10000-(49+26) = 9951 m3/day(here 26 m3 is volume of the secondary Page 3 of 19 sludge wasted) TSS of the secondary effluent: 40 mg/L Sludge washout rate: (10000-49) x 0.04 = 397 kg/day Secondary sludge wasted: (248 + 52 + 358) 397 = 261 kg/day Wasted secondary sludge volume: 261/0.01/1000 = 26 m3/day Biodegradable fraction of secondary sludge = net biomass synthesis / total sludge generated = 248/(248 + 52 + 358) = 37.7% Total STP sludge stabilized in the anaerobic sludge digester Volume of the total sludge: 49 + 26 = 75 m3/day Total quantity of the sludge generated: 2220 kg/day Overall consistency: 29.6 kg/m3 or 2.96% consistency Biodegradable ETP sludge:1960x0.5722+260.1x0.377 = 1220 kg/dayBiodegradable fraction of the ETP sludge: (1220)/(1960+261) = 54.9% Quantity of sludge left after stabilization: (1960+261) x (1-0.549) / 0.9 = 1113 kg/day Biodegradable sludge fraction in the stabilized sludge: 1113 x 0.1 = 111 kg/day Quantity of biodegradable sludge stabilized: 1220 111) = 1109 kg/day or 90.9% of the biodegradable sludge Quantity of COD removed during stabilization in the anaerobic digesterQuantity of primary sludge stabilized in terms of BODu or bCOD: 1960x0.572x0.909x1.07 = 1090 kg/day Quantity of secondary sludge stabilized in terms of BODu or COD: 261x 0.377 x 0.909 x 1.42 = 127 kg/day Total COD removed during stabilization: 1090 + 127 = 1217 kg/day Biogas generation potentialMethane production rate: 0.35 Nm3/kg COD (1 kg COD removed is assumed to produce 0.35 Nm3 of methane) Biogas generation rate: 0.35/0.65 = 0.538 Nm3/kg COD (methane content of the biogas is assumed as 65% by volume) Biogas generation potential: 0.538 x 1217 = 655 Nm3/day Dimensions of the Bheema digester Volume of the hemispherical gas holder: 655x0.5 = 328 m3 Radius of the dome: 5.389 m (Here 4/3 x 22/7 x R3 = 328 m3) Page 4 of 19 Diameter of the digester: 10.779 m (here diameter of the dome = diameter of the digester) Liquid volume of the digester: 75 x 20 = 1500 m3 (here SRT=HRT = 20 days and sludge flow rate is 75 m3/day 49 m3/day primary sludge and 26 m3/day is secondary sludge) Height of the digester: 1500 / (10.7792 x 22/7/4) = 16.438 m 1.2With the help of a schematic diagram show what happens to the biodegradable organic matter of wastewater during the secondary (biological) treatment? List and discuss the bio-kinetics parameters of both aerobic and anaerobic biological treatment?Provide a classification scheme and classify different biological treatment units you have studied under the Unit processes and operations -2 (biological)?

In the absence of oxygen, nitrate nitrogen is used as electron acceptor and in the process nitrate is converted to nitrous oxide and molecular nitrogen through denitrification.. In the absence of oxygen and nitrate nitrogen, sulfate is used as electron acceptor and in the process sulfur/sulfide is formed. When oxygen, nitrate and sulfate are absent, carbon dioxide is used as electron acceptor and methane gas is produced. Hydrolysis, acido-genesis, aceto-genesis and methano-genesis processes generate methane also from the acetic acid and hydrogen formed from the acido-genesis and aceto-genesis. qmax, Ks, Y and kd are the bio-kinetic parameters. The biokinetic parameters are temperature sensitive and need temperature correction. Aerobic treatment processesAnaerobic treatment processes Page 5 of 19 qmax. (maximum substrate utilization rate constant) Units of BODu utilized by a unit of microbial biomass per unit time Typical value is 6/dayIts value can be much lower. Ks (substrate concentration at qmax/2 level of substrate utilization) Mg/l or g/m3 or kg/m3 are the units Typical value is about 40 mg/LIts value can be much higher for anaerobic treatment processes. Y (sludge yield coefficient) A dimensionless parameter defined as ratio of the biomass synthesized to substrate utilized. Typical value is 0.4Value is much lower and typically taken as around 0.1 Kd (decay coefficient or auto-oxidation rate constant) Units of biomass autooxidized per unit of the microbial biomass per unit time. Typical value is 0.1Value is much lower and typically taken as around 0.04 Classification schemes of biological treatment units: Aerobic, anaerobic, and mixed and miscellaneous types Attached growth and suspended growth reactors Single stage or multistage reactors and hybrid reactors Mixed and miscellaneousWaste stabilization pondsAnaerobicpondsFacultativepondsPrimary facultativepondsSecondaryfacultativepondsMaturationpondsVegetated pondsConstructed wetland systemsFree water CWSSub-surfaceflow CWSMultigraderoughing filtersAerobicAttached growthExposedaerobicTricklingfiltersRBCSubmerged aerobicSAFFFABMBBRSuspenededgrothASPMBRAeratedlagoonsAlgalponds (oxidationditches) AnaerobicSuspended growthAnaerobiccontact processUASBAnaerobicbafflereactorMoving bed Attached growthUp-flow Down-flowfiltersExpandedbedFluidizedbedSingle stageMultistageHybrid reactors Biological treatment units for nutrient removal Nitrification units:Page 6 of 19 BOD removal and nitrification together, and BOD removal followed by nitrification. Nitrification in aerobic attached growth and in aerobic suspended growth reactors. Denitrification units: Pre-anoxic and post-anoxic denitrification unitsPhosphorus removal units (BOD removal also occurs): anaerobic fermentation and aerobic treatment and aerobic sludge wastage. Biological treatment units for sludge stabilization Anaerobic digestion KVIC model Janta model Deenbandhu model Bheema digester German model USA model Aerobic sludge digesters Land forming Vermin composting Marks: 2 x 10 = 20 Q.2 2.1Sewage with the following characteristics is treated in a UASB reactor.BOD5 at 20C240 mg/L TSS320 mg/L VSS65% of the TSS Biodegradable VSS80% of the VSS The UASB reactor has 256 m2 area. The reactor zone depth is 3.5 m and upflow velocity of the wastewater is 0.6 m/sec. Treated effluent of the UASB is having 60 mg/L of suspended solids and 80 mg/L of soluble BOD. Net sludge yield coefficient of the anaerobic oxidation is 0.09. Concentration of sludge in the sludge bed - sludge blanket zone is 3%. Sludge is drained out once the sludge bed-sludge blanket depth reaches 1/3rd of the reactor zone depth, and every time only 25% of the accumulated sludge is drained out.Find the frequency of sludge draining and the quantity of sludge drained each time?Assume total hydrolysis of the biodegradable volatile suspended solids and conservation of all other suspended solids. TSS accumulation rate (from the influent) Effluent loading rate: 14.4 m3/m2.day TSS of the wastewater: 0.320 kg/m3 BOD5 at 20C of the wastewater: 0.24 kg/m3 Page 7 of 19 BODu/BOD5 ratio: 1.6 (assumed for the municipal sewage)BODu loading rate: 5.53 kg/m2.day VSS of the wastewater: 65% of TSS or 208 mg/L or 0.208 kg/m3 Biodegradable suspended solids: 80% of VSS or 166.4 mg/L or 0.166 kg/m3 Non-biodegradable suspended solids: 154 mg/L or 0.154 kg/m3:TSS accumulation rate (from the influent): 2.218 kg/m2.day Biosolids accumulation rate Treated effluent BOD (BOD5 at 20C) = 80 mg/L or 0.08 kg/m3 BOD removal efficiency: (0.24-0.08)/0.24 = 66.7% Net yield coefficient: 0.09 Biosolids accumulation rate: BODu loading rate x BOD removal efficiency x Net yield coefficient = 0.332 kg/m2.day Total sludge accumulation rate: 2.218 + 0.332 = 2.55 kg/m2/day Sludge consistency: 3% Sludge volume accumulation rate: 2.55/0.03/1000 = 0.085 m3/m2.day Volume available in the reactor for sludge storage Reactor zone depth: 3.5 m Depth available for sludge storage: 3.5/3 = 1.17 m Depth of sludge accumulation allowed between two successive sludge drainings:0.292 m Time required for sludge accumulation by 0.292 m: 3.435 days Quantity of sludge drained out Reactor area: 256 m2 Draining depth: 0.292 Volume os sludge drained each time: 0.292 x 256 = 74.75 m3/per draining cycle Quantity of sludge drained out: 74.75 x 0.03 x 1000 = 2243 kg/ draining cycle

2.2Discuss the basis and the basic design equations, and also the approach followed for the design of a waste stabilization pond system comprised of an anaerobic pond, a facultative pond and a maturation pond system of 3 ponds connected in series? Waste stabilization pond system includes anaerobic, facultative and maturation ponds. Anaerobic pond: Volumetric organic loading rate is used as the basis of design of an anaerobic pond. The volumetric organic loading rate is temperature dependent and its value may range between 100 to 350 g/m3.day.The equations given below are used for calculating the volumetric organic loading rates.Page 8 of 19 Temp. T ( oC) Volumetric Loading (g/m3 d)25 350For the average ambient air temperature for the coldest month of the year for the site volumetric organic loading rate applicable is calculated. From the known wastewater flow rate and BOD, using the volumetric organic loading rate volume of the anaerobic pond is found. Depth of the pond and aspect ratio of the pond are assumed. Facultative pond Surface organic loading rate is used as the basis of the facultative pond.Surface organic loading rate is temperature (T) or latitude (L) or insolation (So) levels dependent and applicable loading rate is calculated by using one of the following equations: ( )Ts072 . 1 50 = ( )20085 . 1 357 =Ts( )25002 . 0 107 . 1 350 =TsT Ts10 = 90 20 = Ts60 20 = Tso sS = 07 . 1 Ls25 . 6 375 = For the average ambient air temperature for the coldest month of the year for the site surface organic loading rate applicable is calculated. From the known influent flow rate and BOD, using the surface organic loading rate, surface area of the anaerobic pond is found. BOD of the anaerobic pond effluent becomes BOD of the facultative pond influent. It is calculated using the following expressions: Temp. T ( oC) BOD removal (%)25 70Depth of the pond and aspect ratio of the pond are assumed. Maturation pondColiform count reduction to the desired level is used as the basis for the design of the maturation pond. Coliform count reduction is obtained by the following equation: nT BienkNN|.|

\| +=u) (11 ( )( ) 20) (19 . 1 6 . 2=TT BKHere n is number of maturation ponds connected in series. For using this equation coliform count in the influent of the maturation pond should be known and this one is calculated by using the following equation: ( )( )facl T B an T BrwfaclK KNNu u) ( ) (1 1 + +=For using this equation hydraulic retention of both anaerobic pond and facultative pond must be known. Depth of the pond and aspect ratio of the pond are assumed. 2.3An anaerobic digester of 6.0 m diameter and 6.0 m liquid depth and with hemispheric dome (to store the biogas generated) is run at 30 days HRT for stabilizing the sludge. Heat transfer coefficients for the digester floor, walls and (dome) cover are 1.7, 4.7 and 4.0 w/m2 C respectively. For proper sludge stabilization the digester contents are maintained at 35C through feeding the digester with preheated sludge. Average ambient temperature of the site is 23C. Find the temperature to which the sludge should be heated prior to the feeding? Metabolic heat generated by the digester is assumed to compensate 30% of the heat losses occurring from the digester. Liquid volume of the digester: H.Pi D2/4 = 169.65 m3 Sludge flow: 169.65/30 = 5.65 m3/day Heat loss from the digester floor Floor area of the digester: 18.85 m2 Reactor internal temperature: 35C Ambient air temperature: 23C Heat transfer coefficient: 1.7 W/m2.C Heat from the floor: (35-23)x1.7x18.85 = 384.54 W = 7948 Kcal/day Here 1 W is equal to 20.67 Kcal/day (1W x 3600 sec.hr x 24 hr/day / 4.18 J/cal / 1000 J/kJ = 20.67 Kcal/day) Page 10 of 19 (1 W per second is one Joule) Heat loss from the digester walls Wall area of the anaerobic digester: 113.1 m2 Heat transfer coefficient: 4.7 W/m2.C Heat loss from the walls: (35-23)x4.7x113.1 = 6378.84 W = 131850 Kcal/day Heat loss from the digester dome Dome area of the digester: 2 Pi r2 = 4x22/7x32 = 56.55 m2 Heat transfer coefficient: 4.0 W/m2.C Heat loss from the dome: (35-23)x4.0x56.55 = 2714.4 W = 56,098 Kcal/day Metabolic heat generated:30% of the total heat loss = (384.54 + 6378.84 + 2714.4)x0.3 = 2843.334 W = 58772 Kcal/day Heat content of the treated effluent: Flow rate: 5.65 m3/day Temperature: 35C Heat content: 197,750 Kcal/day Heat content of the influent sludge= total heat loss metabolic heat generated= 7948 + 131850 + 56,098 + 197,750 -58772 = 3,34,874 Kcal/day Temperature of the influent sludge: Sludge flow rate: 5.65 m3/day or 5650 L/day Heat content of the influent: 3,34,874 Kcal Temperature of the influent sludge: 3,34,874/5,650 = 59.3C Marks: 3 x 8 = 24 Q.3 3.1Wastewater entering into a 3nd stage RBC system has 17 mg/L of soluble BOD and 21 mg/L ammonical nitrogen. Find out the ammonical nitrogen concentration in the treated effluent coming out from the 3nd stage of the RBC system? Assume use of medium density RBC in the 3rd stage of the RBC system and HRT (retention time) of the 3rd stage standard RBC unit is 1.7 hours. BOD in the treated effluent ) / ( 00974 . 0 2) / ( 00974 . 0 4 1 11Q AS Q ASsn sn + + = Page 11 of 19 Influent BOD: 17 mg/L As is disk surface area for stage-n (in m2) = 11000 m2 Q is flow rate in m3/day: 635.3 m3/day BOD of the treated effluent = 9.47 mg/L Ammonical nitrogen nitrification rate:sBOD Frx = 1 . 0 0 . 1Frx = 1.0 0.1x 9.47 = 0.053 1.5 g/m2.day x Frx = 0.053 x 1.5 = 0.0795 g/m2.day Ammonical-N nitrification rate = 11000x0.0795 = 874.5 g/day Ammonical nitrogen concentration in the effluent = 21 - 874.5/635.3 = 19.6 mg/L 3.2Find ventilation rate for a trickling filter of 6.1 m height and 40 m2 top surface area, when the ambient air temperature is 34C and the sewage temperature is 25C? Take BOD loading rate as 1.5 kg/m3.day. Take specific surface area of the plastic filter medium used as 100 m2/m3. Dair, draft in mm of water column ZT TDh cair|.|

\| =1 1353Z, height of the filter in meters: 6.1 m Tc, cold temperature in K: 298 Th, hot temperature in K: 307 Dair = 0.212 mm or 0.00212 kPa (Np) number of velocity head resistances offered by the filter:( )ALpD N) 10 36 . 1 (5exp 33 . 10=D is depth of packing: 6.1 m BOD loading rate:1.5 kg/m3.day or 0.0625 kg/m3.hr L/A is liquid mass loading rate (kg/m2.hr): 0.381 kg/m2.hr Filter resistance (head loss in filter): 63.01 velocity heads Total head for a plastic filter medium of 100 m2/m3 specific surface area: 63.01 x 1.6 =100.82 velocity heads Air flow velocity: |.|

\|= AgVN PpT22 or 212 .||.|

\| A=PTNg PVP: 0.00212 kPa Page 12 of 19 NpT : 100.82 velocity heads Air flow ventilation velocity: 0.0203 m/sec. Ventilation rate: Filter surface area: 40 m2 Ventilation rate: 2925 m3/hr 3.3A UASB reactor is used for the treatment of 100 m3/day spent wash with 12 kg/m3 COD. If the upflow velocity in the reactor is to be maintained at 0.9 m/hour, if the reactor zone depth of the UASB is 4 m and if the volumetric loading rate is 12 kg/m3/day find the UASB reactor diameter. Further find the recycle ratio of the treated effluent required for the maintenance of the upflow velocity. Assume efficiency treatment of the UASB as 70%?Surface organic loading rate: Strength (COD) of the spent wash: 12 kg/m3Reactorzone depth: 4 m Volumetric loading rate: 12 kg/m3.day Surface organic loading rate: 12 x 4 = 48 kg/day Surface hydraulic loading rate (overflow rate) Upflow velocity in the reactor: 0.9 m/hr Hydraulic loading rate: 0.9 x 24 = 21.6 m3/m2.day or 21.6 m/day Strength of the influent (Ci) to the UASB: (Qf + Qr) Ci = 48 kg/m2.day Qf + Qr = 21.6 m3/m2.day Ci = 48 / 21.6 = 2.222 kg/m3 Strength of the treated effluent (Ce) from the UASB Efficiency of treatment: 70% Ce = (1 - 0.7) x Ci = 0.667 kg/m3 Qf and Qr values and recycle ratioQr = 21.6 Qf----- (from Qf + Qr = 21.6) 12 Qf + 0.667 Qr = 21.6 x 2.22212 Qf +0.667 (21.6 Qf) = 21.6 x 2.222 Qf = 2.96 m3/day Qr = 21.6 2.96 = 18.64 Recycle ratio: Qr / Qf = 18.64 / 2.96 = 6.3 Size of the UASB reactor Spent wash flow rate: 100 m3 Page 13 of 19 Surface loading rate of the spent wash: 2.96 m3/day Surface area of the UASB reactor: 33.8 m2 Diameter of the UASB reactor: 6.56 m 3.4Find the BOD5 : N : P ratio for an activated sludge process operated at 12 day SRT and 6 hour HRT. Take sludge yield coefficient as 0.4 and kd as 0.1/day, and BOD reaction rate constant as 0.23/day. Assume nitrogen and phosphorus levels in the treated (clarified) effluent of the ASP as 0.5 mg/L and 0.3 mg/L respectively. BODu: 100 mg/L --- (assumed value) Net synthesis of microbial biomass:) ( 1) ( .SRT kS S Q YNBSRde i+=Se is taken as negligible and QSi is taken as 100 mg--- (BODu)Yield coefficient: 0.4 Auto-oxidation coefficient: 0.1/day SRT (solids retention time): 12 days Net biomass synthesis: 100 x 0.4 /(1 + 0.1 x 12) = 18.182 mg/L Cell debris generation rate SRT kSRT k S S Q Yf CDGRdd e id. 1. ). ( .+=fd value: 0.15---- assumed value Cell debris generated: 0.15 (100 x 0.4 18.182) = 3.273 mg/l Total microbial biomass: 21.455 mg/L Nitrogen required for 100 mg of BODu Nitrogen content of the biomass: 21.455 x 0.123 = 2.639 mgNitrogen in the treated effluent: 0.5 mg/L Total nitrogen required: 0.5 + 2.639 = 3.139 Phosphorus required for 100 mg of BODu Phosphorus content of the biomass: 21.455 x 0.023 = 0.493 mg Phosphorus in the treated effluent: 0.3 mg/L Total phosphorus required: 0.3+0.493 = 0.793 BOD5 equivalent to BODu:exp(-k.t)} - {1 L BODo t =Page 14 of 19 BOD reaction rate constant: 0.23/day t is time: 5 days--- (for BOD5) BODt = 68.336 mg/L BODt : N : P ratio = 68.336 : 3.139 : 0.793 = 100 : 4.59 : 1.16 Marks: 4 x 6 = 24 Q.4 4.1Write note on a standard RBC unit? Standard RBC unit includes 3.5 m dia HDPE disks of different configurations or corrugation patterns are stacked on a horizontal steel shaft, and placed in a 45 m3 capacity tank.The shaft oriented either perpendicular to or parallel to the wastewater flow Steel shaft of square, round or octagonal shape, 13-30 mm thickness and 8.23 m length (7.23 m length is occupied by disks), with corrosion resistant coating is used. 40% of disc surface is submerged in the wastewaterand the disk surface is alternatively brought in contact with wastewater and atmosphere by rotating at 1 to 1.6 rpm rate either mechanical or pneumatically Mechanical drive of 3.7 or 5.6 kW capacity is used Deep plastic cups are attached to the perimeter of the disks and compressed air is released into the cups for rotation - Air requirement is 5.3 m3/min for standard density shaft and 7.6 m3/min for high density shaft RBC unit is usually provided with an enclosure to Prevents algal growth, to protect discs from sunlight (UV light), to prevent heat loss and to prevent exposure to cold weatherRBC units are of low, medium and high density types Low density or standard type units (9300 m2 disc area) are used for initial stages Medium and high density units (11000 and 16,700 m2 area) are used in the mid and final stages 4.2A trickling filter operating at 65% efficiency of BOD removal is supposed to treat the incoming sewage of 300 mg/L BOD to an outgoing effluent of 30 mg/L BOD. Find the treated effluent recycle ratio required?Recycle ratio: r = Qr/Qf--- definition Page 15 of 19 Recycle stream flow: rQf Feed water flow: Qf Flow to the trickling filter: Qf +rQf Recycle flow strength: 30 mg/L Feed wastewater strength: 300 mg/L Strength of the flow to TFTreated effluent strength: 30 mg/L Treatment efficiency: 65% Strength of the flow to TF: 30/(1-0.65) =85.7 mg/L Recycle ratio 300 Qf + 30 r Qf = 85.7 Qf (1+r) 300 + 30 r = 85.7 +85.7 r r = (300 85.7) / (85.7 30) = 3.85 4.3Draw a properly labeled line diagram and describe the UASB reactors being used in the STPs of Ludhiana? UASB reactor has a reactor zone and a settling zone. The reactor zone is divisible into sludge bed zone, sludge blanket zone and diffused sludge zone. There are provisions for the sampling the reactor contents from different elevations for the treatment process monitoring and control. There are provisions for the draining out of sludge from the sludge bed, sludge blanket and diffused sludge zones.Distribution tubes load sewage at the bottom of the reactor to flow upwards through the sludge bed and sludge blanket. The settling zone has a 3-phase separator comprising of deflectors, biogas collection tunnels and clarified effluent overflow weirs and collection troughs.The wastewater, together with the biogas bubbles generated, flows upward and get into the settling zone. Deflectors direct the biogas bubbles into the biogas collection tunnels. Wastewater free from biogas bubbles enter the settling zone and suspended solids of the wastewater settle on the inclined sufaces and slide down back into the reactor zone. Clarified water overflows into the clarified effluent trough and drained out from the UASB reactor, and Page 16 of 19 scum baffle do not allow washout of the floating scum. A distribution box placed over the reactor with multitude of distribution tubes facilitate the application of sewage to the UASB reactor. 4.4Write note on the fate of the total kjeldhal nitrogen (TKN) of the wastewater during biological treatment of wastewater? During BOD removal organic-N is converted to Ammonical-N. Ammonical-N can be assimilated by microorganisms and transformed back into organic-N Death and decomposition of microbial biomass convert organic-N back into ammonical-N Ammonical-N in excess can be released into atmosphere under favourable conditions Ammonical-N, when enough oxygen is available can be transformed into nitrite and nitrate-N by autotrophic bacteria Nitrate-N can be assimilated by plants and become organic-N. Death and decomposition of plant material can transform the organic-N into ammonical-N and release back. Nitrate-N by heterotrophs can be denitrified into nitrous oxide and molecular nitrogen under anoxic conditions and the formed N2O and N2 escape into the atmosphere. 4.5State BOD kinetics equation, identify the BOD kinetic parameters, and write note on any one method of finding out the BOD kinetics parameters?BOD kinetics equation is exp(-k.t)} - {1 L BODo t =This equation can be used to find BODt provided the values for K and Lo are known. BOD kinetic parameters are Lo and K K is temperature sensitive and its value needs temperature correction which is done through using modified Arhenius equation. KT=K20T-20

There are many methods for finding the BOD kinetic parameters. All the methods require serial BOD test results for the sample in question. The methods include Method of least squares Method of moments (Moore et al. 1950) Log difference method (Fair, 1936) Page 17 of 19 Fugimoto method (Fujimoto, 1961) Daily difference method (Tsivoglou, 1958) Rapid ratio method (Sheehy, 1960) Thomas method (Thomas, 1950) Method of Moments Moores diagram (a nomograph relating K with EBOD/L0 and EBOD/E(BOD.t)) is needed (Moores diagram is different for different n value)Results of serial BOD test for n days are used to find EBOD and EBOD/ E(BOD.t)EBOD/E(BOD.t) value is used to read k value and EBOD/L0 value from the Moores diagram From EBOD/L0, since EBOD is known, L0 is found Marks: 5 x 4 = 20 Q.5 5.1Write note on flow distribution boxes? Flow distribution boxes are used to divide the incoming flow n number of streams, and each of the streams is carried by a distribution tube into the UASB reactor. The incoming flow is made to overflow a weir that is divided into n number of equal length sections. This ensures division of the incoming flow (even if it is highly variable) equally among the outgoing streams. 5.2 Write note on degree-days? In aerobic sludge digestion, stabilization rate of sludge increases with increasing temperature and with increasing solids retention time. The temperature and the SRT are combined together, expressed as degree-days, and used in the design. Typical degree-days used for the aerobic stabilization is about 550. At this the stabilization level will be 38% reduction of the sludge amount. 5.3Write note on poly-hydroxi-butyrate? Page 18 of 19 Poly-hydroxi-butyrate is a fermentation product formed in the anaerobic tank in the phosphate accumulating organisms. In the aerobic reactor, the poly-hydroxi-butyrate is consumed in return to phosphate accumulation in the phosphate accumulating organisms.

5.4Differentiate the pre-anoxic denitrification from the post-anoxic denitrification? Denitrification is anaerobic process. Oxygen limiting conditions, presence of nitrate, and availability of substrate (biodegradable organic matter) facilitate the denitrification process. In the pre-anoxic denitrification process, the incoming wastewater is mixed with nitrified mixed liquor pumped back into a pre-anoxic tank under limited oxygen condition. In the post anoxic tank, the nitrified mixed liquor is allowed to into the post-anoxic tank for the denitrification. Here the mixed liquor may have limited amount of biodegradable organic matter. Either through supplying fresh wastewater or through dosing easily biodegradable substrate like methanol, the substrate limitation is madeup. 5.5Write note on nitrogenous BOD? In the BOD testing nitrification of ammonical nitrogen can occur if DO level is high and if nitrifying bacteria (autotrophs, nitrosomonas and nitrobacter) are present. Source of the amoonical nitrogen could be the incoming wastewater or autooidation of microbial biomass. The nitrification demands oxygen and this demand is known nitrogenous BOD. Using nitrification inhibitors, incubation of sample for