WATER POLLUTION Principle and Control

WASTEWATER TREATMENTPrinciples Biological TreatmentBySumarno

IntroductionBiological Wastewater TreatmentRemoval the organic contaminats in domestic and or industrial wastewater by appropriate treatment process to render the water suitable for discharge to surface waterAnaerobic, Anoxic and Aerobic Processes

In biological treatment the organic contaminants serve as the energy source (electron donor) in the reaction and oxygen, nitrite/nitrate, sulfate or carbon dioxide serve as the electron acceptor.Anaerobic processes use sulfate or carbon dioxide as electron acceptor.Anoxic processes use nitrite/nitrate as their electron acceptor.Aerobic processes use oxygen as electron acceptor Anaerobic Anoxic Aerobic

- 200 mV 0 mV + 200 mV Fig 1. ORP Range (mV) for Wastewater Treatment

Methanogenesis Denitrification Nitrification Activated SludgeAll biological treatment processes are conversion processes in that they convert the readily bio- degradable organic contaminants (soluble/colloidal) into two fraction: (1) a gas which escape from liquid, (2) the excess biomass.

The Dillema of Biomass DisposalThe cell wall of biomass are very complex and therefore quite refractory to further biotransformation.Under aerobic and anaerobic conditions, only 25 40% of resulting biomass which is synthesized may be further biodegraded, the remaining is so refractory that it cannot practically be destroyed (Gosset and Belser 1982)The Excess BiomassAerobic treatment normally produced about 10 times more refractory biomass than anaerobic.The excess biomass presents an greater dillema due to (1) need sludge processing area which is may increasingly be restricted in the future;(2) its possible listing as hazardous waste classification which is necessitates special processing. Treatment Process Selection Anaerobic and Aerobic Processes SelectionAwareness of Aerobic Process Disadvantages Biomass disposal problem caused by large volume of refractory biomass and possibility listing as hazardous wasteAlternative Technology Advantages Capable to biotransformed refractory and or toxic substances anaerobically Nutrient requirement of anaerobic processes only 5 20% of those for aerobic processesMinimal Waste Biomass Anaerobic treatment processes give only 5 20% as much waste biomass compared to aerobic treatment processesRetention Time Requirement Volumetric loading rate anaerobic treatment are often > 10 times than aerobic processes (5 -10 kg/m3 d for anaerobic and 0.5 1.0 kg/m3 d for aerobic) Smaller Reactor VolumeEnergy Requirement and Production No aeration energy requirement for anaerobic processes Energy methane production 12,000 BTU/kg COD biodegradedConstruction Anaerobic processes construction more simple than aerobic processesAchievement of Secondary Effluent Standard Anaerobic processes incapable of producing an effluent which meets standard

Anaerobic Wastewater TreatmentPrinciples of Anaerobic Treatment

Complex Organic Compounds 75% HydrolysisSimple Organic Compounds 5% 10% 30% Acidogenesis 35% 20% Long Chain Fatty Acids 13% 17% H2 and CO2 Acetate 28% Methanogenesis 72% CH4 and CO2Fig 2. Series Metabolisme of Anaerobic Processes

Optimal Anaerobic Treatment Processes Proper pH The pH reactor must range between 6.5 8.2 Adequate macronutrients Nitrogen concentration in the reactor must ranging 40 70 mg/L Sulfide precursor should be added, commonly in the sulfate form Trace metal bioavailability Obligate micronutrients requiremen The obligate requirement for sulfide and trace metal by methanogens Iron to be required in the highest concentration Temperature Commonly anaerobic reactor are operated at mesophilic temperature of 30 370 C

Toxicity acomodation The anaerobic process can accomodate toxicity of various forms in industrial wastewater and even biodegrade certain toxicant e.g. CCL4, tetrachloroethylene, formaldehyde , acrylate, trichloroethylene, chloroform , cyanide Adequate metabolisme time Two measured of time are involved Hydraulid Retention Time (HRT) and Solid Retention Time (SRT) Carbon source for synthesis Electron donor Biodegradable COD Electron acceptor CO2 and/or sulfateOperational Consideration

BOD/COD assay BOD assay has value when applied to anaerobically treated effluent in that it indicate pollutant load for a subsequent treatment unit. A plot log COD vs 1/HRT for treatment processes indicate the non- degradable fraction at the intercept Biochemical Methane Potential Assay Assaying the concentration of organic pollutant in a wastewater which can be aerobically converted to CH4. Evaluating potential anaerobic process efficiency Testing for non-biodegradables remaining after treatment Anaerobic Toxicity Assay A simple assay procedure to evaluate the potential toxicity of a wastewater sample to the anaerobic biomassAerobic Wastewater TreatmentPrinciples of Aerobic TreatmentFermentation AerobicOrganic C + O2 Energy + CO2 + H2O + Residue microbes

New Aerobic + O2 Energy + CO2 + H2O + Residue microbes

New Aerobic + O2 Energy + CO2 + H2O + microbes ResidueNew Cell BiosynthesisTwo kinds of ingredients are required for the biosynthesis of cell components: (1) precursors that provide the carbon, hydrogen, nitrogen, and other elements found in cellular structures, and (2) adenosine triphosphate (ATP) and other forms of chemical energy needed to assemble the precursors into covalently-bonded cellular structure.

microbesSimple precursors C 60H87 N12O23P energy new cellEndogenous RespirationUnder substrate-limited conditions, microbes will feed on each other at a higher rate than new cells can be produced. The aerobic degradation of cellular material is endogenous respiration

aerobic C 60H87 N12O23P CO2 + H2O + NH4+ + Residue microbesOptimal Aerobic Treatment Processes

Temperature The rate of bio-oxidation is a function of temperature. Various microbial species have optimal temperatures for survival and cell synthesis. Psychrophilic microorganisms thrive in a temperature range of -2 to 30C . Optimum temperature is 12 to 18C Mesophilic microorganisms thrive in a temperature range of 20 to 45C . Optimum temperature is 25 to 40C Thermophilic microorganisms thrive in a temperature range of 45 to 75C. Optimum temperature is 55 to 65C.

Food to Microorganisms Ratio (F/M)

Acid Concentration The influent pH has significant impact on wastewater treatment. Benefield and Randall (1985) report that it is possible to treat organic wastewaters over a wide pH range, however the optimum pH for microbial growth is between 6.5 and 7.5.Anoxic Treatment ProcessesPrinciple of Anoxic ProcessesIn the wastewater stabilization pond revealed that the wastewater treatment mechanisms are mainly based on biochemical reaction in the anoxic system

The algal activities in the presence of light proved to be important in the surface layer of the pond and facultative bacteria were working mutually with the algae. Motile flagellate algae (Euglena and Chlamydomonas) were the only species found to exist under anoxic conditions.It was also determined that sulphate-reducing bacteria predominated in the lower volume of the anoxic ponds, rather than acidogenic bacteria, and this caused sulphide and hydrogen sulphide build-up in the ponds contents. The bottom volume of the ponds and the benthic sludge in anoxic ponds contained acid producers and methanogenic bacteria causing the release of biogas.Which one or combined biological treatment processes do you choose is depend on strength of organic pollutants of wastewater, effluent standard and added value from treatment processes.DESIGN OF ACTIVATED SLUDGE PROCESSProcess DescriptionThe activated sludge process is an aerobic growth biological treatment method in which oxygen required substances are removes by biochemical reactions using microorganisms. BacteriaCOHNS + O2 CO2 + NH3 + C5H7NO2 + Other end products Bacteria C5H7NO2 + 5 O2 5 CO2 + 7 H2O + NH3 + energy V, S, X Aeration TankSedimentation TankRecycleQR , S2 , XRQ0 , S0Q0 - QWFig 3. Flow Scheme of Conventional Activated SludgeQW, XW = XRWastageTerminologyACTIVATED SLUDGE floc of microorganisms that form when wastewater is aeratedMIXED LIQUOR mixture of activated sludge and wastewater in the aeration tankMIXED LIQUOR SUSPENDED MATTER (MLSS)measure of the amount of suspended solids in the mixed liquor expressed in mg/lMIXED LIQUOR VOLATILE SUSPENDED MATTER (MLVSS) proportional to the microorganisms concentration in the aeration tankMEAN CELL RESIDENCE TIME (MCRT) the average time a microorganism spends in the treatment processFOOD TO MICROORGANISM RATIO (F/M) ratio of the amount of food expressed as pounds of COD (or BOD) applied per day, to the amount of microorganisms, expressed as the solids inventory in pounds of volatile suspended matter. RETURN ACTIVATED SLUDGE (RAS)settled mixed liquor collected in the clarifier underflow and returned to the aeration basinWASTE ACTIVATED SLUDGE (WAS)excess growth of microorganisms which must be removed to keep the biological system in balance. Various control techniques have been developed to estimate the amount of WAS that must be removed from the processCOMPLETE MIX ACTIVATED SLUDGEan ideal mixing situation where the contents of the aeration tank are at a uniform concentrationPLUG FLOW ACTIVATED SLUDGEan ideal situation where the contents of the aeration tank flows along the length of the tankWASTE ACTIVATED SLUDGE (WAS)excess growth of microorganisms which must be removed to keep the biological system in balance. Various control techniques have been developed to estimate the amount of WAS that must be removed from the processCOMPLETE MIX ACTIVATED SLUDGEan ideal mixing situation where the contents of the aeration tank are at a uniform concentrationPLUG FLOW ACTIVATED SLUDGEan ideal situation where the contents of the aeration tank flows along the length of the tankBACK MIXINGmixing the contents of a tank in the longitudinal or flow oriented directionTRANSVERSE MIXING (or CROSS ROLL) mixing in a direction across the direction of flowSLUDGE REAERATIONpractice of aerating the RAS before it is added to the mixed liquorPROCESS LOADINGorganic loading range as measured by the F/MCONVENTIONAL LOADINGprocess loading of 0.2 to 0.5 lbs BOD applied/lb MLVSS/dayHIGH RATE LOADINGprocess loading of two to three times the conventional loading rateEXTENDED AERATION LOADINGlow rate loading that is one half to one tenth of the conventional loading rateSETTLEABILITYmeasure of the volume occupied by the mixed liquor after settling in a graduated cylinder for 30 minutesgenerally expressed as a percentage based on the ratio of the sludge volume to the supernatant volumeSOLIDS INVENTORY (VOLATILE SOLIDS)amount of volatile suspended solids in the treatment systemSOLUBILITY INDEX ( SI )soluble BOD per total BOD

Aeration SystemAeration is provided by either diffused or mechanical aeration systems. Diffused air systems consist of a blower and a pipe distribution system that is used to bubble air into the mixed liquor. Mechanical aeration systems consist of pumps or mixersDiffused Air SystemDiffused air systems are the most common types of aeration systems used in activated sludge plants. Produce fine or coarse bubbles. Fine Bubble DiffusersFine bubble diffusers are easily clogged by biological growth and by dirty air, resulting in high maintenance costs.The air supply for all fine bubble diffusers should be filtered.

Fig 4. Fine Bubble DiffuserSurface AeratorMechanical Aerator Floating or fixedBrushInjectorOthers

Fig 5. Mechanical AeratorBrush aeratorInjectoraeratorFloatingaeratorDESIGN OF ACTIVATED SLUDGE PROCESSIn the past, designs of biological wastewater treatment processes were based on the empirical parameters developed by experience, which included hydraulic loading, organic loading and retention time.

Nowadays, the design utilizes empirical as well as rational parameters based on biological kinetic equations. Theses equations describe growth of biological solids, substrate utilization rates, food-to-microorganisms ratio, and the mean cell residence time.DESIGN CRITERIA FOR ACTIVATED SLUDGE PROCESSReactor type selectionLoading CriteriaSludge ProductionOxygen Demand and TransferNutrient RequirementControl of Filamentous OrganismEffluents CharacteristicsSelection of Reactor TypeSelection of the reactor type requires several considerations as follows:Reactions kineticOxygen transfer requirementsNature of wastewaterLocal environments conditionsConstruction and maintenance costs

Loading CriteriaThere are 2 main parameter for the design and control of the activated sludge process:The Food to Microorganism Ratio (F/M ratio) F/M = total applied substrate rate/total microbial biomass = Q . S0/V . X = S0 / . XF/M = food to microorganisms ratio, h-1S0 = Influent BOD or COD concentration, mg/LQ = Influent wastewater flowrate, m3/hV = Aeration tank volume, m3 = hydraulic detention time, V/Q, hX = concentration of volatile suspended solid (MLVSS) in aeration tank, mg/L

The Mean Cell Residence Time (c)

c = Vr . X /(Qw . X w + Qe . Xe )

c = the mean cell residence time based on aeration tank, dVr = aeration tank volume, m3X = concentration MLVSS in aeration tank, mg/LQw = excess sludge flowrate, m3/d Xw = concentration of MLVSS in excess sludge, mg/LQe = effluent flowrate, m3/d Xe = concentration of MLVSS in effluent, mg/L

The organic loading parameter and others design parameter for activated sludge processes listed in table 1.

Table1. Design Parameter for Activated Sludge Process (Metcalf et all, 1991)Proses

hrsF/M

Kg BOD/kg MLSS . dVolumetric LoadingKg BOD/m 3 .d MLSS

mg/LV/Q

hrsQ R/QConventional

5-150.2-0.60.32 -0.641500-30004-80.25-0.75Complete Mix5-150.20.40.80-1.922500-40003-50.25-1.0Step Feed5-150.2-0.40.64-0.962000-35003-50.25-0.75Modified Aeration0.2-0.51-5-5.01.2-2.4200-10001.5-3.00.05-0.25Contact Stabilization5-150.2-0.60.96-1.201000-30004000-100000.5-1.03-60.5-1.5Extended Aeration20-300.05-0.150.16-0.403000-600018-360.5-1.5Proses

hrsF/M

Kg BOD/kg MLSS . dVolumetric LoadingKg BOD/m3..dMLSS

mg/LV/Q

hrsQ R/QKrauss Process

5-150.3-0.80.64-1.62000-30004-80.5-1.5High-rate Aeration5-100.4-1.51.6-16.04000-100002-41.0-5.0High Purity Oxygen3-100.25-1.01.6-3.22000-50001-30.25-0.5Oxidation Ditch10-300.05-0.30.08-0.483000-60008-360.75-1.5Microorganism and Substrate BalanceThe term V . MLSS is function of Solid Retention Time (SRT) or c and not Hydraulic Retention Time (HRT) or return sludge ratio, the FM ratio is also function of SRT.Therefore, operation of an activated sludge plant at constant SRT will result in operation in at aconstant FM ratio.The mass balance for microorganism in the entire activated sludge system is expressed as the rate of accumulation of the microorganisms in the inflow plus net growth, minus that in outflow.Mathematically , it is expressed as (Metcalf and Eddy Inc, 1991)

V dX/dt = Q X0 - V rg - (Qw Xw + Qe Xe ) where V volume of aeration tank, m3dX/dt = rate of change of microorganisms concentration (VSS), mg /(l . m3 . d)Q = influent flow, m3/dX0 = microorganisms concentration (VSS) in influent, mg/LX = microorganisms concentration in tank, mg/Lrg = net rate of microorganism growth (VSS), mg/(l . d)

The net rate of bacterial growth is expressed as rg = Y rsu kd Xwhere Y = maximum yield coefficient growth, mg/mg over finite period of logrsu = substrate utilization rate, mg/(m3 d)kd = endogenous decay coefficient, per day

Substituting this equation into above equation, and assuming in the influent is zero and steady-state conditions, this yields

Qw Xw + Qe X e - Y rsu = - kd V X X

1 - Y rsu = - kd c X

The term 1/ is the net specific growth rateThe term rsu can be computed from the following equation r = Q/V (S0 - S) = (S0 - S)/

where S0 - S mass concentration of substrate utilized, mg/LS0 substrate concentration in influent, mg/LS substrate concentration in effluent, mg/L hydraulic retention time, dayEffluent microorganism and substrate concentrationThe mass concentration of microorganism X in the aeration tank Y (S0 - S) m (S0 - S)X = = (1 + kd c) k (1 + kd c)

Wherem = maximum specific growth rate, per dayk = maximum rate of substrate utilization per unit mass of microorganism, per day

The substrate concentration in effluent S can be determination from the substrate mass balance by the following equation

K (1 + kd c)S = c (Y k - kd) 1

WhereS = effluent substrate (soluble BOD5) concentration, mg/LKs half velocity constant, substrate concentration at one half of maximum growth rate, mg/lProcess design and control relationships. In practice, the relationship between specific substrate utilization rate , mean cell residence time (c), and the food to microorganism ratio (F/M) is commonly used for activated sludge process design and process control.

Complete Mix Activated Sludge ProcessIn this modification of activated sludge process, fresh feed and recycled sludge are combined and then introduced in several points in the aeration than from a central channel.Aerated liquor leaves the reactor from effluent channels on both side of the aeration tank (fig. )Oxygen supply and demand are uniform along the tank.Sludge RecycleEffluentWastageInfluentAeration TankClarifierFig 6. Flow Scheme of Complete Mix ProcessContact Stabilization ProcessContact stabilization is another modification of the activated sludge process. A flow diagram for the system is shown in the figure 4. Influent wastewater is mixed with stabilized sludge, and this mixture is aerated in the initial contact tank for which detention time only 20 -40 minute. During initial contact an appreciable fraction of suspended and dissolved BOD is removed by biosorption after contact with well aerated activated sludge. The mixed effluent from the initial contact tank flow to clarifier. Clarified effluent is removed and underflow from the clarifier is taken to stabilized tank, where it is aerated for a period of 1.5 5 hrs.During the this stabilization period, biosorbed organic are broken down by aerobic degradation.Stabilized sludge leaving the stabilization tank is in starved condition and ready for adsorbed organic wasteInitial Contact TankClarifierSludge RecycleStabilized sludgeInfluentEffluentFig 7. Flow Scheme of Contact StabilizationStabilization TankWastageStep Aeration Process ( Step Feed Process )Step aeration is modification of the conventional activated sludge is shown in figure , in which fresh feed is introduced in several point along aeration tank. This arrangement is provide for an equalization F/M ratio along the tank. The aeration tank is divided by baffles into several channel. Each channel constitute of one step of process and the step are linked together in series Aeration TankSedimentation TankRecycle sludgeInfluentEffluentWastageFig 8. Flow Scheme of Step AerationTapered AerationThe purpose of tapered aeration is to match amount of air supply with oxygen demand along the aeration tank.Since at the inlet the oxygen demand is the highest, aerators are space more closely to provide tha higher oxygenation rate. Spacing between aerators increase toward the outlet as oxygen demand decrease. Aeration TankSedimentation TankRecycle sludgeInfluentEffluentWastageFig 9. Flow Scheme of Tapered AerationAerationOxidation DitchFigure show the diagram of the oxidation ditch. An essential part of the system is an aeration ditch provided with an aeration rotor.This rotor has two function : (1) aeration and (2) provision of a flow velocity to mixed liquor in the ditch.Liquid velocity is of the order of 0.3 m/s.The mixture of wastewater and activated sludge repeatedly passed over the aeration rotor at short intervals. A typical rotor has diameter of approximately 75 cm revolved at about 75 rpm.

InfluentAeration rotorClarifierRecycle sludgeWastageEffluentFig 10. Flow Scheme of Oxidation ditchEffluentAeration rotorSludge ProductionDesign of the sludge handling and disposal facilities depends on sludge produced. Quantity of sludge can be estimated by using the equation

Px = Yobs . Q . (S0 - S )

Px = excess sludge (kg/d)Yobs = observed yield (g/g)S0 = Influent BOD or COD concentration, mg/lS = Effluent BOD or COD concentration, mg/lQ = Influent flowrate (m3/d)The observed yield value can be computed using equation Yobs = Y/( 1 + kd . c ) Y = Cell yield coefficient (mg cell produced per mg organic matter removed)kd = endogenous decay coefficient, day-1 c = mean cell residence time, day

Oxygen requirement and transferBOD and wasted sludge per day can be used for estimation of oxygen requirement bacteria C 60H87 N12O23P CO2 + H2O + NH4+ + energy cell

Carbonaceous oxygen demand can be defined asKg O2/d = (total mass of BODL removed, kg/d) (1.42 . mass sludge wasted, kg/d)The BOD of the cell is equal to 1.42 times excess sludgeKg O2/d = (Q . (S0 - S )/F 10-3 mg/kg, kg/d) (1.42 . mass sludge wasted, kg/d)F = conversion factor for converting BOD5 to BODLIf nitrification is desired O2 required to organic nitrogen oxidation should be added to carbonaceous oxygen demand Kg O2/d = (Q . (S0 - S )/F 10-3 mg/kg, kg/d) (1.42 . mass sludge wasted, kg/d) + 4.33 .Q . (N0 - N) 10-3 mg/kg, kg/dN0 = Influent TKN, mg/lN = Effluent TKN, mg/lAir supply must provide :Satisfy the BOD of wastewater,Satisfy the endogenous respiration by sludge organism,Complete mixing,Satisfy dissolved oxygen minimum 2 mg/l.

Example 1.An activated sludge process has a influent BOD concentration of 500 mg/l, influent flow 18,900 m3/d and 56,500 kg of suspended solids under aeration. Calculate the F/M ratio. Assume VSS is 80% of TSS.

SolutionStep 1. Calculate BOD in kg/dBOD = Q . BOD = 18,900 m3/d . 500 mg/l 10-6 kg/mg . 103 l/m3 = 9,450 kg/dStep 2. Calculate the VSS under aerationMLVSS = 56,500 kg . 0.8 = 45,200 kgStep 3. Calculate the F/M ratioF/M = 9,450/45,200 = 0,209 kg BOD/d per kg MLSS

Example 2.Design a complete-mix activated-sludge system.Given:Average design flow 0.32 m3/sPeak design flow 0.80 m3/sRaw wastewater BOD5 240 mg/LRaw wastewater TSS 280 mg/LEffluent BOD5 20 mg/LEffluent TSS 24 mg/LWastewater temperature 300 COperational parameters and biological kinetic coefficients:Design mean cell residence time c = 10 dMLVSS 2400 mg/L (can be 3600 mg/L)VSS/TSS = 0.8TSS concentration in RAS = 9300 mg/LY = 0.5 mg VSS/mg BOD5kd = 0.06/dBOD5 /ultimate BODU = 0.67

Assume:1. BOD (i.e. BOD5) and TSS removal in the primary clarifiers are 33% and 67%, respectively. 2. Specific gravity of the primary sludge is 1.05 and the sludge has 4.4% of solid contents3. Oxygen consumption is 1.42 mg per mg of cell oxidized.

Solution:Step 1. Calculate BOD and TSS loading to the plantDesign flow Q = 0.32 m3/s . 86,400 s/d = 27,648 m3/dSince 1 mg/L = 1 g/m3 = 0.001 kg/m3BOD loading = 0.24 kg/m3 . 27,648 m3/d = 6,636 kg/dTSS loading = 0.28 kg/m3 . 27,648 m3/d = 7741 kg/d

Step 2. Calculate characteristics of primary sludgeBOD removed = 6636 kg/d . 0.33 = 2190 kg/dTSS removed = 7741 kg/d . 0.67 = 5186 kg/dSpecific gravity of sludge = 1.05Solids concentration = 4.4% = 0.044 kg/kgSludge flow rate = 5,186 /[(1.05 . 1000 kg/m3)0.044] = 112 m3/d

Step 3. Calculate flow, BOD, and TSS in primary effluent (secondary influent)Flow = design flow = 27,648 m3/d - 112 m3/d = 27,536 m3/d = Q for Step 6BOD = 6636 kg/d - 2190 kg/d = 4,446 kg/dBOD = (4446 kg/d . 1000 g/kg)/27,536 m3/d = 161.5 g/m3 = 161.5 mg/l = S0TSS = 7741 kg/d - 5186 kg/d = 2,555 kg/d = (2,555 kg/d . 1000 g/kg)/27,536 m3/d = 92.8 g/m3 = 92.8 mg/l

Step 4. Estimate the soluble BOD5 escaping treatment, S, in the effluentUse the following relationshipEffluent BOD = influent soluble BOD escaping treatment,S + BOD of effluent suspended solidsa. Determine the BOD5 of the effluent SS (assuming 63% biodegradable)Biodegradable effluent solids = 24 mg/L . 0.63 = 15.1 mg/LUltimate BODu of the biodegradable effluent solids = 15.1 mg/L . 1.42 mg O2/mg cell = 21.4 mg/LBOD5 = 0.67 BODu = 0.67 . 21.4 mg/L = 14.3 mg/L(b) Solve for influent soluble BOD5 escaping treatment20 mg/L = S + 14.3 mg/LS = 5.7 mg/L

Step 5. Calculate the treatment efficiency E using Eq. E = (S0 - S)/S0 . 100%(a) The efficiency of biological treatment based on soluble BOD is = [(161.5 5.7)/161.5] x 100 = 96.5%(b) The overall plant efficiency including primary treatment is = [(240 20/240] x 100 = 91.7%

Step 6. Calculate the reactor volume using Eq. V = [c Q Y ( S0 - S)] / [X (1 + kd c )c = 10 dQ = 27,536 m3/d (from Step 3)Y = 0.5 mg/mgS0 = 161.5 mg/L (from Step 3)S = 5.7 mg/L (from Step 4b)X = 2400 mg/Lkd = 0.06 d-1 (10 d) (27,536 m3/d) (0.5) (161.5 5.7) mg/l V = (2,400 mg/l) ( 1 + 0.006 day-1 . 10 days)

= 5,586 m3Step 7. Determine the dimensions of the aeration tankProvide 4 rectangular tanks with common walls. Use width-to-length ratio of 1:2 and water depth of 4.4 m with 0.6 m freeboardw . 2 w . (4.4 m) . 4 = 5586 m3 w = 12.6 mwidth = 12.6 m and length = 25.2 mwater depth = 4.4 m (total tank depth 5.0 m)

Step 8. Calculate the sludge wasting flow rate from the aeration tank total mass SS in reactor Vr X c = = SS wasting rate Qw Xw + Qe Xe (5,586 m3/d) (2,400 mg/l)10 days = Qw (9,300 mg/l) + (27,536 m3/d) (24 mg/l . 0.8) Qw = 270 m3/d

Step 9. Estimate the quantity of sludge to be wasted daily(a) Calculated observed yield Y 0.5Yobs = = = 0.3125 1 + kd c 1 + 0.06 (10 days) (b) Calculate the increased in the mass of MLVSSpx = Yobs Q (S0 - S) . (1 kg/1000 g) = 0.3125 . 27,536 m3/d . (161.5 5.7) g/m3 . 0.001 kg/g = 1341 kg/d

(c) Calculate the increased in MLSS (or TSS)pss = (1341 kg/d)/0.8 = 1676 kg/d

(d) Calculate lost of TSS in effluentpe = (27,536 270) m3/d . 24 g/m . Kg/1000 g = 654 kg/dNote: Flow is less sludge wasting rate from Step 8.

(e) Calculate the amount sludge that should be wastedWastewater sludge = pss - pe = (1676 654) kg/d = 1022 kg/d

Step 10. Estimate return activated sludge rateUsing a mass balance of VSS, Q and Qr are the influent and RAS flow rates, respectively. VSS in aerator = 2400 mg/LVSS in RAS = 9300 mg/L . 0.8 = 7440 mg/L2400 (Q + Qr) = 7440 . QrQr /Q = 0.476Qr = 0.4762 . 27,536 m3/d = 13,110 m3/d = 0.152 m3/s

Step 11. Check hydraulic retention time (HRT = ) = V/Q = 5586 m3/(27,536 m3/d) = 0.203 d = 4.87 hNote: The preferred range of HRT is 515 h.

Step 12. Check F/M ratio using U in Eq. S0 - S (161.5 - 5.7) mg/lU = = = 0.32 day-1 X (0.203 day) (2,400 mg/l)Step 13. Check organic loading rate and mass of ultimate BODu utilized Q S0 27,536 m3/d . 161.5 g/m3 Loading = = V 5586 m3 . 1000 g/kg

= 0,80 kg BOD5/m3 d

BOD5 = 0.67 BODu (given)BODu used = Q (S0S)/0.67 = [27,536 m3/d. (161.5 - 5.7) g/m3]/0.67 = 6403 kg/d

Step 14. Compute theoretical oxygen requirementsThe theoretical oxygen required is calculated from Eq. Q (S0 S) O2 = 1000 g/kg F = 6403 kg/d (from Step 13) - 1.42 . 1341 kg/d (from Step 9b) = 4499 kg/d

Step 15. Compute the volume of air requiredAssume that air density 1.202 kg/m3 and contains 23.2% mass oxygen, the oxygen transfer efficiency for the aeration equipment is 8% and a safety factor 2 is used to determined the actual volume for sizing the blower.(a) The theoretical air required is 4499 kg/dAir = = 16,200 m3/d 1.202 kg/m3 . 0.232 g O2/g air

(b) The actual air required at an 8% oxygen transfer efficiencyAir = (16,200 m3/d)/0,08 = 02,000 m3/d = 140 m/min

(c) The design air required (with a factor of safety 2) isAir = 140 m3/min x 2 = 280 m3/mDesign Based on Biokinetic Equations.Microbial GrowthExponential Phase:dX/dt = XX = cell concentration (mg dry mass or VSS/L)dX/dt = volumetric cell production rate (mg/L d) = specific growth rate (h-1 or d-1 )rg = dX/dt = Xrg = volumetric cell production rate (g VSS m-3 d-1)X = cell concentration (g VSS L-1) = the specific growth rate (d-1)Likewise: rSU = dS/dt = - q Sr SU = volumetric substrate consumption rate (g S m-3 d-1 )S = substrate concentration (g S m-3)q = maximum specific rate of substrate degradation, d-1Y = True Yield Coefficient (gVSS/gS) = biomass produced/substrate consumedY = (X X0)/(S0 S) rg = - Y rSU or rSU = - X/YMonod kinetic model = m S/(KS + S)

rg = X = [m/(KS + S)]Xm = maximum specific growth rate (d-1)Ks = saturation or Monod constant (g l-1)S = limiting substrate (g l-1)rSU = -rg/Y = X/Y = m S X/[Y(Ks +S)]

The cell growth rate (and therefore the substrate removal rate) increases with the substrate concentration, up to a certain level when it stabilize at m.

If the limiting substrate concentration is low: conditions of slow growth.If part of the biomass produced is degraded with endogenous decay: rg = -Y rSU kd XWith kd = endogenous coefficient decay (d-1)rg =Volumetric biomass production rate (g VSS/m3 d)-Y rSU = Rate of biomass production from substrate consumptionkd X = Rate of biomass consumption by endogenous respirationAnd = m S/(Ks + S) kdrSU = - rg /Y kd X/Y = -Y-1 ( X + kd X) = - m S X/[Y(Ks + S)]Continuous Treatment in CSTRUnder a steady state in a well mixed reactor (CSTR) X = Xr and S = Sr (X and S = concentrations) and dX/dt = dS/dt = 0V , Xr , SrQ, X0, S0 Q, X, S This also means:Q (X X0) = PX = biomass production rateQ (S0 S) = PS = substrate consumption rate

V dXr/dt = cell in - cell out + cell produced = Q X0 Q X + rg VFor simplification, X0 = 0 and by definition rg = X X V = X Q = Q/VBy definition D = dilution rate = Q/V = = 1/HRT (Hydraulic Retention Time): The growth rate is dictated by the dilution rate.Monod model: = m S/(KS +S)Under steady state: = D, therefore D = m S/(KS +S)Solving this equation S = D KS/(m D)S mass balance: V dS/dt = 0 = Q S0 Q S + rSU V rSU V = - Q (S0 - S)Since rSUV= (-rg/Y) V = - V X/Y = - D V X/Y = - Q X/Y (since D = Q/V)Q (S0 S) = Q X/Y and X = Y(S0 S)Maximum Dilution Rate: DmaxCSTR: = D = Q/VAt washout condition: S = S0 = Dmax KS/(mDmax)Dmax = m S0/(KS + S0) = m/(1 + KS/S0)

Cell wash-out occurs at too high dilution rates (D >Dmax) and is especially sensitive at low initial substrate concentration.Influence of endogenous decayNow = m S/(KS + S) kdS: The biomass balance is identical, hence: D = = m S/(KS + S) kd. Solving this equation gives: S = KS (D + kd)/(m D kd)S mass balance: V dS/dt = 0 = Q S0 Q S + rSU VrSU = -rg/Y kd X/Y = - X/Y-kd X/Y = - X(D + kd)/YSolving this equation gives: X = Y D(S0 S)/(D + kd) = Y(S0 S)/(1 + kd/D)Dmax is obtained for S = S0 = KS (D + kd)/(m D kd)Solving this equation gives Dmax = m/(1 + KS /S0) - kdInfluence of non-biodegradable VSS (nbVSS )An amount of non biodegradable, also called inert, VSS is introduced into the reactor in the wastewater.This amount is not degraded biologically and therefore, at a steady state, the nbVSS concen- trations in the effluent and reactor are similar to the nbVSS concentration in the influent (X0,i)The total mass of VSS in the Bioreactor includes the biomass produced (rg), the nbVSS introduced (X0,i) and the debris released from the endogenous decay:rVSS = total VSS production rate = rg + Q X0,i/V + fd (kd) Xrg = -Y rSU kdX = biomass production from bCOD (-Y rSU) minus endogenous decay (kd X)fd (kd) X = rate of cell debris production with fd = fraction of biomass remaining as cell debris. The cell debris production is directly proportional to thebiomass concentration and kdQ X0.i/V = Amount of nbVSS in the influent Q = influent flow rate, X0,i = influent nbVSS and V = reactor volume).Solid Retention Time (SRT)The SRT is defined as the average time the solids stay inside the aeration tank. When VSS = active cells (X), the SRT is also called Mean Cell Retention Time (MCRT) with:SRT = Amount of active biomass = VX (kg)The production rate can be obtained from the biomass mass balance under steady state as Xout Xin = Production, with Xout = Qe Xe + Qw Xw , Xin = Q0 X0

Note Xr = XwOften Xw >> Xe and Qw Xw + Qe Xe Qw XwOther name of SRT: Sludge AgeExpression of SBiomass balance:V dX/dt = 0 = Q0 X0 - (Qe Xe + Qw Xw) + V rg(Qe Xe + Qw Xw) Q0 X0 = V XFrom the definition of SRT:(Qe Xe + Qw Xw) Q0 X0 = (Qe Xe + Qw Xw)= V X/SRT1/SRT = = m S/(Ks + S) kdS = KS (1 + (kd) SRT)/[SRT(m kd) 1]Expression of XX can be obtained from S mass balance:V dS/dt = Q0 S0 (Qe Se + Qw Sw) + rSU V = 0With the assumptions that S = Sw = Se = Sr (S = bsCOD) and since (Qe + Qw ) = Q0, this becomes:Q0 (S0 S) = - rSU V

Since rg = -Y rSU kd X; rg = X = X/SRT and HRT = V/QX = (SRT/HRT) Y(S0 S)/(1 + (kd)SRT)Expression of U 1 S0 - S = Y U - kd = Y - kdSRT X

X KS 1 1 1 = + =S0 - S k S K U

where:SRT: Solids retention time, dY: Biomass yield, mg VSS/mg sCODU : Substrate utilization rate, mg sCOD/mg VSS.dkd : Endogenous decay coefficient, 1/dS0 : Influent substrate concentration, mg sCOD/lS : Effluent substrate concentration, mg sCOD/lX : Biomass concentration, mg VSS/l : Hydraulic retention time, dKS : Half-velocity constant, mg sCOD/lk : Maximum rate of substrate utilization, mg sCOD/mg VSS . dPlotting 1/SRT versus U, the biokinetic coefficients Y can be determined from the slope and kd from the intercept of the equation.Plotting 1/U or X/(S0 - S) versus 1/S, the biokinetic coefficients KS can be determined from the slope and k from the intercept of the equation.

Equations describing the performance of the system are the mass balance equations of both the biomass and substrate. The biomass balance can be expressed by:[rate of change of biomass in the reactor] = [rate of increase due to growth] [rate of loss due to endogenous respiration] [delibarate wastage]which can be mathematically expressed as:V dX/dt = X V kd X V Qw X where V = reactor volume (l); X = biomass concentration in the reactor (mg/l); kd = biomass decay coefficient (day-1 );Qw = wastage flow rate (s-1 ); t = time (s).

At steady-state conditions, dX/dt = 0, hence, Eq. abovebecomes: = kd + QW V (a)Since the solid retention time (SRT) is defined as: total mass of organisms in the reactorSRT = total mass of organisms leaving the system/daySRT = V X/QW X = V/QW (b)Substituting Eq. (b) into Eq. (a) results in: = kd + 1/SRT (c)Substituting for the value of from Eq. (c) into Monod model: = m S/(KS +S) yields the following equation that describes the steady-state condition for substrate concentration in the reactor:

SRT KS 1 1 = + (d) 1 + SRT x kd m S m

.Substituting the value of KS and kd in Eq. (d) and plotting SRT/[1 + (SRT x kd)] versus 1/S, the biokinetic coefficients, m can be determined from the Y-intercept of the equationIn continuous-flow and completely-mixed reactor, determination of the biokinetic coefficients KS, k, m , Y, and kd is usually achieved by collecting data from lab-scale or pilot-scale experimental setups operated at various hydraulic retention times (HRTs) and/or at various sludge retention times (SRTs), and by allowing steady-state condition to prevail for each HRT or SRT under investigation.

Determination of m