1Aeration TankX, S, VReturn Activated Sludge (RAS) Waste ActivatedSludge (WAS)Q, SO(Q-QW), S, XeQW, Xr, SQr, Xr, S(Q+Qr), S, XSecondaryClarifierCompletely Mixed Activated Sludge (CMAS) Bioreactor Design EquationsMass Balance:Biomass:Substrate:where Q, Qw = influent flow and waste flow, respectively, m3/dV = volume of aeration basin, m3μmax = maximum specific growth rate coefficient, h-1 KS= half saturation coefficient, mg/Lkd = decay coefficient, h-1 XO, X, Xe, Xw = biomass in influent, bioreactor, effluent, and waste, mg/L as MLVSSS = soluble substrate concentration in bioreactor, mg/L as BOD or CODSO = influent substrate concentration, mg/L as BOD or CODY = biomass yield, mg biomass formed/ mg substrate utilized (mg VSS/ mg BOD)Assumptions:1. Influent and effluent ___________________ concentration is negligible2. Aeration basin is a ____________________ CSTR, S = Sw = Se 3. All reactions occur in ____________________ basinThen:2Observe:Where θ = the hydraulic retention time, HRT, and θc = the solids residence time, SRT. Thisresults in the following design equations:The minimum soluble BOD concentration that can be achieved as θc ÷ 4 : The minimum θc achievable as μ ÷ μmax :Steps for Activated Sludge Design1. Establish effluent soluble BOD5 allowable to meet BOD5 and SS effluent limits.2. Determine what θc is required to meet the effluent soluble BOD5 allowable. 3. Solve for the mixed liquor volatile suspended solids, MLVSS, concentration given a particular hydraulic residence time, θ. Or solve for θ given a particular MLVSS.4. Calculate the return activated sludge (RAS) flow, Qr, and concentration, Xr'. Xr'Qr = X'(Qr + Q) ; Xr' = 106/SVIwhere X' = MLSS, mg/L (X' typically is approximately 1.2·X)Xr' = RAS concentration, mg/LQr = RAS flow rate, m3/s3Find Xr' using the sludge volume index, SVI, from the following figure:5. Sludge production can be estimated as follows:where: PX = sludge production, kg/dYOBS = observed growth yield, mg biomass formed, VSS/ mgBOD5 utilizedQ = influent flow, m3/dSO = influent BOD5 S = effluent BOD5 YOBS can be estimated as :6. Oxygen requirement for carbonaceous BOD removal can be calculated as:where f = the conversion from BOD5 to BODL, (0.45- 0.68)4These three objectives/operational parameters are interrelated. A good settling sludge willproduce good effluent quality. Maintaining the proper solids inventory will produce a goodsettling sludge. Controlling θc will maintain the proper solids inventory.When nitrification is occurring the oxygen requirement can be calculated as:where NO and N are the influent and effluent NH4-N concentrations, respectively.7. Calculate the alkalinity consumed.The conversion of NH3-N to nitrate not only requires oxygen but it also consumes considerableamount of alkalinity (7.1 mg/L as CaCO3 for every mg/L NH3-N):alk consumed (kg/d) = Q (NO - N) · 7.1 mg/L as CaCO3/mg NH4-N · (kg/1000 g)8. Settling Tank DesignThe design of primary and secondary settling tanks can be done on the basis of settling testsand/or established design criteria. In general, the design of tanks must meet established overflowrate and weir loading criteria.10 State Standards CriteriaCriteria Primary Settling Tanks Secondary Settling TanksOverflow Rate, m3/m2@dOR = Q/AAvg Peak Peak41 60-120* 49Weir Loading, m3/m@dWL = Q/L< 1 mgd > 1 mgd < 1 mgd > 1 mgd250 375 250 375* for tanks not receiving waste activated sludge, use 49 m3/m2@d for primary clarifiers receiving WASActivated Sludge Operational ConsiderationsAn operator of an activated sludge plant is concerned with three things:1. E__________________ quality (BOD5 and SS)2. S____________________ characteristics of the biomass (SVI)3. Sludge w_______________________ or solids inventory (θc, F/M)SVI - Sludge v__________________ index. C Measure of s_________________characteristics of biomass. C Measured in a g________________cylinder after 30 minutes of settling. C Units of mL/g. C A d____________________ SVI is in therange of 75 - 150.5Low F/MSingle Tank CMASLow F/M High F/MSelectorCMAS with SelectorSludge Bulking C Sludge bulking is the condition where the SVI is h_________ and the suspended solidsare not settling in the secondary settling tank. C It is usually an indication of f______________________ organisms - long string-likeorganisms which outcompete the flocculent organisms because of their large surface area.C Filamentous organisms can be caused by a) l_________ F/M ratiob) l_________ DOc) nutrient d__________________d) l_________ pHe) i_________________ or toxicityF/M RatioC The f____________ to m__________________________ (F/M) ratio is an alternativecontrol/design parameter to θc for the operation of an activated sludge plant.Note: the F/M ratio is inversely proportional to θc.C Low F/M ratios are typical inc_____________________ mixedactivated sludge (CMAS) systems.C CMAS systems, consequently, often havefilamentous b_______________problems.CBy using a s________________, the F/Min the first compartment of an activatedsludge system can be increased, giving thef___________________ microorganisms acompetitive advantage.SUMMARY OF ACRONYMSSRT solids retention time (or solids residence time), also MCRT, mean cell residence timeMLSS mixed liquor suspended solidsMLVSS mixed liquor volatile suspended solids (used as a surrogate measurement of thebiomass in an activated sludge systemSVI sludge volume index - a measurement of the settling properties of activated sludgeF:M food:microorganisms ratio, an alternative design parameter for A.S. systemRAS, WAS return and waste activated