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Activated Sludge Process Design 2 Information Checklist 1. Select the type of biological treatment process. 2. Conduct a material mass balance and determine expected range of flows (minimum, average, and peak) and loadings (COD, TSS, nutrients, etc.). 3. Determine biological kinetic coefficients (lab studies). 4. Develop a preliminary site plan, piping layout, and location of collection boxes, return sludge pumps, etc. 5. Obtain design criteria. 6. Obtain effluent quality criteria (BOD 5 , TSS, TN and TP). 7. Develop data on settling characteristics of the biological solids. 8. Obtain list of equipment manufacturers and provide equipment selection guide. 3 Biological Reactor Design Criteria 1. Design a BNR process with anaerobic, anoxic, and aerobic reactor. 2. Anaerobic will receive influent and returned sludge from clarifier. Anoxic will receive effluent from anaerobic and recycle from aerobic. 3. The effluent will have BOD 5 /TSS/TN/NO 3 - -N/NH 4 + -N/TP of 10/10/10/8/1/1 mg/L or better, respectively. 4. Provide four independent process trains in parallel. 5. Anaerobic and anoxic must be deep, square and θ>1.5 hr. 6. Provide equipment for measuring raw wastewater flow, return activated sludge, waste sludge, and air supply. 7. Blowers shall be capable of delivering max. air requirements considering the largest single unit out of service. 8. Aeration equipment shall provide complete mixing of MLSS and shall be capable of maintaining a min. of 2.0 mg/L DO. 4 Design Criteria - continued 9. Diffusers and piping shall be capable of delivering 150% of the average air requirements. 10. The sludge pump and piping for RAS shall be designed to provide capacity up to 150% of average design flow. 11. Internal recycle between aeration and anoxic basins will have a capacity 200% of the average design flow. 12. All sidestreams from the sludge-handling facilities (thickeners, digesters, and dewatering units) shall be returned to the aeration basin. 13. The sludge wasting shall be achieved from the common collection box containing the effluent MLSS from aeration basins. 14. The basin hydraulics shall be checked at peak design flow plus the design return sludge flow when only three basins are in operation. 5 Design Criteria - continued 12. The biological kinetic coefficients and operational parameters shall be determined from laboratory studies. Sludge age, θ c = 12 days Y= 0.6 mg VSS/mg BOD 5 ; k d = 0.06 day -1 Y N = 0.2 mg VSS/mg NH 4 + -N; k d,N = 0.05 day -1 (N=nitrification) MLVSS = 3000 mg/L; MLVSS/MLSS = 0.8 RAS concentration = 10,000 mg TSS/L BOD 5 = 0.68 BOD L Biological solids are 65% biodegradable. 13. Other conditions Q peak = 1.321 m 3 /sec, Q ave = 0.44 m 3 /sec BOD 5 , S 0 = 250 mg/L, Org-N= 17 mg/L, NH 4 + -N= 19 mg/L, NO 3 - -N= 0 mg/L, TN= 36 mg/L, TP= 6 mg/L TSS, X 0 = 260 mg/L; primary sludge VSS/TSS = 0.74 Anaerobic digester solids content = 0.06 Schematic Flow and Piping Arrangement 6 7 Design Strategy 1. Conduct a material mass balance 2. Calculate θ of anaerobic zone 3. Calculate θ c , biomass increase, and θ of anoxic zone 4. Calculate θ c , biomass increase, and θ of aerobic zone 5. Calculate the dimensions of anaerobic, anoxic, and aerobic reactors 6. Design the required mixing equipment 7. Design aeration system 8. Select influent and effluent structure, develop hydraulic profile 8 Design Calculations First iteration 1. Calculate influent flow (stream 1) Flow = 0.44 m 3 /sec × 86,400 sec/d = 38,016 m 3 /d BOD 5 = 38,016 m 3 /d×250 g/m 3 ×kg/1,000 g = 9,504 kg/d TSS = 38,016 m 3 /d×260 g/m 3 ×kg/1,000 g = 9,884 kg/d Org-N= 646 kg/d, NH 4 + -N= 722 kg/d, NO 3 - -N= 0 kg/d, TN= 1,369 kg/d, TP= 228 kg/d 2. Calculate primary sludge characteristics (stream 3) BOD 5 (34% removal) = 9,504 kg/day × 0.34 = 3,231 kg/d TSS (63% removal) = 9,884 kg/day × 0.63 = 6,227 kg/d Org-N (30% removal)= 194 kg/d NH 4 + -N (0.36% removal)= 2.6 kg/d NO 3 - -N= 0 kg/d TN (14.4% removal)= 197 kg/d TP (16% removal)= 37 kg/d Solids concentration = 4.5% Specific gravity = 1.03 Sludge flow rate = 6,227 kg/d × 1,000 g/kg ÷ 0.045 g/g ÷1.03 ÷ 1 g/cm 3 ÷ 10 6 m 3 /m 3 = 134 m 3 /day 9 Design Calculations - continued 3. Calculate primary treated effluent (stream 4) Q = 38,016 m 3 /d - 134 m 3 /d = 37,882 m 3 /d BOD 5 = 9,504 kg/ - 3,231 kg/d = 6,273 kg/day = 6,273 kg/d ÷ 37,822 m 3 /d × 1000 g/kg = 166 mg/L TSS = 9,884 kg/d - 6,227 kg/d = 3,657 kg/day = 3,657 kg/d ÷ 37,822 m 3 /d × 1000 g/kg = 97 mg/L Org-N = 452 kg/d → 12 mg/L NH 4 + -N = 720 kg/d → 19 mg/L NO 3 - -N= 0 kg/d TN = 1172 kg/d → 31 mg/L TP = 192 kg/d → 5 mg/L 4. The characteristics of streams 2 and 5 are the same as those for streams 1 and 4, respectively. 5. Effluent standards (stream 6) were given as design criteria 6. Calculate WAS (stream 7) Eff. BOD 5 = Eff. sol. BOD 5 + BOD 5 of eff. biological solids Eff. sol. BOD 5 = 10 mg/L – 10 mg/L × 0.65 ×1.42 × 0.68 = 3.7 mg/L 10 Design Calculations - continued Increase in TVSS due to overall BOD 5 removal = Y/(1+k d ·θ c ) (S 0 - S 1 ) Q = 0.6/(1+0.06×12) (166 – 3.7) g/m 3 × 37,882 m 3 /d× kg/1000 g = 2,144 kg/d Increase in TVSS due to nitrification = Y N /(1+k d , N ·θ c ) (S 0,N - S 1,N ) Q = 0.2/(1+0.05×12) (19* – 1**) g/m 3 × 37,882 m 3 /d× kg/1000 g = 85 kg/d *NH 4 + -N in stream 4 **Design criteria Increase in TSS = (2,144+85) kg/d ÷ 0.8 = 2,789 kg/d TSS in WAS = Increase in TSS - TSS lost in the effluent = 2,789 kg/d - [10 g/m 3 × (37,882 - Q WAS ***) m 3 /d ×kg/1000 g] = 2,410 kg/d MLSS concentration = 3,000 mg/L ÷ 0.8 = 3,750 mg/L Q WAS = 2,410 kg/d ÷ 3,750 mg/L × 10 6 kg/mg × 10 -3 L/m 3 = 643 m 3 /d ***first iteration assumes that Q WAS =0 BOD 5 in WAS =2,410 kg/d × 0.65 g/g × 1.42 g/g × 0.68 g/g = 1,513 kg/d Sol. BOD 5 = 3.7 g/m 3 × 643 m 3 /d × kg/1000 g = 2.4 kg/d Total BOD 5 in WAS = 1,513 kg/d + 2.4 kg/d = 1,515 kg/d