Biological Treatment Activated Sludge Activated Sludge Influent GC PST Aeration Tank Secondary settling Tank BS The aeration tank contains mixed liquor suspended solids (MLSS), a combination of influent wastewater and return (recycled) activated sludge from the settling tank The MLSS includes: • Bacteria (both aerobes and facultative) • Slime: usually in flocs composed of extracellular polymeric substances, live bacterial cells, cell debris (dead, lysed) • Free cells • Protozoa • Nondegradable organic matter • Inert suspended solids Activated sludge floc with slime • MLSS rapidly (20-45 min) absorbs organic matter in wastewater influent • Bacteria then solubilize and oxidize organic matter • The food (organic matter) to microorganisms ratio (F/M) dictates character of bacteria and flocs Design Considerations Activated sludge process is defined by: Aeration period = detention time= 3-30 hrs BOD loading = (BOD entering per day)/ (V=volume of aeration tank) F/M= (Q*BODin)/(V*MLSS) Sludge age = mean cell residence time t sludge = MLSS *V SS e * Qe + SS w * Qw Aeration Tank SSe= suspended solids in the effluent SSw= suspended solids in waste sludge Qe = effluent flow rate Qw= sludge flow rate Qw, SSw SST Qe, SSe Types of Activated Sludge Systems Conventional aeration Step feed to aeration High-purity oxygen Extended aeration Loading and operation procedure for different aeration processes Process BOD load (lb BOD/d per 1000 ft3) MLSS mg/l F/M lb BOD/d per lb MLSS tsludge days Aeration period hr R% % BOD removal Conventional 20-40 1000-3000 0.2-0.5 5-15 4-7.5 20-40 80-90 Step aeration 40-60 1500-3500 0.2-0.5 5-15 4-7 30-50 80-90 Extended aeration 10-20 2000-8000 0.05-0.2 >20 20-30 50-100 85-95 High-purity oxygen >120 4000-8000 0.6-1.5 3-10 1-3 30-50 80-90 Example Consider the following activated sludge system Calculate: The aeration period BOD loading F/M ratio SS and BOD removal efficiency tsludge Return activated sludge rate What is the required QR if the MLSS is to be 2200 mg/l MLSS=2600 mg/l Q=2.14 mgd BOD=185 mg/l SS=212mg/l QR=0.85mgd Qw=39,000 gpd SSw=8600 mg/l Aeration: units each 40x40ft2 15.5 ft deep BOD=15 mg/l SSe=15 mg/l Settling tank Solution V= 4*40*40*15.5* (7.481/106)= 0.74 million gallon t=V/Q=(0.74/2.99)*(24)=5.9 hr BOD loading= [2.14*185*8.34]/[4*40*40*15.5/1000]= 33.3 lb BOD/d/1000 ft3 Note that the BOD loading from re-circulated sludge is negligible F /M = 2.14 mgd × 185 mg / l × 8.34 = 0.21 lb BOD / day per lb MLSS 0.74 mil gal × 2600 mg / l × 8.34 SS removal= (212-15)/212= 0.93= 93% BOD removal= (185-15)/185= 0.92= 92% t sludge = 0.74 mil gal × 2600 mg / l × 8.34 = 5.2 days (2.14 mgd ×15 mg / l × 8.34) + (0.039 mgd × 8600 mg / l × 8.34) Return sludge ratio (R)= 0.85/2.14=0.40 =40% To find QR for a MLSS =2200, we apply mass balance on suspended solids at the entrance of the aeration tank 2.14*212 + QR*8600 = (2.14+ QR)*2200 Solve for QR= 0.66 mgd