Home Lecture Quiz Design Example Design of Completely Mixed Activated Sludge System Design a completely mixed activated sludge system to serve 60000 people that will give a final effluent that is nitrified and has 5-day BOD not exceeding 25 mg/l. The following design data is available. Sewage flow = 150 l/person-day = 9000 m3/day BOD5 = 54 g/person-day = 360 mg/l ; BODu = 1.47 BOD5 Total kjeldahl nitrogen (TKN) = 8 g/person-day = 53 mg/l Phosphorus = 2 g/person-day = 13.3 mg/l Winter temperature in aeration tank = 18°C Yield coefficient Y = 0.6 ; Decay constant Kd = 0.07 per day ; Specific substrate utilization rate = (0.038 mg/l)-1 (h)-1 at 18°C Assume 30% raw BOD5 is removed in primary sedimentation, and BOD5 going to aeration is, therefore, 252 mg/l (0.7 x 360 mg/l). Design: (a) Selection of qc, t and MLSS concentration: Considering the operating temperature and the desire to have nitrification and good sludge settling characteristics, adopt qc = 5d. As there is no special fear of toxic inflows, the HRT, t may be kept between 3-4 h, and MLSS = 4000 mg/l. (b) Effluent BOD5: Substrate concentration, S =  1 (1/qc + kd)=         1     (1/5 + 0.07)                                          qY                (0.038)(0.6) S = 12 mg/l. Assume suspended solids (SS) in effluent = 20 mg/l and VSS/SS =0.8. If degradable fraction of volatile suspended solids (VSS) =0.7 (check later), BOD5 of VSS in effluent = 0.7(0.8x20) = 11mg/l. Thus, total effluent BOD5 = 12 + 11 = 23 mg/l (acceptable). (c) Aeration Tank: VX = YQqc(SO - S) where X = 0.8(4000) = 3200 mg/l          1+ kdqc or 3200 V = (0.6)(5)(9000)(252-12)                        [1 + (0.07)(5)]       V = 1500 m3 Detention time, t = 1500 x 24 = 4h                                9000 F/M = (252-12)(9000) = 0.45 kg BOD5 per kg MLSS per day           (3200) (1500) Let the aeration tank be in the form of four square shaped compartments operated in two parallel rows, each with two cells measuring 11m x 11m x 3.1m (d) Return Sludge Pumping: If suspended solids concentration of return flow is 1% = 10,000 mg/l R =        MLSS     = 0.67       (10000)-MLSS Qr = 0.67 x 9000 = 6000 m3/d (e) Surplus Sludge Production: Net VSS produced QwXr = VX =  (3200)(1500)(103/106) = 960 kg/d                                      qc                   (5) or SS produced =960/0.8 = 1200 kg/d If SS are removed as underflow with solids concentration 1% and assuming specific gravity of sludge as 1.0, Liquid sludge to be removed = 1200 x 100/1 = 120,000 kg/d                                                               = 120 m3/d (f) Oxygen Requirement: For carbonaceous demand, oxygen required = (BODu removed) - (BODu of solids leaving)                        = 1.47 (2160 kg/d) - 1.42 (960 kg/d)                        = 72.5 kg/h For nitrification, oxygen required = 4.33 (TKN oxidized, kg/d) Incoming TKN at 8.0 g/ person-day = 480 kg/day. Assume 30% is removed in primary sedimentation and the balance 336 kg/day is oxidized to nitrates. Thus, oxygen required                        = 4.33 x 336 = 1455 kg/day = 60.6 kg/h Total oxygen required         = 72.5 + 60.6 = 133 kg/h = 1.0 kg/kg of BODu removed. Oxygen uptake rate per unit tank volume = 133/1500                       = 90.6 mg/h/l tank volume (g) Power Requirement: Assume oxygenation capacity of aerators at field conditions is only 70% of the capacity at standard conditions and mechanical aerators are capable of giving 2 kg oyxgen per kWh at standard conditions. Power required =   136   = 97 kW (130 hp)                         0.7 x 2                       = (97 x 24 x 365) / 60,000 = 14.2 kWh/year/person