1. The questions relate to the following data for a completely mixed activated s
ID: 1714061 • Letter: 1
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1. The questions relate to the following data for a completely mixed activated sludge system Qin 5MGD (flow from primary) CMFR volume- 4.5 MG BOD in WW from primary treatment BODin 200 mg/l BOD in effluent from final clarifier BODou 30mg/l MLSS 3500mg/l VM-70% SS in waste sludge11,000 mg/L Waste sludge flow 0.125 MGD S.G. of sludge -1.01 SV- 318 mLL a) The biomass or MO concentration in the tank is closest to A) 2500 mg/l B) 3500 mg/l C) 5000mg/l. D) Cannot be determined from given data. b) The lbs. of solids under aeration is closest to A) 50,000 B) 70,000 C) 110,000 D) 130,000 c) The F/M ratio in lb BOD/lb MLVSS-day for this system is closest to: A) 0.05 B) 0.07 C) 0.09 D) 0.11Explanation / Answer
(a) Biomass concentration in tank is measured as suspended solids (mg/L). Hence, it is also called MLSS.
Biomass concentration in tank = MLSS = 3500 mg/L
Therefore, answer is (B) 3500 mg/L
(b) Total solids under aeration = Volume of reactor x MLSS in tank
= 4.5 MG x 3500 mg/L
= 4.5 x 106 Gallons x 3500 mg/L
= 4.5 x 106 x 3.78541 x 3500 mg
= 59620207500 mg
= 131440.06 lbs
Therefore, answer is (D) 130,000
(c) F/M ratio = Daily BOD load applied to aerator (in lb) / Total microbial mass in aerator (in lb)
= Daily BOD load applied to aerator (in mg) / Total microbial mass in aerator (in mg)
= Qin (BODin - BODout) / (Volume of reactor x MLSS in tank)
= 5 MG x (200 mg/L - 30 mg/L) / (4.5 MG x 3500 mg/L)
= 5 x 106 x 3.78541 x 170 / (4.5 x 106 x 3.78541 x 3500)
= 0.054
Therefore, answer is (A) 0.05
(d) Hydraulic retention time = Volume of reactor / Inflow sewage discharge
= CMFR volume / Qin
= 4.5 MG / 5 MGD
= 0.9 days
= 0.9 x 24 hr
= 21.6 hr
Therefore, answer is (D) 22 hr
(e) Mass of suspended solids wasted = SS in waste sludge x Waste sludge flow
= 11000 mg/L x 0.125 MGD
= 11000 x 0.125 x 106 x 3.78541 mg/d
= 5204938750 mg/d
= 11474.93 lb/d
Therefore, answer is (A) 10,500
(f) MCRT = Sludge retention time = Mass of MLSS in system / Mass of solid leaving the system or wasted per day
= 131440.06 lbs / 11474.93 lb/d
= 11.45 days
Therefore, answer is (C) 11 d
(h) Settled volume of sludge = 318 mL/L
MLSS = 3500 mg/L = 3.5 g/L
SVI = Sludge Volume Index = Volume occupied (in mL) by one gram of solids in mixed liquor after settling for 30 min
= 318 mL/L / 3.5 g/L
= 90.86
Therefore, answer is (C) 90
(g) MLSS concentration in return sludge = 106 / SVI = 106 / 90.86 = 11006.3 mg/L
Sludge recirculation ratio = Initial MLSS / (MLSS concentration in return sludge - Initial MLSS)
= 3500 / (11006.3 - 3500)
= 0.4663
= 46.63%
Therefore, answer is (D) 50%
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