Question

2. (40 pts) A wastewater treatment plant has an activated sludge system. The aeration tank is 252 ft long, 25 ft wide, and 18 ft deep. An engineer will model the steady-state performance as first order (k= 8.3 d ?) as a series of 9 equal-volume CFSTRs. The flow rate is 3.1 MGD and the influent BODS is 165 mg/L. a) If the engineer's model is correct, what is the outlet BOD5 concentration (mg/L)? b) If the model is correct, what would the BOD5 concentration be in a sample collected from the tank at a point 168 ft from the inlet end (mg/L)? c) If the reactor was a true PFR, what would the outlet BOD5 concentration be (mg/L)?

Answer 1

la) V of Aeration tank *18 ht 252 ft x 25 ft 113,400 ft B 39 11-13 m m3 0. 3048 m m] K = 8.3 ㅓ 8 = 3.1 MGD 3.1810 gallons /day 11734. 77 m3/day [ gallon = 0.0037 m3] •day со ing/L Since the engineer has modeled it as 9 equal CFSTR then each CESTR has volume => 356.79 m² 9 =165 - v 2. 3 4 5 6 구 8 9 He 4 ta Ja totes xco xico 7 =6 X X 168 ft + 28th 2 ft c = effluent concentration after the 1st CESTR 252 x Ca effluent concentration after the 9th CF STR final effluent BODs concentration

L Consider the 1st CFSTR only Shaving volume v'] Since its a Continuous Flow Stirred Tank Reactor that L Cont Not Concentration batch same at all points of stirring Also Steady State has been achieved. . Accumulation So rate balance eg for the 1st CFSTR => 20 mass - AR IR OR + RR 1 V Accumulation Rate =0 Output Input Rate = 2,9 Rate Reaction Rate =48 -ker [rue ? BOD is decaying >> o=68-cig - KC, v C, Q+ KC, V = Cog C = Cog 8 + kv Similarly: C ag &+kv co Coo Stkv o 9+kV 2 le su Q g+ KV. 2 2 C₂ = Ce Q Stkv Coro Skv 8 HTV 3 co 9+ KV

9 ... so co la ) ( 9 165 mg/L х 11734.77 11734.77 m2 / day m3/day + 8.3 x 356.79 3 dayt 1 21. 77 mg/L of 28 ft = 9 (b) length each CESTR 952 ft 28 ft x 6 = 168 ft : Sample collected from the tank at point 16 8 ft from the inlet end will be the effluent of 6th CFSTR i a C = ? 6 coloring Q+kv 6 1 165 ng/L 11734. 71 11734.77 + 8.3 x 356.79 42.76 mg/L C) there are same egs separate as for blug flow reactor equations hence we reactor with t = time (etention) for batch Be length velocity 02 t = v 8

: t= v = >lor 32 11.13 m3 11734.77 m² Laboresco 0.27 days I day For 1st order; the eg for batch reactor C = Co exp(-kt) = 165 mg / x exp(-8.3 x day mg / x 0.27 days = 17.02