Question

Environmental Engineering 3&4

So 3. An activated sludge plant treats 5 MGD and the average influent BOD is 200 mg/L. The WWTP has 2 aeration basins, each 100 ft x 25 ft x 15 ft deep and 2 final clarifiers with diameters of 20 ft. About 35% of BOD is removed in primary clarifiers. Determine the BOD in the primary effluent, hydraulic retention time, F/M ratio, BOD exiting the aeration basin and mass of solids wasted (1b/d). The operation parameters are: MLVSS = 2000 mg/L Ks = 200 mg/L Yield coefficient (Y) = 0.60 max = 4/day Bacterial decay rate (kd) = 0.060 day 1 Mean cell residence time (Oc) = 10 days Answers: BOD in primary effluent = 130 mg/L HRT = 2.7 h F/M = 0.58 BOD exiting aeration basin = 5 = 8.3 mg/L Mass solids wasted = QwXw - 467 lb/d 4. A sample of wastewater is incubated for five days at a temperature of 20 deg. C. After this incubation period, the BOD is found to be 211 mg/L. Assume that the reaction rate constant is 0.14 days? What is the ultimate BOD? Answer: ultimate BOD = 417 mg/L BODY Kp = KBOD * 1,135

Answer 1

Page L Date ubhash Son Sos influent BOD = 200 mgle. PR BOD is removed. as given in the question Sf = - So - (85% of s.) 200 - 35 x 200 effluent = Sf = 130 ngle. BOD. HRT: higdrallic retention time : 2X 100 X 25 X 15 X 24 ft Exhrs 5x 106 x 0.1336 ft 2.7 hrs. ART = Qo So E sx1080.1335 x 200 2X100X25 x 15 x24 x ROOD 0:58 E VX Qw Xu 100 x 25 x 15 ft x 2000 mg Qualy = 467 lb/d.

Datel 211 = la clé SX0-14) 2012 (1-0.496) 211= lo x 0.504 lo z 418.65 ngle.