Question

9.9 Design an aerated grit chamber system to treat a 1-day sustained peak hourly flow of 1.6 m/s with an average flow of 0.65 m/s. See Table 9.4 for design guidance. Determine: a. The grit chamber volume (assuming two chambers will be used) b. The dimensions of the two grit chambers. c. The average hydraulic retention time in each grit chamber. d. Air requirements, assuming 0.20 m of air per m length of tank per minute. e. The quantity of grit removed at peak flow, assuming a typical value of 0.20 m' of grit per one thousand m of untreated wastewater. 9.11 A wastewater treatment plant will receive a flow of 35,000 m/day. Calculate the surface area (m“), diameter (m), volume (m'), and hydraulic retention time of a 3 m deep circular, primary clarifier that would remove 50 percent of suspended solids. Assume the surface overflow rate used for the design is 60 m/m -day.

Answer 1

Given, Peak hourly flow of güt charmbet = 16 m3) sec Average flow = 0.65 m3/src We will take peak flow a, Grit Chamber volume :- Assume a Total Wetention time = 3 min Guit chamber volume (each) = 1; / x 3 min x 60S 2 min depth = 151.2003 b. Dimensions of Two Grit Chambers – Using a typical width to depth ratio and Assume width to depth ratio = 12:1 Eepth =3m a = width = 1.213) = 3-6 m b= hength = Volume 1512 Width & Wepth 3.6*3 = hength = 14 m s The Average hydraulic Retention time in each Fretention Time = 151.2 m 3 0.65 (125) - 2326153 Hc - 3.87 min

d, Air Requirements - Assume, 0.2 l m/ min-m of hength of Tank for Frequired (hemoth basis) = 14 mx 0.2 m/in-on = 2.8 m/min for each . Crit Chamber Total air supply required = 2.872 = 5-6 m3/min & Quality of Grit Thermoved at peak flow: Assuming a value of 0.2 m3/03 m² volume Curit [016 m3/s)[86,400 s/d)(0.20 mo/103 mm)] = 27.648 mild Quantity of Grit removed = 27.648 m3/day

910) * - 35000 m3/day, Depth = 3m Circular efficiency = 50% Surface Overflow rate = 60 m2%n- day Sustaice Area = Q = 35000 7.583.33 m2 que surface Leading 60 » Surface Area = π d ²9 d =. 83.33 = 27.2537" A volume = Surface Area x depth = 583.33 x 3=1450 m Hydraulic retention time = = = 15 00 = the day = 2 = 12 hour
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