Question

Design an aerated grit chamber to treat a municipal wastewater for the following conditions. Two chambers will be required for cleaning while others in operation. Each chamber will receive one-half of the flow.

   Qavg = 0.6 m3/s, peaking factor = 2.5

   Detentionn time, t = 3 min

   Width/depth ratio = 1.5 : 1

   Depth, d = 3 m

   Number of chambers, n = 2

   Air supply rate = 0.3 m3/min-m of length

   Quantity of grit = 0.06 m3/1000 m3 of wastewater

Answer 1

- As per Given Data: Q-0.6 m/s (Average) Peak factor=215 is a peak = 0 6x2.5 = 1.5 m3/s. Detention tion time, t = 8 min - time,t = 3 min = 3X60-180 sec # of chamber, n=2 so Volum required per Basin n = .5 X 180 2 = 135 m3 Given [/= 135 m3/basin Now Depth of basin as per given dota=3m. in width, B = 115*3=4.5m su 4 - 1 So, Surface area of basin. A=I - 135 - 45m2 or LXB = 45m2 in L = 25 = 25 = 10m So, dimention of basin will be 4.5810X 3 to: or 4.5X10X3.5 (boord free) (0.5 m each each grit chomber at average Now, detention time in fbw, Detention time grit chombo = 450 sec - 135 m3 x2 tention time = o 6 m/s 7.5 min

Now, Air Supply requirement. Anstand that Given that 0.3/min-m. of length will be adequote. Air required (length basis) = 10x0.3 m3/min-m - 3 m3/min for each grit bosin Now is Total Air Supply required - 3x2 = 6 m in. stimate the quantity of grit at peak flow, Given. that a value of 0.06 m3/1000 m3 at waste water Volume of grit = 115 (m3/s) x 86400 / x 0.06 m² / 13m3 - 7.70 m3/h Ans