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3 June, 21:26

g A primary sedimentation basin is designed for an average flow of 0.3 m3/s. The TSS concentration in the influent is 240 mg/L. The average solids removal efficiency of the basin is 60 percent. The sludge has an average solids concentration of 4 percent and a specific gravity of 1.025. Calculate (a) the quantity and volume of sludge produced, (b) the effluent flow rate, and (c) the pump cycle time if the pumping rate is 570 L/min.

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  1. 3 June, 21:35
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    (a) 0.243 m3/day

    (b) 96 mg/l

    (c) 0.426 m3/min

    Explanation:

    The sludge has an average solids concentration of 4 percent and considering TSS concentration in the influent of 240 mg/L then solids in sludge will be 0.04*240 = 9.6 mg/L

    Considering the average flow of 0.3 m3/s then mass of sludge per day will be given by 0.3*1000*9.6*60*60*24/1000000=248.832 kg/day

    To get volume, considering specific gravity given as 1.025 and taking density of water as 1000 kg/m3 then density of sludge is 1025 kg/m3

    Volume is mass/density hence 248.832/1025=0.2427629268292 m3/day

    Approximately, the volume of sludge is 0.243 m3/day.

    (b)

    Efficiency of 60 percent is equivalent to 0.6

    Efficiency = (influent concentration - flow rate) / influent concentration

    0.6 = (240-flow rate) / 240

    Flow rate = 96 mg/l

    (c)

    Cycle time = 0.243/0.57=0.4263157894736 m3/min

    Rounded off, cycle time is 0.426 m3/day
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