Question

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.

Answer 1

Answer:

(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