Question

A 75,000 ft3 clarifier is to be used to treat wastewater. The recycle ratio is 50%, the sludge volume index (SVI) is 125, and the return activated sludge concentration is 8000 mg/L. The biomass concentration is 3500 mg/L. The combined design flow rate of the primary and secondary clarifiers is 2.5 MGD. After primary treatment, the wastewater has an influent BOD concentration of 200 mg/L and an influent suspended solids concentration of 200 mg/L. Two secondary clarifiers, each 28 ft in diameter, are then used. After secondary treatment, the effluent BOD concentration is 15 mg/L, and the effluent suspended solids concentration is 20 mg/L. The volume of sludge produced is 0.5 MGD. What is most nearly the solids residence time

Answer 1

11 hours approximately

Step-by-step explanation:

We are to calculate mean cell residence time mcrt

= Mass of solid in reactor/mass of solid wasted in a day

Q = Qe + We

Q = 2.5

Qw = 0.5

Qe = 2.5 - 0.5

= 2 MGD

10⁶/svi

= 10⁶/125

= 8000

X = 3500

Xe = 20mg/

1MGD = 0.1337million

Mcrt = 75000x3500/[0.5*8000*10⁶+2*20*10⁶] x 0.1337

= 262500000/[4000000000+40000000} x 0.1337

= 262500000/574800000

= 0.45668 days

= 0.45668 x 24 hours

= 10.9603 hours

Approximately 11 hours