Design volume of the aeration basin - sewer construction:

Problem:

What is the design volume of the aeration basin?

Solution

Assuming that 30% of the plant influent BOD5 is removed during primary sedimentation, this means that S_o = 444 mg/l × 0.70 = 310 mg/l.

Q_c = 1/t

Thus  1/Qc = [(Q₀×Y)/(V×X)(S₀ - S)] -k_d

¼ = [days]=33.75×10⁶ [-1/d]× 55/100[g/g]×(310[mg/l]-2 [mg/l])

V[l]×4300[mg/l]-0.05[1/day]

Solving for V, we get V = 5007558 l = 5,00,000 l.