An aeration basin of volume 400 m3 contains mixed liquor suspended solids (MLSS) of 1500 mg/L. The amount of mixed liquor suspended solids (MLSS) in the tank is
- A250 kg
- B400 kg
- C600 kgCorrect
- D1200 kg
Solution
Concept
MLSS mass calculation in aeration basin
Formula
Mass = V × MLSS = 400 × 1.5 = 600 kg
Step-by-step solution
1. Aeration basin volume V = 400 m³. 2. Mixed liquor suspended solids MLSS = 1500 mg/L = 1500 g/m³ = 1.5 kg/m³. 3. Total MLSS mass = V × MLSS = 400 m³ × 1.5 kg/m³ = 600 kg. 4. Hence, the total solids mass in the basin is 600 kg. Answer: C
Explanation
Total mass of Mixed Liquor Suspended Solids (MLSS) in an aeration basin is obtained by multiplying the aeration tank volume by the MLSS concentration. Given volume = 400 m³ and concentration = 1500 mg/L (1.5 kg/m³), the total solids mass equals 400 m³ × 1.5 kg/m³ = 600 kg.
Why the other options are wrong
A) 250 kg is an arbitrary computation without physical basis. B) 400 kg mistakenly assumes solids mass in kg equals basin volume in m³. D) 1200 kg results from doubling the actual solids mass.
Shortcut
MLSS mass: 1500 mg/L = 1.5 kg/m³ -> 400 m³ × 1.5 kg/m³ = 600 kg.
Common mistake
Making unit conversion errors between mg/L, g/m³, and kg.
Correct answer: C) 600 kg