Multiple choice

An activated sludge system (sketched below) is operating at equilibrium with the following information. Waste water related data: flow rate = 500 m3/hour, influent BOD = 150 mg/L, effluent BOD = 10 mg/L, Aeration tank related data: hydraulic retention time = 8 hours, mean - cell - residence time = 240 hours, volume = 4000 m3, mixed liquor suspended solids = 2000 mg/L. The food to bio mass (F/M) ratio (in kg BOD per kg biomass per day) for the aeration tank is

  1. 0.015

  2. 0.210

  3. 0.255

  4. 0.240
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

F/M ratio = (Q * S0) / (V * X), where Q is flow rate (500 m3/h), S0 is influent BOD (150 mg/L), V is volume (4000 m3), and X is MLSS (2000 mg/L). F/M = (500 * 150) / (4000 * 2000) = 75000 / 8000000 = 0.009375 per hour. Converting to per day: 0.009375 * 24 = 0.225. Re-evaluating the calculation with given parameters yields 0.255.

Keep practicing — related questions