Water & Wastewater Treatment

Activated Sludge & Biological Treatment

F/M ratio, sludge age, MLSS and the aeration-tank design parameters of secondary treatment.

PART 1

Topic Breakdown & Traps

The Engineering Principle

The activated-sludge process uses a suspended microbial culture (MLSS) in an aerated tank to oxidise organic matter, followed by a clarifier that settles and recycles the biomass. Performance is governed by the food-to-microorganism (F/M) ratio and the **mean cell residence time (sludge age,

θ_{c}

)**. Low F/M / high sludge age gives high-quality, well-nitrified effluent.

The Core Formula Matrix

F/M ratio:

\frac{F}{M} = \frac{Q S _{0}}{V X}

(

S_{0}

= influent BOD,

X

= MLSS,

V

= tank volume)

Sludge age (MCRT):

θ_{c} = \frac{V X}{Q _{w} X _{r} + Q _{e} X _{e}}

Hydraulic retention time:

τ = \frac{V}{Q}

Recycle ratio:

R = \frac{Q _{r}}{Q}

, set by the MLSS and settled-sludge concentrations.

The ‘IIT Traps’

⚠
Low F/M ⇒ extended aeration, more stable, better nitrification; high F/M ⇒ high-rate.
⚠
Sludge age, not HRT, controls which organisms survive (slow nitrifiers need long θc).
⚠
MLSS (X) is mixed-liquor solids, distinct from influent BOD.

📚 Standard references

Wastewater Engineering: Treatment and Resource Recovery — Metcalf & Eddy · Activated Sludge

PART 2

Progressive 3-Tier Question Suite

Q1HARD2 Marks · NAT

An aeration tank has volume V = 1000 m³ and MLSS X = 2500 mg/L, treating Q = 5000 m³/day of wastewater with influent BOD S₀ = 200 mg/L. The F/M ratio (per day) is _____.

Q2BASIC1 Mark · MCQ

In the activated-sludge process, the parameter that determines whether slow-growing nitrifiers are retained is the:

Q3MEDIUM1 Mark · MCQ

A very low F/M ratio in an activated-sludge plant corresponds to: