Settling basins following air-activated sludge reactors

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

0 formulas

10 exam-style examples

~45 min

All Environmental Engineering lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Settling basins following air-activated sludge reactors within Environmental Engineering. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what settling basins following air-activated sludge reactors describes physically and when it applies.
State every one of the 0 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: mg/L × MGD × 8.34 = lb/day is the single most used conversion.

Lecture

Why this section exists. Settling basins following air-activated sludge reactors is the part of Environmental Engineering that lets you connect a treatment unit or receiving water body to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a mass balance across one reactor or one unit process. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. mg/L × MGD × 8.34 = lb/day is the single most used conversion. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Aeration basin at a wastewater treatment plant with churning aerated water and walkways. — Photo 1. Where this shows up in practice: settling basins following air-activated sludge reactors.
Capstone Studio instructional photograph

Environmental Engineering — Settling basins following air-activated sludge reactors: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a treatment unit or receiving water body. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 0 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

All configurations EXCEPT extended aeration 400−700 16−28 1,000−1,200 40−64 19−29 4−6 38 8 2 12−15
Extended aeration 200−400 8−16 600−800 24−32 5−24 1−5 34 7 2 12−15
Settling basins following chemical flocculation reactors 800−1,200 2
Metcalf and Eddy; AECOM, Wastewater Engineering: Treatment and Resource Recovery, 5th ed., New York: McGraw-Hill, 2014, p. 190, Table 8-34.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Hydraulic detention time in a tank — Settling basins following air-activated sludge reactors

A treatment tank holds 91,449 gal and treats 4.0 MGD. Find the hydraulic detention time.

Given

V = 91,449 gal
Q = 4.0 MGD

Find

Detention time θ

Start with the thinking

θ = V/Q with consistent volume units.
Express the answer in hours for design comparison.

Step-by-step solution

Detention
$θ = V / Q$
Flow in gal/day
$4.0 \times 1 0^{6} = 4, 000, 000 g a l / d a y$
Substituting
$θ = 91449/4, 000, 000 = 0.0229 d a y$
Convert
$θ = 0.0229 (24) = 0.55 h r$

Answer: θ ≈ 0.55 hr

Why the other options are there

0.023 hr (day/hour conversion missed)
43.74 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Settling basins following air-activated sludge reactors

Example 2

Converting ppm to mg/m³ — Settling basins following air-activated sludge reactors

A stack gas contains 25.5 ppm of a compound with molecular weight 46 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 25.5 ppm
MW = 46 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 25.5 (46) /24.45$
Evaluate — 47.98 mg/m³
Reverse check
$pp m = 47.98 (24.45) /46 = 25.5 ✓$

Answer: ≈ 48.0 mg/m³

Why the other options are there

13.6 mg/m³ (ratio inverted)
52.4 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Settling basins following air-activated sludge reactors

Example 3

First-order removal in a CSTR versus a plug-flow reactor — Settling basins following air-activated sludge reactors

A reactor of volume 1,786 m³ treats 0.15 m³/s carrying 89 mg/L of a contaminant that decays first-order with k = 0.10 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Given

V = 1,786 m³
Q = 0.15 m³/s
C₀ = 89 mg/L
k = 0.10 h⁻¹

Find

τ, CSTR effluent and PFR effluent

Start with the thinking

The mass balance for steady state is: in − out − reaction = 0.
For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

Formula
$τ = V / Q$
Substituting
$τ = 1786/0.15 = 11, 907 s = 3.307 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 89/ (1 + 0.10 \times 3.307) = 66.88 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 89 e^{(} - 0.10 \times 3.307) = 63.94 m g / L$
Comparison — plug flow removes 28.2% versus 24.9% for the CSTR

Answer: τ = 3.31 h; C_CSTR = 66.9 mg/L, C_PFR = 63.9 mg/L

Why the other options are there

59.6 mg/L (linear decay assumed)
66.9 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Settling basins following air-activated sludge reactors

Example 4

Activated sludge F/M ratio, aeration time and sludge production — Settling basins following air-activated sludge reactors

An activated sludge plant treats 14,109 m³/d with an influent BOD of 366 mg/L, effluent BOD 15 mg/L, MLSS 1,975 mg/L, and an aeration basin volume of 3,599 m³. With Y = 0.65 mg VSS/mg BOD, k_d = 0.07 d⁻¹ and an SRT of 9 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 14,109 m³/d
S₀ = 366 mg/L, S = 15 mg/L
X = 1,975 mg/L, V = 3,599 m³
Y = 0.65, k_d = 0.07 d⁻¹, SRT = 9 d

Find

F/M ratio, detention time τ and sludge wasted per day

Start with the thinking

F/M compares the food entering daily with the microbial mass held in the basin; conventional plants run near 0.2–0.5 d⁻¹.
Endogenous decay reduces net sludge yield as the SRT lengthens.

Step-by-step solution

Formula
$F / M = Q S_{0} / (V X)$
Substituting
$F / M = 14109 (366) / (3599 \times 1975) = 0.726 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 3599/14109 = 0.2551 d = 6.12 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.65 (14109) (366 - 15) / (1 + 0.07 \times 9) m g / d$
Evaluate
$P_{x} = 1, 975 k g V S S / d a y$

Answer: F/M = 0.73 d⁻¹, τ = 6.1 h, sludge = 1,975 kg/d

Why the other options are there

F/M = 2,615 (basin volume omitted)
P_x = 3,219 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Settling basins following air-activated sludge reactors

Example 5

Circular clarifier overflow rate, detention time and Stokes settling — Settling basins following air-activated sludge reactors

A circular clarifier 33 m in diameter and 3.5 m deep treats 26,846 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 100 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 26,846 m³/d
D = 33 m, depth = 3.5 m
d_p = 100 µm, ρ_s = 2650 kg/m³
μ = 0.001 Pa·s

Find

SOR, detention time, weir loading and the Stokes settling velocity

Start with the thinking

An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

Surface area
$A = (π /4) (33)^{2} = 855.3 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 26846/855.3 = 31.39 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 2, 994/26846 = 2.68 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 26846/(π × 33) = 258.9 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (1.0 e - 4)^{2} / (18 \times 0.001) = 777.0 m / d$
Compare
$v_{s} = 777.0 m / d > S O R = 31.39 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 31.4 m/d, τ = 2.7 h, weir loading = 258.9 m³/m·d, v_s = 777.0 m/d

Why the other options are there

SOR = 74.0 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Settling basins following air-activated sludge reactors

Example 6

Hydraulic detention time in a tank — Settling basins following air-activated sludge reactors (2)

A treatment tank holds 117,919 gal and treats 2.7 MGD. Find the hydraulic detention time.

Given

V = 117,919 gal
Q = 2.7 MGD

Find

Detention time θ

Start with the thinking

θ = V/Q with consistent volume units.
Express the answer in hours for design comparison.

Step-by-step solution

Detention
$θ = V / Q$
Flow in gal/day
$2.7 \times 1 0^{6} = 2, 700, 000 g a l / d a y$
Substituting
$θ = 117919/2, 700, 000 = 0.0437 d a y$
Convert
$θ = 0.0437 (24) = 1.05 h r$

Answer: θ ≈ 1.05 hr

Why the other options are there

0.044 hr (day/hour conversion missed)
22.90 hr (ratio inverted)

Reference: FE Reference Handbook — Environmental Engineering → Settling basins following air-activated sludge reactors

Example 7

Converting ppm to mg/m³ — Settling basins following air-activated sludge reactors (2)

A stack gas contains 51.0 ppm of a compound with molecular weight 44 g/mol at 25 °C and 1 atm. Express the concentration in mg/m³.

Given

C = 51.0 ppm
MW = 44 g/mol
Molar volume = 24.45 L/mol

Find

Concentration in mg/m³

Start with the thinking

ppm by volume needs the molar volume to become a mass concentration.
24.45 L/mol applies at 25 °C; use 22.4 at 0 °C.

Step-by-step solution

Conversion
$m g / m^{3} = pp m \times M W /24.45$
Substituting
$m g / m^{3} = 51.0 (44) /24.45$
Evaluate — 91.78 mg/m³
Reverse check
$pp m = 91.78 (24.45) /44 = 51.0 ✓$

Answer: ≈ 91.8 mg/m³

Why the other options are there

28.3 mg/m³ (ratio inverted)
100.2 mg/m³ (0 °C molar volume used)

Reference: FE Reference Handbook — Environmental Engineering → Settling basins following air-activated sludge reactors

Example 8

First-order removal in a CSTR versus a plug-flow reactor — Settling basins following air-activated sludge reactors (2)

A reactor of volume 1,565 m³ treats 0.50 m³/s carrying 141 mg/L of a contaminant that decays first-order with k = 0.30 h⁻¹. Compute the hydraulic residence time and the effluent concentration if the tank behaves as a CSTR and as a plug-flow reactor.

Given

V = 1,565 m³
Q = 0.50 m³/s
C₀ = 141 mg/L
k = 0.30 h⁻¹

Find

τ, CSTR effluent and PFR effluent

Start with the thinking

The mass balance for steady state is: in − out − reaction = 0.
For the same volume, plug flow always outperforms a single completely mixed tank for first-order kinetics.

Step-by-step solution

Formula
$τ = V / Q$
Substituting
$τ = 1565/0.50 = 3, 130 s = 0.869 h$
Formula (CSTR)
$C = C_{0} / (1 + k τ)$
Substituting
$C = 141/ (1 + 0.30 \times 0.869) = 111.8 m g / L$
Formula (PFR)
$C = C_{0} e^{(} - k τ)$
Substituting
$C = 141 e^{(} - 0.30 \times 0.869) = 108.6 m g / L$
Comparison — plug flow removes 23.0% versus 20.7% for the CSTR

Answer: τ = 0.87 h; C_CSTR = 111.8 mg/L, C_PFR = 108.6 mg/L

Why the other options are there

104.2 mg/L (linear decay assumed)
111.8 mg/L for both reactors

Reference: FE Reference Handbook — Environmental Engineering → Settling basins following air-activated sludge reactors

Example 9

Activated sludge F/M ratio, aeration time and sludge production — Settling basins following air-activated sludge reactors (2)

An activated sludge plant treats 10,935 m³/d with an influent BOD of 246 mg/L, effluent BOD 13 mg/L, MLSS 2,660 mg/L, and an aeration basin volume of 3,982 m³. With Y = 0.70 mg VSS/mg BOD, k_d = 0.08 d⁻¹ and an SRT of 12 days, compute the F/M ratio, hydraulic detention time and daily sludge production.

Given

Q = 10,935 m³/d
S₀ = 246 mg/L, S = 13 mg/L
X = 2,660 mg/L, V = 3,982 m³
Y = 0.70, k_d = 0.08 d⁻¹, SRT = 12 d

Find

F/M ratio, detention time τ and sludge wasted per day

Start with the thinking

F/M compares the food entering daily with the microbial mass held in the basin; conventional plants run near 0.2–0.5 d⁻¹.
Endogenous decay reduces net sludge yield as the SRT lengthens.

Step-by-step solution

Formula
$F / M = Q S_{0} / (V X)$
Substituting
$F / M = 10935 (246) / (3982 \times 2660) = 0.254 d^{- 1}$
Formula
$τ = V / Q$
Substituting
$τ = 3982/10935 = 0.3642 d = 8.74 h$
Formula
$P_{x} = Y Q (S_{0} - S) / (1 + k_{d} \cdot S R T)$
Substituting
$P_{x} = 0.70 (10935) (246 - 13) / (1 + 0.08 \times 12) m g / d$
Evaluate
$P_{x} = 909.9 k g V S S / d a y$

Answer: F/M = 0.25 d⁻¹, τ = 8.7 h, sludge = 909.9 kg/d

Why the other options are there

F/M = 1,011 (basin volume omitted)
P_x = 1,783 kg/d (decay term ignored)

Reference: FE Reference Handbook — Environmental Engineering → Settling basins following air-activated sludge reactors

Example 10

Circular clarifier overflow rate, detention time and Stokes settling — Settling basins following air-activated sludge reactors (2)

A circular clarifier 29 m in diameter and 3.0 m deep treats 20,539 m³/d. Compute the surface overflow rate, the detention time and the weir loading rate, then check whether a 40 µm particle (SG = 2.65) settles out by Stokes' law.

Given

Q = 20,539 m³/d
D = 29 m, depth = 3.0 m
d_p = 40 µm, ρ_s = 2650 kg/m³
μ = 0.001 Pa·s

Find

SOR, detention time, weir loading and the Stokes settling velocity

Start with the thinking

An ideal settling basin removes every particle whose settling velocity exceeds the overflow rate — depth does not matter for removal, only for detention.
Stokes' law applies to the laminar (small particle) regime.

Step-by-step solution

Surface area
$A = (π /4) (29)^{2} = 660.5 m^{2}$
Formula
$S O R = Q / A$
Substituting — SOR = 20539/660.5 = 31.10 m³/m²·d
Formula — τ = V/Q = A·h/Q
Substituting
$τ = 1, 982/20539 = 2.32 h$
Formula
$w e i r l o a d in g = Q / (π D)$
Substituting — 20539/(π × 29) = 225.4 m³/m·d
Formula
$v_{s} = g (ρ_{s} - ρ) d^{2} / (18 μ)$
Substituting
$v_{s} = 9.81 (1650) (4.0 e - 5)^{2} / (18 \times 0.001) = 124.3 m / d$
Compare
$v_{s} = 124.3 m / d > S O R = 31.10 m / d \to p a r t i c l e i sr e m o v e d$

Answer: SOR = 31.1 m/d, τ = 2.3 h, weir loading = 225.4 m³/m·d, v_s = 124.3 m/d

Why the other options are there

SOR = 75.1 m/d (side area used)
v_s computed with the diameter unsquared

Reference: FE Reference Handbook — Environmental Engineering → Settling basins following air-activated sludge reactors

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a treatment unit or receiving water body, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Settling basins following air-activated sludge reactors contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: a mass balance across one reactor or one unit process.
Unit rule: mg/L × MGD × 8.34 = lb/day is the single most used conversion.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

mg/L × MGD × 8.34 = lb/day is the single most used conversion
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.