Wastewater Collection & Treatment

A. Wastewater collection systems

Collection Systems, Flow Rates and Inflow/Infiltration

Sizing sanitary sewers with Manning for self-cleansing velocity, converting population to peak design flow, and the inflow/infiltration and wet-well cycling that drive lift-station design.

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B. Wastewater treatment systems

BOD Loading, Removal Efficiency and Treatment Trains

Converting concentrations to mass loads with the 8.34 factor, tracking BOD and TSS removal unit-by-unit through the treatment train, and meeting characteristic secondary effluent limits.

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C. Preliminary treatment

Preliminary Treatment: Screening and Grit Removal

Bar-screen velocity and head loss, horizontal and aerated grit-chamber design, and flow equalization that smooths the load before primary and secondary treatment.

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D. Primary treatment

Primary Treatment and Clarifier Design

Sizing primary clarifiers by surface overflow rate and detention time, predicting BOD/TSS removal, checking weir loading, and the scraper and sludge withdrawal that close the mass balance.

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E. Secondary treatment

Activated Sludge: F/M, SRT and Secondary Clarifiers

The F/M ratio, mean cell residence time (SRT), MLSS mass balance, hydraulic retention time, RAS/WAS flows, and the coupled secondary clarifier that together control a biological treatment plant.

Activated sludge is the heart of modern wastewater treatment and the densest source of secondary-treatment points on the PE-CIV-WRE exam. The idea is elegant: grow a concentrated population of microorganisms (the 'activated' floc) in an aerated reactor, let them eat the dissolved and colloidal BOD that primary treatment could not settle, then separate the biomass in a clarifier and recycle most of it to keep the population high. The whole process is controlled by just two ratios — the F/M ratio and the solids retention time — and a mass balance that ties the reactor to its clarifier. Master those and you can answer almost anything in this area. The biological relations are in the NCEES PE Civil Reference Handbook §6.8.5.3 Activated Sludge Treatment; the secondary-clarifier overflow and solids-loading design split follows standard Metcalf & Eddy practice.

The system: reactor, clarifier, recycle, and waste

Picture four flows. Influent (

Q_0

, BOD

S_0

) enters the aeration basin (volume

V

, holding a microbial concentration

X

= MLSS). The mixed liquor flows to the secondary clarifier, where biomass settles. Clarified effluent (

Q_e

, BOD

S_e

) leaves the top. Settled sludge is split: most returns as RAS (

Q_r

, concentration

X_r

) to reseed the reactor, and a small waste stream (WAS,

Q_w

,

X_w

) is removed to hold the population steady. Every design quantity below is a statement about one of these flows or a balance among them.

Q_0 + Q_r \to \text{aeration basin }(V,X) \to \text{clarifier} \to \{Q_e,X_e\}\ +\ \{Q_r,X_r\}\ +\ \{Q_w,X_w\}

The F/M ratio: feeding the bugs

The food-to-microorganism ratio is the rate of BOD fed divided by the mass of organisms available to eat it — pounds of BOD per day per pound of MLSS. It is the operator's throttle. A high F/M (overfed,

>0.5

) grows dispersed, poorly-settling biomass; a low F/M (underfed,

<0.2

) drives the population into endogenous respiration, producing dense, well-settling floc but more aeration demand. Conventional plants target

0.2\text{–}0.4\ \text{day}^{-1}

. Because the numerator and denominator are both masses-per-time and mass, the

8.34

factors cancel and you can compute F/M directly from concentrations.

\frac{F}{M} = \frac{Q_0 S_0}{V\,X}

Solids retention time: the master variable

Mean cell residence time — SRT, sludge age, or

\theta_c

— is the average number of days a microbe stays in the system. It is the total mass of solids in the reactor divided by the mass of solids leaving per day (in the waste plus the effluent). SRT is ·the· design and control variable: it sets which organisms can survive (slow-growing nitrifiers need

\theta_c\gtrsim8\text{–}10\ \text{days}

), the effluent quality, the sludge settleability, and the waste rate. You adjust SRT primarily by changing the waste flow

Q_w

. Conventional plants run

\theta_c\approx5\text{–}15\ \text{days}

; extended aeration

20\text{–}30

.

\theta_c = \frac{V\,X}{Q_w X_w + Q_e X_e}\ \approx\ \frac{V\,X}{Q_w X_w}\ \text{(when effluent solids are small)}

MLSS, HRT, and the kinetic link

Two more quantities round out the reactor. The hydraulic retention time

\theta=V/Q_0

is how long the ·liquid· stays in the basin (hours), in sharp contrast to the SRT (days) of the solids — decoupling those two is the whole point of recycling sludge. The MLSS concentration

X

that the basin actually carries is fixed by the kinetics: it rises with SRT and with the BOD removed, and falls with longer HRT and higher decay. This equation ties the biological growth constants (

Y

,

k_d

) to the solids the plant must manage.

X = \frac{\theta_c\,Y\,(S_0 - S_e)}{\theta\,(1 + k_d\,\theta_c)},\qquad \theta=\frac{V}{Q_0}

RAS and WAS: holding the population

Return activated sludge (

Q_r

) reseeds the reactor; its rate is set by a solids balance at the reactor inlet. Treating the underflow concentration

X_r

as the maximum the clarifier can thicken to, the recycle ratio follows directly, and the limiting

X_r

comes from the sludge volume index:

X_{r,\max}=10^6/\text{SVI}

. Waste activated sludge (

Q_w

) is the deliberate removal of biomass that holds SRT constant — wasting from the concentrated RAS line needs a much smaller flow than wasting dilute mixed liquor. Get the recycle ratio from

X

and

X_r

; get

Q_w

from the SRT you want.

R = \frac{Q_r}{Q_0} = \frac{X}{X_r - X},\qquad X_{r,\max}=\frac{10^6}{\text{SVI}}

The coupled secondary clarifier

The clarifier must do two jobs at once, and each has its own loading limit. As a clarifier it must not let solids escape upward, so its surface overflow rate is checked on the influent wastewater flow

Q_0

ALONE — the recycle does not leave over the weir, so it does not count toward overflow. As a thickener it must move the solids down to the underflow, so its solids loading rate is checked on

Q_0+Q_r

at the MLSS concentration. This split is the single most-tested subtlety of secondary clarification: overflow on

Q_0

, solids loading on

Q_0+Q_r

. A clarifier that passes overflow but fails solids loading will thicken poorly and 'lose its blanket,' sending solids over the weir.

v_0 = \frac{Q_0}{A}\quad(\text{influent only}),\qquad \text{SLR}=\frac{8.34\,(Q_0+Q_r)\,X}{A}

Exam strategy

Label the four flows on the figure before computing anything — most errors are flow-bookkeeping errors. For F/M, plug concentrations straight in; the

8.34

s cancel. For SRT, mass-in-reactor over mass-wasted; if effluent solids are given, include

Q_eX_e

, otherwise the RAS-line waste

Q_wX_w

dominates. Keep HRT (hours, on

Q_0

, the liquid) rigidly separate from SRT (days, the solids). For the clarifier, write the two checks side by side: overflow rate on

Q_0

only, solids loading on

Q_0+Q_r

. Verify SRT lands in the right band for the process (and above ~

8\text{–}10\ \text{days}

if nitrification is required).

Key equations

Food-to-microorganism ratio

\frac{F}{M}=\frac{Q_0 S_0}{V X}

$Q_0$ influent flow, $S_0$ influent BOD (mg/L), $V$ aeration volume, $X$ MLSS (mg/L). Units day $^{-1}$ . Conventional target $0.2\text{–}0.4$ .

Solids retention time (SRT,

\theta_c

)

\theta_c=\frac{V X}{Q_w X_w + Q_e X_e}

Sludge age in days = solids inventory ÷ solids wasted per day. Master control variable; nitrification needs $\theta_c\gtrsim8\text{–}10\ \text{d}$ .

MLSS from kinetics

X=\frac{\theta_c\,Y(S_0-S_e)}{\theta(1+k_d\theta_c)}

$Y$ = yield (mg VSS/mg BOD, typical range 0.4–1.2), $k_d$ = endogenous decay (day $^{-1}$ , ~0.06), $\theta=V/Q_0$ = HRT. Predicts the MLSS the basin carries.

Hydraulic retention time

\theta=\frac{V}{Q_0}

Liquid residence time (hours). Distinct from SRT: HRT is on influent flow; SRT is on the solids. Conventional aeration $\theta\approx4\text{–}8\ \text{hr}$ .

Process efficiency

E=\frac{S_0-S_e}{S_0}\times100\%

BOD removal across secondary; conventional activated sludge $85\text{–}95\%$ .

Recycle (RAS) ratio

R=\frac{Q_r}{Q_0}=\frac{X}{X_r-X}

From a solids balance at the reactor inlet. $X$ = MLSS, $X_r$ = RAS concentration. Typical $R\approx0.25\text{–}1.0$ .

SVI and limiting RAS concentration

\text{SVI}=\frac{SV\times1000}{\text{MLSS}},\qquad X_{r,\max}=\frac{10^6}{\text{SVI}}

SVI (mL/g) from the 30-min settled volume $SV$ (mL/L). Good sludge: $80\text{–}120$ . Sets the maximum thickening of the underflow.

Secondary clarifier overflow rate

v_0=\frac{Q_0}{A}

Surface overflow rate uses INFLUENT flow only ( $Q_0$ ), NOT $Q_0+Q_r$ . RAS does not leave over the weir. Limit ~400–800 gpd/ft $^2$ avg.

Secondary clarifier solids loading rate

\text{SLR}=\frac{8.34\,(Q_0+Q_r)\,X}{A}

Thickening check uses $Q_0+Q_r$ at MLSS $X$ . Units lb/day-ft $^2$ ; $Q$ in MGD, $X$ in mg/L. Typical avg ~19–29.

Clarifier solids mass balance

(Q_0+Q_r)X = Q_e X_e + Q_r X_r + Q_w X_w

Steady-state solids in = solids out. Closes the system; solve for an unknown flow or concentration.

Solids (system) loading mass

SL=8.34\,Q\,X

Mass of suspended solids per day (lb/day); $Q$ in MGD, $X$ in mg/L. General form of the $8.34$ conversion for solids.

Worked examples

F/M ratio and loadings

Problem. An aeration basin of $V=1.0\ \text{MG}$ treats $Q_0=5.0\ \text{MGD}$ with influent $\text{BOD}_5=200\ \text{mg/L}$ at an MLSS of $X=2{,}500\ \text{mg/L}$ . Find the F/M ratio and the hydraulic retention time.

Solution. F/M:

\dfrac{F}{M}=\dfrac{Q_0 S_0}{V X}=\dfrac{5.0\times200}{1.0\times2{,}500}=\dfrac{1000}{2500}=0.400\ \text{day}^{-1}

. HRT:

\theta=V/Q_0=1.0/5.0=0.200\ \text{day}\times24=4.80\ \text{hr}

. Final:

\text{F/M}=\mathbf{0.400\ day^{-1}}

,

\theta=\mathbf{4.80\ hr}

. Sanity check via masses: BOD load

=8.34(5.0)(200)=8{,}340\ \text{lb/d}

; MLSS mass

=8.34(1.0)(2{,}500)=20{,}850\ \text{lb}

; ratio

=8{,}340/20{,}850=0.400

— identical, confirming the

8.34

s cancel. Both values are squarely conventional.

\frac{F}{M}=\frac{5.0(200)}{1.0(2500)}=0.400\ \text{day}^{-1}

SRT and the waste sludge rate

Problem. For the same basin ( $V=1.0\ \text{MG}$ , $X=2{,}500\ \text{mg/L}$ , $Q_0=5.0\ \text{MGD}$ ), the plant wastes from the RAS line at $X_w=10{,}000\ \text{mg/L}$ , and the effluent carries $X_e=15\ \text{mg/L}$ at $Q_e\approx Q_0$ . What waste flow $Q_w$ holds the SRT at $10\ \text{days}$ ?

Solution. Rearrange

\theta_c=\dfrac{VX}{Q_wX_w+Q_eX_e}

for

Q_w

:

Q_w=\dfrac{VX/\theta_c - Q_eX_e}{X_w}=\dfrac{(1.0)(2{,}500)/10 - (5.0)(15)}{10{,}000}=\dfrac{250-75}{10{,}000}=0.0175\ \text{MGD}

. That is

17{,}500\ \text{gpd}=12.2\ \text{gpm}

. Final:

Q_w=\mathbf{0.0175\ MGD}

(about

12\ \text{gpm}

). Sanity check: back-substitute —

\theta_c=\dfrac{(1.0)(2500)}{(0.0175)(10{,}000)+(5.0)(15)}=\dfrac{2500}{175+75}=10.0\ \text{days}

. The tiny waste flow relative to

Q_0

is expected because WAS is drawn from concentrated RAS.

Q_w=\frac{VX/\theta_c - Q_eX_e}{X_w}=\frac{250-75}{10{,}000}=0.0175\ \text{MGD}

RAS ratio from SVI

Problem. The mixed liquor in the same plant has an SVI of $120\ \text{mL/g}$ at $X=2{,}500\ \text{mg/L}$ . Estimate the maximum RAS concentration, the required recycle ratio, and the RAS flow.

Solution. Limiting RAS concentration:

X_r=10^6/\text{SVI}=10^6/120=8{,}330\ \text{mg/L}

. Recycle ratio:

R=\dfrac{X}{X_r-X}=\dfrac{2{,}500}{8{,}330-2{,}500}=\dfrac{2{,}500}{5{,}830}=0.429

. RAS flow:

Q_r=R\,Q_0=0.429\times5.0=2.14\ \text{MGD}

. Final:

X_r\approx\mathbf{8{,}330\ mg/L}

,

R\approx\mathbf{0.43}

,

Q_r\approx\mathbf{2.14\ MGD}

. Sanity check:

R=0.43

sits in the typical

0.25\text{–}1.0

recycle band, and an SVI of

120

is a fair-to-good settling sludge, consistent with a moderate recycle.

R=\frac{X}{X_r-X}=\frac{2{,}500}{8{,}330-2{,}500}=0.429

Secondary clarifier: overflow vs. solids loading

Problem. Size-check a $90\text{-ft}$ diameter secondary clarifier for the plant: $Q_0=5.0\ \text{MGD}$ , MLSS $X=2{,}500\ \text{mg/L}$ , recycle $Q_r=1.67\ \text{MGD}$ ( $R=0.33$ ). Compute the surface overflow rate and the solids loading rate, and compare to typical limits.

Solution. Area:

A=\pi D^2/4=\pi(90)^2/4=6{,}360\ \text{ft}^2

. Overflow rate (INFLUENT only):

v_0=Q_0/A=5.0\times10^6/6{,}360=786\ \text{gpd/ft}^2

— at the top of the

400\text{–}800

average band, OK. Solids loading rate (uses

Q_0+Q_r

):

\text{SLR}=\dfrac{8.34(Q_0+Q_r)X}{A}=\dfrac{8.34(5.0+1.67)(2{,}500)}{6{,}360}=\dfrac{139{,}000}{6{,}360}=21.9\ \text{lb/day-ft}^2

— within the typical

19\text{–}29

average range. Final:

v_0=\mathbf{786\ gpd/ft^2}

,

\text{SLR}=\mathbf{21.9\ lb/day\text{-}ft^2}

— both acceptable. Sanity check: had we (wrongly) put

Q_0+Q_r

in the overflow rate it would read

1{,}050\ \text{gpd/ft}^2

, over the limit — exactly the error the influent-only rule prevents.

v_0=\frac{Q_0}{A}=786\ \tfrac{\text{gpd}}{\text{ft}^2},\qquad \text{SLR}=\frac{8.34(6.67)(2500)}{6360}=21.9\ \tfrac{\text{lb}}{\text{day-ft}^2}

Common pitfalls

•Including RAS in the clarifier OVERFLOW rate. Surface overflow uses $Q_0$ only; the recycle leaves through the underflow, not over the weir. Solids loading is what uses $Q_0+Q_r$ .
•Confusing SRT (sludge age, days, on the solids) with HRT (liquid time, hours, on $Q_0$ ). They are deliberately decoupled by recycling — never substitute one for the other.
•Computing F/M with the $8.34$ factor left in. The factors cancel; F/M $=Q_0S_0/(VX)$ uses concentrations directly (or masses on both top and bottom).
•Wasting from mixed liquor vs. RAS without changing $X_w$ : $Q_w$ from the dilute reactor ( $X\approx2{,}500$ ) is several times larger than from the thick RAS line ( $X_r\approx8{,}000\text{–}10{,}000$ ) for the same SRT.
•Forgetting nitrification needs a long SRT. If the permit requires ammonia removal, $\theta_c$ must be roughly $\ge8\text{–}10\ \text{days}$ — a short-SRT high-rate design will not nitrify.
•SVI sign error: $X_{r,\max}=10^6/\text{SVI}$ , so a HIGH SVI (bulking sludge) gives a LOW maximum RAS concentration and forces a high recycle ratio.
•Mixing F/M units: it is per day (mass BOD per day per unit mass MLSS). Volumetric loading (mass BOD per unit reactor volume) is a different parameter — do not interchange them.

References

NCEES PE Civil Reference Handbook — §6.8.5.3 Activated Sludge Treatment (F/M, SRT, MLSS, RAS, SVI, solids loading)
Recommended Standards for Wastewater Facilities (Ten States Standards), 2014 — secondary clarifier overflow and solids loading criteria
Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery — activated-sludge kinetics, design tables, and the secondary-clarifier overflow vs. solids-loading split

F. Nutrient removal

Nutrient Removal: Nitrification, Denitrification and Phosphorus

Nitrification oxygen and alkalinity demand, anoxic denitrification, biological (A2O) and chemical phosphorus removal, and the SRT and stoichiometry that hit effluent N and P limits.

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G. Solids treatment, handling, and disposal

Sludge Production, Digestion and Solids Handling

Compute dry-solids mass from TSS and BOD removal, convert to wet-sludge volume at a percent solids, and size thickening, digestion, gas production, and biosolids disposal.

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H. Disinfection

Wastewater Disinfection (Chlorine and UV)

Set chlorine dose from demand and residual, size contact using CT and the baffling factor, dechlorinate, and apply UV dose and transmittance to meet coliform limits.

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I. Advanced treatment

Advanced Treatment: Filtration, AOP and Reuse

Tertiary effluent filtration, membrane bioreactors and reverse osmosis for reuse, advanced oxidation, and the multi-barrier quality criteria that govern water reuse.

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