Water and Wastewater

A. Water and wastewater characteristics

Water and Wastewater Characteristics and Loading

Read water quality the way a treatment plant does: organic, solids, and nutrient measures, then convert concentrations into mass loads and removal efficiencies.

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C. Physical processes

Coagulation, Sedimentation, and Filtration

The clarification train as loading-rate arithmetic: overflow rate versus settling velocity, detention and weir loading, and rapid-filter rates and backwash.

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D. Chemical processes

Chemical Treatment: Coagulation and Lime–Soda Softening

Coagulant chemistry and alkalinity demand, the velocity gradient G that sizes mixing, hardness as CaCO3, the lime–soda dose, and the meq/L bar chart.

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E. Biological processes

Activated Sludge: F/M, SRT, and Clarifiers

The workhorse of biological wastewater treatment, sized by F/M and solids retention time, balanced across the aeration basin and coupled secondary clarifier.

Activated sludge is the central process of modern wastewater treatment: a suspension of microorganisms in an aerated basin consumes the organic load (BOD), and a downstream secondary clarifier settles that biomass so it can be returned to the basin and kept working. The whole design lives or dies on two control parameters — the food-to-microorganism ratio (

\text{F/M}

) and the solids retention time (

\text{SRT}

, or mean cell residence time

\theta_c

) — and on recognizing that the basin and clarifier are one coupled system. The governing relations appear in the NCEES FE Reference Handbook — Environmental Engineering under Activated Sludge and the accompanying design-parameter table.

The F/M ratio

The food-to-microorganism ratio is the organic load applied per unit of biomass per day — the 'how much are we feeding each microbe' parameter. Food is the influent BOD load

Q_0 S_0

; microorganisms are the mass of mixed-liquor suspended solids in the basin,

V\,X_A

. A high

\text{F/M}

means a heavily loaded, fast-growing, young sludge; a low

\text{F/M}

means a lightly loaded, well-stabilized sludge. Conventional plants run

\text{F/M}\approx 0.2

–

0.4\,\text{kg BOD/(kg MLSS·d)}

.

\text{F/M} = \frac{Q_0\,S_0}{V\,X_A}

MLSS and the aeration basin mass balance

The biomass concentration in the basin, the mixed-liquor suspended solids

X_A

, results from a balance between growth on substrate and loss by decay and wasting. Combining the steady-state biomass and substrate balances gives the design relation linking basin concentration to the kinetic constants, the retention times, and the substrate removed. This is what couples the biological kinetics to the physical sizing.

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

Hydraulic retention time versus solids retention time

Two residence times govern the process, and confusing them is the classic error. Hydraulic retention time

\theta = V/Q

is how long the water stays in the basin — hours. Solids retention time

\theta_c

(SRT, sludge age) is how long the biomass stays in the system — days — because settled solids are recycled rather than flushed out with the flow. The recycle is precisely what decouples

\theta_c

from

\theta

, letting a basin with a

6

-hour hydraulic time hold a

10

-day-old sludge.

\theta = \frac{V}{Q}, \qquad \theta_c = \frac{V\,X_A}{Q_w X_w + Q_e X_e}

SRT, wasting, and process control

Solids retention time is the master control variable: it sets the sludge age, the effluent quality, and whether slow-growing organisms (like nitrifiers) survive. You control

\theta_c

by wasting — deliberately removing biomass at

Q_w

,

X_w

from the system each day. Because the wasted and effluent solids equal the daily biomass production at steady state, increasing the wasting rate lowers the sludge age and

\text{F/M}

moves inversely to

\theta_c

.

\frac{1}{\theta_c} = \frac{Q_w X_w + Q_e X_e}{V\,X_A}

Sludge recycle (RAS) and the coupled clarifier

The secondary clarifier thickens settled biomass to a concentration

X_r

, and most of it is returned as return activated sludge (RAS) at flow

Q_R

to maintain the basin MLSS. A solids mass balance around the clarifier ties everything together: what flows in from the basin must leave in the effluent, the recycle, and the waste streams. Neglecting the small effluent and waste solids, the recycle ratio needed to hold a target MLSS follows directly from the basin and underflow concentrations.

(Q_0 + Q_R)\,X_A = Q_e X_e + Q_R X_r + Q_w X_w

Settleability and clarifier loading

The clarifier can only return solids it can settle, so its loading must match the sludge's settling character. The sludge volume index (

\text{SVI}

) — the volume one gram of settled solids occupies after

30

minutes — gauges settleability; below about

100\,\text{mL/g}

settles well, above

150\,\text{mL/g}

signals bulking. A secondary clarifier is checked on two loadings: the surface overflow rate (a settling criterion) and the solids loading rate (a thickening criterion), the latter being what usually controls because the unit must thicken a heavy recycled solids flux.

\text{SVI} = \frac{V_{30}\,(\text{mL/L})\times 1000}{X_{\text{MLSS}}\,(\text{mg/L})}, \qquad \text{SLR} = \frac{(Q_0+Q_R)\,X_A}{A}

Exam strategy

Identify the loading type before computing:

\text{F/M}

and volumetric loading use the BOD mass load over basin mass or volume;

\theta

uses volume over flow;

\theta_c

uses basin solids inventory over the daily solids leaving. Keep

\theta

(hours, water) and

\theta_c

(days, solids) rigorously separate. For wasting, solve the SRT definition for

Q_w

; for recycle, write the clarifier solids balance and, if effluent/waste solids are negligible, use

R = X_A/(X_r - X_A)

. Convert all concentrations to consistent mass units (kg/m³ or mg/L) before substituting, and sanity-check answers against the handbook's design-parameter ranges.

Key equations

Food-to-microorganism ratio

\text{F/M} = \dfrac{Q_0\,S_0}{V\,X_A}

Applied BOD load per unit biomass per day (kg BOD/(kg MLSS·d)). $Q_0$ =influent flow, $S_0$ =influent BOD, $V$ =basin volume, $X_A$ =MLSS. Conventional 0.2–0.4.

Solids retention time (SRT)

\theta_c = \dfrac{V\,X_A}{Q_w X_w + Q_e X_e}

Mean cell residence time / sludge age (d). Basin solids inventory divided by solids leaving in waste + effluent.

Hydraulic retention time

\theta = \dfrac{V}{Q}

Liquid residence time in the basin (h). Typically 4–8 h for conventional activated sludge; distinct from SRT.

Biomass concentration (kinetics)

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

Design MLSS from kinetics. $Y$ =yield (kg VSS/kg BOD), $k_d$ =decay rate (d⁻¹), $S_e$ =effluent BOD.

Clarifier solids mass balance

(Q_0 + Q_R)\,X_A = Q_e X_e + Q_R X_r + Q_w X_w

Steady-state solids around the secondary clarifier. $X_r$ =RAS (underflow) solids, $X_w$ =waste solids, $X_e$ =effluent SS.

Recycle ratio

R = \dfrac{Q_R}{Q_0} = \dfrac{X_A}{X_r - X_A}

Return-sludge ratio to hold target MLSS (right form neglects effluent/waste solids). Typically 0.25–1.0.

Sludge volume index

\text{SVI} = \dfrac{V_{30}\,(\text{mL/L})\times 1000}{X_{\text{MLSS}}\,(\text{mg/L})}

Volume (mL) of 1 g settled solids after 30 min. <100 good settling, >150 bulking. Underflow X_r ≈ 10⁶/SVI.

Solids loading rate

\text{SLR} = \dfrac{(Q_0 + Q_R)\,X_A}{A}

Thickening criterion for the secondary clarifier (kg/(m²·d) or lb/(ft²·d)); often controls over overflow rate.

Volumetric organic loading

L_v = \dfrac{Q_0\,S_0}{V}

BOD load per basin volume (kg BOD/(m³·d)). Conventional ~0.3–0.6.

Worked examples

F/M, HRT, and volumetric loading

Problem. An aeration basin of $V = 2{,}500\,\text{m}^3$ treats $Q_0 = 10{,}000\,\text{m}^3/\text{d}$ of primary effluent with $S_0 = 150\,\text{mg/L}$ BOD, holding $X_A = 2{,}500\,\text{mg/L}$ MLSS. Find the F/M ratio, hydraulic retention time, and volumetric loading, and compare to conventional ranges.

Solution. F/M ratio (mg/L cancels in the ratio):

\text{F/M} = \frac{Q_0 S_0}{V X_A} = \frac{10{,}000\times150}{2{,}500\times2{,}500} = \frac{1.5\times10^{6}}{6.25\times10^{6}} = 0.24\,\text{d}^{-1}.

Hydraulic retention time:

\theta = \frac{V}{Q_0} = \frac{2{,}500}{10{,}000} = 0.25\,\text{d} = 6.0\,\text{h}.

Volumetric loading (with

S_0 = 0.150\,\text{kg/m}^3

):

L_v = \frac{Q_0 S_0}{V} = \frac{10{,}000\times0.150}{2{,}500} = 0.60\,\text{kg BOD/(m}^3\text{·d)}.

Sanity check:

\text{F/M}=0.24

(range

0.2

–

0.4

),

\theta=6\,\text{h}

(range

4

–

8\,\text{h}

),

L_v=0.60

(range

0.3

–

0.6

) — all consistent with a conventional plant. Final:

\text{F/M}=0.24\,\text{d}^{-1}

,

\theta=6.0\,\text{h}

,

L_v=0.60\,\text{kg/(m}^3\text{·d)}

.

Solids retention time and wasting rate

Problem. For the same basin ( $V = 2{,}500\,\text{m}^3$ , $X_A = 2{,}500\,\text{mg/L}$ ), the plant targets $\theta_c = 10\,\text{d}$ . Waste sludge is at $X_w = 10{,}000\,\text{mg/L}$ ; the effluent carries $X_e = 15\,\text{mg/L}$ at $Q_e \approx 10{,}000\,\text{m}^3/\text{d}$ . Find the required waste flow $Q_w$ .

Solution. From the SRT definition, the daily solids leaving must equal the inventory over

\theta_c

:

Q_w X_w + Q_e X_e = \frac{V X_A}{\theta_c} = \frac{2{,}500\times2{,}500}{10} = 625{,}000\ (\text{mg/L·m}^3/\text{d}).

Effluent solids:

Q_e X_e = 10{,}000\times15 = 150{,}000

. Waste solids:

Q_w X_w = 625{,}000 - 150{,}000 = 475{,}000

.

Q_w = \frac{475{,}000}{10{,}000} = 47.5\,\text{m}^3/\text{d}.

Sanity check: neglecting effluent losses would give

Q_w\approx 62.5\,\text{m}^3/\text{d}

; the real value is lower because the effluent already carries some solids out.

Q_w\approx 0.5\%

of plant flow is a realistic wasting rate. Final:

Q_w \approx 47.5\,\text{m}^3/\text{d}

.

Settleability, recycle ratio, and solids loading

Problem. The mixed liquor ( $X_A = 2{,}500\,\text{mg/L}$ ) gives a $30$ -minute settled volume of $250\,\text{mL/L}$ . The secondary clarifier has surface area $A = 400\,\text{m}^2$ . Find the SVI, estimate the RAS (underflow) concentration, the recycle ratio needed to hold the MLSS (neglecting effluent/waste solids), and the solids loading rate for $Q_0 = 10{,}000\,\text{m}^3/\text{d}$ .

Solution. SVI:

\text{SVI} = \frac{V_{30}\times1000}{X_{\text{MLSS}}} = \frac{250\times1000}{2{,}500} = 100\,\text{mL/g} \ (\text{good settling}).

Underflow estimate:

X_r \approx 10^{6}/\text{SVI} = 10^{6}/100 = 10{,}000\,\text{mg/L}

. Recycle ratio (effluent/waste solids neglected):

R = \frac{X_A}{X_r - X_A} = \frac{2{,}500}{10{,}000 - 2{,}500} = \frac{2{,}500}{7{,}500} = 0.333,\quad Q_R = 0.333\times10{,}000 = 3{,}330\,\text{m}^3/\text{d}.

Solids loading rate (with

X_A = 2.5\,\text{kg/m}^3

):

\text{SLR} = \frac{(Q_0 + Q_R)X_A}{A} = \frac{(10{,}000 + 3{,}330)(2.5)}{400} = \frac{33{,}325}{400} = 83.3\,\text{kg/(m}^2\text{·d)} = 3.47\,\text{kg/(m}^2\text{·h)}.

Sanity check:

\text{SVI}=100\,\text{mL/g}

implies

X_r\approx10{,}000\,\text{mg/L}

, internally consistent;

R=0.33

sits in the

0.25

–

1.0

band;

\text{SLR}\approx3.5\,\text{kg/(m}^2\text{·h)}

is under the usual

\sim4

–

6

limit. Final:

\text{SVI}=100\,\text{mL/g}

,

X_r\approx10{,}000\,\text{mg/L}

,

R=0.333

,

\text{SLR}\approx83.3\,\text{kg/(m}^2\text{·d)}

.

Common pitfalls

•Confusing hydraulic retention time θ (hours, follows the water) with solids retention time θc (days, follows the recycled biomass). They differ by the recycle — that is the whole point of activated sludge.
•Computing F/M with effluent BOD or with the wrong solids. Food is the influent BOD load Q₀S₀; microorganisms are the basin inventory V·X_A (MLSS, sometimes MLVSS) — not the recycle solids.
•Dividing influent flow by effluent flow inconsistently in the clarifier balance. Use one steady-state solids balance: (Q₀+Q_R)X_A = Q_eX_e + Q_RX_r + Q_wX_w.
•Sizing the secondary clarifier on overflow rate alone. For activated sludge the solids loading (thickening) rate usually controls, because the unit must handle the heavy recycled solids flux.
•Forgetting the effluent solids in the SRT denominator. Neglecting Q_eX_e overstates the required wasting rate Q_w — include it when effluent SS is given.
•Reading SVI backwards: a HIGH SVI (>150 mL/g) means poor, bulking sludge that settles badly; a low SVI settles and thickens well.
•Mixing concentration units mid-problem. Keep MLSS, RAS, and BOD all in mg/L or all in kg/m³ before substituting; 2,500 mg/L = 2.5 kg/m³.

References

NCEES FE Reference Handbook — Environmental Engineering
Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery — activated-sludge kinetics, SRT/F:M, clarifier coupling, RAS/WAS
Davis & Cornwell, Introduction to Environmental Engineering — F/M, sludge age, secondary clarifier loading

Fixed-Film Processes and Nutrient Removal

Trickling filters and RBCs, Monod growth kinetics, and biological/chemical nitrogen and phosphorus removal — the attached-growth half of secondary treatment.

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F. Sludge treatment and handling

Disinfection and Sludge Treatment

The CT concept and Chick-Watson inactivation, chlorine chemistry and breakpoint, UV dose, and the thickening-digestion-dewatering train that turns sludge into biosolids.

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G. Water conservation and reuse

Water Conservation and Reuse

Demand management and graywater, reclaimed-water classes, membrane treatment (MF/UF/NF/RO recovery and rejection) and advanced oxidation, and indirect potable reuse.

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