Wastewater: Sources, Sewers, Sewage Treatment, Sludge and Industrial Effluents
1. Sources of wastewater and population forecasting
A point source discharges at an identifiable outlet — a sewer outfall, an industrial drain, a treatment plant — so it can be measured, permitted and treated. A non-point (diffuse) source enters over a wide area and intermittently, usually with rain: runoff from farmland carrying fertiliser, pesticides and sediment, urban storm runoff with oil and litter, and seepage from septic tanks and dumps. Point sources are controlled by treatment and discharge standards; non-point sources by land-use practices — buffer strips, contour farming, controlled fertiliser use, detention ponds and constructed wetlands. The domestic sewage flow is normally taken as about 80% of the water supplied, the rest being consumed or lost, with an allowance for infiltration into the sewers.
Sewers and plants are designed for the population 20–30 years ahead, so the census record is extrapolated. The arithmetic increase method adds the average increase per decade, P_n = P + n x̄, and suits old, stable cities. The geometric increase method applies the average percentage growth, P_n = P(1 + r)ⁿ, and suits young, fast-growing ones. The incremental increase method adds the average increment and the average increment of the increments, P_n = P + n x̄ + [n(n + 1)/2]ȳ, a compromise between the two. The logistic curve recognises that growth must level off at a saturation population; from three censuses P₀, P₁, P₂ at equal intervals, P_s = [2P₀P₁P₂ − P₁²(P₀ + P₂)]/(P₀P₂ − P₁²). Graphical extension and comparison with similar cities complete the set.
2. Sanitary and storm sewers, and sewer appurtenances
A separate system carries sewage and storm water in different pipes; a combined system carries both, with overflows that discharge diluted sewage to the river in heavy rain. Sanitary sewers are designed for the peak flow — the average sewage flow times a peak factor that is larger for smaller populations — running partly full, and they are checked at both ends of the velocity range. The minimum is the self-cleansing velocity, which must move the grit that enters, about 0.6–0.9 m/s at design flow (Camp’s relation links it to the particle size and specific gravity); the maximum, about 2.5–3 m/s, protects the pipe from scour. Flow is computed by Manning’s equation, with the proportional depth, velocity and discharge of a partly full circular pipe read from the partial-flow curves. Sewers follow the ground where possible, and below a certain gradient the flow is pumped. Storm sewers are sized by the rational method, Q = CiA/360 with Q in m³/s, i in mm/h and A in hectares, using a weighted runoff coefficient C = ΣC_iA_i/ΣA_i and the intensity for a duration equal to the time of concentration (inlet time plus time of flow in the drain).
Sewer appurtenances are the structures that make the network work and let it be maintained. Manholes give access at every junction, change of direction, gradient or diameter, and at intervals on straight runs; a drop manhole carries an incoming sewer that arrives well above the outgoing one down a vertical pipe, so the sewage does not fall into the chamber and splash the worker. Lamp holes and clean-outs admit inspection or rods. Street inlets and catch basins admit storm water, the latter with a sump that traps grit. An inverted siphon (depressed sewer) carries sewage under a river, railway or valley in a pipe that runs full under pressure, usually as several pipes so that low flows still reach self-cleansing velocity. Flushing tanks clean flat upper reaches, grease and oil traps protect sewers from hotel and garage waste, storm-water regulators and overflows divert excess flow in combined systems, and ventilating shafts release the gases — including H₂S, which corrodes concrete crowns and is dangerous to workers.
3. Preliminary, primary, secondary and tertiary treatment
Preliminary treatment protects the plant: bar screens and comminutors remove rags and large solids, grit chambers settle sand and grit at a controlled horizontal velocity of about 0.3 m/s (held constant by a proportional-flow weir or a Parshall flume) so that the lighter organics stay in suspension, and skimming tanks remove oil and grease. Primary treatment is plain sedimentation, which removes about 60% of the suspended solids and 30–35% of the BOD at an overflow rate of a few tens of m³/m²·d. Secondary treatment removes the dissolved and colloidal organics biologically. In the activated sludge process the settled sewage is aerated with a suspended culture (the mixed liquor), and the solids are settled in a secondary clarifier and mostly returned. Its design and control parameters are the food-to-microorganism ratio F/M = QS₀/(VX), the hydraulic retention time θ = V/Q, the solids retention time (mean cell residence time) θ_c = VX/(Q_wX_r + (Q − Q_w)X_e), the sludge volume index SVI = (settled volume in mL/L after 30 min) × 1000/MLSS (mg/L), in mL/g, and the return ratio R = Q_r/Q, which from a solids balance on the clarifier is R = X/(X_r − X) with X_r ≈ 10⁶/SVI.
Monod kinetics in a completely mixed reactor with recycle give the steady-state results: the effluent substrate S = K_s(1 + k_dθ_c)/[θ_c(Yk − k_d) − 1], which depends only on θ_c, and the biomass X = θ_cY(S₀ − S)/[θ(1 + k_dθ_c)]. Conventional plants run at F/M about 0.2–0.4 d⁻¹ and θ_c about 5–15 d; extended aeration uses a very low F/M and long θ_c, producing little, well-stabilised sludge. An SVI below about 100–150 mL/g means good settling; a high SVI means bulking, usually from filamentous organisms, and rising sludge comes from denitrification in the clarifier. Attached-growth processes hold the biomass on media: the trickling filter (low- or high-rate, with recirculation), whose efficiency the NRC equation estimates as E = 100/[1 + 0.4432√(W/(VF))] for W the BOD load in kg/d, V the medium volume in m³ and recirculation factor F = (1 + R)/(1 + 0.1R)²; and the rotating biological contactor. Waste stabilisation ponds rely on the algae–bacteria symbiosis; the UASB reactor treats strong wastes anaerobically in a granular sludge blanket. Tertiary (advanced) treatment targets what remains: nitrogen by nitrification (autotrophs, needing long θ_c and alkalinity) followed by denitrification in an anoxic zone with a carbon source; phosphorus by chemical precipitation with alum, ferric salts or lime, or biologically by phosphate-accumulating organisms cycled through anaerobic and aerobic zones; suspended solids by filtration; and pathogens by disinfection. Membrane bioreactors combine activated sludge with membrane separation.
| Parameter | Definition | Typical conventional value | Controls |
|---|---|---|---|
| F/M | QS₀/(VX) | 0.2–0.4 d⁻¹ | Loading; settleability |
| θ_c (SRT) | VX/(solids wasted per day) | 5–15 d | Effluent quality; nitrification; sludge yield |
| θ (HRT) | V/Q | 4–8 h | Tank volume |
| SVI | Settled mL/L × 1000/MLSS | Below about 150 mL/g | Clarifier performance; bulking |
| R | Q_r/Q = X/(X_r − X) | 0.25–0.5 | MLSS held in the aeration tank |
4. Sludge generation, processing and disposal; sewage farming
Primary and waste activated sludge are mostly water — 95–99% — so the first task is to remove water. The volume of a sludge of dry-solids mass M_s, solids fraction P_s and specific gravity S_sl is V = M_s/(ρ_w S_sl P_s), and because the solids stay the same, V₁/V₂ = (100 − p₂)/(100 − p₁) for moisture contents p₁ and p₂ in per cent: thickening from 98% to 95% moisture cuts the volume to 2/5. Thickening (gravity or dissolved-air flotation) is followed by stabilisation, which destroys pathogens and odour-producing organics: anaerobic digestion (usually mesophilic, about 35 °C) converts organics to biogas of roughly 60–70% methane, aerobic digestion oxidises them, and lime stabilisation raises the pH. The theoretical methane yield follows from its COD: CH₄ + 2O₂ → CO₂ + 2H₂O, so 1 mol of CH₄ (22.4 L at STP) is worth 64 g of COD and each kg of COD removed yields 0.35 m³ of methane. Dewatering on sand drying beds, belt presses, filter presses or centrifuges gives a cake, which is disposed of by land application as a soil conditioner (subject to metals and pathogens), composting, landfill or incineration.
Sewage farming applies sewage, usually after at least primary treatment, to land for irrigation: the soil filters and oxidises the organics and the crops use the nutrients. It needs suitable, permeable soil, rest periods and crops not eaten raw. Applied too heavily or continuously, the soil pores clog with organic solids and the land becomes waterlogged and anaerobic — sewage sickness — which is prevented by pre-treatment, intermittent application, rotation and underdrains. Salinity, heavy metals from industrial inflow and the health of farm workers are the limits on the practice.
5. Industrial effluents, CETPs, wastewater recycling and zero liquid discharge
Industrial effluents differ from sewage in strength, composition and variability, and each industry has a signature. The strength of an effluent is often expressed as a population equivalent, its BOD load divided by the BOD contributed per person per day. Treatment begins with equalisation (to damp variations in flow and quality) and neutralisation, then removes the specific contaminants — oil by gravity separators, metals by precipitation, cyanide by alkaline chlorination, colour by coagulation, adsorption or oxidation — before biological treatment where the waste is biodegradable.
| Industry | Characteristic pollutants |
|---|---|
| Distillery | Spent wash with very high BOD and COD, dark colour, low pH |
| Tannery | Chromium, sulphide, high salinity, organic load |
| Textile dyeing | Colour, high pH, COD, dissolved salts |
| Pulp and paper | Lignin, colour, suspended fibre, chlorinated organics |
| Dairy and food | High, readily biodegradable BOD; fats |
| Electroplating | Heavy metals, cyanide, acids |
| Refinery and petrochemical | Oil, phenols, sulphides |
Small and medium units in an industrial estate often cannot afford, run or staff a treatment plant each. A Common Effluent Treatment Plant (CETP) treats their combined effluent centrally, with the economies of scale and professional operation that brings: member units pre-treat to meet an inlet standard (removing, for instance, metals or extreme pH that would upset the biology), the effluent is conveyed by pipeline or tanker, and the CETP treats it to the discharge standard, with the costs shared by a formula. Wastewater recycling returns treated effluent to cooling towers, boilers, process water, flushing or irrigation after tertiary treatment. Zero liquid discharge (ZLD) is its limit: no liquid effluent leaves the site at all. The usual train is pretreatment and biological treatment, then ultrafiltration and reverse osmosis to recover most of the water, then the RO reject concentrated in multiple-effect evaporators and crystallisers until only solid salt remains for disposal. ZLD saves water and ends discharge, but it is energy-intensive and turns a liquid problem into a solid-waste one.
Key takeaways
- Point sources are treated at an outlet; non-point sources are controlled by land use. Forecast by arithmetic (P + n x̄), geometric (P(1 + r)ⁿ), incremental (+ [n(n + 1)/2]ȳ) or logistic (P_s from three equally spaced censuses), with n in decades.
- Sanitary sewers are checked for a self-cleansing minimum and a scouring maximum velocity; storm sewers use Q = CiA/360 with A in hectares and a weighted C. Drop manholes, inverted siphons and ventilating shafts each solve a specific problem.
- F/M = QS₀/(VX); θ_c = VX/(solids wasted per day); SVI = settled mL/L × 1000/MLSS; R = X/(X_r − X) with X_r ≈ 10⁶/SVI; X = θ_cY(S₀ − S)/[θ(1 + k_dθ_c)].
- NRC trickling filter E = 100/[1 + 0.4432√(W/(VF))], F = (1 + R)/(1 + 0.1R)². Tertiary treatment removes N by nitrification–denitrification and P chemically or by PAOs.
- Sludge V = M_s/(ρ_wS_slP_s) and V₁/V₂ = (100 − p₂)/(100 − p₁); 0.35 m³ CH₄ per kg COD at STP. CETPs serve clusters of small units; ZLD recovers water by RO and evaporation and leaves only solid salt.
Practice questions (19)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Which of the following are non-point sources of water pollution?
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Answer: A — Fertiliser runoff from farmland after rain; C — Storm runoff from city streets carrying oil and litter; D — Seepage from many scattered septic tanks
(a), (c) and (d) enter over a wide area and intermittently, with no single outlet to treat, so they are diffuse sources controlled through land use and drainage practice. (b) discharges at one identifiable outfall — the definition of a point source.The populations of a town in four successive censuses were 40 000, 46 000, 52 000 and 58 000. By the arithmetic increase method, what will the population be two decades after the last census?
Numerical answer — type the value.
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Answer: 70000
The increase is 6000 per decade, so P₂ = 58 000 + 2 × 6000 = 70 000. Arithmetic increase assumes a constant absolute growth, which suits a large, established city. Enter it as 70000.A town’s population grew from 50 000 to 60 000 to 72 000 over two decades. By the geometric increase method, what will its population be two decades after it reached 72 000?
Numerical answer — type the value.
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Answer: 103680
The growth rate is 20% per decade in both decades, so P = 72 000 × 1.2² = 72 000 × 1.44 = 103 680. The arithmetic method would give only 72 000 + 2 × 11 000 = 94 000, because it ignores compounding. Enter it as 103680.A city recorded 100 000, 120 000, 145 000 and 175 000 in four successive censuses. By the incremental increase method, what will its population be two decades after the last census?
Numerical answer — type the value.
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Answer: 240000
Increases 20 000, 25 000, 30 000: x̄ = 25 000. Incremental increases 5000, 5000: ȳ = 5000. P₂ = 175 000 + 2 × 25 000 + (2 × 3/2) × 5000 = 175 000 + 50 000 + 15 000 = 240 000. Using n²/2 in place of n(n + 1)/2 gives 235 000. Enter it as 240000.Three censuses at equal intervals recorded 40 000, 100 000 and 130 000. Fitting a logistic curve, what is the saturation population?
Numerical answer — type the value.
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Answer: 137500
P_s = [2P₀P₁P₂ − P₁²(P₀ + P₂)]/(P₀P₂ − P₁²) = [2 × 4 × 10⁴ × 10⁵ × 1.3 × 10⁵ − 10¹⁰ × 1.7 × 10⁵]/(5.2 × 10⁹ − 10¹⁰) = (1.04 × 10¹⁵ − 1.70 × 10¹⁵)/(−4.8 × 10⁹) = 137 500. Both numerator and denominator are negative here; a sign slip gives a negative or absurd answer. Enter it as 137500.A 30 ha residential area has a runoff coefficient of 0.6. For a design rainfall intensity of 50 mm/h, what peak storm-water flow must its drain carry, in m³/s?
Numerical answer — type the value.
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Answer: 2.5
With A in hectares, Q = CiA/360 = 0.6 × 50 × 30/360 = 2.5 m³/s. Using the km² form (÷3.6) without converting 30 ha to 0.3 km² gives 250 m³/s, a hundred times too large.A sewer must cross under a river. The appurtenance used is:
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Answer: A — an inverted siphon, running full under pressure, often as several pipes
An inverted (depressed) siphon dips below the obstacle and runs full; several pipes of different sizes let low flows keep a self-cleansing velocity in the smallest. A drop manhole joins sewers at different levels, a catch basin traps grit from street inlets, and a flushing tank cleans flat upper reaches.An aeration tank of 3000 m³ holds an MLVSS of 2500 mg/L and receives 10 000 m³/d of settled sewage with a BOD of 200 mg/L. What is the F/M ratio, in d⁻¹? Give the answer to two decimal places.
Numerical answer — type the value.
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Answer: 0.27
F/M = QS₀/(VX) = 10 000 × 200/(3000 × 2500) = 2 × 10⁶/7.5 × 10⁶ = 0.27 d⁻¹, within the conventional 0.2–0.4 d⁻¹. The concentrations cancel their units, so no conversion to kg is needed as long as both are in mg/L.The plant of the previous question (V = 3000 m³, X = 2500 mg/L) wastes 100 m³/d of return sludge at 10 000 mg/L. Neglecting solids in the effluent, what is the mean cell residence time, in days?
Numerical answer — type the value.
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Answer: 7.5
θ_c = VX/(Q_wX_r) = 3000 × 2500/(100 × 10 000) = 7.5 × 10⁶/10⁶ = 7.5 d. The hydraulic retention time is only V/Q = 0.3 d = 7.2 h; recycling is what keeps the cells 25 times longer than the water.A 1-litre sample of mixed liquor with an MLSS of 2500 mg/L shows 250 mL of settled sludge after 30 minutes. What is its sludge volume index, in mL/g?
Numerical answer — type the value.
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Answer: 100
SVI = settled volume (mL/L) × 1000/MLSS (mg/L) = 250 × 1000/2500 = 100 mL/g, a well-settling sludge. Omitting the factor 1000 gives 0.1, which is in mL/mg.An activated-sludge plant is to hold an MLSS of 3000 mg/L, and its sludge has an SVI of 120 mL/g. Taking the return-sludge concentration as 10⁶/SVI, what recirculation ratio Q_r/Q is needed? Give the answer to two decimal places.
Numerical answer — type the value.
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Answer: 0.56
X_r = 10⁶/120 = 8333 mg/L. A solids balance on the tank, neglecting influent solids and growth, gives Q_rX_r = (Q + Q_r)X, so R = X/(X_r − X) = 3000/(8333 − 3000) = 0.5625 ≈ 0.56. Writing R = X/X_r = 0.36 forgets that the return flow also dilutes the tank.A completely mixed activated-sludge reactor has θ_c = 10 d and θ = 6 h. With Y = 0.6, k_d = 0.06 d⁻¹ and a BOD removal S₀ − S = 190 mg/L, what is the steady-state biomass concentration X, in mg/L?
Numerical answer — type the value.
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Answer: 2850
X = θ_cY(S₀ − S)/[θ(1 + k_dθ_c)] = 10 × 0.6 × 190/[0.25 × (1 + 0.6)] = 1140/0.4 = 2850 mg/L. Leaving θ in hours (6) instead of days (0.25) gives 118.75 mg/L, twenty-four times too small.A single-stage trickling filter of 1000 m³ receives a BOD load of 1000 kg/d with a recirculation ratio of 1. Using the NRC equation, E = 100/[1 + 0.4432√(W/(VF))] with F = (1 + R)/(1 + 0.1R)², what is its BOD removal efficiency, in per cent? Give the answer to one decimal place.
Numerical answer — type the value.
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Answer: 74.4
F = (1 + 1)/(1 + 0.1)² = 2/1.21 = 1.653. W/(VF) = 1000/(1000 × 1.653) = 0.605, whose square root is 0.778. E = 100/(1 + 0.4432 × 0.778) = 100/1.345 = 74.4%. Without recirculation (F = 1) it would be 100/1.4432 = 69.3%.Which of the following statements about secondary and tertiary treatment are correct?
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Answer: A — Nitrification requires a solids retention time long enough to keep the slow-growing nitrifiers; B — Denitrification needs an anoxic zone and a carbon source; D — Biological phosphorus removal cycles the sludge through anaerobic and aerobic zones
(a), (b) and (d) are correct: nitrifiers are autotrophs with low growth rates; denitrifiers use nitrate only when oxygen is absent and need organic carbon as electron donor; phosphate-accumulating organisms release P anaerobically and take up more aerobically. (c) is false: a high SVI means a bulky, poorly settling sludge.A plant produces 1000 kg of dry solids per day as a sludge containing 4% solids, with a specific gravity of 1.02. What volume of sludge is produced, in m³/d? Give the answer to one decimal place.
Numerical answer — type the value.
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Answer: 24.5
V = M_s/(ρ_w S_sl P_s) = 1000/(1000 × 1.02 × 0.04) = 1000/40.8 = 24.5 m³/d. Taking 4 in place of 0.04 gives 0.245 m³/d, a hundredfold error from using the percentage as a fraction.100 m³ of sludge at 98% moisture is thickened to 95% moisture. What is its new volume, in m³?
Numerical answer — type the value.
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Answer: 40
The solids are conserved: V₂ = V₁(100 − p₁)/(100 − p₂) = 100 × 2/5 = 40 m³. Reasoning from the water fraction, 100 × 95/98 = 96.9 m³, is the tempting error.An anaerobic digester removes 1000 kg of COD per day. Using the COD equivalence of methane (1 mol CH₄ = 64 g COD, and 22.4 L per mol at STP), what volume of methane is produced at STP, in m³/d? Give the answer to the nearest whole number.
Numerical answer — type the value.
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Answer: 350
CH₄ + 2O₂ → CO₂ + 2H₂O, so one mole of methane (22.4 L) carries 64 g of COD, i.e. 0.35 L CH₄ per g COD = 0.35 m³ per kg. 1000 kg COD/d × 0.35 = 350 m³/d. At 35 °C the same methane occupies about 12.8% more volume.Which of the following statements about industrial effluent management are correct?
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Answer: A — A CETP lets a cluster of small units share one treatment plant; B — Member units of a CETP may have to pre-treat to meet its inlet standard; C — Zero liquid discharge typically uses reverse osmosis followed by evaporators and crystallisers
(a), (b) and (c) are correct. (d) is false: ZLD eliminates the liquid effluent by driving the dissolved salts out as a solid, which must itself be stored or landfilled — and the evaporation step consumes a great deal of energy.An industry discharges 5000 m³/d of effluent with a BOD of 400 mg/L. Taking the per-capita BOD contribution as 80 g/d, what is the population equivalent of the effluent?
Numerical answer — type the value.
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Answer: 25000
BOD load = 5000 m³/d × 400 g/m³ = 2 × 10⁶ g/d = 2000 kg/d. PE = 2 × 10⁶/80 = 25 000. Using 400 mg/L as 0.4 g/m³ instead of 400 g/m³ loses a factor of 1000. Enter it as 25000.