Wastewater: Sewerage, Primary and Secondary Treatment, Sludge and Reuse

Wastewater treatment inverts the question the previous chapter answered. There the standard was what a person may safely drink; here it is what a river may safely receive, and the difference changes every design decision. A drinking-water plant removes what is present; a sewage works must remove what is present and stabilise what it removes, because the organic matter it takes out of the water is still organic matter and has to go somewhere. That is why the chapter has two halves — a liquid train and a sludge train — and why candidates who learn only the first find that half the marks are in the second. The liquid train is a sequence of increasingly biological steps: screens and grit chambers take out what would damage the plant, primary sedimentation takes out what settles, and secondary treatment takes out what does not settle by feeding it to bacteria and then settling the bacteria. The genuinely clever idea in the whole subject is in that last clause. Activated sludge works because the solids are recycled: the settled biomass goes back to the head of the aeration tank, so the microbes stay in the plant far longer than the water does — a sludge age of nearly ten days against a hydraulic retention of under seven hours in the worked example below. Decoupling those two times is what lets a tank of a manageable size treat a city, and almost every question about secondary treatment is a question about that ratio.
Overview

How much sewage, and why a sewer is designed on a gradient

2 min read · 1 part

The quantity of domestic wastewater is conventionally taken as 70 to 80 per cent of the water supplied, the balance being what is drunk, evaporated, used on gardens or lost. So a town of 50,000 at 135 lpcd draws 6.75 MLD and generates about 5.4 MLD = 5400 m³/day. To that is added infiltration through joints and manholes, which is small per metre and large per kilometre, and in a combined system the storm flow, which is larger than everything else put together and is the reason separate systems are now the default. The sewer is then sized not on the average but on the peak: a peak factor of 3 is usual for a small town, falling towards 2 for a large one because the peaks of many sub-catchments do not coincide. At a factor of 3 this town’s sewer must carry 16.2 MLD = 0.1875 m³/s = 187.5 L/s.

🎯 A sewer has a minimum velocity as well as a maximum
A water main only needs to be big enough. A sewer has a lower bound too, and it is the constraint that decides the design. Sewage carries grit and organic solids; below about 0.6 m/s those settle out, the deposit reduces the section, the velocity falls further, and the sewer silts itself shut — a failure that compounds. Above about 3 m/s the same grit abrades the invert. So a sewer must run between the two at the flows it will actually see, and since the diameter is fixed by the peak flow, the free variable is the gradient. Work a 300 mm sewer running full, n = 0.013, R = D/4 = 0.075 m: at 1 in 300 it runs at 0.790 m/s, at 1 in 500 at 0.612 m/s — only just self-cleansing — and at 1 in 1000 at 0.433 m/s, which will silt. The flat-terrain problem in Indian towns is exactly this: there is no gradient to be had, so the sewer must either be laid progressively deeper, at rising excavation cost, or pumped. And the self-cleansing check must be made at the minimum flow, not the peak, because a sewer sized for the peak of a town that has not been built yet runs part-full and slow for years.

The quantity of domestic wastewater is conventionally taken as 70 to 80 per cent of the water supplied, the balance being what is drunk, evaporated, used on gardens or lost. So a town of…

Preliminary and primary treatment

1 min read · 3 parts
  1. Screens — bar screens at 20 to 50 mm remove rags, plastics and anything that would foul a pump. They protect the plant rather than treat the sewage, which is why they are called preliminary.
  2. Grit chamber — designed to settle sand and silt at about 0.15 to 0.3 mm while keeping organics in suspension, which is done by holding the velocity near 0.3 m/s: fast enough to keep organic matter moving, slow enough to drop mineral grit.
  3. Primary sedimentation — the same overflow-rate design as a water-treatment clarifier, removing roughly 60% of suspended solids and 30 to 35% of BOD. It removes what settles, and nothing else.

Notice what primary treatment cannot do. A BOD removal of a third is nowhere near an effluent standard, because most of the oxygen demand in domestic sewage is dissolved or colloidal and gravity has no purchase on it. That is the gap secondary treatment exists to close, and it closes it by a different mechanism entirely — not by settling the organic matter but by eating it.

Notice what primary treatment cannot do. A BOD removal of a third is nowhere near an effluent standard, because most of the oxygen demand in domestic sewage is dissolved or colloidal and…

Secondary treatment, and the two clocks that must differ

3 min read · 2 parts

In the activated sludge process the settled sewage is aerated in a tank holding a dense suspension of bacteria — the mixed liquor suspended solids, MLSS — which oxidise the dissolved organics and grow. The mixture then goes to a secondary clarifier where the biomass settles, and the crucial step is that most of it is pumped back to the head of the aeration tank as return sludge, with only the surplus wasted. That recycle is what separates the two clocks: the water passes through in a hydraulic retention time of hours, while the bacteria stay for a sludge retention time — the sludge age — of days.

A worked activated sludge plant: 5000 m³/day, S₀ = 250 mg/L, MLSS = 3000 mg/L, F/M = 0.3/day
QuantityRelationValue
Aeration tank volumeV = QS₀ / ((F/M)·X)1388.9 m³
Hydraulic retention timeHRT = V/Q6.67 hours
Sludge age (Y = 0.6, k_d = 0.06/day, S_e = 25 mg/L)1/SRT = Y(S₀ − S_e)/(HRT·X) − k_d9.80 days
BOD removal(250 − 25)/25090%
🧠 Sludge age is a ratio of masses, not of times
The definition that makes every activated-sludge question tractable is SRT = (mass of solids in the system) / (mass of solids leaving per day), that is VX / (Q_w X_r). Read that way, three otherwise mysterious operating rules become obvious. Wasting more sludge shortens the sludge age, because it increases the denominator — and a plant is controlled by how much it wastes, not by how much it aerates. A short sludge age selects fast-growing organisms and gives a poorly settling, dispersed sludge; a long one starves them and gives a fine, pin-point floc that also settles badly, so both extremes fail in the clarifier rather than in the aerator. And nitrification needs a long sludge age, because nitrifying bacteria grow slowly: if the sludge age is shorter than their doubling time they are literally washed out of the plant faster than they can reproduce, and no amount of aeration will produce nitrification. That last point is why a plant designed for BOD removal alone cannot be persuaded to nitrify by running the blowers harder.

The alternative family is the attached-growth processes — the trickling filter and its modern relatives — where the biomass grows as a film on media over which the sewage is distributed, rather than as a suspension. They need less energy and far less operator skill, and they handle shock loads better because the biomass cannot be washed out; they give lower efficiency for the same footprint, and they lose performance in cold weather. The NRC formula sizes one: at 1000 kg BOD/day into 500 m³ of media with a recirculation factor of 1, the efficiency is 100/(1 + 0.4432√(W/VF)) = 61.5% — respectable, and clearly below the 90% the activated sludge plant above reached on the same sewage. Which of the two is right for a town is a question about operating capacity as much as about engineering, and stating that plainly is more useful than pretending one dominates.

In the activated sludge process the settled sewage is aerated in a tank holding a dense suspension of bacteria — the mixed liquor suspended solids, MLSS — which oxidise the dissolved…

Sludge, and the arithmetic that surprises everyone once

2 min read · 1 part

Everything removed from the liquid arrives as sludge, and sludge is mostly water — typically 95 to 99 per cent of it. The treatment train for sludge is therefore about removing water, in four steps: thickening (gravity or flotation), stabilisation (usually anaerobic digestion, which destroys volatile solids and produces methane worth burning), dewatering (centrifuge, filter press or drying beds), and disposal or use — landfill, incineration, or application to land where the standards permit it. Digestion is the step worth understanding: it is the same biology as secondary treatment run without oxygen, so instead of ending in carbon dioxide and new cells it ends in methane, and a large works can run much of its own aeration on the gas its digesters produce.

⚠️ Five percentage points of water can halve the volume
The solids in a sludge do not change when you thicken it — only the water does. So the volume goes as the reciprocal of the solids fraction: V₂/V₁ = (100 − p₁)/(100 − p₂) where p is the moisture percentage. Put numbers to it and the result is startling every time. Thickening 100 m³ of sludge from 95% to 90% moisture halves it to 50 m³, and taking it on to 80% quarters it to 25 m³. Five percentage points of moisture — which sounds like a trivial change and reads on a meter as almost nothing — removed half the tankers. The intuition that misleads is treating moisture as a linear scale; it is the solids fraction that doubles, from 5% to 10%. This is why dewatering dominates the cost of sludge handling and why a plant fights for every point of dryness before the sludge goes on a lorry: the haulage bill is proportional to volume, and volume is a reciprocal, not a difference.

Finally, reuse. Treated sewage is a resource in a water-short country, and the standard it must meet depends entirely on what it is reused for — the syllabus asks for reuse "for different applications" precisely because there is no single number. Irrigation of non-edible crops and landscape watering need coliform control and little else; industrial cooling needs low solids and low hardness rather than low BOD; groundwater recharge and any indirect potable route need the full drinking-water treatment train applied a second time. The engineering point worth carrying is that reuse quality is set by exposure, not by the sewage: the same effluent is entirely safe for a golf course and entirely unacceptable for a vegetable field eaten raw.

Everything removed from the liquid arrives as sludge, and sludge is mostly water — typically 95 to 99 per cent of it. The treatment train for sludge is therefore about removing water, in…

Key takeaways

  • Sewage is taken as 70–80% of water supply; the sewer is sized on the peak, typically 3× the average for a small town.
  • A sewer needs a minimum velocity of about 0.6 m/s as well as a maximum, and the free variable is the gradient.
  • Primary sedimentation removes ~60% of SS but only ~33% of BOD, because most of the demand is dissolved or colloidal.
  • Activated sludge works by recycling solids, so sludge age ≫ hydraulic retention — 9.8 days against 6.67 hours here.
  • Nitrification needs a long sludge age: too short and the slow-growing nitrifiers are washed out faster than they reproduce.
  • Sludge volume goes as 1/(100 − moisture %): 95% → 90% moisture halves the volume, 95% → 80% quarters it.

Practice questions (8)

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.

  1. An activated sludge plant treats 5000 m³/day of sewage with a BOD of 250 mg/L. The MLSS is 3000 mg/L and the design F/M ratio is 0.3 per day. The aeration tank volume in m³ is ______ (round to one decimal place)

    Numerical answer — type the value.

    Show answer

    Answer: 1388.9

    1388.9 m³. The food-to-microorganism ratio is F/M = QS₀/(VX) — food arriving per day divided by the mass of microbes holding it. Rearranged, V = QS₀/((F/M)·X) = (5000 × 250)/(0.3 × 3000) = 1,250,000/900 = 1388.9 m³. The units look wrong at first glance and are not: mg/L and m³/day multiply to grams per day on both sides of the ratio, so they cancel and F/M comes out per day. What makes this worth more than a formula substitution is the second number it gives you free. The hydraulic retention time is V/Q = 1388.9/5000 = 0.278 day = 6.67 hours, which sounds far too short to treat sewage biologically — and it would be, if the bacteria left with the water. They do not: the secondary clarifier settles them and the return line brings them back, so the sludge age works out to about 9.8 days on standard kinetics. The plant holds its microbes fourteen times longer than its water, and that decoupling is what makes a tank of this modest size sufficient. A candidate who reports the HRT as the treatment time has understood the arithmetic and missed the process.

  2. 100 m³ of sludge at 95% moisture content is thickened to 90% moisture content. Assuming no solids are lost, its new volume in m³ is ______

    Numerical answer — type the value.

    Show answer

    Answer: 50

    50 m³ — exactly half. The solids are conserved, so equate them: at 95% moisture the solids fraction is 5%, giving 100 × 0.05 = 5 m³ of solids; at 90% moisture the solids fraction is 10%, so the new volume is 5/0.10 = 50 m³. The general relation is V₂/V₁ = (100 − p₁)/(100 − p₂), and the reason it surprises is that moisture looks like a linear scale while the volume depends on its complement in a reciprocal. Going from 95% to 90% does not remove 5% of the water — it doubles the solids concentration, from 5% to 10%, and so halves the volume. Push it one step further and the effect compounds: at 80% moisture the same solids occupy 5/0.20 = 25 m³, a quarter of the original. This single relation is why dewatering dominates the economics of sludge handling. Haulage, storage and disposal are all charged by volume, so a plant that moves its cake from 80% to 75% moisture cuts its lorry movements by 20% — and why the last few points of dryness are the expensive ones, because the same reciprocal that rewards you early demands progressively more energy as the water left is progressively more tightly bound.

  3. An activated sludge plant designed for BOD removal is not nitrifying. The operator increases the air supply substantially and nitrification still does not occur. The most likely reason is that

    1. the sludge age is shorter than the doubling time of the nitrifying bacteria, so they are washed out
    2. nitrification is an anaerobic process and additional air suppresses it
    3. the BOD is too high, and nitrifiers cannot function until it is below 5 mg/L
    4. the MLSS is too high, and nitrifiers are inhibited by crowding
    Show answer

    Answer: A — the sludge age is shorter than the doubling time of the nitrifying bacteria, so they are washed out

    The sludge age is too short and the nitrifiers are washed out. This is the cleanest illustration of why sludge age, and not aeration, is the master control on an activated sludge plant. Nitrifying bacteria — Nitrosomonas and Nitrobacter — are autotrophs with a low growth rate, doubling in days rather than hours, whereas the heterotrophs that remove carbonaceous BOD double in hours. Now read the definition of sludge age as what it physically is: SRT = (solids in the system)/(solids leaving per day) is the average time an organism spends in the plant. If that time is shorter than the nitrifiers’ doubling time, they are being removed faster than they can reproduce, and their population runs down to nothing no matter what else is done. It is a washout, and it is irreversible while the wasting rate is unchanged. Aeration cannot fix it, because oxygen was never the limit — the population was. The correction is to waste less sludge, which lengthens the sludge age, and in a plant designed for carbonaceous removal alone that usually also means a larger aeration tank, since holding more solids at the same F/M requires more volume. Option (b) is simply wrong — nitrification is strictly aerobic; it is denitrification that is anoxic. Option (c) inverts a real effect: a high BOD favours heterotrophs in competition, but it does not impose a threshold. Option (d) invents an inhibition that does not exist.
Show the other 5 questions

  1. A 300 mm diameter sewer runs full at a gradient of 1 in 500. Taking Manning n = 0.013, the velocity of flow in m/s is ______ (round to three decimal places)

    Numerical answer — type the value.

    Show answer

    Answer: 0.612

    0.612 m/s. For a circular pipe running full the hydraulic radius is R = A/P = (πD²/4)/(πD) = D/4 = 0.075 m. Manning gives v = (1/n)R^(2/3)S^(1/2) = (1/0.013) × 0.075^(2/3) × (1/500)^(1/2) = 76.92 × 0.17795 × 0.044721 = 0.612 m/s. Now do the thing the question is really testing, which is to judge the number rather than report it. The self-cleansing velocity for a sanitary sewer is about 0.6 m/s, so this design passes — but only just, with about 2% to spare, and that margin is at the full flow. In the early years of a scheme the sewer runs part-full at a fraction of the design flow, and the actual velocity is lower, which is why a sewer is checked at the minimum anticipated flow and not the design flow. Two comparisons put the gradient in context: at 1 in 300 the same pipe runs at 0.790 m/s, comfortably clean; at 1 in 1000 it runs at 0.433 m/s and will silt up, the deposit will reduce the section, the velocity will fall further, and the failure compounds until it blocks. Note also what does not rescue a flat sewer: increasing the diameter raises R but reduces nothing about the slope, and at constant flow a larger pipe runs shallower and slower. On flat terrain the only real options are to lay progressively deeper or to pump.

  2. Which of the following are correct comparisons of the activated sludge process with a trickling filter?

    1. The activated sludge process generally achieves higher BOD removal for the same land area
    2. The trickling filter tolerates shock loads better, because its biomass is attached rather than suspended
    3. The activated sludge process needs more power and more skilled operation
    4. The trickling filter produces no sludge and therefore needs no secondary clarifier
    Show answer

    Answer: A — The activated sludge process generally achieves higher BOD removal for the same land area; B — The trickling filter tolerates shock loads better, because its biomass is attached rather than suspended; C — The activated sludge process needs more power and more skilled operation

    (a), (b) and (c). The three correct statements are the trade that decides which process a town gets. (a) is right and the worked numbers show the size of it: on the same 250 mg/L sewage, the activated sludge plant reached 90% BOD removal, while the NRC formula gives a 500 m³ trickling filter loaded at 1000 kg BOD/day only 61.5%. Suspension puts the biomass in contact with every part of the flow; a film only contacts what runs over it. (b) is right for a reason worth stating precisely: an attached film cannot be washed out of the plant, so a sudden hydraulic surge dilutes a trickling filter but does not destroy its biomass, whereas the same surge can carry MLSS over the weir of a secondary clarifier and cripple an activated sludge plant for days. (c) is right and is often decisive in practice — continuous aeration is the single largest power demand on a sewage works, and the process must be actively controlled through its wasting rate, which needs an operator who understands sludge age. A trickling filter with a rotating distributor can run on gravity and inattention. (d) is wrong, and it is the one that catches people. A trickling filter continuously sloughs its biofilm as the film thickens and the innermost layer loses its oxygen supply; that sloughed solids stream is real, it is substantial, and it must be settled in a secondary clarifier exactly as activated sludge must. Every biological process converts dissolved BOD into cells, and cells are sludge — there is no biological treatment that produces none.

  3. A town of 50,000 people has a water supply of 135 litres per capita per day. Taking sewage as 80% of the water supplied and a peak factor of 3, the design flow for the sewer in litres per second is ______ (round to one decimal place)

    Numerical answer — type the value.

    Show answer

    Answer: 187.5

    187.5 L/s. Work it in one chain and the units take care of themselves. Water supplied = 50,000 × 135 = 6,750,000 L/day = 6.75 MLD. Sewage = 0.80 × 6.75 = 5.4 MLD = 5400 m³/day, the average dry weather flow. Design flow = 3 × 5.4 = 16.2 MLD, and 16.2 × 10⁶ L/day ÷ 86,400 s = 187.5 L/s, which is 0.1875 m³/s. Three things in that chain are worth holding. The 80% is a convention, not a measurement: it is water supplied minus what is drunk, evaporated, or put on gardens, and a town with heavy garden use returns less while one with high groundwater returns more through infiltration. The peak factor of 3 is for a small town; it falls towards 2 for a large one, because the peaks of many sub-catchments do not coincide — the bigger the sewershed, the smoother its hydrograph, which is the same statistical smoothing that governs flood peaks in the hydrology chapter. And the design flow is what fixes the diameter, but not the gradient: the gradient is then chosen so that the velocity stays between the self-cleansing minimum of about 0.6 m/s and the abrasion maximum of about 3 m/s, and it must be checked at the minimum flow the sewer will actually see rather than at this peak.

  4. Primary sedimentation removes about 60% of suspended solids but only about one third of the BOD of domestic sewage. The reason is that

    1. much of the oxygen demand is dissolved or colloidal, and gravity acts on neither
    2. the settled solids continue to exert oxygen demand from the tank floor
    3. BOD is measured after five days, by which time settling has reversed
    4. primary tanks are designed for a shorter detention time than BOD removal needs
    Show answer

    Answer: A — much of the oxygen demand is dissolved or colloidal, and gravity acts on neither

    Much of the demand is dissolved or colloidal. Suspended solids and BOD are different quantities measured on different principles, and a settling tank can only act on one of them. Gravity separation works on particles with mass and size; the sugars, organic acids, proteins and detergents that make up a large share of domestic oxygen demand are in solution, and a colloid is effectively in solution as far as a settling tank is concerned — the previous chapter’s arithmetic showed a 1 μm particle needing 39 days to fall 3 m. So primary sedimentation takes out the settleable fraction of the BOD and leaves the rest untouched, which is why the two removal figures diverge so sharply. Everything downstream follows from that gap. Secondary treatment exists precisely because primary treatment cannot reach dissolved organics, and it reaches them by a completely different mechanism — feeding them to bacteria, which converts dissolved BOD into settleable cells, and then settling the cells. Read that way the whole train is coherent: the secondary clarifier is not a second attempt at primary sedimentation, it is settling a substance the biology created for the purpose. Option (b) describes something real in a poorly designed tank but it is not why the two percentages differ. Option (c) misdescribes the BOD test, which is run on a sample in a bottle. Option (d) blames the design, when even an infinitely long tank would leave dissolved BOD exactly where it was.

  5. A trickling filter of volume 500 m³ receives an organic load of 1000 kg BOD per day with a recirculation factor F = 1. By the NRC formula, efficiency = 100/(1 + 0.4432√(W/VF)), where W is in kg/day and V in m³. The BOD removal efficiency in per cent is ______ (round to two decimal places)

    Numerical answer — type the value.

    Show answer

    Answer: 61.47

    61.47%. Substitute directly: W/VF = 1000/(500 × 1) = 2.0, √2 = 1.41421, 0.4432 × 1.41421 = 0.62678, so efficiency = 100/1.62678 = 61.47%. The number is more useful placed beside its alternative than on its own. The activated sludge plant worked earlier in this chapter treated sewage of the same strength to 90% removal; this filter reaches 61.5%. That difference is the honest reason activated sludge dominates in large cities and is not the automatic answer everywhere: a trickling filter needs no blowers, tolerates a shock load without losing its biomass, and can be run by staff who do not have to understand sludge age — real advantages for a small or remote works, and the reason the process is far from obsolete. Two features of the formula are examinable in their own right. It is empirical, fitted to military-installation data, so it carries units inside it and cannot be non-dimensionalised or rearranged freely. And efficiency depends on W/V, the organic loading rate, not on W alone: doubling the filter volume at the same load takes W/VF to 1.0 and lifts efficiency to 69.3%, a gain of under eight points for twice the media. That flattening is why filters are staged in series rather than built larger — a second stage attacks a much weaker sewage and earns more per cubic metre than enlarging the first.