An ideal rectangular settling tank is redesigned to be twice as deep, keeping the plan area and the flow rate unchanged. What happens to the size of the smallest particle that is completely removed?

It is unchanged — depth does not enter the removal criterion at all

It halves, because the detention time has doubled

It falls by a factor of √2, because settling velocity goes as d²

It doubles, because the particle now has further to fall

Which of the following statements about the drinking-water treatment train are correct?

Coagulation is needed because colloidal particles will not settle in any practicable tank

Flocculation requires gentle mixing, because vigorous stirring breaks the floc it is meant to build

Chlorine is preferred over ozone mainly because it leaves a residual in the distribution system

Coagulation is needed because colloidal particles will not settle in any practicable tank

Flocculation requires gentle mixing, because vigorous stirring breaks the floc it is meant to build

Chlorine is preferred over ozone mainly because it leaves a residual in the distribution system

A rapid sand filter removes bacteria reliably enough that disinfection can be omitted

On a breakpoint chlorination curve, the combined residual falls as the applied chlorine dose is increased over a certain range. The reason is that

the added chlorine oxidises the chloramines already formed, destroying them

the chlorine escapes as a gas once its solubility limit is exceeded

the pH rises with the dose and shifts the equilibrium away from residual chlorine

the measurement method under-reads at high dose and the fall is an artefact

Water Quality and Treatment: Parameters, Demand, Settling and the Treatment Train

Environmental Engineering is Section 5 of the Civil Engineering paper, and water treatment is the one design problem in that paper where the specification is a public health standard rather than a load, and the whole subject follows from that inversion. A structural engineer is given the load and asked for a section; here you are given a required outcome — so many organisms, so much turbidity, so little of each dissolved species — and asked what sequence of processes reaches it from whatever the source happens to be. That sequence is the treatment train, and it is not arbitrary: each unit exists because the one before it cannot do that job. Screening removes what floats and what is large. Coagulation exists for one reason and one reason only, which a single number makes obvious — a one-micron clay colloid settling under gravity takes thirty-nine days to fall three metres, so it will never settle out of a tank a city can afford to build, and the only way to remove it is to make it bigger. Sedimentation then removes what coagulation has made settleable, and its design rule is the counter-intuitive one in this chapter: a settling tank is sized by its surface area, not its volume, because the critical settling velocity equals the surface overflow rate Q/A and the depth cancels out. Filtration polishes what sedimentation leaves. Disinfection kills what filtration passes. Read in that order the train is a chain of failures being covered, which is the honest way to see it and the way the numerical questions are set.

Quality parameters, and the three that are really tests

The syllabus asks for physical, chemical and biological parameters and a water quality index. The physical ones — turbidity, colour, taste, odour, temperature — are what a user notices, and the standards for them are set by acceptability rather than by toxicity. The chemical ones split again: some are health-based limits (arsenic, fluoride, nitrate, lead), and some are operational (pH, alkalinity, hardness, chloride, iron). The biological ones are indicator-based, and that word is the whole idea — nobody tests drinking water for every pathogen, because you would need a separate test for each. Instead coliforms are counted, because they are abundant wherever faecal contamination is present, easy to culture, and harmless themselves. A coliform count is not a measure of danger; it is a measure of the route by which danger arrives.

The parameters that carry arithmetic in the exam
Parameter	What it measures	The relation you must be able to use
BOD	Oxygen microbes will consume degrading the organics	`BOD_t = L₀(1 − e^(−kt))`, and `k_T = k₂₀ · 1.047^(T−20)`
COD	Oxygen a strong oxidant consumes — biodegradable or not	Always ≥ BOD. The ratio BOD/COD is the biodegradability, not the strength
Hardness	Divalent cations, chiefly Ca²⁺ and Mg²⁺	As CaCO₃: `mg/L × 50 / (equivalent weight)`
DO	Oxygen actually dissolved, against a saturation value	Deficit `D = DO_sat − DO`; saturation FALLS as temperature rises

⚠️ The five-day test never sees the whole demand

BOD₅ is not the BOD. At the standard k = 0.23/day, 1 − e^(−0.23 × 5) = 0.6834, so the five-day test measures about 68% of the ultimate demand and never converges on it — the exponential has no last day. An ultimate BOD of 300 mg/L reports as BOD₅ = 205 mg/L, and a question that gives you one and asks for the other is asking whether you know that. Two further traps ride on the same relation. First, k is quoted to two different bases in different books: k base e and K base 10, with k = 2.303K, so a value of 0.1 means 0.23 if it is base 10. Second, temperature raises k but lowers the saturation DO — the demand goes up exactly when the supply goes down, which is why a summer discharge is the design case rather than a winter one.

Water requirement, forecasting, and the distribution system

Demand is a population times a rate, and both halves are estimates. The rate is the per-capita demand, conventionally around 135 litres per capita per day for a fully sewered Indian town under the CPHEEO manual, plus institutional, commercial, industrial and fire demand and an allowance for losses. The population is the harder half, because it must be forecast to the design period — 30 years for a headworks that cannot be enlarged, far less for a pump that can. And the forecast is not one method but a family of them, which is the honest difficulty: the same two censuses give materially different answers depending on which curve you fit.

Arithmetic increase — a constant number added per decade. Fits a large, old, saturated city. From 50,000 (2010) to 80,000 (2020) it projects 110,000 for 2030.
Geometric increase — a constant percentage per decade. Fits a young, growing town. The same two figures give r = 0.60 and project 128,000 — 18,000 more people, a 16.4% difference from the same data.
Incremental increase — arithmetic plus the average of the increments of the increments; a compromise, and the one most syllabuses treat as the default.
Logistic — an S-curve with a saturation population. The only one that refuses to grow for ever, and therefore the only one honest about a bounded city.

The distribution system then carries that demand, and its layouts trade the same two things everywhere: dead-end (or tree) systems are cheap, easy to design and easy to isolate, but a break upstream cuts everyone below it and the far ends stagnate; grid (or ring) systems cost more and need network analysis — Hardy Cross, which is the pipe-network method of the Water Resources chapter — but water reaches any point by more than one path, so a break isolates a segment rather than a suburb, and nothing stagnates. Storage sits between supply and demand for the same reason a battery sits between a generator and a load: the source runs at the average rate and the city draws at the peak rate, so the service reservoir absorbs the difference and lets the treatment plant be sized for the average.

Settling, and the depth that cancels out

A discrete particle falling through still water reaches a terminal velocity when drag balances buoyant weight. In the laminar range that is Stokes’ law, v_s = g(G − 1)d²/18ν, and the d² in it is the single most consequential exponent in this chapter — halving the particle size quarters the settling velocity. Put three real sizes through it and the treatment train explains itself.

Why the train has the units it has (G = 2.65, ν = 1.01 × 10⁻⁶ m²/s)
Particle	Settling velocity	Time to fall 3 m	So it is removed by
0.05 mm fine sand	2.226 mm/s = 192.3 m/day	22.4 minutes	A grit chamber, before anything else
0.01 mm silt	7.69 m/day	9.4 hours	Plain sedimentation, if you are patient
1 μm clay colloid	0.0769 m/day	39 days	Nothing gravitational. It must be coagulated first

🎯 The overflow rate is a velocity wearing different units

Design an ideal rectangular tank and the algebra does something surprising. A particle enters at the top and must reach the floor before the flow carries it out: it has the detention time t = V/Q = LWH/Q to fall the depth H, so it is fully removed if v_s ≥ H/t = HQ/(LWH) = Q/(LW) = Q/A. The depth H cancels. The critical settling velocity equals the surface overflow rate Q/A, and a tank is therefore sized by its plan area alone — a deeper tank of the same footprint removes exactly the same particles, it just holds them longer on the way. For Q = 4000 m³/day at an overflow rate of 40 m³/m²/day, A = 100 m², and the critical velocity is 40 m/day = 0.463 mm/s. A particle settling slower than that is not entirely lost: it is removed in the ratio v_s/v_c, so a 20 m/day particle is 50% removed. This is why real plants add plates or tubes — they multiply area without multiplying volume, which is the only variable that matters.

Coagulation, filtration and disinfection

A clay colloid does not settle because it is small and because it is charged — the particles carry like charges, repel one another, and so never aggregate into anything heavy. Coagulation attacks the charge: a trivalent salt, usually alum Al₂(SO₄)₃·18H₂O, hydrolyses and compresses the double layer, and rapid mixing distributes it in seconds because the reaction is faster than the mixer. Flocculation is then the opposite of rapid mixing — slow, gentle, long stirring that lets destabilised particles collide and grow without the shear tearing them apart again. It is the one unit process in the plant where stirring harder makes the result worse, which is why its velocity gradient G is specified with an upper limit rather than a lower one. Alum also consumes alkalinity, so a soft, low-alkalinity water needs lime added alongside it or the pH falls out of the effective range and the dose stops working.

Rapid against slow sand filtration — a factor of 25 in land
	Rapid sand	Slow sand
Rate of filtration	~5 m³/m²/h	~0.2 m³/m²/h
Area for 4000 m³/day	33.3 m²	833.3 m²
Needs pre-treatment?	Yes — coagulation and sedimentation first	No — it can take raw water of low turbidity
How it is cleaned	Backwashing, in minutes	Scraping the schmutzdecke, in hours
Removes bacteria how?	Mechanically — disinfection still essential	Biologically, in the surface mat — genuinely high removal

Disinfection closes the train, and chlorine dominates it for a reason worth naming: it is the only common disinfectant that leaves a residual. Ozone and ultraviolet both kill more effectively in the contact tank and neither protects the water in the pipe afterwards, so a distribution system with a long residence time and any chance of ingress is chlorinated whatever else is done to it. The dose splits into the chlorine demand — what is consumed by ammonia, organics and reduced species — and the residual that survives it: a 3 mg/L dose leaving 0.5 mg/L means a demand of 2.5 mg/L, and 4000 m³/day at 3 mg/L is 12 kg of chlorine a day. Add ammonia and the curve becomes the breakpoint curve, whose falling limb is the part candidates misread: past the hump, more chlorine gives less residual, because it is destroying the chloramines it earlier formed. Only beyond the breakpoint is the residual free chlorine, and only free chlorine disinfects quickly.

Hardness, and the arithmetic of expressing everything as CaCO₃

Hardness is the sum of divalent cations, chiefly calcium and magnesium, and it is reported as CaCO₃ for one purely practical reason: the species are chemically different but interchangeable in effect, so converting them to a common currency lets them be added. The conversion is mg/L as CaCO₃ = (mg/L of ion) × 50 / (equivalent weight of ion), where 50 is the equivalent weight of CaCO₃ itself. Calcium at 40 mg/L becomes 40 × 50/20.04 = 99.80; magnesium at 24.3 mg/L becomes 24.3 × 50/12.15 = 100.00 exactly. Their total, 199.8 mg/L as CaCO₃, is very hard water — and note that magnesium contributed as much as calcium while weighing 40% less, because its equivalent weight is smaller. That is the whole reason for the convention.

🧠 Carbonate hardness is the smaller of two numbers

Hardness splits into carbonate (temporary, removable by boiling or by lime) and non-carbonate (permanent, needing soda ash or ion exchange), and the split is decided by the alkalinity, not by the hardness. The rule is a one-liner worth memorising: carbonate hardness = min(total hardness, total alkalinity), and non-carbonate hardness is whatever is left. So water with 200 mg/L hardness and 150 mg/L alkalinity has 150 carbonate and 50 non-carbonate; water with 200 hardness and 250 alkalinity has 200 carbonate, 0 non-carbonate, and 50 mg/L of alkalinity that is not associated with hardness at all. Candidates lose this mark by subtracting in the wrong direction and reporting a negative non-carbonate hardness, which cannot exist.

Key takeaways

BOD₅ is about 68% of the ultimate BOD at k = 0.23/day, never all of it: L₀ = 300 reports as BOD₅ = 205.
A settling tank is sized by surface overflow rate Q/A; the depth cancels out of the removal calculation entirely.
A particle slower than the critical velocity is removed in the ratio v_s/v_c — not lost entirely.
Coagulation exists because a 1 μm colloid needs 39 days to fall 3 m. Nothing gravitational can remove it.
Hardness as CaCO₃ = ion mg/L × 50 / equivalent weight; carbonate hardness = min(hardness, alkalinity).
Chlorine is used because it leaves a residual; past the breakpoint hump, more chlorine gives less residual.

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.

A wastewater has an ultimate BOD of 300 mg/L. Taking the first-order rate constant as k = 0.23 per day (base e), its 5-day BOD in mg/L is ______ (round to the nearest whole number)
Numerical answer — type the value.
Show answer
Answer: 205
205 mg/L. BOD₅ = L₀(1 − e^(−kt)) = 300 × (1 − e^(−0.23 × 5)) = 300 × (1 − e^(−1.15)) = 300 × (1 − 0.31664) = 300 × 0.68336 = 205.01 mg/L. The number worth carrying out of this is not 205 but 0.683: at the standard rate constant the five-day test sees a little over two-thirds of the demand, and it is the same fraction whatever L₀ is, because the exponential does not know how much organic matter is present. That is why BOD₅ and BOD are different quantities and why a question is entitled to give you either. Two ways to lose the mark. If a paper quotes K = 0.1 per day it almost certainly means base 10, and k = 2.303 × 0.1 = 0.23 — using 0.1 in the base-e formula gives 118 mg/L, which is wrong by 40%. And if the test temperature is not 20 °C, correct the rate first with k_T = k₂₀ × 1.047^(T−20): at 27 °C that is 0.23 × 1.047⁷ = 0.317/day, and BOD₅ rises to 238 mg/L on the same sample.
An ideal rectangular settling tank is redesigned to be twice as deep, keeping the plan area and the flow rate unchanged. What happens to the size of the smallest particle that is completely removed?
1. It is unchanged — depth does not enter the removal criterion at all
2. It halves, because the detention time has doubled
3. It falls by a factor of √2, because settling velocity goes as d²
4. It doubles, because the particle now has further to fall
Show answer
Answer: A — It is unchanged — depth does not enter the removal criterion at all
Unchanged. This is the result that makes sedimentation design work, and it survives exactly because two effects cancel. Doubling the depth doubles the detention time — the particle has twice as long to settle — but it also doubles the distance it must fall. Write it out: the particle is removed if v_s ≥ H/t, and t = LWH/Q, so H/t = HQ/(LWH) = Q/(LW) = Q/A. H cancels. The criterion is v_s ≥ Q/A, the surface overflow rate, and it contains no depth at all. Two consequences follow, and both are examinable. First, a settling tank is a plan-area device, so the way to improve one is to add plates or tubes that multiply effective area inside the same volume — a tube settler is not a clever material, it is this algebra turned into hardware. Second, depth is chosen for entirely different reasons — sludge storage, scour velocity, the risk of the sludge blanket being re-entrained — and the standard 3 to 4 m has nothing to do with the particles being caught. Option (b) is the trap: detention time is a real quantity, it is just not the criterion. Option (c) inverts the question, applying Stokes to a change that Stokes never sees.
A sedimentation tank treats 4000 m³/day at a surface overflow rate of 40 m³/m²/day. A particle with a settling velocity of 20 m/day enters uniformly across the depth at the inlet. The percentage of these particles removed is ______
Numerical answer — type the value.
Show answer
Answer: 50
50%. The critical settling velocity is the overflow rate itself, v_c = Q/A = 40 m/day, so a 20 m/day particle is slower than critical and is not completely removed — but it is not completely lost either, and that is the point of the question. Over the detention time it falls a fixed distance; a particle that starts low enough in the tank reaches the floor within that distance, and one that starts higher does not. Since particles enter uniformly across the depth, the fraction removed is the fraction of the depth that a particle can cover: v_s/v_c = 20/40 = 0.50, so 50%. The general rule is removal = v_s/v_c for v_s < v_c, and 100% for v_s ≥ v_c. Note what is not needed: the flow, the area and the depth all drop out once you have the ratio — the 4000 m³/day is there to let you check that A = 100 m², not because the answer depends on it. The assumption that does matter is the uniform inlet distribution; if the tank were fed only at the surface, none of these particles would be removed at all.
A water sample contains 40 mg/L of Ca²⁺ (atomic weight 40.08) and 24.3 mg/L of Mg²⁺ (atomic weight 24.31). Its total hardness in mg/L as CaCO₃ is ______ (round to one decimal place)
Numerical answer — type the value.
Show answer
Answer: 199.8
199.8 mg/L as CaCO₃. Both ions are divalent, so their equivalent weights are half their atomic weights: Ca 20.04, Mg 12.155. CaCO₃ has a molecular weight of 100 and is also divalent, so its equivalent weight is 50. Convert each ion with mg/L × 50 / (equivalent weight): calcium gives 40 × 50/20.04 = 99.80, magnesium gives 24.3 × 50/12.155 = 99.96 ≈ 100.00. Adding, 199.8 mg/L as CaCO₃ — very hard water on any published scale. The instructive part is the comparison: magnesium was present at 40% less mass than calcium and contributed the same hardness, because its equivalent weight is smaller and hardness is about charge rather than weight. That is exactly why the convention exists — without a common currency the two numbers cannot be added at all, and reporting "64.3 mg/L of hardness" by summing the raw concentrations is meaningless. The follow-up a question usually asks is the split: carbonate hardness is min(total hardness, total alkalinity), so if this water had 150 mg/L alkalinity as CaCO₃, it would carry 150 carbonate and 49.8 non-carbonate.
Which of the following statements about the drinking-water treatment train are correct?
1. Coagulation is needed because colloidal particles will not settle in any practicable tank
2. Flocculation requires gentle mixing, because vigorous stirring breaks the floc it is meant to build
3. Chlorine is preferred over ozone mainly because it leaves a residual in the distribution system
4. A rapid sand filter removes bacteria reliably enough that disinfection can be omitted
Show answer
Answer: A — Coagulation is needed because colloidal particles will not settle in any practicable tank; B — Flocculation requires gentle mixing, because vigorous stirring breaks the floc it is meant to build; C — Chlorine is preferred over ozone mainly because it leaves a residual in the distribution system
(a), (b) and (c). Take them in order. (a) is correct and is the quantitative justification for the whole coagulation stage: Stokes gives a 1 μm clay colloid a settling velocity of 0.077 m/day, so it needs 39 days to fall three metres. No city builds a tank with a 39-day detention time, so the particle cannot be settled — it has to be made bigger first, which is what coagulation and flocculation do. (b) is correct and is the one unit process in the plant where working harder makes the result worse: rapid mixing distributes the coagulant in seconds, but flocculation then needs a low velocity gradient over 20 to 40 minutes, because the shear that helps particles collide will also tear apart the floc they form. That is why G for flocculation is specified with an upper limit. (c) is correct and is the honest reason chlorine survives against better disinfectants: ozone and UV both inactivate organisms more effectively in the contact tank, and neither protects the water afterwards. A pipe network with hours of residence time and any chance of ingress needs a residual, and chlorine is the common disinfectant that leaves one. (d) is wrong, and it is the dangerous one. A rapid sand filter removes bacteria only mechanically and incompletely; unlike a slow sand filter, it has no biological layer doing the work. Disinfection after rapid filtration is not a polish — it is the barrier.
A town had 50,000 people in 2010 and 80,000 in 2020. Using the geometric increase method, its projected population in 2030 is ______
Numerical answer — type the value.
Show answer
Answer: 128000
128,000. The geometric method assumes a constant percentage growth per decade. From 50,000 to 80,000 the ratio is 80/50 = 1.60, so r = 0.60 per decade, and the next decade multiplies again: 80,000 × 1.60 = 128,000. Compare it with the arithmetic method on the same two data points: there the increase is a constant 30,000 per decade, giving 80,000 + 30,000 = 110,000. The two answers differ by 18,000 people, 16.4%, from identical data — and that gap is the reason the choice of method is examined at all rather than being a formality. The rule of thumb is about the stage of the town: geometric suits a young, rapidly growing settlement and will over-predict a mature one, because nothing compounds for ever; arithmetic suits a large, saturated city and will under-predict a booming one. Where a question does not say which to use, the incremental increase method is the usual default, since it sits between the two by construction. And a design period matters here in a way candidates skip: over-forecasting builds capacity that sits idle and is paid for by today’s users, while under-forecasting builds a headworks that cannot be enlarged.
On a breakpoint chlorination curve, the combined residual falls as the applied chlorine dose is increased over a certain range. The reason is that
1. the added chlorine oxidises the chloramines already formed, destroying them
2. the chlorine escapes as a gas once its solubility limit is exceeded
3. the pH rises with the dose and shifts the equilibrium away from residual chlorine
4. the measurement method under-reads at high dose and the fall is an artefact
Show answer
Answer: A — the added chlorine oxidises the chloramines already formed, destroying them
The added chlorine destroys the chloramines it earlier formed. The breakpoint curve has three regions and the middle one is the counter-intuitive part. At low dose, chlorine added to a water containing ammonia forms chloramines — combined residual — and the residual rises roughly with the dose. Past a peak, further chlorine no longer forms chloramines but oxidises them, largely to nitrogen gas, so the measured residual falls even as more chlorine is added. The minimum of that falling limb is the breakpoint. Only beyond it does the residual start to rise again, and now it is free chlorine — hypochlorous acid and hypochlorite — which is the fast, effective disinfectant. Three practical consequences follow, and each has been examined. Dosing to the rising limb before the hump gives a residual that looks adequate on a test but is combined chlorine, which disinfects far more slowly. Dosing into the falling limb wastes chlorine and can leave less residual than a smaller dose would. And the taste-and-odour complaints associated with chlorination are largely from chloramines rather than free chlorine, so going past the breakpoint often improves the taste rather than worsening it. Option (b) is chemically wrong at these concentrations, (c) has the pH dependence backwards, and (d) invents an instrument fault to avoid the chemistry.
A plant treats 4000 m³/day. The rapid sand filters are designed for a filtration rate of 5 m³/m²/h. The required filter area in m² is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 33.33
33.33 m². The only real step is the unit conversion candidates skip: the flow is per day and the rate is per hour. Q = 4000/24 = 166.67 m³/h, so A = 166.67/5 = 33.33 m². Dividing 4000 by 5 directly gives 800 m², wrong by a factor of 24, and it is by far the commonest way to lose this mark. The number is worth holding beside its alternative. A slow sand filter runs at about 0.2 m³/m²/h, which for the same town needs 166.67/0.2 = 833.3 m² — twenty-five times the land. That single ratio explains why almost every new Indian plant is rapid-sand: land near a growing town costs more than the coagulation and sedimentation stages that rapid filtration requires upstream. The trade is real in both directions, though. Slow sand filtration needs no chemicals, no coagulation, no skilled dosing, and its biological surface layer genuinely removes bacteria, which a rapid filter does not — so it remains the right answer for a small, remote supply with clean raw water and no chemist. In practice a plant also builds one extra filter beyond the calculated area, because backwashing takes a unit out of service and the design flow does not pause while it does.