Water Quality, Lakes and Rivers, Reactor Kinetics, and Water Treatment
1. Water and wastewater quality parameters, and BOD kinetics
Quality parameters are physical (turbidity in NTU, colour, taste and odour, temperature, and solids — total solids split by filtration into suspended and dissolved, and by ignition at 550 °C into volatile and fixed), chemical (pH, alkalinity, hardness, chloride, fluoride, nitrogen as organic, ammonia, nitrite and nitrate, phosphorus, metals, dissolved oxygen, and the oxygen demands BOD, COD and TOC) and biological (coliforms and pathogens, treated in Section 3). Hardness is the sum of the divalent cations, chiefly Ca²⁺ and Mg²⁺, expressed as CaCO₃: each ion’s mg/L is multiplied by 50/(its equivalent weight), so 40 mg/L Ca²⁺ (eq. wt 20) is 100 and 24 mg/L Mg²⁺ (eq. wt 12) is 100 mg/L as CaCO₃. The part of hardness matched by alkalinity is carbonate (temporary) hardness, removable by boiling; the rest is non-carbonate (permanent). For drinking water, IS 10500:2012 sets, for example, pH 6.5–8.5, TDS 500 mg/L (2000 where no other source exists), total hardness 200 mg/L as CaCO₃ (600), fluoride 1.0 mg/L (1.5), nitrate 45 mg/L and arsenic 0.01 mg/L.
Biochemical oxygen demand is the oxygen microbes use to oxidise the biodegradable organic matter, measured over 5 days at 20 °C (BOD₅). Its exertion is modelled as first order: the BOD remaining is L_t = L₀e^(−kt), so the BOD exerted is BOD_t = L₀(1 − e^(−kt)), where L₀ is the ultimate carbonaceous BOD and k the rate constant (base e, typically 0.1–0.3 d⁻¹; in base 10, k₁₀ = k/2.303). The rate rises with temperature by the van ’t Hoff–Arrhenius form k_T = k₂₀θ^(T−20), with θ ≈ 1.047 for BOD; the ultimate BOD itself does not change. In a dilution test, BOD = (D₁ − D₂)/P, where D₁ and D₂ are the initial and final DO of the diluted sample and P the fraction of sample in the bottle. After about 8–10 days nitrifiers begin to exert a nitrogenous BOD, which is why the carbonaceous test is read at 5 days or run with a nitrification inhibitor. COD oxidises almost all organics with dichromate in hot acid, so COD ≥ BOD, and a BOD₅/COD ratio above about 0.5 marks a readily biodegradable waste.
2. Eutrophication, thermal stratification, and the oxygen sag curve in rivers
Eutrophication is the enrichment of a water body with nutrients, chiefly phosphorus and nitrogen, which fuels algal growth. Lakes pass from oligotrophic (clear, nutrient-poor, oxygenated to the bottom) through mesotrophic to eutrophic (turbid, algal blooms, anoxic bottom water). In most fresh waters phosphorus is the limiting nutrient — Liebig’s law of the minimum — so control concentrates on P from sewage, detergents and fertiliser runoff. The damage comes after the bloom: dead algae sink and decompose, consuming the oxygen of the deep water, releasing ammonia, iron, manganese and H₂S, and some cyanobacteria release toxins. Thermal stratification sets the stage. In summer a warm, light epilimnion floats on a cold, dense hypolimnion, separated by the metalimnion in which the temperature falls steeply (the thermocline). The hypolimnion is cut off from the air and its oxygen is not replaced. In autumn the surface cools to the density of the deep water and the lake turns over, mixing oxygen down and nutrients up; in temperate lakes a second overturn follows in spring after the ice melts.
A river receiving an organic load loses oxygen to deoxygenation by BOD and gains it by reaeration from the air in proportion to the deficit. With the deficit D = DO_sat − DO, the Streeter–Phelps equation is dD/dt = k₁L − k₂D, whose solution is D_t = [k₁L₀/(k₂ − k₁)](e^(−k₁t) − e^(−k₂t)) + D₀e^(−k₂t). The deficit rises to a maximum — the critical deficit — at the critical time t_c = [1/(k₂ − k₁)] ln{(k₂/k₁)[1 − D₀(k₂ − k₁)/(k₁L₀)]}, where D_c = (k₁/k₂)L₀e^(−k₁t_c), and then recovers. The ratio f = k₂/k₁ is the self-purification factor. Distance follows from time as x = ut. The initial conditions come from mixing at the outfall: L₀ = (Q_wL_w + Q_rL_r)/(Q_w + Q_r), and the same for DO and temperature. Along the sag the river passes through zones of degradation, active decomposition and recovery to clear water.
3. Kinetics and reactor design: mass balance, order, batch, CMFR and PFR
Every unit is analysed with a mass balance on a control volume: accumulation = in − out + generation (reaction). An energy balance is written the same way for heat, as in a digester or an incinerator. The rate of a reaction r = −kCⁿ has order n: zero order (C falls linearly, t½ = C₀/2k), first order (C = C₀e^(−kt), t½ = ln 2/k, independent of C₀) and second order (1/C = 1/C₀ + kt). The order is found by plotting the data: C, ln C or 1/C against t, whichever is straight. In a batch reactor nothing enters or leaves, and the concentration simply follows the rate law in time.
A completely mixed flow reactor (CMFR, CSTR) has the same concentration everywhere as in its outflow. At steady state with first-order decay, QC₀ = QC + kCV, so C/C₀ = 1/(1 + kτ) with τ = V/Q. A plug-flow reactor (PFR) moves fluid through in order without mixing along its length; every element spends exactly τ inside, like a batch reactor in transit, so C/C₀ = e^(−kτ). For any positive order the PFR needs a smaller volume for the same removal, because the CMFR works throughout at the low outlet concentration. CMFRs in series approach plug flow: n equal tanks give C/C₀ = 1/(1 + kτ/n)ⁿ, which tends to e^(−kτ) as n grows. The CMFR’s virtue is that it dilutes a shock load or a toxic spike at once, which is why aeration tanks are often completely mixed. Real tanks lie between the two ideals, with dead zones and short-circuiting that a tracer test reveals.
| Reactor | C/C₀ | τ for 90% removal, k = 0.2 h⁻¹ | Removal at kτ = 2 |
|---|---|---|---|
| Batch or PFR | e^(−kτ) | ln 10/0.2 = 11.5 h | 1 − e^(−2) = 86.5% |
| One CMFR | 1/(1 + kτ) | 9/0.2 = 45 h | 1 − 1/3 = 66.7% |
| n CMFRs in series | 1/(1 + kτ/n)ⁿ | Between the two | Two tanks: 1 − 1/4 = 75% |
4. Water treatment: screening, sedimentation, coagulation, filtration, desalination and disinfection
Screening at the intake removes floating debris and fish with coarse bars and finer travelling screens. Plain sedimentation removes discrete particles that settle at their terminal velocity; for small spheres Stokes’ law gives v_s = g(ρ_s − ρ)d²/(18μ). In an ideal settling tank (Hazen–Camp theory) a particle is removed completely if v_s ≥ the surface overflow rate v₀ = Q/A_s, and a slower one in the fraction v_s/v₀ — so removal depends on the plan area and not on the depth. The depth sets the detention time t = V/Q (typically 2–4 h for plain settling). Coagulation adds a coagulant — alum, Al₂(SO₄)₃·14H₂O, or ferric salts — to destabilise the negatively charged colloids that will not settle by compressing the double layer, neutralising charge and sweep-floc enmeshment; the optimum dose and pH are found by the jar test. Alum consumes alkalinity: each 594 mg of alum reacts with alkalinity equivalent to 300 mg of CaCO₃. Flocculation then grows the floc by slow mixing, designed on the velocity gradient G = √(P/(μV)), typically 20–80 s⁻¹, and the product Gt.
Filtration removes the floc that escapes the clarifier. A slow sand filter (about 0.1–0.4 m/h) works mainly biologically in the schmutzdecke on its surface and is cleaned by scraping; a rapid sand filter (about 5–15 m/h) works by depth straining and adhesion after coagulation and is cleaned by backwashing, which fluidises the bed. The filter area is Q divided by the filtration rate. Desalination removes dissolved salts: reverse osmosis pushes water through a semipermeable membrane against the osmotic pressure π = iCRT (about 24 atm for 0.5 M NaCl, near sea water), so the applied pressure must exceed it; electrodialysis moves ions through ion-exchange membranes in an electric field and suits brackish water; thermal distillation (multi-stage flash, multi-effect) boils and condenses. All produce a concentrated brine that must be disposed of. Disinfection is most often by chlorine: Cl₂ + H₂O → HOCl + HCl, and HOCl ⇌ H⁺ + OCl⁻ with pK_a ≈ 7.5, so the fraction as HOCl is 1/(1 + 10^(pH − 7.5)). HOCl is some eighty times more effective than OCl⁻, so chlorine works better at low pH. Chlorine first reacts with reducing agents and ammonia to form chloramines, and past the breakpoint every further dose appears as free residual; a residual of at least 0.2 mg/L is kept at the consumer’s tap. Ozone and UV (about 254 nm) disinfect without a lasting residual; chlorination of water rich in organics forms trihalomethanes.
5. Water distribution and storage
Treated water reaches consumers by gravity, by pumping, or by pumping with storage — the usual arrangement, in which pumps run at a steady rate and elevated reservoirs absorb the difference between that rate and the hourly demand, while keeping a residual pressure in the mains. The network is laid out as a dead-end (tree) system, simple but with stagnant tails and no alternative path during repairs; a grid-iron system of interconnected loops, analysed by Hardy Cross; a ring system round a district; or a radial system fed from zonal reservoirs. Storage is sized for three components: the balancing (equalising) volume, from a mass curve of demand against the pumping schedule; a fire reserve; and an emergency reserve for breakdowns. Demand varies with the hour and season, so mains are designed for peak factors on the average daily demand, and pipe materials, valves (sluice, air, scour and pressure-relief), hydrants and meters complete the system. Leakage control and pressure management are the largest levers for water conservation in an existing network.
- Dead-end: simple and cheap; stagnant water at the tails and no supply during repairs.
- Grid-iron: every point fed from more than one direction; more valves and pipe, and needs network analysis.
- Ring and radial: rings suit well-planned towns; radial zones fed from local reservoirs keep pressures even.
Key takeaways
- Hardness as CaCO₃ = Σ mg/L × 50/eq. wt; carbonate hardness is the part matched by alkalinity. BOD_t = L₀(1 − e^(−kt)), and k_T = k₂₀θ^(T−20) with θ ≈ 1.047 changes the rate, not L₀.
- Phosphorus usually limits algae in fresh water; summer stratification isolates the hypolimnion, which goes anoxic under a bloom, until autumn overturn mixes the lake.
- Streeter–Phelps: t_c = [1/(k₂ − k₁)] ln{(k₂/k₁)[1 − D₀(k₂ − k₁)/(k₁L₀)]}, D_c = (k₁/k₂)L₀e^(−k₁t_c); initial L₀ and D₀ come from mixing at the outfall.
- First order: batch and PFR give e^(−kτ), one CMFR 1/(1 + kτ), n CMFRs 1/(1 + kτ/n)ⁿ; the PFR is smaller for the same removal, the CMFR damps shocks.
- Settling removal is v_s/(Q/A); detention time V/Q; G = √(P/μV); RO pressure must beat π = iCRT; HOCl fraction 1/(1 + 10^(pH − 7.5)); free residual appears only past the breakpoint.
Practice questions (20)
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 water contains 40 mg/L of Ca²⁺ and 24 mg/L of Mg²⁺. What is its total hardness, in mg/L as CaCO₃? (Ca = 40, Mg = 24)
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Answer: 200
Ca²⁺: 40 × 50/20 = 100; Mg²⁺: 24 × 50/12 = 100; total hardness = 200 mg/L as CaCO₃. Adding the mg/L directly (64) ignores that each ion must be converted by its own equivalent weight.The water of the previous question — total hardness 200 mg/L as CaCO₃ — has an alkalinity of 150 mg/L as CaCO₃. What is its non-carbonate hardness, in mg/L as CaCO₃?
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Answer: 50
When alkalinity is less than total hardness, carbonate hardness equals the alkalinity, 150, and non-carbonate hardness is the remainder: 200 − 150 = 50 mg/L as CaCO₃. Had alkalinity exceeded hardness, all the hardness would be carbonate and non-carbonate hardness would be zero.A wastewater has an ultimate carbonaceous BOD of 250 mg/L and a BOD rate constant (base e) of 0.23 d⁻¹ at 20 °C. What is its 5-day BOD, in mg/L? Give the answer to the nearest whole number.
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Answer: 171
BOD₅ = L₀(1 − e^(−kt)) = 250(1 − e^(−1.15)) = 250 × 0.6834 = 170.8 ≈ 171 mg/L. Using the formula for BOD remaining, 250e^(−1.15) = 79 mg/L, gives what is left to exert, not what has been exerted.A waste has a BOD₅ at 20 °C of 200 mg/L, with k₂₀ = 0.23 d⁻¹ (base e). Taking θ = 1.047, what is its 3-day BOD when incubated at 30 °C, in mg/L? Give the answer to one decimal place.
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Answer: 194.5
L₀ = 200/(1 − e^(−0.23 × 5)) = 200/0.6834 = 292.7 mg/L, unchanged by temperature. k₃₀ = 0.23 × 1.047¹⁰ = 0.23 × 1.583 = 0.364 d⁻¹. BOD₃ at 30 °C = 292.7(1 − e^(−0.364 × 3)) = 292.7 × 0.6645 = 194.49 ≈ 194.5 mg/L. Correcting BOD₅ by 1.047¹⁰ instead gives 317, above the ultimate BOD, which is impossible.Which of the following statements about lakes are correct?
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Answer: A — In summer stratification the hypolimnion is cut off from reaeration; B — Phosphorus is usually the limiting nutrient for algae in fresh water; D — Autumn overturn occurs when the surface water cools to the density of the deeper water
(a), (b) and (d) are correct. (c) is false: an oligotrophic lake produces too little organic matter to exhaust the oxygen of its deep water; an anoxic hypolimnion is the mark of a eutrophic lake, where decaying algae consume it.A sewage outfall discharges 1 m³/s with a BOD of 100 mg/L into a river flowing at 10 m³/s with a BOD of 2 mg/L. What is the BOD of the river just below the outfall after complete mixing, in mg/L? Give the answer to two decimal places.
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Answer: 10.91
C = (Q_rC_r + Q_wC_w)/(Q_r + Q_w) = (10 × 2 + 1 × 100)/11 = 120/11 = 10.91 mg/L. Averaging the two concentrations, 51 mg/L, ignores that the river carries ten times the flow.Below an outfall a river has an initial ultimate BOD of 10 mg/L and an initial DO deficit of 1 mg/L. The deoxygenation and reaeration rate constants (base e) are 0.2 d⁻¹ and 0.4 d⁻¹. Using the Streeter–Phelps model, what is the critical DO deficit, in mg/L? Give the answer to two decimal places.
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Answer: 2.78
t_c = [1/(k₂ − k₁)] ln{(k₂/k₁)[1 − D₀(k₂ − k₁)/(k₁L₀)]} = 5 ln{2 × [1 − 1 × 0.2/(0.2 × 10)]} = 5 ln 1.8 = 2.94 d. D_c = (k₁/k₂)L₀e^(−k₁t_c) = 0.5 × 10 × e^(−0.588) = 5 × 0.5556 = 2.78 mg/L. Substituting t = 2.94 d in the full sag equation gives the same 2.78, a check worth making.For the river of the previous question (L₀ = 10 mg/L, D₀ = 1 mg/L, k₁ = 0.2 d⁻¹, k₂ = 0.4 d⁻¹), at what time after the outfall does the critical deficit occur, in days? Give the answer to two decimal places.
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Answer: 2.94
t_c = [1/(0.4 − 0.2)] ln{(0.4/0.2)[1 − 1 × (0.4 − 0.2)/(0.2 × 10)]} = 5 ln(2 × 0.9) = 5 × 0.5878 = 2.94 d. Leaving out the initial deficit term gives 5 ln 2 = 3.47 d, which is right only when D₀ = 0.A first-order contaminant (k = 0.2 h⁻¹) must be reduced by 90%. What hydraulic retention time, in hours, does a single completely mixed flow reactor need?
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Answer: 45
C/C₀ = 1/(1 + kτ) = 0.1, so kτ = 9 and τ = 9/0.2 = 45 h. A plug-flow reactor needs only ln 10/0.2 = 11.5 h: the CMFR is nearly four times larger because it operates entirely at the outlet concentration.Two equal completely mixed reactors in series each have kτ = 1 for a first-order reaction. The overall fractional removal is:
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Answer: A — 75%
Each tank passes C/C₀ = 1/(1 + 1) = 0.5, so the pair passes 0.5 × 0.5 = 0.25 and removes 75%. A single tank of the combined volume (kτ = 2) removes only 66.7%, and a plug-flow reactor of that volume 1 − e^(−2) = 86.5% (option C): tanks in series lie between the two ideals.A discrete particle of diameter 0.02 mm and specific gravity 2.65 settles in water at 20 °C (μ = 1.0 × 10⁻³ Pa·s, ρ = 1000 kg/m³). What is its Stokes settling velocity, in mm/s? Give the answer to two decimal places. (g = 9.81 m/s²)
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Answer: 0.36
v_s = g(ρ_s − ρ)d²/(18μ) = 9.81 × 1650 × (2 × 10⁻⁵)²/(18 × 10⁻³) = 6.474 × 10⁻³/18 = 3.60 × 10⁻⁴ m/s = 0.36 mm/s. Using 2650 in place of the density difference 1650 gives 0.58 mm/s.A settling tank of plan area 200 m² and volume 600 m³ treats 10 000 m³/d. What is its detention time, in hours? Give the answer to two decimal places.
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Answer: 1.44
t = V/Q = 600/10 000 d = 0.06 d = 1.44 h. The overflow rate of the same tank is Q/A = 10 000/200 = 50 m³/m²·d; the two are linked through the depth, 3 m, since t = depth/overflow rate.In the tank of the previous question (overflow rate 50 m³/m²·d), what percentage of discrete particles with a settling velocity of 0.4 mm/s is removed, under ideal-settling theory? Give the answer to one decimal place.
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Answer: 69.1
v_s = 0.4 × 10⁻³ × 86 400 = 34.56 m/d, below v₀ = 50 m/d, so the fraction removed is v_s/v₀ = 34.56/50 = 0.691, i.e. 69.1%. Comparing mm/s with m/d without converting would suggest either no removal or complete removal.A plant treating 20 ML/d doses alum, Al₂(SO₄)₃·14H₂O (molar mass 594), at 30 mg/L. How much natural alkalinity does the alum consume, in mg/L as CaCO₃? Give the answer to two decimal places.
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Answer: 15.15
Al₂(SO₄)₃·14H₂O + 3Ca(HCO₃)₂ → 2Al(OH)₃ + 3CaSO₄ + 6CO₂ + 14H₂O: 594 mg of alum uses alkalinity equivalent to 3 × 100 = 300 mg of CaCO₃. So 30 × 300/594 = 15.15 mg/L as CaCO₃. The daily alum demand is 30 mg/L × 20 ML/d = 600 kg/d.A flocculation basin of volume 1000 m³ receives 500 W of useful mixing power. With μ = 1.0 × 10⁻³ Pa·s, what is the mean velocity gradient, in s⁻¹? Give the answer to one decimal place.
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Answer: 22.4
G = √(P/(μV)) = √(500/(10⁻³ × 1000)) = √500 = 22.4 s⁻¹, within the usual 20–80 s⁻¹ for flocculation. Leaving out the square root gives 500, which would shear the floc apart.Which of the following statements about filtration and desalination are correct?
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Answer: A — A slow sand filter removes impurities largely by biological action in its surface layer; B — A rapid sand filter is cleaned by backwashing; C — In reverse osmosis the applied pressure must exceed the osmotic pressure of the feed
(a), (b) and (c) are correct. (d) is false: every desalination process splits the feed into product water and a brine that carries the salts, and its disposal is a design problem in its own right.Taking pK_a = 7.5 for hypochlorous acid, what fraction of the free chlorine is present as HOCl at pH 8.5? Give the answer to three decimal places.
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Answer: 0.091
[HOCl]/([HOCl] + [OCl⁻]) = 1/(1 + 10^(pH − pK_a)) = 1/(1 + 10¹) = 1/11 = 0.091. One pH unit above pK_a only about 9% remains as the far stronger disinfectant HOCl, which is why chlorination is less effective in alkaline water.In breakpoint chlorination, the free chlorine residual first appears:
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Answer: A — after the breakpoint, once the chloramines have been oxidised
The first chlorine is used by reducing agents; the next forms chloramines, so the residual rises as combined chlorine (the hump); further chlorine oxidises the chloramines to N₂ and the residual falls to the breakpoint. Only beyond it does added chlorine persist as free residual, rising one-for-one with the dose.What is the osmotic pressure of a 0.5 M NaCl solution at 298 K, in atm, assuming complete dissociation? Give the answer to one decimal place. (R = 0.08206 L·atm/(mol·K))
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Answer: 24.5
π = iCRT = 2 × 0.5 × 0.08206 × 298 = 24.45 ≈ 24.5 atm, since NaCl gives two ions. Forgetting the van ’t Hoff factor halves it to 12.2 atm and would under-size the high-pressure pumps of a reverse-osmosis plant.Which of the following statements about water distribution and storage are correct?
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Answer: A — A dead-end system can leave stagnant water at its tails; B — A grid-iron system can supply a point from more than one direction during repairs; C — Balancing storage is found from a mass curve of demand against the pumping schedule
(a), (b) and (c) are correct. (d) is false: elevated reservoirs exist partly to maintain a residual pressure in the mains, which keeps contaminated groundwater from entering through leaks and delivers water to upper floors.