Question 4 of 5: Filtration mechanisms; chlorine residuals; break-point

Question 4: Filtration mechanisms; chlorine residuals; break-point (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. For the illustrative numbers used below: a rapid sand filter of effective size 0.5 mm and bed depth 0.75 m at a filtration rate of 5 m/h and porosity 0.40, treating water at 15 °C carrying 10 $\mu$m floc of specific gravity 1.01; a free chlorine residual of 1.0 mg/L at pH 7.5; and a water carrying 2.0 mg/L of total ammonia nitrogen.

Find. (a) Four capture mechanisms in a granular filter, with evidence that each matters; (b) the definitions of and distinction between free and combined residual chlorine; (c) the break-point process.

Part (a) — Approach. Separate transport (what brings a particle from the bulk flow to a grain surface) from attachment (what keeps it there), and show numerically that the mechanism a student expects — sieving — is the least important of them.

1. Part (a), mechanism 1 — straining. A particle larger than the constriction between grains cannot pass and is retained at the surface. For approximately uniform spheres the constriction is about $0.155\,d$, so for 0.5 mm sand $$d_{\text{pore}}\approx 0.155\times 500=\boxed{77.5\ \mu\text{m}}$$ A Giardia cyst is about 5 $\mu$m and a typical clay colloid 1 $\mu$m, fifteen to seventy times smaller than the opening. Straining alone therefore removes only the coarsest floc and grit, and if it were the dominant mechanism the filter would blind at the surface within minutes. It matters most as the failure mode: poorly coagulated water strains at the top 50 mm, headloss climbs steeply and the bed depth is wasted.
2. Mechanism 2 — sedimentation within the pores. Each pore behaves as a very shallow settling basin whose "surface" is the upper hemisphere of the grain below. For a 10 $\mu$m floc of specific gravity 1.01 at 15 °C, $$v_s=\frac{g(\rho_p-\rho)d^2}{18\mu} =\frac{9.81\times 10.9\times (10\times 10^{-6})^2}{18\times 1.139\times 10^{-3}} =5.2\times 10^{-7}\ \text{m/s}\ (0.045\ \text{m/d})$$ which sounds negligible until it is compared with the time available. The residence time in the pores is $$t=\frac{L\varepsilon}{v}=\frac{0.75\times 0.40}{5/3600}=216\ \text{s}, \qquad \text{settling distance}=5.2\times 10^{-7}\times 216 =\boxed{113\ \mu\text{m}}$$ That distance exceeds the 77.5 $\mu$m pore, so a particle of this size can cross an entire pore and reach a grain by settling alone. The Reynolds number is $4.6\times 10^{-6}$, comfortably within the Stokes regime.
3. Mechanism 3 — interception and diffusion. A particle following a streamline that passes within one particle radius of the grain touches it: this is interception, and it grows with the ratio of particle to grain diameter. Below roughly 1 $\mu$m, Brownian motion becomes the dominant transport instead, carrying particles across streamlines to the grain by diffusion. The two mechanisms bracket a minimum in filter efficiency near 1 to 3 $\mu$m — too large to diffuse effectively, too small to intercept or settle — which is, uncomfortably, exactly the size range of Cryptosporidium oocysts and is the reason filtration is credited only 2 to 2.5 log for protozoa and must be paired with disinfection.
4. Mechanism 4 — attachment (adsorption). Transport is necessary but not sufficient: a particle that touches a grain will re-entrain unless it adheres. Attachment is governed by London–van der Waals attraction working against electrostatic repulsion between two similarly charged surfaces, so it is controlled by the chemistry upstream. Effective coagulation, which neutralises the particle charge, raises the attachment efficiency $\alpha$ from near 0.01 to near 1 and is worth two orders of magnitude in filtrate turbidity; a filter aid polymer does the same job at the filter. The practical statement is the one that answers the question: filtration is a chemical process disguised as a physical one, which is why filtered-water turbidity tracks coagulant dose far more closely than it tracks filtration rate.

Part (b) — free and combined residual chlorine. The total chlorine residual is everything that will oxidise iodide at pH 4 in the standard DPD test. It is made of two fractions with quite different behaviour.

Free available residual chlorine is the sum of hypochlorous acid and hypochlorite ion, $[\text{HOCl}]+[\text{OCl}^{-}]$, produced when chlorine hydrolyses in water that has no ammonia left to react with. It is a powerful, fast-acting germicide whose active fraction is set entirely by pH through the p$K_a$ of 7.54 established in Question 2: at pH 7.5 a 1.0 mg/L free residual is $$\alpha_{\text{HOCl}}\times 1.0=\boxed{0.52\ \text{mg/L as HOCl}}$$ It is also unstable — consumed by sunlight, organics, iron and manganese — and it is the fraction that generates trihalomethanes.

Combined available residual chlorine is chlorine bound to nitrogen as the chloramines, formed by $$\text{NH}_3+\text{HOCl}\rightarrow\text{NH}_2\text{Cl} +\text{H}_2\text{O}\quad\text{(monochloramine)}$$ $$\text{NH}_2\text{Cl}+\text{HOCl}\rightarrow\text{NHCl}_2 +\text{H}_2\text{O},\qquad \text{NHCl}_2+\text{HOCl}\rightarrow\text{NCl}_3+\text{H}_2\text{O}$$ Which chloramine dominates is decided by pH and by the applied chlorine-to- ammonia ratio: monochloramine above about pH 7, dichloramine below it, nitrogen trichloride only in acid water or at high ratios, and the last two are responsible for the swimming-pool odour and taste complaints.

The differences that matter in practice are four. Strength: the combined residual is a far weaker oxidant, needing on the order of $$\frac{CT_{\text{chloramine}}}{CT_{\text{free}}}=\frac{1850}{104} \approx 18$$ times the $CT$ for the same 3-log Giardia inactivation at 10 °C. Persistence: the combined residual is much more stable and survives to the far reaches of a large distribution system, which is why many Canadian utilities disinfect with free chlorine at the plant and convert to chloramine for distribution. By-products: chloramines form far fewer THMs and HAAs, but bring nitrosamines and the risk of nitrification in the mains. Measurement: DPD gives the free residual on the first reading and the total after adding iodide, and the combined residual is the difference, which is why an operator who reports only "total chlorine" has not demonstrated compliance with a free-chlorine $CT$ requirement.

Part (c) — break-point chlorination. Break-point chlorination is the deliberate addition of chlorine beyond the point at which all ammonia has been oxidised, so that a free residual can be established in a water that contains ammonia nitrogen. Following the curve as dose increases:

1. Part (c), zone 1 — immediate demand. The first increment of chlorine is consumed by reduced iron, manganese, sulphide and readily oxidisable organics. No residual of any kind appears.
2. Zone 2 — chloramine formation. Chlorine now reacts with ammonia to form monochloramine, one mole for one mole, so the combined residual rises almost linearly with dose. On a mass basis the ratio is $$\frac{M_{\text{Cl}_2}}{M_{\text{N}}}=\frac{70.9}{14.0}=5.06:1$$ so for 2.0 mg N/L the residual peaks near a dose of about 10 mg/L as Cl$_2$. This is the "hump", and a plant that stops here is chloraminating.
3. Zone 3 — destruction. Further chlorine oxidises the chloramines rather than adding to them, and the residual falls even as the dose rises — the counter-intuitive part of the curve. The overall stoichiometry is $$2\,\text{NH}_3+3\,\text{Cl}_2\rightarrow\text{N}_2\uparrow +6\,\text{HCl}$$ giving a theoretical requirement of $$\frac{3\times 70.9}{2\times 14.0}=7.6\ \text{mg Cl}_2\ \text{per mg N}, \qquad D_{\text{bp}}=7.6\times 2.0=\boxed{15.2\ \text{mg/L as Cl}_2}$$
4. Zone 4 — the break-point and beyond. At the minimum the ammonia is gone and the residual is at its lowest; every further increment now appears as free residual, rising at unit slope. Because side reactions with organic nitrogen consume chlorine that the stoichiometry does not account for, plants design on a practical ratio of 8:1 to 10:1, so $$D_{\text{practical}}\approx 10\times 2.0=\boxed{20\ \text{mg/L as Cl}_2}$$ and confirm it by a chlorine-demand curve run on the actual water.

Break-point chlorination is used to remove taste and odour caused by chloramines, to establish a free residual for $CT$ credit, and occasionally as an ammonia-removal process in its own right. Its costs are equally definite: a high chemical dose, roughly 14.3 mg/L of alkalinity destroyed per mg/L of ammonia nitrogen oxidised, elevated chloride and total dissolved solids, and — because a large free residual is being created in the presence of organic matter — a strong tendency to form THMs. In a Canadian plant it is therefore normally a corrective or seasonal measure, with biological nitrification preferred where the ammonia load is continuous.