Question 4 of 5: Filtration, Sedimentation and Hardness Removal

Question 4: Filtration, Sedimentation and Hardness Removal (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.

(a) Schmutzdecke, headloss and backwash (9 marks)

Schmutzdecke. The term — German for “dirt blanket” — describes the biologically active mat of algae, bacteria, protozoa and trapped debris that forms on the surface of a slow sand filter after a few days of operation. In a slow sand filter the schmutzdecke does most of the work: it is a biological as well as a physical barrier, achieving genuine removal of bacteria and protozoan cysts by predation and straining, and the filter is not considered mature until it has developed. It is worth stating plainly that a schmutzdecke does not develop in a rapid sand filter, and its absence is the defining difference between the two processes. A rapid filter runs at 5–15 m/h against 0.1–0.3 m/h, is preceded by coagulation and sedimentation rather than replacing them, removes particles throughout the depth of the bed rather than at the surface, and is cleaned by backwashing in minutes rather than by scraping the top 20 mm every few weeks. What a rapid filter does form at its surface, if it is run without adequate pretreatment, is a surface cake that causes rapid headloss development and short runs — the failure mode that the schmutzdecke represents by design in the slow filter.

Filter headloss. A clean bed already resists flow; the Carman–Kozeny (or Rose) equation gives that clean-bed loss as a function of porosity, grain size and sphericity, filtration rate and bed depth. As the run proceeds, deposited solids reduce the pore volume, headloss climbs, and the run ends when either the available head is exhausted or the effluent turbidity breaks through — a well-designed filter reaches both limits at about the same time. The subtlety worth marks is negative head: because headloss accumulates preferentially in the top few centimetres, the pressure part-way down the bed can fall below atmospheric. Dissolved gases then come out of solution and lodge in the pores, a condition called air binding, which further increases headloss, disrupts the flow distribution, and can lift the media during the next backwash.

Given. A rapid filter with bed depth 0.75 m, effective size 0.55 mm, sphericity 0.85, porosity 0.42, run at 5 m/h; water at 20 °C. The filter carries 1.8 m of submergence and is taken off line at 3.0 m of headloss.

Find. The clean-bed headloss, and whether air binding is expected at the terminal condition.

1. Bed Reynolds number. With \(v = 5/3600 = 1.389\times10^{-3}\) m/s, \(\mathrm{Re} = \rho\varphi dv/\mu = 0.647\), so the flow is in the laminar range and the viscous term of the friction factor dominates: \(f = 150(1-\varepsilon)/\mathrm{Re} + 1.75 = 136.3\).
2. Clean-bed headloss. The Carman–Kozeny equation gives \[h_L = f\,\frac{1-\varepsilon}{\varepsilon^3}\,\frac{L}{\varphi d}\,\frac{v^2}{g} = 136.3\left(\frac{0.58}{0.0741}\right)\left(\frac{0.75}{4.68\times10^{-4}}\right) \frac{(1.389\times10^{-3})^2}{9.81} \;\Longrightarrow\; \boxed{\ h_L = 0.34\ \text{m}\ }\]
3. Air binding. The pressure head at the bottom of the bed is the submergence plus the bed depth less the accumulated loss: \(1.8 + 0.75 - 3.0 = -0.45\) m. The negative value confirms sub-atmospheric pressure within the bed at the terminal headloss, so this filter is air-binding-prone and either the submergence must be increased or the run terminated earlier.

Filter backwash. Cleaning reverses the flow upward at a rate high enough to fluidise the bed — typically 0.5–0.7 m/min, or 10–12 L/m²·s — expanding it 20–50 per cent so that the grains separate and abrade against one another, releasing the trapped floc which is carried out over the wash troughs. Fluidisation alone is a poor scrubber, because in a fluidised bed the grains move together with little relative motion, so modern practice adds air scour beneath or before the water wash, or uses a simultaneous air-and-water wash, to shear the deposits off. Backwash also re-stratifies a dual-media bed, the coarse light anthracite settling above the fine heavy sand. Two operational points close the answer: after a wash the filter passes a brief burst of turbidity while it re-ripens, which is why filter-to-waste is provided; and the wash water, typically 2–4 per cent of production, must be recovered and treated because it carries the concentrated pathogen load the filter removed.

For the filter above, washing at 0.6 m/min for 8 minutes uses 4.8 m³ per m² against 180 m³/m² produced in a 36-hour run, or 2.7 per cent of production.

(b) Surface overflow rate and settling efficiency (8 marks)

The relationship is the central result of ideal-settling (Hazen–Camp) theory, and its most useful consequence is counter-intuitive: removal in a primary sedimentation tank depends on the plan area alone and is independent of the depth.

Consider an ideal rectangular tank of length \(L\), width \(W\) and depth \(H\) passing a flow \(Q\), in which flow is horizontal and uniform at \(v_h = Q/(WH)\), and particles settle at their own terminal velocity \(v_s\) independently of one another. A particle entering at the very top of the inlet is removed only if it reaches the floor within the detention time, that is if \(H/v_s \le L/v_h\). Substituting \(v_h\) and rearranging, \[v_s \ge \frac{QH}{WHL} = \frac{Q}{LW} = \frac{Q}{A} \equiv v_c,\] and the depth has cancelled. The critical settling velocity \(v_c\) is numerically the surface overflow rate, \(Q/A\), which is why the SOR (in m³/m²·d, i.e. m/d) is quoted as a velocity. Every particle with \(v_s \ge v_c\) is completely removed; a slower particle is removed only if it enters low enough, and since the inlet concentration is uniform over the depth its fractional removal is exactly \(v_s/v_c\). The efficiency–SOR relationship is therefore a rising line to complete removal:

Given. Two primary tanks, each 25 m × 6 m × 3.0 m deep, treating 12 000 m³/d. Primary solids have a density of 1050 kg/m³; water at 20 °C.

Find. The surface overflow rate, the smallest particle completely removed, the removal of a particle settling at 20 m/d, and the effect of adding a third tank instead of deepening the two.

1. Overflow rate and detention. \(A = 2(25)(6) = 300\) m², so \(\text{SOR} = 12\,000/300 = 40\ \text{m}^3/\text{m}^2\!\cdot\!\text{d}\). The volume is 900 m³ and the detention time \(900/12\,000 = 0.075\) d = 1.8 h, consistent with the identity \(\text{SOR} = H/\tau = 3.0/0.075 = 40\) m/d.
2. Smallest particle completely removed. Setting the Stokes velocity equal to \(v_c = 40/86\,400 = 4.63\times10^{-4}\) m/s, \[d = \sqrt{\frac{18\mu v_c}{g(\rho_p-\rho)}} = \sqrt{\frac{18(1.002\times10^{-3})(4.63\times10^{-4})}{9.81(51.8)}} \;\Longrightarrow\; \boxed{\ d = 128\ \mu\text{m}\ }\] The particle Reynolds number is 0.059, so Stokes’ law is legitimately applied.
3. Partial removal. A particle settling at 20 m/d is removed to the extent \(v_s/v_c = 20/40 = 50\) per cent.
4. Third tank versus a deeper tank. A third identical tank raises the area to 450 m² and halves the SOR to 20 m/d; the smallest fully removed particle shrinks by \(\sqrt{2}\) to 91 µm, and the 20 m/d particle goes from 50 to 100 per cent removal. Deepening the existing two tanks to 4.5 m adds exactly the same volume, lengthens the detention time to 2.7 h — and changes the overflow rate not at all, so it removes not one additional particle.

(c) Permanent hardness and ion exchange (8 marks)

Permanent hardness, more properly non-carbonate hardness, is the portion of the total hardness that is not associated with bicarbonate and carbonate alkalinity — the calcium and magnesium paired with sulfate, chloride and nitrate. The name is historical and physically apt: temporary (carbonate) hardness precipitates when the water is boiled, because bicarbonate decomposes to carbonate and carbon dioxide and the calcium drops out as CaCO₃, while the sulfate and chloride hardness remains in solution however long the water is boiled. Operationally, \(\text{TH} = \text{CH} + \text{NCH}\) with \(\text{CH} = \min(\text{alkalinity}, \text{TH})\), all expressed as CaCO₃. The distinction governs the choice of process: lime alone removes only carbonate hardness, whereas removing non-carbonate hardness requires soda ash as well — or ion exchange, which is indifferent to the anion entirely.

Principle of ion exchange. A strong-acid cation resin is a cross-linked polystyrene bead carrying fixed sulfonate groups, each holding a mobile counter-ion. On the sodium cycle the exchange is \[2\,\text{R--SO}_3\text{Na} + \text{Ca}^{2+} \rightleftharpoons (\text{R--SO}_3)_2\text{Ca} + 2\,\text{Na}^+,\] with the identical reaction for magnesium. The resin’s affinity for divalent calcium and magnesium exceeds its affinity for monovalent sodium, so at the low ionic strength of a natural water the equilibrium lies far to the right and the hardness ions are held while sodium is released. Because the exchange is on the cation, the anion is irrelevant — which is exactly why ion exchange removes carbonate and non-carbonate hardness alike, and why it is the process of choice when permanent hardness dominates. Water leaves a fresh bed at essentially zero hardness. When the exchange front reaches the bottom of the bed, hardness breaks through and the resin is regenerated by flooding it with a concentrated brine, typically 10 per cent NaCl: the high sodium concentration reverses the equilibrium by mass action, stripping the calcium and magnesium into a waste stream. Three practical consequences follow. Softened water is essentially zero-hardness and would be corrosive, so part of the raw flow is bypassed and blended to a target of about 80–100 mg/L as CaCO₃. Sodium is added to the finished water in exchange for the hardness, about 0.46 mg of sodium per mg of hardness as CaCO₃, which matters for sodium-restricted consumers. And the spent brine is a concentrated chloride waste whose disposal is often the binding constraint on using the process at all.

Given. The very hard groundwater of Question 2 — total hardness 303 mg/L as CaCO₃, alkalinity 180 mg/L as CaCO₃ — treated at 300 m³/d. Resin volume 2.0 m³ at an operating capacity of 50 kg CaCO₃/m³, regenerated with 120 kg NaCl per m³ of resin. Target finished hardness 85 mg/L as CaCO₃.

Find. The split between temporary and permanent hardness, the bypass fraction, the run length between regenerations, and the regenerant efficiency.

1. Split the hardness. Carbonate hardness is the lesser of alkalinity and total hardness, so \(\text{CH} = 180\) mg/L and \(\text{NCH} = 303 - 180 = 123\ \text{mg/L as CaCO}_3\) — 41 per cent of the hardness is permanent, and lime softening alone could not touch it.
2. Blend to the target. With the softened stream at essentially zero hardness, the bypass fraction follows from a mass balance on the blend point: \(x = 85/303 = 0.281\), so 28 per cent of the raw flow bypasses the vessel and 215.8 m³/d is softened.
3. Run length. The hardness load on the resin is \(215.8(303)/1000 = 65.3\) kg CaCO₃/d against a bed capacity of \(2.0(50) = 100\) kg, so \[\boxed{\ t_{run} = \frac{100}{65.3} = 1.53\ \text{d} \approx 37\ \text{h}\ }\]
4. Regenerant efficiency. Stoichiometrically the 100 kg of CaCO₃ removed is \(100\,000/50.04 = 1998\) equivalents, requiring \(1998(58.44)/1000 = 117\) kg of NaCl. Applying 240 kg gives an efficiency of \(117/240 = 49\) per cent, which is normal — the excess brine is the price of driving an unfavourable equilibrium backwards, and it goes to waste.