16-Civ-B5 Water Supply and Wastewater Treatment · May 2016

Question 3 of 5: Effluent Limits on cBOD, Ammonia and Phosphorus, and the Sludge Train

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examination, May 2016 — 98-Civ-B5 Water Supply and Wastewater Engineering. Three hours; closed book, one two-sided aid sheet and an approved calculator permitted. Question 1 is compulsory and the candidate attempts any three of Questions 2–5; every question carries 25 marks, so a complete paper is 100 marks. All five questions are worked here, because this set is a study resource rather than a timed sitting.

Reference texts.

• Metcalf & Eddy / AECOM, Wastewater Engineering: Treatment and Resource Recovery, 5th ed. — activated sludge kinetics, secondary clarifier loading, nitrogen chemistry, sludge processing.
• Crittenden, Trussell, Hand, Howe & Tchobanoglous, MWH's Water Treatment: Principles and Design, 3rd ed. — chlorine chemistry and breakpoint, granular media filtration, UV disinfection.
• Davis & Cornwell, Introduction to Environmental Engineering, 5th ed. — water-quality indicators, hardness chemistry, sedimentation theory, residuals management.
• Mihelcic & Zimmerman, Environmental Engineering: Fundamentals, Sustainability, Design, 3rd ed. — mass balances, reaction kinetics, nutrient cycling.
• APHA/AWWA/WEF, Standard Methods for the Examination of Water and Wastewater — Methods 2130 (turbidity), 2340 (hardness), 5210 (BOD), 9222 (membrane-filter coliforms).
• Health Canada, Guidelines for Canadian Drinking Water Quality; CCME, Canadian Environmental Quality Guidelines; Environment and Climate Change Canada, Wastewater Systems Effluent Regulations (SOR/2012-139).

Check: representative design data. Questions 1 to 4 are discussion questions and the source paper prints no numbers at all in them. Every number that appears in those four answers is a representative Canadian municipal value chosen by the solver so that each definition or mechanism can be made concrete and checkable; each is labelled where it is used, and all of them. Only Question 5(b) uses data given by the examiner. The graded content of Questions 1–4 is the reasoning, not the arithmetic.

Question 3: Effluent Limits on cBOD, Ammonia and Phosphorus, and the Sludge Train (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. A representative secondary plant and its receiving water are used so that each limit can be argued from a number rather than from an assertion.

Given data — representative secondary plant and receiving water

Quantity	Value
Treated effluent flow	10 000 m3/d
Effluent cBOD5, total ammonia nitrogen, total phosphorus	25 mg/L; 20 mg N/L; 1.0 mg P/L
Receiving stream flow and upstream cBOD5	2.0 m3/s; 2.0 mg/L
Combined primary and secondary sludge	1000 m3/d at 0.8 % dry solids
Thickened and dewatered target solids concentrations	4 % after thickening; 22 % cake

Find. (a) The physical mechanism each of the three limits is written to prevent, quantified for this plant; (b) what stabilisation, thickening and dewatering each accomplish, and the volume and mass consequences of the sludge train.

Approach. Treat each effluent parameter as the driver of a distinct failure mode in the receiving water — oxygen depletion, direct toxicity plus a second oxygen demand, and nutrient-driven biomass production — and put a number on each mode; then follow the sludge as a conserved mass of dry solids through three processes that change its volume, its stability and its handling properties but not its mass.

(a) Why cBOD₅, ammonia nitrogen and phosphorus are limited (15 marks)

1. cBOD₅ — the immediate oxygen debt. Biodegradable organic matter discharged to a stream is oxidised by the same heterotrophic bacteria that would have consumed it in a treatment plant, and the oxygen comes out of the water column. The result is the classical dissolved-oxygen sag, in which deoxygenation competes with reaeration: $$\frac{dD}{dt} = k_d L - k_2 D$$ where $D$ is the oxygen deficit and $L$ the remaining ultimate carbonaceous demand. If the sag drives dissolved oxygen below roughly 5 mg/L, cold-water fisheries are lost; below about 2 mg/L the reach becomes septic. The Wastewater Systems Effluent Regulations therefore set a national secondary-treatment standard of 25 mg/L average cBOD₅, matched by a 25 mg/L suspended-solids limit.
2. Quantify the loading this plant imposes. The receiving stream carries $$Q_r = 2.0\ \mathrm{m^3/s}\times 86\,400\ \mathrm{s/d} = 172\,800\ \mathrm{m^3/d}$$ so a completely mixed conservative balance at the end of the mixing zone gives $$C_{\text{mix}} = \frac{Q_r C_{\text{up}} + Q_e C_e}{Q_r + Q_e} = \frac{172\,800\times 2.0 + 10\,000\times 25}{182\,800} = \boxed{3.26\ \text{mg/L cBOD}_5}$$ Even at the regulated 25 mg/L the discharge raises the stream's organic load by 63 per cent above background, which is why a limit that looks generous at the end of the pipe is not generous at all in a small river, and why provincial permits routinely tighten the national number.
3. Ammonia nitrogen — a toxicant and a second, larger oxygen demand. The toxicity argument was made in Question 1: it is the un-ionised fraction that kills, it rises tenfold per pH unit, and the CCME guideline of 0.019 mg/L NH₃-N is easily exceeded by an effluent whose total ammonia looks modest. The Canadian regulations additionally require that the effluent not be acutely lethal to rainbow trout, and un-ionised ammonia is the single most common cause of failing that test. The second argument is stoichiometric: nitrification consumes oxygen, $$\mathrm{NH_4^{+} + 2\,O_2 \rightarrow NO_3^{-} + 2\,H^{+} + H_2O}, \qquad \frac{2\times 31.998}{14.007} = 4.57\ \frac{\mathrm{g\ O_2}}{\mathrm{g\ N}}$$ so the 20 mg N/L in this effluent carries $$\text{NOD} = 4.57\times 20 = \boxed{91.4\ \text{mg/L of oxygen demand}}$$ which is nearly four times the entire permitted carbonaceous demand. A plant can meet its cBOD₅ limit comfortably and still deoxygenate the river downstream, because the nitrogenous demand is exerted more slowly and therefore further along the reach.
4. Phosphorus — the limiting nutrient, and the oxygen demand it creates later. Phosphorus is limiting in almost all Canadian fresh waters, so it is phosphorus, not nitrogen or carbon, that decides how much algal biomass a lake can grow. The Redfield stoichiometry of algal protoplasm, $\mathrm{C_{106}H_{263}O_{110}N_{16}P}$, has a molar mass of 3553 g/mol against 30.97 g/mol of phosphorus, so $$\frac{3553}{30.97} = \boxed{114.7\ \text{g of algae per g of P}}$$ and when that biomass dies and decays it consumes, from $\mathrm{C_{106}H_{263}O_{110}N_{16}P + 138\,O_2 \rightarrow 106\,CO_2 + 16\,HNO_3 + H_3PO_4 + 122\,H_2O}$, $$\frac{138\times 31.998}{30.97} = \boxed{142.6\ \text{g of O}_2\ \text{per g of P}}$$ One gram of phosphorus therefore represents roughly 140 times the eventual oxygen demand of one gram of residual cBOD, delivered to the hypolimnion of a lake weeks after the discharge. That leverage is why effluent phosphorus limits are written at 1.0 mg/L or, in sensitive watersheds, at 0.1–0.3 mg/L, concentrations that would be absurd for any other parameter.
5. Read the three limits as three different failure modes. They are not interchangeable expressions of “cleanliness”. The cBOD₅ limit protects dissolved oxygen in the first few kilometres; the ammonia limit protects fish immediately and dissolved oxygen further downstream; and the phosphorus limit protects the trophic state of a lake or estuary over a season. A plant can satisfy any one of them while violating the intent of the others, which is why modern Canadian permits specify all three and increasingly add total nitrogen and acute-lethality testing as well.

(b) Sludge stabilisation, thickening and dewatering (10 marks)

6. Fix the mass that must be handled. The three processes change volume and character, not mass, so start from the dry solids: $$M_{\mathrm{DS}} = 1000\ \mathrm{m^3/d}\times 0.008 \times 1000\ \mathrm{kg/m^3} = \boxed{8000\ \text{kg/d of dry solids}}$$ Every step below moves this same 8000 kg/d into a smaller and less offensive volume.
7. Stabilisation — making the solids biologically inert. Raw sludge is 70–80 per cent volatile, putrescible, malodorous and rich in pathogens; stabilisation destroys the volatile fraction so the material can be stored, transported and used without nuisance or risk. Anaerobic digestion is the dominant route at municipal scale: mesophilic digesters at 35 °C with a solids retention time of 15–20 days destroy roughly half the volatile solids through hydrolysis, acidogenesis, acetogenesis and methanogenesis, and recover the energy as biogas. On this plant, with 75 per cent volatile solids and 55 per cent destruction, $$\text{VS destroyed} = 8000\times 0.75\times 0.55 = 3300\ \mathrm{kg/d}, \qquad \text{biogas} \approx 3300\times 0.95 = \boxed{3135\ \mathrm{m^3/d}}$$ at roughly 60–65 per cent methane, enough to heat the digesters and usually to co-generate as well. The alternatives are aerobic digestion (simpler, no gas recovery, higher power cost, favoured at small plants), lime stabilisation to pH 12 for two hours (fast, chemical-intensive, adds mass), and composting. In Canada the target of stabilisation is defined by biosolids quality criteria — the CCME guidance and provincial instruments such as British Columbia's Organic Matter Recycling Regulation — which set pathogen and vector-attraction-reduction requirements for Class A and Class B material.
8. Thickening — removing free water cheaply. Thickening raises the solids concentration while the sludge is still a pumpable liquid, and it is the highest-return step in the whole train because digester volume, heating energy and pumping all scale with volume. Gravity thickeners suit primary sludge; dissolved-air flotation, gravity belt thickeners and rotary drums suit the light, flocculent waste activated sludge. Conserving dry solids, $$V_2 = V_1\frac{P_1}{P_2} = 1000\times\frac{0.8}{4.0} = \boxed{200\ \mathrm{m^3/d}}$$ an 80 per cent reduction in volume for a change in solids concentration that a casual reader would call minor.
9. Dewatering — producing a solid that can be handled. Dewatering takes the stabilised sludge past the point where it behaves as a liquid and produces a spadeable cake, because haulage and land application are priced by wet tonne. Mechanical devices — belt filter presses, decanter centrifuges, plate-and-frame presses and increasingly screw presses — are preceded by chemical conditioning with a cationic polymer to build a shear-resistant floc and release bound water. At a 22 per cent cake, $$V_3 = V_1\frac{P_1}{P_3} = 1000\times\frac{0.8}{22} = \boxed{36.4\ \mathrm{m^3/d}}$$ a 96.4 per cent overall reduction from the 1000 m³/d that left the clarifiers. Note that both filtrate and centrate return to the head of the plant carrying high ammonia and phosphorus, a recycle load of 15–25 per cent of the plant's nitrogen that must be accounted for in the aeration design rather than discovered in operation.

Final results — Question 3

Quantity	Result
Receiving-stream flow	172 800 m3/d
Mixed cBOD5 below the outfall	3.26 mg/L (from 2.0 mg/L upstream)
Nitrogenous oxygen demand of 20 mg N/L	91.4 mg/L O2 (at 4.57 g O2/g N)
Algal biomass grown per gram of phosphorus	114.7 g algae/g P
Eventual oxygen demand per gram of phosphorus	142.6 g O2/g P
Dry solids in the sludge train	8000 kg/d
Volatile solids destroyed and biogas produced	3300 kg/d; 3135 m3/d
Sludge volume: raw / thickened / cake	1000 / 200 / 36.4 m3/d (96.4 % reduction)

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