Question 2 of 5: Oxygen-Demand Definitions and Water-Quality Parameters

Question 2: Oxygen-Demand Definitions and Water-Quality Parameters (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) BOD₅, cBOD₅ and ultimate BOD (10 marks)

All three are measured in the same bottle test — a diluted, seeded, sealed sample incubated in the dark at 20 °C, with dissolved oxygen measured before and after — and they differ only in what is counted and for how long.

BOD₅ is the oxygen consumed in five days by an uninhibited sample. It is an arbitrary but universal index: five days was chosen historically because it approximates the travel time of British rivers to the sea, and it has survived because the whole regulatory apparatus is calibrated to it. Its defect is that it counts everything that respires, including nitrifying bacteria if they happen to be active, so a well-nitrified effluent can report a BOD₅ far higher than its true organic content.

cBOD₅ removes that ambiguity. A nitrification inhibitor — typically allylthiourea or 2-chloro-6-(trichloromethyl)pyridine — is added so that only the carbonaceous demand is exerted. This is now the standard regulatory parameter for treated effluent in Canada and the United States, precisely because a plant that does its job well and nitrifies should not be penalised for it. For raw sewage, where nitrifiers are scarce and outcompeted, BOD₅ and cBOD₅ are practically identical.

Ultimate BOD, \(L_0\) or BOD_u, is the total oxygen the sample would consume if the reaction ran to completion. It is not measured directly but extracted from the first-order model \(y_t = L_0(1 - e^{-kt})\) by fitting a series of incubation times. Ultimate BOD is the quantity that actually belongs in a receiving-water model, because the river does not stop at five days; the sag calculation in Question 1 used \(L_0\), not BOD₅.

Given. A settled sewage with \(\text{cBOD}_5 = 200\) mg/L, first-order rate constant \(k = 0.23\ \text{d}^{-1}\) (base e, 20 °C), and TKN of 25 mg N/L.

Find. The carbonaceous ultimate BOD, the ultimate nitrogenous demand, and what fraction of the true total demand a five-day test actually captures.

1. Carbonaceous ultimate BOD. Inverting the first-order model, \[L_0 = \frac{y_5}{1 - e^{-k(5)}} = \frac{200}{1 - e^{-1.15}} = \frac{200}{0.683} \;\Longrightarrow\; \boxed{\ L_0 = 293\ \text{mg/L}\ }\] so the five-day test recovers only 68 per cent of the carbonaceous demand.
2. Ultimate nitrogenous demand. At 4.57 g O₂ per g of nitrogen oxidised, \(\text{NOD}_u = 4.57(25) = 114\) mg/L.
3. Total ultimate demand. \(293 + 114 = 407\) mg/L, of which nitrification is 28 per cent. A five-day cBOD result of 200 mg/L therefore represents under half of the oxygen this wastewater will ultimately demand of a receiving water.

(b) Fluorides, nitrates, sulfates and hardness (15 marks)

Fluoride is the parameter with a genuine optimum rather than a maximum. Below roughly 0.5 mg/L, dental caries rates rise measurably; between about 0.7 and 1.2 mg/L the caries benefit is realised without cosmetic harm, which is why Health Canada gives an optimal target near 0.7 mg/L for communities that choose to fluoridate; above roughly 1.5 mg/L, the maximum acceptable concentration, prolonged exposure causes dental fluorosis and at much higher concentrations skeletal fluorosis. Natural fluoride is common in granitic and volcanic groundwaters, so in a Canadian context the design question is usually removal — by activated alumina, bone char, or reverse osmosis — and not addition. Fluoride is significant because it is one of the very few drinking-water parameters where both too little and too much are defects.

Nitrate matters for one acute reason and one chronic one. In infants under six months, nitrate reduced to nitrite in the gut oxidises haemoglobin to methaemoglobin, which cannot carry oxygen — methaemoglobinaemia, or blue-baby syndrome. The Health Canada maximum acceptable concentration is 45 mg/L expressed as nitrate, which is exactly 10 mg/L expressed as nitrogen: converting, \(10 \times 62.0/14.0 = 44.3 \approx 45\) mg/L, and confusing the two bases by the factor 4.43 is the single commonest error in reporting nitrate. Chronically, elevated nitrate is a reliable indicator of agricultural or septic-field contamination of an aquifer, and it signals that the same source may be delivering pathogens and pesticides. Nitrate is not removed by conventional treatment; it requires ion exchange, reverse osmosis, or biological denitrification.

Sulfate is regulated on aesthetics and comfort rather than toxicity. Above roughly 500 mg/L it imparts a bitter taste and acts as a laxative on unacclimatised consumers, giving the aesthetic objective. Its engineering significance is larger than its health significance: sulfate attacks Portland cement through the formation of expansive ettringite, so high-sulfate groundwater dictates sulfate-resisting cement in buried structures; in sewers and anaerobic digesters, sulfate-reducing bacteria convert it to hydrogen sulfide, which is odorous, toxic, and oxidises on the crown of the pipe to sulfuric acid, causing crown corrosion. High sulfate also interferes with the taste of the water and can accelerate lead and copper release.

Hardness, the sum of the multivalent cations expressed as CaCO₃ and dominated by calcium and magnesium, has no health-based guideline — there is some epidemiological evidence that hard water is mildly protective against cardiovascular disease. Its significance is economic and operational. Hard water consumes soap, forms curd, and deposits scale in kettles, water heaters, boilers and heat exchangers, where a millimetre of calcium carbonate can cost several per cent in heat-transfer efficiency. Very soft water, on the other hand, is aggressive and corrodes distribution piping, dissolving lead and copper. The practical optimum for a municipal supply is around 80–100 mg/L as CaCO₃, achieved by softening a hard source and then blending back.

Given. A groundwater with 80 mg/L Ca²⁺ and 25 mg/L Mg²⁺.

Find. The total hardness as CaCO₃ and its classification.

1. Convert each cation. Multiplying by the ratio of equivalent weights, \(50.04/20.04 = 2.497\) for calcium and \(50.04/12.15 = 4.118\) for magnesium, \[\text{TH} = 80(2.497) + 25(4.118) = 199.8 + 102.9 \;\Longrightarrow\; \boxed{\ \text{TH} = 303\ \text{mg/L as CaCO}_3\ }\]
2. Classify. On the conventional scale — soft below 60, moderately hard 60–120, hard 120–180, very hard above 180 mg/L — this water is very hard and would be softened before distribution.