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16-Civ-B5 Water Supply and Wastewater Treatment · December 2019

Question 2 of 5: pH and Disinfection Efficiency; Indicator Organisms

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

Notes on this paper

Paper format. National Examination, December 2019 — 16-Civ-B5 Water Supply and Wastewater Treatment. Three hours; closed book with one aid sheet written on both sides; an approved Casio or Sharp calculator is permitted. Question 1 is compulsory; attempt any three of the remaining four. All five questions carry 25 marks, so the paper is marked out of 100. Every question is solved here, because the complete set is the study resource.

Reference texts for this subject. Metcalf & Eddy / Tchobanoglous, Stensel, Tsuchihashi & Burton, Wastewater Engineering: Treatment and Resource Recovery, 5th ed. (McGraw-Hill) — the primary reference for Q1(ii)–(iv), Q3 and Q5. Crittenden et al., MWH's Water Treatment: Principles and Design, 3rd ed. (Wiley) — coagulation, disinfection and filtration for Q1(i), Q1(v), Q2 and Q4(a). Davis, Water and Wastewater Engineering: Design Principles and Practice (McGraw-Hill) and Mihelcic & Zimmerman, Environmental Engineering: Fundamentals, Sustainability, Design (Wiley) — distribution systems and sewer hydraulics for Q4(b) and Q5. Canadian regulatory frame: Guidelines for Canadian Drinking Water Quality (Health Canada), the Canadian Environmental Quality Guidelines (CCME) for ammonia, and the federal Wastewater Systems Effluent Regulations (SOR/2012-139).

Check — assumptions declared once, used throughout. The paper supplies no water-quality data for Q1–Q4, so every illustrative number below is the solver's own representative value, clearly labelled where it is introduced; the marks lie in the definitions and the reasoning, and the numbers are there to make each distinction concrete. Free-chlorine speciation uses pKa = 7.54 at 25 °C; ammonia speciation uses the Emerson relation. Q5 is solved from the supplied partial-flow curves and independently from the exact circular-segment geometry.

Question 2: pH and Disinfection Efficiency; Indicator Organisms (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.

Part (a) — pH and disinfection efficiency (15 marks)

When chlorine gas is added to water it hydrolyses essentially completely and then dissociates as a weak acid:

$$\mathrm{Cl_2} + \mathrm{H_2O} \rightleftharpoons \mathrm{HOCl} + \mathrm{H^+} + \mathrm{Cl^-}$$ $$\mathrm{HOCl} \rightleftharpoons \mathrm{H^+} + \mathrm{OCl^-}, \qquad K_a = 2.9\times10^{-8}\ \text{at }25\,{}^{\circ}\mathrm{C},\quad \mathrm{p}K_a = 7.54$$

The sum HOCl + OCl− is the free available chlorine, and analytical methods report that sum. But the two species are not equally effective. Hypochlorous acid is a small, electrically neutral molecule that diffuses through the lipid cell wall and oxidises intracellular enzymes and nucleic acids; the hypochlorite ion carries a negative charge, is repelled by the negatively charged cell surface, and is therefore roughly 80 to 100 times less germicidal. Disinfection efficiency at a fixed measured residual is thus governed by how the free chlorine is speciated, and speciation is governed entirely by pH.

The fraction present as HOCl follows directly from the dissociation equilibrium:

$$\alpha_{\mathrm{HOCl}} = \frac{[\mathrm{HOCl}]}{[\mathrm{HOCl}]+[\mathrm{OCl^-}]} = \frac{1}{1+10^{(\mathrm{pH}-\mathrm{p}K_a)}}$$
56789100.00.20.40.60.81.0pKa = 7.54pH 7.3: 63.5 % HOClpH 8.5: 9.9 % HOClpHFraction of free chlorineHOCl (hypochlorous acid)OCl⁻ (hypochlorite ion)Free-chlorine speciation, 25 °C
Figure 2.1 — Free-chlorine speciation against pH. At the plant's original pH of 8.5 only about a tenth of the measured free chlorine is the potent HOCl species; lowering the pH to 7.3 moves the operating point to the steep part of the curve, where nearly two-thirds is HOCl.

Given. Post-filtration water whose pH is reduced from 8.5 to 7.3; free-chlorine pKa = 7.54 at 25 °C; HOCl taken as 80 times as germicidal as OCl−; the plant currently applies 2.0 mg/L of free chlorine to achieve its target CT. Find. The predicted effect on the chlorine dose required for the same level of inactivation, and the reason for it.

Approach. Evaluate the HOCl fraction at each pH, convert the two fractions into a germicidal-equivalent residual using the potency ratio, and rescale the dose so that the product of germicidal-equivalent concentration and contact time is unchanged.

  1. Speciation at the original pH. $\alpha_{8.5} = 1/(1+10^{(8.5-7.54)}) = 1/(1+9.12) = 0.099$ — only 9.9 per cent of the measured free chlorine is HOCl.
  2. Speciation at the new pH. $\alpha_{7.3} = 1/(1+10^{(7.3-7.54)}) = 1/(1+0.575) = 0.635$ — now 63.5 per cent is HOCl. The HOCl concentration at a given total residual therefore rises by a factor of $0.635/0.099 = 6.4$.
  3. Germicidal-equivalent residual. Crediting the hypochlorite ion with 1/80 of the potency of HOCl, the effective residual per unit of measured free chlorine is $\alpha + (1-\alpha)/80$: it is 0.110 at pH 8.5 and 0.639 at pH 7.3. Dividing, the same measured residual is now $$\boxed{5.8\ \text{times as germicidal at pH 7.3 as at pH 8.5}}$$
  4. Dose required for the same inactivation. Holding CT and contact time constant, the required applied free chlorine falls in the same proportion: $$C_{7.3} = C_{8.5}\times\frac{0.110}{0.639} = 2.0 \times 0.172 = 0.34\ \text{mg/L}$$ so on disinfection grounds alone the dose could be cut to about one-sixth. In practice the dose is set by chlorine demand plus the residual to be maintained, so the realistic prediction is a substantially reduced dose — commonly 30 to 50 per cent less — or, at unchanged dose, a large increase in the safety margin on Giardia and virus inactivation.

The same conclusion appears in the regulatory CT tables, which is the form an operator would cite: the tabulated CT for 3-log Giardia inactivation rises by roughly 50 per cent for each unit increase in pH across this range, and the tables stop at pH 9 precisely because free chlorine has almost ceased to work by then. Three consequences should be stated with the answer. Lowering pH raises the corrosivity of the finished water, so the reduction has to be reconciled with the corrosion-control target and the lead-and-copper requirements — often by re-raising pH after the contact tank. Lower pH also shifts THM formation modestly and increases haloacetic acid formation. And where the plant chloraminates, the argument does not apply at all: monochloramine formation and stability improve at higher pH, so a chloraminating plant deliberately runs the opposite way.

Question 2(a) — effect of lowering post-filtration pH from 8.5 to 7.3
QuantitypH 8.5pH 7.3Change
HOCl fraction of free chlorine9.9 per cent63.5 per cent× 6.4
Germicidal-equivalent residual factor0.1100.639× 5.8
Free-chlorine dose for equal inactivation2.0 mg/L0.34 mg/L− 83 per cent

Part (b) — Indicator organisms in the biological testing of water (10 marks)

An indicator organism is a micro-organism whose presence in water is taken as evidence that faecal contamination has occurred and that enteric pathogens may therefore be present. Its significance is entirely practical. The pathogens that matter — Salmonella, Campylobacter, enteric viruses, Cryptosporidium, Giardia — are numerous in kind, are present intermittently and in small numbers, and are individually difficult, slow and expensive to culture. Testing a distribution sample for each of them would be impossible as a routine. An indicator collapses that impossible battery into one cheap, fast, standardised test that can be run daily on every sample, and a utility's entire microbiological compliance programme rests on it.

The basis of selection is a set of criteria that any candidate organism must satisfy. It should be present whenever the pathogens are present and absent when they are absent, which restricts the choice to organisms of exclusively or predominantly faecal origin. It should be more numerous than the pathogens, so that the indicator is detected before the pathogen becomes a risk — there are of the order of 109 coliforms per gram of human faeces. It should be at least as persistent as the pathogens in the aquatic environment and at least as resistant to the treatment barriers applied, so that its removal is a conservative proxy for theirs. It must not multiply in the water, or its density would no longer be a measure of contamination. It should be harmless to the analyst, and detectable by a simple, rapid, inexpensive and reproducible method applicable to all types of water. Finally, its density should bear some quantitative relationship to the degree of contamination.

No single organism satisfies every criterion, which is why practice uses a small hierarchy. Total coliforms are the broadest group; because some members are environmental rather than faecal, a positive is treated in the Guidelines for Canadian Drinking Water Quality as a signal to investigate the integrity of the system rather than as proof of faecal contamination, and the guideline is none detectable per 100 mL. Escherichia coli is essentially exclusively faecal and is the definitive indicator of recent faecal contamination in drinking water, with the same guideline of none detectable per 100 mL; it is the parameter on which a boil-water advisory is issued. Enterococci survive longer in saline and cold water and are the preferred indicator for marine recreational waters. Clostridium perfringens spores, being chlorine-resistant and long-lived, act as a conservative surrogate for protozoan cysts and oocysts, and coliphages serve the same role for enteric viruses. The limitation of the whole approach must also be stated: Cryptosporidium is far more resistant to chlorine than E. coli is, so an absence of E. coli does not demonstrate the absence of oocysts — which is why the multi-barrier approach adds turbidity and filter performance as continuously monitored surrogates rather than relying on the indicator alone.