Question 2 of 5: pH and disinfection; municipal flow projection

Question 2: pH and disinfection; municipal flow projection (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. Part (a): a free-chlorine residual in a water whose pH is the operating variable, at 20 °C. Part (b): population $P = 5\,000$; average daily water demand $q = 450$ L per capita per day; all other quantities to be assumed and stated.

Find. (a) The chemistry by which pH governs the germicidal strength of a free chlorine residual, and its practical consequence for operating dose; (b) the average and the peak wastewater flow.

Part (a) — Approach. Write the two reactions chlorine undergoes in water, identify which product does the disinfecting, evaluate the acid–base equilibrium across the operating pH range, and convert the speciation into the quantity an operator actually controls, namely $CT$.

1. Part (a) — hydrolysis puts chlorine into the water as hypochlorous acid. Gaseous chlorine hydrolyses essentially completely above pH 3: $$\text{Cl}_2+\text{H}_2\text{O}\rightarrow\text{HOCl}+\text{H}^{+} +\text{Cl}^{-}$$ and sodium hypochlorite arrives at the same place from the other side, $\text{NaOCl}+\text{H}_2\text{O}\rightarrow\text{HOCl}+\text{Na}^{+} +\text{OH}^{-}$. Note already that gas chlorine depresses pH while hypochlorite raises it, so the choice of chemical is itself a pH decision in a poorly buffered water.
2. The acid–base equilibrium is the whole answer. Hypochlorous acid is weak and dissociates to the hypochlorite ion: $$\text{HOCl}\rightleftharpoons\text{H}^{+}+\text{OCl}^{-},\qquad K_a=\frac{[\text{H}^{+}][\text{OCl}^{-}]}{[\text{HOCl}]},\qquad \mathrm{p}K_a=7.54\ \text{at}\ 20\,{}^{\circ}\text{C}$$ Both species together are the free available chlorine residual, but they are not equally effective: the neutral HOCl molecule crosses the cell membrane, whereas the negatively charged OCl$^{-}$ ion is repelled by the negatively charged cell wall and is roughly 80 times weaker. Rearranging, $$\alpha_{\text{HOCl}}=\frac{[\text{HOCl}]}{[\text{HOCl}]+[\text{OCl}^{-}]} =\frac{1}{1+10^{\,\mathrm{pH}-\mathrm{p}K_a}}$$
3. Evaluate the speciation across the operating range. Substituting pH values into the expression above gives the germicidal share of the residual: $$\alpha_{\text{HOCl}}(7.0)=\frac{1}{1+10^{-0.54}}=\frac{1}{1.2884} =0.7762$$ and, repeating, $$\boxed{\alpha_{\text{HOCl}}=97.2,\ 77.6,\ 52.3,\ 25.7,\ 3.4\ \text{per cent at pH}\ 6.0,\ 7.0,\ 7.5,\ 8.0,\ 9.0}$$ A drift of one pH unit from 7 to 8 therefore destroys two-thirds of the active species even though the measured total residual has not moved.
Free-chlorine speciation against pH at 20 °C (pK_a = 7.54). The germicidal species HOCl collapses from 97 per cent at pH 6 to 3 per cent at pH 9.
4. Convert speciation into required contact time. Weighting the two species at an 80:1 potency ratio gives a germicidal-equivalent factor $\Phi=\alpha+(1-\alpha)/80$, and holding the kill constant means holding $\Phi\,CT$ constant: $$\frac{CT_{\mathrm{pH}\,8.0}}{CT_{\mathrm{pH}\,7.0}} =\frac{\Phi(7.0)}{\Phi(8.0)}=\frac{0.7790}{0.2667}=\boxed{2.92}$$ So the same log-inactivation at pH 8.0 needs nearly three times the residual, three times the contact tank, or some combination of the two. This is the number that matters operationally, and it is slightly smaller than the raw HOCl ratio of 3.01 precisely because OCl$^{-}$ is weak rather than inert.
5. Consequences worth stating. Temperature moves p$K_a$ (7.54 at 20 °C, about 7.75 at 5 °C), so cold Canadian water actually holds slightly more HOCl at a given pH, partly offsetting the slower kinetics. A lime-softened water leaving at pH 10 has almost no germicidal capacity in its free residual and must be recarbonated before the contact tank, not after. Conversely, raising pH to about 8.3 for lead and copper corrosion control — now common practice in Canada — buys distribution-system benefits at a real disinfection cost that has to be paid back in $CT$. A chloraminating plant runs the argument in reverse, deliberately trading germicidal power for residual stability and lower THM formation.

Part (b) — Approach. Convert per-capita water demand to an average daily wastewater flow through a stated return factor, apply the Harmon peaking factor to get peak dry-weather flow, add a stated extraneous allowance for infiltration and inflow, and check the low end with the Babbitt minimum factor.

1. Part (b) — average daily water demand. $$Q_{\text{water}}=\frac{P\,q}{1000}=\frac{5000\times 450}{1000} =2250\ \text{m}^3/\text{d}$$
2. Convert to average daily wastewater flow. Not all water supplied reaches the sewer: lawn watering, car washing, leakage and process losses are consumed. Taking a return factor $f_r=0.80$, $$Q_{\text{avg}}=f_r\,Q_{\text{water}}=0.80\times 2250 =\boxed{1800\ \text{m}^3/\text{d}\ (20.8\ \text{L/s})}$$
3. Peaking factor and peak dry-weather flow. With $P=5$ thousand, the Harmon factor evaluated in Question 1 is 3.245, so $$Q_{\text{peak,DWF}}=\mathrm{PF}\times Q_{\text{avg}}=3.245\times 1800 =\boxed{5841\ \text{m}^3/\text{d}\ (67.6\ \text{L/s})}$$
4. Add the extraneous flow the pipe must also carry. Sanitary sewers leak inward. Taking a peak infiltration and inflow allowance of 90 L per capita per day, $$Q_{\text{I/I}}=\frac{5000\times 90}{1000}=450\ \text{m}^3/\text{d}, \qquad Q_{\text{design}}=5841+450 =\boxed{6291\ \text{m}^3/\text{d}\ (72.8\ \text{L/s})}$$ This, not the sanitary peak, is the flow the collection system is sized on.
5. Check the low end, because velocity governs there. The Babbitt minimum-flow relation gives $$\frac{Q_{\min}}{Q_{\text{avg}}}=0.2\,P^{0.2}=0.2\times 5^{0.2}=0.276, \qquad Q_{\min}=0.276\times 1800=497\ \text{m}^3/\text{d}\ (5.8\ \text{L/s})$$ The design flow is therefore 12.7 times the minimum. A sewer sized on 72.8 L/s runs at a small fraction of its depth at 5.8 L/s, which is precisely the condition in which solids deposit — hence the rule that the grade must still deliver about 0.6 m/s, or a boundary shear of roughly 1.5 Pa, at minimum flow.