Question 1 of 5: Define and Differentiate — Five Pairs from Water and Wastewater Practice

Question 1: Define and Differentiate — Five Pairs from Water and Wastewater Practice (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. The question supplies no data, so one representative data set is carried through all five parts. Each definition closes with a numerical illustration drawn from it, which is what converts a definition into a markable engineering answer rather than a recital.

Find. For each pair: a definition that says what the quantity is, the physical or chemical basis on which the two members differ, and one number from the table above that makes the distinction concrete.

1. Part (i) — Temporary versus permanent hardness.

   Hardness is the sum of the multivalent cations in solution, overwhelmingly Ca²⁺ and Mg²⁺, expressed on a common basis of mg/L as CaCO₃. The division into temporary and permanent is a division by anion, not by cation: temporary (carbonate) hardness is that portion of the hardness balanced by carbonate and bicarbonate alkalinity, and permanent (non-carbonate) hardness is the remainder, balanced by sulphate, chloride and nitrate.

   The names come from what happens on boiling. Heating drives the bicarbonate equilibrium to the right,

   $$\mathrm{Ca}^{2+} + 2\,\mathrm{HCO}_3^{-} \;\xrightarrow{\ \Delta\ }\; \mathrm{CaCO}_3(s)\downarrow + \mathrm{CO}_2\uparrow + \mathrm{H}_2\mathrm{O}$$

   so the carbonate fraction precipitates as scale and is "temporarily" removed; the sulphate and chloride salts stay dissolved and are "permanent". Converting the tabulated cations to a common basis,

   $$\text{TH} = C_{\mathrm{Ca}}\frac{M_{\mathrm{CaCO_3}}}{M_{\mathrm{Ca}}} + C_{\mathrm{Mg}}\frac{M_{\mathrm{CaCO_3}}}{M_{\mathrm{Mg}}} = 80\times\frac{100.09}{40.08} + 24\times\frac{100.09}{24.31}$$

   which gives 199.8 + 98.8 mg/L. Because the alkalinity (220) is less than the total hardness, all of the alkalinity is carbonate hardness and the balance is non-carbonate:

   $$\boxed{\ \text{TH} = 298.6,\quad \text{carbonate (temporary)} = 220,\quad \text{non-carbonate (permanent)} = 78.6\ \text{mg/L as CaCO}_3\ }$$

   The distinction is a design decision, not a curiosity: lime alone precipitates the carbonate fraction, whereas the 78.6 mg/L of non-carbonate hardness must be given its own carbonate ion, so it needs soda ash as well in a lime–soda process, or a different technology altogether (ion exchange, nanofiltration).

2. Part (ii) — Total ammonia nitrogen versus free ammonia.

   Total ammonia nitrogen (TAN) is the analytical quantity: the sum of the ionised and un-ionised species reported as nitrogen, TAN = NH₄⁺-N + NH₃-N. It is what a distillation–titration or an ion-selective electrode measures, and what a discharge permit normally names. Free (un-ionised) ammonia is only the NH₃ molecule — the neutral, lipid-soluble species that crosses gill membranes and is therefore the toxic one.

   The two are linked by the acid dissociation of the ammonium ion, which is strongly dependent on pH and temperature:

   $$\mathrm{NH}_4^{+} \rightleftharpoons \mathrm{NH}_3 + \mathrm{H}^{+},\qquad f_{\mathrm{NH_3}} = \frac{1}{1 + 10^{\,(\mathrm{p}K_a - \mathrm{pH})}}$$

   with the Emerson correlation \(\mathrm{p}K_a = 0.09018 + 2729.92/T\) for T in kelvin, giving pK_a = 9.40 at 20 °C. Substituting pH = 8.0 gives f = 0.0381, so of the 20 mg N/L reported,

   $$\boxed{\ \mathrm{NH_3\text{-}N} = 0.0381 \times 20 = 0.761\ \mathrm{mg/L}\ }$$

   That is roughly 40 times the CCME guideline of 0.019 mg/L un-ionised NH₃-N for the protection of freshwater aquatic life, even though only 3.8 % of the measured ammonia is in that form. A TAN number quoted without pH and temperature is therefore uninterpretable for toxicity.

3. Part (iii) — Self-cleansing versus scouring velocity in sewers.

   Both are lower-bound design velocities, and they differ in what they are required to move. The self-cleansing velocity is the velocity that must be reached at least once each day so that the solids arriving with the sewage never settle out in the first place; Canadian and US practice sets it at about 0.6–0.75 m/s in sanitary sewers, equivalent to a boundary shear stress of roughly 1.5 Pa on a 1 mm grit particle. The scouring velocity is the higher velocity needed to re-entrain material that has already deposited and consolidated, typically 3–4 Pa of shear, and it is what a flushing or peak-flow event must deliver. (The upper limit — about 3 m/s — is a separate criterion, set by abrasion of the pipe wall, and is a maximum, not a minimum.)

   The common physics is the tractive force. On a uniform-flow reach,

   $$\tau_0 = \rho g R S = 1000 \times 9.81 \times 0.1117 \times 0.005$$

   using the hydraulic radius R = 0.1117 m computed for this sewer in Question 5. Because Manning gives \(v \propto \sqrt{S}\) while \(\tau_0 \propto S\), velocity scales with the square root of shear, so the two criterion velocities follow from the design velocity v = 1.262 m/s directly:

   $$\boxed{\ \tau_0 = 5.48\ \mathrm{Pa};\quad v_{\text{self-cleansing}} = 0.66\ \mathrm{m/s};\quad v_{\text{scour}} = 1.08\ \mathrm{m/s}\ }$$

   The sewer of Question 5 runs at 1.26 m/s, so it clears both thresholds and would also re-suspend a deposit left by an earlier low-flow period. The practical lesson is that the self-cleansing criterion binds at minimum night flow, while the scouring criterion is checked at peak flow.

4. Part (iv) — Combined versus free residual chlorine.

   Free residual chlorine is the sum of the species formed by chlorine and water alone, HOCl and OCl⁻. Combined residual is chlorine bound to nitrogen — the chloramines — formed when ammonia is present:

   $$\mathrm{NH}_3 + \mathrm{HOCl} \rightarrow \mathrm{NH}_2\mathrm{Cl} + \mathrm{H}_2\mathrm{O} \qquad \mathrm{NH}_2\mathrm{Cl} + \mathrm{HOCl} \rightarrow \mathrm{NHCl}_2 + \mathrm{H}_2\mathrm{O}$$

   They differ in strength and in persistence, and the trade is deliberate. Free chlorine is the far stronger disinfectant — free-chlorine CT values are one to two orders of magnitude below monochloramine CT values for the same log removal — but it decays quickly and forms trihalomethanes with natural organic matter. Combined residual is weak but stable, which is why it is the residual of choice for a long distribution system and for biofilm control.

   Which one exists at the tap is decided by the dose relative to the ammonia present. Monochloramine formation consumes chlorine at the mass ratio \(M_{\mathrm{Cl_2}}/M_{\mathrm{N}} = 70.90/14.01 = 5.06\) mg Cl₂ per mg NH₃-N, and the breakpoint — where the chloramines are oxidised to nitrogen gas — occurs near 1.5 mol Cl₂ per mol N, i.e. 7.59 mg/mg. With 0.50 mg N/L present, the breakpoint demand is 3.80 mg/L, so a 5.0 mg/L dose passes it:

   $$\boxed{\ \text{dose to breakpoint} = 3.80\ \mathrm{mg/L};\quad \text{free residual at a 5.0 mg/L dose} \approx 1.20\ \mathrm{mg/L}\ }$$

   Below 3.80 mg/L the plant would leave a combined residual only, and the operator who reads "1.2 mg/L residual" without distinguishing the two forms may be reporting a residual with almost no disinfecting power.

5. Part (v) — Coagulation versus flocculation.

   Coagulation is the chemical step: destabilisation of colloids by adding a hydrolysing metal salt or polymer, which compresses or neutralises the diffuse double layer so that particles are no longer kept apart by electrostatic repulsion. It is complete in seconds, which is why it is carried out in a small, violently mixed chamber. Flocculation is the physical step that follows: gentle, sustained stirring that provides the velocity gradients needed to bring the now-destabilised particles into contact often enough to grow into settleable flocs, over 20–40 minutes.

   The Camp–Stein velocity gradient quantifies both and shows how sharply they differ:

   $$G = \sqrt{\frac{P}{\mu V}} \quad\Longrightarrow\quad P = G^2 \mu V = 700^2 \times 1.002\times10^{-3} \times 5.0$$

   A rapid mix at G = 700 s⁻¹ for 30 s and a flocculator at G = 40 s⁻¹ for 25 min give

   $$\boxed{\ P_{\text{rapid mix}} = 2.45\ \mathrm{kW};\quad Gt = 21\,000 \ \text{(mix)} \ \text{versus}\ 60\,000\ \text{(floc)}\ }$$

   The two Gt products are of the same order but arrive at that value in opposite ways — huge G for a moment, or modest G for a long time. Running a flocculator at rapid-mix intensity does not make bigger flocs; it shears them apart, because floc strength sets a ceiling on G of roughly 70 s⁻¹.