16-Civ-B5 Water Supply and Wastewater Treatment · May 2018
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Paper format. National Examination, May 2018 — 16-Civ-B5 Water Supply and Wastewater Engineering. Three hours, closed book, one two-sided aid sheet and an approved Casio or Sharp calculator permitted. Question 1 is compulsory and candidates attempt any three of Questions 2–5; every question carries 25 marks. Marks are shown at the end of each question and the paper explicitly invites candidates to state any assumptions they make. All five questions are worked below, because the set is a study resource rather than a three-hour sitting.
Reference texts.
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. No data are supplied; each part is a definition worth five marks. To show that the definitions are operational rather than verbal, a small representative calculation is attached to each one, using the values stated in that part.
Find. A defensible definition of each parameter or process, and a statement of why a plant operator or designer cares about it.
Approach. Define the quantity, give its governing expression, work one illustrative number, and close with the operating decision the number drives.
Backwashing is the periodic reversal of flow through a granular medium filter that fluidises the bed and carries the accumulated floc, precipitated iron and manganese and biological growth out of the filter to waste. During a normal filter run, solids are stored within the depth of the bed; head loss climbs and, eventually, either the available driving head is exhausted or the stored solids begin to break through into the filtrate. Backwashing restores the clean-bed condition so the next run can begin. A complete wash sequence on a rapid sand filter is normally air scour first (at sub-fluidisation water rates, so that grain-to-grain abrasion can scrub the attached material off the media), then air plus water, then a high-rate water wash at 0.5–0.9 m/min that expands the bed by 20–50 per cent and carries the released solids to the wash-water troughs.
Its significance is threefold. It is what makes granular filtration a sustainable unit process rather than a disposable one; it consumes product water, so the wash regime is a direct operating cost; and it governs filtrate quality immediately after the wash, because a freshly washed bed passes a turbidity spike until it ripens. Taking a filter loaded at 5 m/h for a 48 h run and washed at 0.6 m/min for 10 minutes, the wash water is
$$\text{wash fraction}=\frac{v_{bw}t_{bw}}{v_f t_{run}} =\frac{0.6\times 10}{5\times 48}=\boxed{2.5\ \%\ \text{of production}}$$which is the number a designer uses to size the wash-water equalisation tank and to decide whether recovering the spent wash water to the head of the works is worth the recycled solids load it brings with it.
The food-to-microorganism ratio is the mass of biodegradable organic matter applied per day divided by the mass of active biomass held in the aeration tank,
$$\frac{F}{M}=\frac{Q\,S_0}{V\,X_v} \qquad\left[\frac{\text{kg BOD}_5/\text{d}}{\text{kg MLVSS}}\right]$$where Q is the influent flow, S0 the influent BOD, V the aeration-tank volume and Xv the mixed-liquor volatile suspended solids. It is the loading parameter that sets which region of the growth curve the culture is operating in. For a plant treating 10 000 m3/d at 200 mg/L BOD5 in a 2500 m3 tank at 3000 mg/L MLVSS, the applied food is 2000 kg BOD/d against 7500 kg of biomass, so
$$\frac{F}{M}=\frac{2000}{7500}=\boxed{0.27\ \text{kg BOD}_5\,\text{kg MLVSS}^{-1}\,\text{d}^{-1}}$$which sits in the conventional range of 0.2–0.4 and predicts a well-flocculated, readily settling sludge. Its significance is that both extremes misbehave in recognisable ways: a high F/M starves the culture of contact time, leaves soluble BOD in the effluent and encourages dispersed growth, while a very low F/M (extended aeration, below about 0.05) produces a highly mineralised but pin-floc sludge that can carry solids over the clarifier weirs. F/M is the inverse companion of sludge age, and in practice operators control it by wasting sludge to hold MLVSS at target.
Mixed liquor suspended solids (MLSS) is the total suspended-solids concentration of the aerated mixture of incoming wastewater and recycled biomass in the aeration tank, typically 2000–4000 mg/L in a conventional plant. Its volatile fraction (MLVSS, usually 70–80 per cent of MLSS) is the surrogate for active biomass. Return activated sludge (RAS) is the thickened sludge continuously pumped from the secondary clarifier underflow back to the head of the aeration tank; it is what maintains MLSS at a concentration far above anything the influent could supply, and so decouples the solids retention time from the hydraulic retention time.
The two are linked by a solids balance on the clarifier. Neglecting the solids leaving over the weirs, everything entering the clarifier must leave in the underflow, so
$$(Q+Q_R)X=Q_R X_R\quad\Rightarrow\quad R=\frac{Q_R}{Q}=\frac{X}{X_R-X}$$For MLSS of 3000 mg/L and a RAS concentration of 8000 mg/L, the required recycle ratio is R = 3000/(8000 − 3000) = 0.60, so a 10 000 m3/d plant returns
$$Q_R=R\,Q=0.60\times 10\,000=\boxed{6000\ \text{m}^3/\text{d}}$$Their significance is operational: MLSS is the knob an operator turns (through wasting) to set treatment capacity, and the RAS rate is the knob that keeps the clarifier sludge blanket at a workable depth. Return too little and the blanket rises until solids escape; return too much and the clarifier is hydraulically overloaded by its own recycle while the RAS thins out.
Break-point chlorination is the practice of applying free chlorine beyond the dose at which all ammonia and oxidisable nitrogenous matter have been destroyed, so that any further chlorine appears as a free available residual. Chlorine added to a water containing ammonia first reacts to form chloramines,
$$\mathrm{NH_3+HOCl\rightarrow NH_2Cl+H_2O}$$and the measured residual rises as combined chlorine. Past a Cl2:N mass ratio of roughly 5:1 the added chlorine begins to oxidise the chloramines it has just made, and the residual falls,
$$\mathrm{2\,NH_3+3\,HOCl\rightarrow N_2\uparrow+3\,H^{+}+3\,Cl^{-}+3\,H_2O}$$The minimum in the residual curve is the break-point. The stoichiometry of the destruction reaction fixes where it falls:
$$\frac{\text{Cl}_2}{\text{NH}_3\text{-N}} =\frac{3\times 70.91}{2\times 14.01}=7.59\ \frac{\text{g Cl}_2}{\text{g N}} \quad\Rightarrow\quad \text{dose}=7.59\times 2.0=\boxed{15.2\ \text{mg/L}}$$for a water carrying 2.0 mg/L of ammonia nitrogen (side reactions and other chlorine demand push the practical dose to roughly 8–10 g Cl2 per gram of N). Its significance is that free chlorine is a far stronger and faster disinfectant than the combined residual, and that the ammonia and the taste-and-odour producing chloro-organics are destroyed in the process; the cost is a large chemical dose and an elevated risk of forming trihalomethanes from any natural organic matter still present.
The Sludge Volume Index is the volume in millilitres occupied by one gram of suspended solids after a mixed-liquor sample has settled quiescently for 30 minutes in a one-litre cylinder,
$$\mathrm{SVI}=\frac{SV_{30}\,(\text{mL/L})\times 1000}{\mathrm{MLSS}\,(\text{mg/L})} \qquad[\text{mL/g}]$$For a mixed liquor at 3000 mg/L that settles to 250 mL/L,
$$\mathrm{SVI}=\frac{250\times 1000}{3000}=\boxed{83\ \text{mL/g}}$$Values below about 100 mL/g indicate a dense, well-settling sludge; 100 to 150 is acceptable; and above roughly 150 mL/g the sludge is bulking, usually because filamentous organisms are dominant. Its significance is that SVI is the single cheapest early-warning test in an activated-sludge plant. A rising SVI predicts a rising sludge blanket and eventual solids carry-over days before the effluent turbidity shows it, and it also caps the achievable RAS concentration — the underflow cannot be thickened much beyond roughly 106/SVI mg/L, which for an SVI of 83 is about 12 000 mg/L.
| Part | Parameter | Representative calculation | Result |
|---|---|---|---|
| i | Backwash water use | (0.6 × 10) / (5 × 48) | 2.5 % of production |
| ii | F/M ratio | 2000 kg BOD/d ÷ 7500 kg MLVSS | 0.27 kg BOD kg−1 d−1 |
| iii | RAS flow | R = 3000/(8000 − 3000) = 0.60 | 6000 m3/d |
| iv | Break-point dose | 7.59 g Cl2/g N × 2.0 mg/L | 15.2 mg/L Cl2 |
| v | SVI | 250 × 1000 / 3000 | 83 mL/g (good settling) |
Check: the five illustrative datasets are the solver's own representative values — the examination question supplies no numbers. They are included because a five-mark "define and describe the significance" answer is stronger when the definition is shown to produce a number an operator would act on. The break-point ratio of 7.59 g Cl2/g N is the theoretical stoichiometric value; texts quote 7.6:1 theoretical and 8–10:1 in practice.
