Question 3 of 5: Distribution systems; load-based effluent limits

Question 3: Distribution systems; load-based effluent limits (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) — requirements of an adequate distribution system. A distribution system is judged against six requirements, and a design that fails any one of them fails as a whole.

Quantity and pressure. It must deliver the peak hourly domestic demand, or the maximum daily demand plus the fire flow, whichever is greater, while holding a residual pressure of roughly 275 to 550 kPa in normal service and not less than about 140 kPa anywhere during a fire draw. Quality. It must preserve the quality of water leaving the plant: a disinfectant residual to the far point, no cross-connection or backflow path, materials and linings that neither leach nor foster biofilm, and short enough residence times that nitrification and DBP formation do not develop. Reliability. No single break, valve failure or power outage should leave a district unsupplied; that means redundant paths, sectionalising valves close enough together that a repair isolates one block rather than one district, and storage sized for equalisation, fire and emergency. Hydraulic integrity. Velocities of roughly 0.6 to 1.5 m/s, controlled headloss, and no dead legs in which water ages. Operability and maintainability. Hydrants, blow-offs, air valves, meters, and district metered areas that make leakage detectable. Economy and expandability. Least life-cycle cost, with a grid that can be extended without re-laying trunk mains.

The grid-iron (looped or interlaced) layout carries mains in both directions around each block so every node is fed from at least two sides. Its advantages are directly the requirements above: water can reach any point by alternative routes, so a break or a repair isolates a short length instead of a district; flow arrives at a node from two directions, so headloss and the required pipe diameter are lower for a given demand; hydrant flows are higher and better sustained; and continuous circulation prevents the stagnation that costs a disinfectant residual and generates taste, odour and coliform regrowth. Its disadvantages are cost and complexity: more pipe, many more valves and fittings per unit of service area, and a network that cannot be solved by inspection, so Hardy Cross or a hydraulic model is needed to determine flows and pressures. Leak location is also harder, because a leak is fed from several directions at once.

The dead-end (branching or tree) system runs a trunk with submains and laterals terminating in closed ends. Its advantages are economy and simplicity: the least pipe length and the fewest valves, flows that are determinate by simple addition so the design can be done by hand, and a natural fit to ribbon development along a highway or valley. Its disadvantages are serious in a modern setting: every terminal is stagnant, so the residual decays, sediment accumulates and water quality complaints concentrate at the ends; a single break cuts off everything downstream of it; fire flow at a terminal is limited by one pipe; and the system needs a routine flushing programme, which wastes treated water. In Canadian practice the grid is the default for a built-up service area, dead ends are tolerated only for short cul-de-sacs and in early phases of a subdivision, and where they are unavoidable they are fitted with automatic flushing devices and looped as soon as the next phase is built.

Part (b) — Given.

Find. The cBOD$_5$, TP and TAN concentration limits that apply at 15 000 m$^3$/d, and the consequence for the aeration system.

Approach. Every condition in this question is written on mass per day, so the whole calculation is one conversion applied three times: compute the permitted load at the present rated capacity, hold it, and divide it back out over the expanded flow.

1. Part (b) — convert the present concentration limits into the loads the permit actually protects. With $C$ in mg/L and $Q$ in m$^3$/d, and using 1 mg/L = 1 g/m$^3$, $$L=\frac{C\,Q}{1000}$$ $$L_{\text{BOD}}=\frac{10\times 12000}{1000}=120\ \text{kg/d},\qquad L_{\text{TP}}=\frac{0.5\times 12000}{1000}=6.0\ \text{kg/d}$$ The TAN limit needs no conversion: the regulator has stated it directly as 60 kg/d.
2. Hold the load and re-express it at the expanded flow. Inverting the same relation, $$C_2=\frac{1000\,L}{Q_2}$$ $$C_{2,\text{BOD}}=\frac{1000\times 120}{15000}=\boxed{8.0\ \text{mg/L cBOD}_5}$$ $$C_{2,\text{TP}}=\frac{1000\times 6.0}{15000}=\boxed{0.40\ \text{mg/L TP}}$$ $$C_{2,\text{TAN}}=\frac{1000\times 60}{15000}=\boxed{4.0\ \text{mg/L TAN}}$$
3. Check the structure of the answer. Because the load is held, every concentration limit is divided by the capacity ratio $$\frac{Q_2}{Q_1}=\frac{15000}{12000}=1.25,\qquad \frac{C_1}{C_2}=\frac{10}{8.0}=\frac{0.5}{0.40}=1.25$$ so the two derived limits must tighten by exactly 20 per cent and nothing else. The classic wrong answer moves the concentration in the same direction as the flow and reports 12.5 mg/L cBOD$_5$ — a relaxation of the permit in the name of a condition written to prevent one. The plant is being allowed more water, not more pollution.
4. Comment on aeration — carbonaceous demand first. Assume a raw cBOD$_5$ of 200 mg/L and TKN of 40 mg/L, $Y=0.40$ kg VSS per kg BOD$_5$, $k_d=0.06$ d$^{-1}$, an SRT of 10 d and $f=\text{BOD}_5/\text{BOD}_u =0.68$. The carbonaceous demand after expansion is $$\Delta S\,Q=\frac{(200-8.0)\times 15000}{1000}=2880\ \text{kg cBOD}_5/\text{d}, \qquad P_{x,\text{bio}}=\frac{Y\,\Delta S\,Q}{1+k_d\theta_c} =\frac{0.40\times 2880}{1.6}=720\ \text{kg VSS/d}$$ $$R_{o,\text{C}}=\frac{\Delta S\,Q}{f}-1.42P_{x,\text{bio}} =\frac{2880}{0.68}-1.42\times 720=4235-1022=3213\ \text{kg O}_2/\text{d}$$
5. The TAN limit is what changes the plant. Nitrogen leaving in the cell mass is about 0.12 g N per g VSS, so $$N_{\text{ox}}=\frac{(40-4.0)\times 15000}{1000}-0.12\times 720 =540-86.4=453.6\ \text{kg N/d}$$ and at the stoichiometric 4.57 g O$_2$ per g N, $$R_{o,\text{N}}=4.57\times 453.6=2073\ \text{kg O}_2/\text{d},\qquad R_o=3213+2073=\boxed{5286\ \text{kg O}_2/\text{d}}$$ Against the present duty — carbon only, at 12 000 m$^3$/d and a 10 mg/L limit, giving 2544 kg O$_2$/d — the requirement rises by a factor of $$\frac{5286}{2544}=\boxed{2.08}$$ while the flow rises only 1.25. That is the answer to the question asked: the aeration system must roughly double, and almost none of the increase is the extra hydraulic capacity. At a field transfer of 2.0 kg O$_2$/kWh the blower duty goes from about 53 kW to about 110 kW.
6. Three consequences that follow from the same numbers. First, nitrification needs aerobic solids retention time, not just air: at a Canadian winter mixed-liquor temperature near 12 °C the autotroph growth rate is roughly 0.44 d$^{-1}$, so a washout SRT near 2.3 d and a safety factor of 2.5 put the design SRT at about 6 d — more tank, not just more blower. Second, nitrification destroys alkalinity at 7.14 g as CaCO$_3$ per g N: $$\Delta\text{Alk}=7.14\times 453.6=3239\ \text{kg CaCO}_3/\text{d} \equiv 216\ \text{mg/L}$$ which will collapse the pH of a soft receiving-water catchment unless alkalinity is added or denitrification recovers about half of it. Third, the 0.40 mg/L TP limit is below what biological removal alone reliably achieves, so chemical polishing is implied: taking a 4.0 mg/L secondary effluent and a 2:1 Al:P molar dose, $$D_{\text{alum}}=\frac{4.0-0.40}{30.97}\times 594.4=69\ \text{mg/L} \equiv 1037\ \text{kg/d}$$ with the corresponding chemical sludge. The revised permit is therefore not a paper change: it converts a carbon-removal plant into a nitrifying plant with chemical phosphorus removal.