Completely-Mixed Activated Sludge Design: 22-Agric-B11 Principles of Waste Management, Undated paper, Q3

Question 3: Completely-Mixed Activated Sludge Design (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.

Quantity	Value
Design flow, Q	10,000 m³/d
Influent BOD₅, S₀	300 mg/L
Influent TSS / VSS	100 mg/L / 0 mg/L (simplified)
Influent NH₄⁺-N / TP	40 mg/L / 5 mg/L
Effluent SS (85% biodegradable)	10 mg/L
SRT (pilot-tested)	6 d
Y, k, K_s, b, f_d	0.60 mg VSS/mg BOD₅; 5.0 d^-1; 50 mg/L; 0.06 d^-1; 0.15

Check: "other information in the textbook" is taken as the standard Metcalf & Eddy design values — BOD₅/BOD_u ratio f = 0.68 (used to convert the biodegradable-solids bCOD into an equivalent BOD₅, and to convert the substrate-utilization oxygen demand from a BOD₅ to a BOD_u basis); cell COD/VSS content 1.42 mg O₂/mg VSS; mixed-liquor VSS concentration X = 2,500 mg/L (typical CMAS design range 1,500–4,000 mg/L); VSS/TSS ratio of the wasted sludge = 0.85; aeration-tank length:width ratio 2:1 at 4.5 m side water depth.

Find. Effluent BOD₅; aeration tank dimensions; daily sludge production rate; oxygen utilization rate; and a recommendation for reaching ≥95% ammonia removal.

Approach. Use Monod/Lawrence-McCarty kinetics with the pilot-tested SRT to find the soluble effluent BOD₅, add the particulate BOD₅ contributed by biodegradable effluent solids to get the TOTAL effluent BOD₅; size the aeration tank and biomass/sludge production from the standard CMAS mass-balance equations; compute the oxygen utilization rate from the substrate removed net of the oxygen credited to synthesized biomass; then evaluate whether the design SRT is adequate for reliable nitrification.

1) Soluble effluent BOD₅. At steady state, $\dfrac{1}{\text{SRT}} = \dfrac{Yk S}{K_s+S} - b$. Solving for the soluble substrate concentration $S$ with $\text{SRT}=6\ \text{d}$: $$\left(\frac{1}{6}+0.06\right)(50+S) = (0.60)(5.0)\,S \;\Rightarrow\; S = \boxed{4.09\ \text{mg/L (soluble BOD}_5\text{)}}$$ This alone is already well under the 10 mg/L target, but the effluent also carries biodegradable suspended solids that exert their own BOD₅ once discharged.
Particulate BOD₅ from effluent solids. Of the 10 mg/L effluent SS, 85% (8.5 mg/L) is biodegradable. Converting biodegradable VSS to an oxygen-demand basis via the cell COD content (1.42 mg O₂/mg VSS) and then to a 5-day basis via the standard BOD₅/BOD_u ratio (f = 0.68): $$\text{BOD}_{5,particulate} = (8.5)(1.42)(0.68) = \boxed{8.21\ \text{mg/L}}$$
Total effluent BOD₅. $$\text{BOD}_{5,total} = S + \text{BOD}_{5,particulate} = 4.09+8.21 = \boxed{12.3\ \text{mg/L}}$$
2) Biomass (sludge) production. With $\Delta S = S_0-S = 300-4.09=295.9\ \text{mg/L}$, the standard biomass-plus-debris production formula gives $$P_{x,VSS} = \frac{Y\,Q\,\Delta S}{1+b\,\text{SRT}}\left[1+f_d\,b\,\text{SRT}\right] = \frac{(0.60)(10{,}000)(295.9)}{1+(0.06)(6)}\big[1+(0.15)(0.06)(6)\big] = \boxed{1{,}376\ \text{kg VSS/d}}$$ Converting to a total-solids (TSS) basis at an assumed 0.85 VSS/TSS ratio for the wasted sludge, $$P_{x,TSS} = \frac{1{,}376}{0.85} = \boxed{1{,}619\ \text{kg TSS/d}}$$
Aeration tank volume and dimensions. From the SRT definition $\text{SRT}=\dfrac{X\,V}{P_{x,VSS}}$, with an assumed mixed-liquor concentration $X=2{,}500\ \text{mg/L}$: $$V = \frac{P_{x,VSS}\times\text{SRT}}{X} = \frac{(1{,}376{,}000\ \text{g/d})(6\ \text{d})}{2{,}500\ \text{g/m}^3} = \boxed{3{,}302\ \text{m}^3}$$ (hydraulic retention time check: $\tau=V/Q=3{,}302/10{,}000\times24=7.9\ \text{h}$, squarely inside the typical 4–8 h CMAS range — a useful sanity check on the assumed $X$.) At an assumed 4.5 m side water depth and a 2:1 length:width ratio, the plan area is $A=3{,}302/4.5=733.9\ \text{m}^2$, so $$W=\sqrt{A/2}=\boxed{19.2\ \text{m}}, \qquad L=2W=\boxed{38.3\ \text{m}}$$
Oxygen utilization rate. $$R_o = \frac{Q\,\Delta S}{f} - 1.42\,P_{x,VSS} = \frac{(10{,}000)(295.9)}{0.68} - 1.42(1{,}376{,}000\ \text{g/d}) = \boxed{2{,}398\ \text{kg O}_2/\text{d}}\ (\approx 99.9\ \text{kg O}_2/\text{h})$$

Quantity	Result
Soluble effluent BOD₅	4.09 mg/L
Total effluent BOD₅	12.3 mg/L
Daily biomass production (VSS)	1,376 kg/d
Daily sludge production rate (TSS)	1,619 kg/d
Aeration tank volume	3,302 m³ (τ ≈ 7.9 h)
Aeration tank dimensions	L = 38.3 m × W = 19.2 m × depth 4.5 m
Oxygen utilization rate	2,398 kg O₂/d (≈ 99.9 kg O₂/h)

3) Facility upgrades for ≥95% ammonia removal (5 marks). Checking whether nitrification is even kinetically feasible at the design SRT: using a typical 20°C nitrifier (ammonia-oxidizing bacteria) maximum growth rate $\mu_{max}\approx0.75\ \text{d}^{-1}$, the minimum SRT for nitrification is $\theta_{c,min}=1/\mu_{max}\approx1.33\ \text{d}$, and applying a standard design safety factor of 2–2.5 gives a recommended design SRT of only $\approx3.3\ \text{d}$ — comfortably BELOW the current 6-day SRT. This means the existing basin is theoretically large enough to nitrify under ideal, warm, well-buffered conditions, so simply confirming that nitrification is occurring is not, by itself, the fix; the risk is RELIABILITY, not basin size. Recommended facility upgrades, in order of priority: (1) alkalinity supplementation — nitrification consumes roughly 7.14 mg alkalinity as CaCO₃ per mg NH₄⁺-N oxidized, and without added alkalinity (lime or soda ash dosing) the process pH can crash and self-inhibit nitrification well before 95% removal is reached; (2) dedicated DO control in the aerobic zone (maintain DO ≥ 2 mg/L), since nitrifiers are far more oxygen-sensitive than heterotrophs and are the first population to be inhibited if DO sags during peak loading; (3) additional basin volume or a separate nitrification stage sized with a winter-temperature safety factor, since $\mu_{max}$ for nitrifiers falls sharply as temperature drops (a summer-adequate SRT margin can disappear entirely in cold weather); and (4) a secondary clarifier with return-activated-sludge (RAS)/waste-activated-sludge (WAS) capability, which the process description does not yet include but is structurally required to decouple SRT from hydraulic retention time and hold a target MLSS at all — without RAS/WAS there is no way to control SRT independently of flow at all, so this is the most fundamental "upgrade" of the four.

Topic: CMAS kinetics for effluent quality, biomass/sludge production, oxygen demand, and nitrification-reliability design.

Key relations: $1/\text{SRT}=YkS/(K_s+S)-b$; $P_{x,VSS}=\dfrac{Y\,Q\,\Delta S}{1+b\,\text{SRT}}[1+f_d\,b\,\text{SRT}]$; $\text{SRT}=XV/P_{x,VSS}$; $R_o=Q\Delta S/f-1.42P_{x,VSS}$; $\theta_{c,min}=1/\mu_{max}$ for nitrification.

Why this works: total effluent BOD₅ from a biological process is never just the residual soluble substrate — the biomass itself that escapes the clarifier exerts an oxygen demand once discharged, which is why regulatory permits are written against TOTAL effluent BOD₅/TSS rather than soluble substrate alone. Similarly, "is the tank big enough" and "is ammonia removal reliable" are two different engineering questions — kinetic feasibility (SRT > θc,min) is necessary but not sufficient; alkalinity, DO, and temperature margin are what actually determine whether a 95% removal GUARANTEE can be met.

Common pitfall: reporting only the soluble-substrate BOD₅ from the Monod equation as "the" effluent BOD₅ and ignoring the particulate contribution from effluent solids; and assuming that because the design SRT exceeds the bare minimum for nitrification, no further ammonia-removal upgrade is needed — kinetic feasibility under ideal summer conditions says nothing about winter reliability, alkalinity depletion, or DO excursions.

Source: Tchobanoglous, Burton & Stensel, Metcalf & Eddy Wastewater Engineering Ch. 8 (CMAS design, nitrification kinetics and alkalinity demand).

Question 3 of 5: Completely-Mixed Activated Sludge Design

Question 3: Completely-Mixed Activated Sludge Design (25 marks)