Aerated Lagoon Design: 22-Agric-B11 Principles of Waste Management, May 2014, Q5

Question 5: Aerated Lagoon Design (20 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. $Q=0.30\ \text{mgd}$, $\text{BOD}_{5,0}=500\ \text{mg/L}$, target removal $\ge75\%$; $k_{20^\circ\text{C}}=0.68\ \text{d}^{-1}$, $\theta=1.047$, temperature $10$–$25^\circ\text{C}$; aerator transfers 2.5 lb $\text{O}_2$/hp-hr at standard conditions; 10-hp aerator units available.

Quantity	Value
Flow, $Q$	0.30 mgd
Influent BOD5	500 mg/L
Target removal	≥75%
$k_{20^\circ\text{C}}$ / $\theta$	0.68 d⁻¹ / 1.047
Temperature range	10–25°C
Aerator O2 transfer	2.5 lb O2/hp·hr

Find. Lagoon number/volume/configuration, excess sludge production and oxygen demand per day, and the number of 10-hp aerators required.

Figure 4. Two aerated-lagoon cells in series, sized for the winter (worst-case, slowest-kinetics) detention time.

Approach. Correct the rate constant to the coldest (worst-case) design temperature, size the completely-mixed lagoon volume from first-order BOD removal kinetics, then build the sludge yield and oxygen demand from the BOD actually removed, and finally divide the oxygen requirement by the aerator's rated transfer capacity.

Temperature-corrected rate constant. Winter (10°C) governs because it is the slowest kinetics and hence needs the largest volume to still hit 75% removal: $$k_{10^\circ\text{C}} = 0.68\times1.047^{(10-20)} = \boxed{0.430\ \text{d}^{-1}}$$
Detention time and volume. For a completely-mixed lagoon with first-order BOD removal, $S_e/S_0=1/(1+kt)$. Requiring $S_e/S_0=0.25$ (75% removal): $$t = \dfrac{(1/0.25)-1}{0.430} = \boxed{6.98\ \text{days}\approx7.0\ \text{d}}$$ Converting flow, $Q=0.30\ \text{mgd}\times3{,}785.4\ \text{m}^3/\text{MG}=1{,}135.6\ \text{m}^3/\text{day}$, so $$V = Qt = 1{,}135.6\times6.98 = \boxed{7{,}930\ \text{m}^3}$$ Check: no design depth is given; a typical aerated-lagoon depth of 3.0 m (Metcalf & Eddy) is assumed, giving a total surface area of ≈2,640 m². Configuration: two equal cells in series (≈3,965 m³ each) — the first cell carries most of the BOD removal, the second polishes the effluent and improves reliability against short-circuiting, a standard aerated-lagoon layout.
BOD removed. $S_e = 0.25\times500=125\ \text{mg/L}$, so the removed load is $$\Delta BOD = Q(S_0-S_e) = 1{,}135.6\times(500-125)/1000 = \boxed{425.9\ \text{kg/day}\ (938.9\ \text{lb/day})}$$
Excess sludge production. Aerated lagoons run without solids recycle, so the mean cell residence time equals the hydraulic detention time, $\theta_c=t=6.98\ \text{d}$. Using typical activated-sludge kinetic coefficients (Check: not given in the source; $Y=0.6\ \text{kg VSS/kg BOD}_5$ removed, $k_d=0.06\ \text{d}^{-1}$, Metcalf & Eddy): $$P_x = \dfrac{Y\,\Delta BOD}{1+k_d\theta_c} = \dfrac{0.6\times425.9}{1+0.06(6.98)} = \boxed{180\ \text{kg VSS/day}}$$
Oxygen supply required. Using a typical aerated-lagoon field oxygen-requirement factor of 1.5 kg $\text{O}_2$/kg $\text{BOD}_5$ removed (Check, Metcalf & Eddy typical range 0.7–1.5): $$O_{2,req} = 1.5\times425.9 = \boxed{638.8\ \text{kg/day}\ (1{,}408\ \text{lb/day})}$$
Number of 10-hp aerators. At the manufacturer's 2.5 lb $\text{O}_2$/hp·hr transfer rate, the daily oxygen demand requires $$\text{hp-hr/day} = \dfrac{1{,}408}{2.5} = 563.3\ \text{hp-hr/day} \Rightarrow \text{continuous power} = \dfrac{563.3}{24} = 23.5\ \text{hp}$$ $$n = \left\lceil \dfrac{23.5}{10} \right\rceil = \boxed{3\ \text{aerators}\ (10\ \text{hp each})}$$

Quantity	Result
Detention time / volume	7.0 days / 7,930 m³
Configuration	2 cells in series, ≈3,965 m³ each, ≈3.0 m depth
Excess sludge production	180 kg VSS/day
Oxygen supply required	638.8 kg/day (1,408 lb/day)
Aerators required	3 × 10-hp units

Topic: Aerated-lagoon detention time sizing, biological sludge yield, and aerator selection.

Key relations: $S_e/S_0=1/(1+kt)$ (completely-mixed, first-order); $k_T=k_{20}\theta^{T-20}$; $P_x=Y\Delta BOD/(1+k_d\theta_c)$; $n_{aerators}=\lceil O_{2,req}/(transfer\ rate\times24\times hp_{unit})\rceil$.

Why this works: The lowest expected temperature always governs detention-time sizing because a first-order rate constant falls with temperature, so the coldest month needs the most volume to still hit the removal target; sludge yield and oxygen demand then both scale directly off the BOD mass actually removed by that sized reactor.

Common pitfall: Sizing the lagoon on the summer (warm, fast-kinetics) rate constant undersizes the reactor and misses the removal target every winter; forgetting to convert the rate constant to lagoon temperature before applying $S_e/S_0=1/(1+kt)$ silently uses the wrong $k$ altogether.

Source: Metcalf & Eddy/Tchobanoglous, Wastewater Engineering: Treatment and Resource Recovery, aerated-lagoon and activated-sludge design chapters.

Question 5 of 6: Aerated Lagoon Design

Question 5: Aerated Lagoon Design (20 marks)