22-Agric-B11 Principles of Waste Management · May 2015
Question 4 of 5: Aerated Lagoon Design
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 04-Agric-B11, Principles of Waste Management — May 2015. 3-hour duration, open-book exam. Answer Question 1 plus any three of Questions 2 to 5; all five questions are answered below as a complete study resource.
Reference texts: Tchobanoglous, Burton & Stensel, Metcalf & Eddy Wastewater Engineering: Treatment and Resource Recovery; MWPS-18, Livestock Waste Facilities Handbook (MidWest Plan Service); Rynk et al., On-Farm Composting Handbook (NRAES-54); Sommer & Christensen (eds.), Animal Manure Recycling: Treatment and Management.
Check: lagoon depth = 3 m and 2 equal aerated cells in series (typical aerated-lagoon configuration, Metcalf & Eddy); field oxygen-requirement factor = 1.5 kg O2/kg BOD5 removed; biomass synthesis yield Y = 0.6 kg VSS/kg BOD5 removed and endogenous decay $k_d$ = 0.06 d-1 (typical aerated-lagoon values, no solids recycle so MCRT = HRT); settled-sludge specific gravity 1.03 and 3% solids content as removed; sludge storage allowance = 20% of lagoon volume before desludging is required — none of these are stated by the question.
Find. Lagoon volume and cell dimensions; oxygen requirement per hour; sludge production and desludging frequency.
Two-cell aerated-lagoon train sized for the coldest (winter, 10 °C) design condition.
Approach. Correct the rate constant to the coldest (winter) lagoon temperature since a lower $k$ always governs the largest detention time, size the lagoon volume from a first-order complete-mix BOD removal model, then compute oxygen demand from the mass of BOD removed and sludge production from a standard biomass-yield relation.
Temperature-corrected rate constant. Sizing on the coldest expected condition (10 °C) always governs, since a lower $k$ needs a longer detention time for the same removal: $$k_T = k_{20}\,\theta^{(T-20)} = 0.68 \times 1.037^{(10-20)} = \boxed{0.473\ \text{d}^{-1}}$$
Detention time and lagoon volume. For a complete-mix, first-order, no-recycle lagoon, $S_e/S_0 = 1/(1+k_T t)$. At the minimum 80% removal, $S_e/S_0=0.20$, so $$t = \frac{S_0/S_e - 1}{k_T} = \frac{5-1}{0.473} = \boxed{8.46\ \text{d}}$$ giving a total volume $$V = Q\,t = 5{,}000 \times 8.46 = \boxed{42{,}300\ \text{m}^3}$$ Split into 2 equal cells in series (typical practice) at an assumed 3 m operating depth, each cell has surface area $A = (V/2)/3 = 7{,}050\ \text{m}^2$; taking a 2:1 length:width rectangle, $W=\sqrt{A/2}=59.4$ m and $L=2W=118.7$ m, so each cell is roughly 119 m × 59 m × 3 m deep.
Oxygen requirement. Mass of BOD5 removed: $S_e = 400(0.20) = 80$ mg/L, so $$\text{BOD removed} = Q(S_0-S_e) = 5{,}000\ \text{m}^3/\text{d} \times (400-80)\ \text{g/m}^3 \times 10^{-3} = \boxed{1{,}600\ \text{kg/d}}$$ At an assumed field oxygen factor of 1.5 kg O2/kg BOD5 removed, $$\dot m_{O_2} = 1.5 \times 1{,}600 = 2{,}400\ \text{kg/d} = \boxed{100\ \text{kg O}_2/\text{hr}}$$
Excess sludge production. Using the standard biomass-yield relation with no solids recycle (MCRT = HRT = 8.46 d): $$P_x = \frac{Y \cdot \text{BOD removed}}{1+k_d\,\theta_c} = \frac{0.6 \times 1{,}600}{1+0.06(8.46)} = \boxed{636.8\ \text{kg dry solids/d}}$$
Desludging frequency. At an assumed 3% solids content and SG 1.03 for the settled sludge as removed, the daily sludge volume is $$V_{sludge} = \frac{636.8/0.03}{1.03\times1{,}000} = 20.6\ \text{m}^3/\text{d} \Rightarrow 7{,}520\ \text{m}^3/\text{yr}$$ Allowing 20% of the total lagoon volume (8,460 m³) for sludge accumulation before it must be removed, $$\text{years to fill} = \frac{8{,}460}{7{,}520} = \boxed{1.1\ \text{years}}$$ so the lagoon should be desludged on roughly an annual basis.
