18-Env-A4 Water and Wastewater Engineering · December 2017
Question 5 of 5: Activated Sludge – Waste Sludge Volume and Aeration Tank Sizing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2017 — 04-Env-A4 / Water and Wastewater Engineering. 3 hours duration; closed book with one double-sided aid sheet; approved calculator permitted. The paper instructs that Question 1 is compulsory and any three of the remaining four questions are required; all five are solved below for completeness.
Reference texts. Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.) — activated-sludge kinetics, solids/hydraulic retention time, nitrogen and phosphorus forms; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.) — discrete settling theory, indicator organisms, coagulation chemistry; MWH’s Water Treatment: Principles and Design (3rd ed.) — process selection, softening, rapid and slow sand filtration; Guidelines for Canadian Drinking Water Quality (Health Canada).
Question 5: Activated Sludge – Waste Sludge Volume and Aeration Tank Sizing (25 marks)
Given. The plant flow, primary effluent $BOD_5$, observed yield, clarifier underflow concentration and target SRT below:
Quantity
Symbol
Value
Plant flow
$Q$
10,000 m³/d
Primary effluent $BOD_5$
$S_0$
120 mg/L
Observed TSS yield
$Y_{obs}$
0.75 kg TSS/kg $BOD_5$
Clarifier underflow (WAS) concentration
$X_u$
8,000 mg/L
Target solids retention time
$SRT$
10 d
Find. (i) the volumetric flow of waste activated sludge per day, $Q_w$; (ii) the aeration-tank volume $V$ needed to hold that biomass at a 10-day SRT.
Approach. Convert the $BOD_5$ load to a daily WAS solids-production rate via the given observed yield, divide by the underflow concentration for $Q_w$, then use the steady-state solids balance $SRT=(MLSS\cdot V)/P_x$ (mass of biomass in the tank turns over once per SRT through wasting) to size the aeration volume.
Daily $BOD_5$ load and WAS solids production. With essentially complete biological removal of the primary effluent $BOD_5$ assumed (no effluent $BOD_5$ is given, and the yield is quoted as an observed/net figure that already reflects endogenous decay), the load applied is
$$Q\,S_0=10{,}000\ \tfrac{\text{m}^3}{\text{d}}\times0.120\ \tfrac{\text{kg}}{\text{m}^3}=1200\ \text{kg}\ BOD_5/\text{d}.$$
The waste-sludge TSS production rate follows from the given yield:
$$\boxed{P_x=Y_{obs}\,Q\,S_0=0.75\times1200=900\ \text{kg TSS/d}}.$$
WAS volumetric flow. Dividing the daily solids mass by the clarifier underflow concentration ($X_u=8{,}000\ \text{mg/L}=8.0\ \text{kg/m}^3$):
$$\boxed{Q_w=\frac{P_x}{X_u}=\frac{900}{8.0}=112.5\ \text{m}^3/\text{d}}.$$
Aeration-tank volume from the SRT solids balance. At steady state the mass of biomass held in the aeration tank ($MLSS\times V$) must turn over once every SRT through wasting at rate $P_x$ (effluent solids taken as negligible), so
$$SRT=\frac{MLSS\cdot V}{P_x}\ \Longrightarrow\ V=\frac{P_x\cdot SRT}{MLSS}.$$
No aeration-tank MLSS is stated in the question, so a typical conventional activated-sludge design value $MLSS=3{,}000\ \text{mg/L}=3.0\ \text{kg/m}^3$ is adopted (flagged below).
Solve for $V$.
$$\boxed{V=\frac{P_x\cdot SRT}{MLSS}=\frac{900\times10}{3.0}=3{,}000\ \text{m}^3}.$$
Check: assumes (a) essentially complete $BOD_5$ removal across the aeration tank, consistent with the yield being quoted as an observed/net figure; (b) effluent suspended solids are negligible in the SRT solids balance; (c) aeration-tank $MLSS=3{,}000\ \text{mg/L}$, a typical conventional activated-sludge design value (Metcalf & Eddy design-range table, 1,500–4,000 mg/L) since the question does not state one — $V$ scales inversely with whatever MLSS is actually specified or selected.
