Activated Sludge — Waste Sludge Production and Aeration Tank Volume: 18-Env-A4 Water and Wastewater Engineering, December 2018, Q5

Question 5: Activated Sludge — Waste Sludge Production and Aeration Tank Volume (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.

Process data
Quantity	Symbol	Value
Wastewater flow	$Q$	20,000 m³/d
Raw sewage $BOD_5$	$BOD_{5,raw}$	150 mg/L
Raw sewage TSS	$TSS_{raw}$	200 mg/L
Primary clarifier TSS removal	$\eta_{TSS}$	65%
Primary clarifier $BOD_5$ removal	$\eta_{BOD}$	40%
Observed sludge (TSS) yield	$Y$	0.7 kg TSS/kg $BOD_5$
Secondary clarifier underflow (WAS) concentration	$X_u$	8,000 mg/L
Solids retention time	$\theta_c$	10 d

Find. (a) The mass of solids wasted per day and the corresponding WAS volumetric flow. (b) The aeration tank volume required for a 10-day SRT.

Approach. Find the $BOD_5$ load entering the secondary (biological) process from the primary clarifier's removal efficiency; apply the observed sludge yield to that load to get the daily waste solids production $P_x$; divide $P_x$ by the underflow concentration for WAS volume (part a); then use the SRT definition $\theta_c=XV/P_x$ with an assumed design MLSS to size the aeration tank, and cross-check with the resulting F/M ratio and hydraulic retention time (part b).

Check: two assumptions are required because the exam does not supply them. (1) Part (a) treats the $BOD_5$ entering the secondary process (90 mg/L, after the primary clarifier) as essentially fully removed there — no secondary effluent $BOD_5$ is given, so $P_x=Y\,Q\,BOD_{5,\text{to secondary}}$ is used rather than $Y\,Q\,(S_0-S)$ with a nonzero effluent $S$; $P_x$ scales down proportionally if a nonzero effluent $BOD_5$ is later specified. (2) Part (b) requires the aeration-tank MLSS concentration $X$, which the exam also does not supply; a standard conventional-activated-sludge design value of $X=3{,}000$ mg/L is assumed (Metcalf & Eddy conventional plug-flow design range, 1,500–3,000 mg/L). The resulting HRT (5.0 h) and F/M ratio (0.14 kg $BOD_5$/kg MLSS·d, shown in Step 5 to be independent of the MLSS assumed) both fall within the expected range for a conventional design, supporting the assumption.

Part (a) — $BOD_5$ load to the secondary process. The primary clarifier removes 40% of the raw $BOD_5$, so the $BOD_5$ entering the aeration tank is $$BOD_{5,\text{to secondary}}=BOD_{5,raw}(1-\eta_{BOD})=150(1-0.40)=90\text{ mg/L}.$$
Waste solids production rate, $P_x$. $$P_x=Y\times Q\times BOD_{5,\text{to secondary}}=0.7\times20{,}000\ \text{m}^3/\text{d}\times90\ \text{g/m}^3=1{,}260{,}000\text{ g/d}=\boxed{1{,}260\text{ kg TSS/d}}.$$
Volume of waste activated sludge, $Q_{WAS}$. Dividing the daily solids mass by the underflow concentration ($X_u=8{,}000$ mg/L $=8.0$ kg/m³): $$Q_{WAS}=\frac{P_x}{X_u}=\frac{1{,}260}{8.0}=\boxed{157.5\text{ m}^3/\text{d}}.$$
Part (b) — aeration tank volume from the SRT definition. At steady state, the solids retention time equals the total biomass inventory in the aeration tank divided by the daily solids wasting rate (effluent solids assumed negligible): $$\theta_c=\frac{X\,V}{P_x}\quad\Rightarrow\quad V=\frac{\theta_c\,P_x}{X}.$$ Using the assumed design MLSS $X=3{,}000$ mg/L $=3.0$ kg/m³: $$V=\frac{10\times1{,}260}{3.0}=\boxed{4{,}200\text{ m}^3}.$$
Cross-checks — HRT and F/M ratio. Hydraulic retention time: $t=V/Q=4{,}200/20{,}000=0.21\text{ d}=\boxed{5.0\text{ h}}$, within the typical 4–8 h range for conventional activated sludge. Food-to-microorganism ratio: $$\frac{F}{M}=\frac{Q\,BOD_{5,\text{to secondary}}}{V\,X}=\frac{20{,}000\times0.090}{4{,}200\times3.0}=\boxed{0.14\text{ kg }BOD_5/\text{kg MLSS}\cdot\text{d}}.$$ Because $V\propto1/X$ by construction ($V=\theta_c P_x/X$), the product $VX=\theta_c P_x$ is fixed regardless of which MLSS value is assumed, so this F/M result (and the conclusion that the design is self-consistent) would be unchanged had a different MLSS in the 1,500–3,000 mg/L range been assumed instead — only the tank volume and HRT would shift.

Question 5 — final results
Quantity	Value
$BOD_5$ entering secondary process	90 mg/L
Waste solids production, $P_x$	1,260 kg TSS/d
Waste activated sludge volume, $Q_{WAS}$	157.5 m³/d
Aeration tank volume (assumed MLSS 3,000 mg/L)	4,200 m³
Hydraulic retention time	5.0 h
F/M ratio	0.14 kg $BOD_5$/kg MLSS·d

Topic: Activated-sludge waste solids production, WAS volume, and aeration tank sizing from the SRT (mean cell residence time) definition.

Key relations: $P_x=Y\,Q\,(S_0-S)$ (observed-yield solids production); $Q_{WAS}=P_x/X_u$; $\theta_c=XV/P_x\Rightarrow V=\theta_c P_x/X$; $F/M=Q\,S_0/(VX)$.

Why this works. The observed sludge yield $Y$ already lumps biomass synthesis and endogenous decay into one empirical coefficient, so $P_x$ follows directly from the substrate load without needing separate $Y$ and $k_d$ kinetic coefficients. The SRT (solids retention time) definition is a steady-state mass balance — at steady state, whatever solids mass is produced per day must equal whatever is wasted per day, and $\theta_c=XV/P_x$ simply restates that as "total solids inventory / daily wasting rate."

Common pitfall. Confusing hydraulic retention time (a plug-flow transit-time concept) with solids retention time (a solids-inventory concept) — in an activated-sludge system with return sludge recycle, $\theta_c\gg t$ (here 10 d vs. 5 h) precisely because RAS keeps recycling biomass back into the tank instead of letting it wash out with the flow; using the raw (pre-primary) $BOD_5$ rather than the $BOD_5$ actually entering the biological process when computing $P_x$.

Source: Metcalf & Eddy §8 (activated-sludge process design, SRT, observed yield).

Question 5 of 5: Activated Sludge — Waste Sludge Production and Aeration Tank Volume

Question 5: Activated Sludge — Waste Sludge Production and Aeration Tank Volume (25 marks)