18-Env-A4 Water and Wastewater Engineering · December 2013
Question 5 of 6: Activated Sludge — WAS Volume and Aeration Tank Sizing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 — 04-Env-A4 / Water and Wastewater Engineering. 3 hours duration; closed book with one double-sided aid sheet; approved calculator permitted. The paper instructs candidates to attempt any two questions from Part A and any two from Part B (100 marks); all six are solved below for completeness.
Reference texts. Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.) — activated-sludge kinetics, nitrification, aeration, anaerobic digestion; 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 sand filtration; Guidelines for Canadian Drinking Water Quality (Health Canada).
Question B2: Activated Sludge — WAS Volume and Aeration Tank Sizing (25 marks)
120 mg/L (assumed ≈ complete removal to a low secondary effluent)
VSS (observed) yield
$Y$
0.65 kg VSS/kg BOD₅ removed
Clarifier underflow (WAS) concentration
$X_u$
8,000 mg/L
Find. The volume of waste activated sludge produced per day, $V_w$.
Approach. Convert the BOD₅ load removed to a net biomass (VSS) production rate using the given observed yield, then divide that mass rate by the underflow solids concentration to get a daily volumetric wasting rate.
Net biomass (VSS) produced per day. With no secondary effluent BOD given, essentially all of the 120 mg/L is treated as removed across the aeration tank:
$$P_x=Y\,Q\,\Delta S=0.65\times20{,}000\ \tfrac{\text{m}^3}{\text{d}}\times0.120\ \tfrac{\text{kg}}{\text{m}^3}=\boxed{1560\ \text{kg VSS/d}}.$$
Waste activated sludge volume. Dividing the mass production rate by the underflow solids concentration ($X_u=8{,}000\ \text{mg/L}=8.0\ \text{kg/m}^3$) gives the daily wasted volume:
$$V_w=\frac{P_x}{X_u}=\frac{1560\ \text{kg/d}}{8.0\ \text{kg/m}^3}=\boxed{195\ \text{m}^3/\text{d}}.$$
Check: this assumes essentially complete BOD₅ removal (a reasonable design target for a nitrifying activated-sludge plant, and consistent with the yield being described as an "observed"/net figure that already reflects whatever endogenous decay occurs) and that WAS is wasted directly from the clarifier underflow rather than from the aeration tank mixed liquor.
Quantity
Value
Net VSS production, $P_x$
1,560 kg VSS/d
WAS volume, $V_w$
195 m³/d
(II) Aeration Tank Volume for Consistent Nitrification
Achieving reliable nitrification means designing the solids retention time (SRT) long enough that the slow-growing nitrifying bacteria are not washed out of the system faster than they can reproduce — even under the coldest design temperature and with a safety margin for peak loads. This requires first estimating the minimum SRT from nitrifier growth kinetics, applying a safety factor, and then sizing the aeration tank volume from the standard SRT mass-balance using the biomass production rate already found in part (I).
Given.
Given / assumed design data
Quantity
Symbol
Value
Nitrifier max. growth rate at 20°C
$\mu_{n,\max,20}$
0.75 d-1 (typical published value)
Temperature coefficient (growth)
$\theta_\mu$
1.123 (typical)
Nitrifier decay rate at 20°C
$k_{dn,20}$
0.08 d-1 (typical)
Temperature coefficient (decay)
$\theta_{kdn}$
1.04 (typical)
Design temperature
$T$
10°C (Canadian winter design condition)
Design safety factor
$SF$
2.0
Design MLVSS
$X$
3,000 mg/L (typical value)
Find. The design SRT and the corresponding aeration tank volume $V$.
Approach. Temperature-correct the nitrifier kinetic coefficients to the winter design condition, find the minimum SRT at which the net nitrifier growth rate is positive, apply a safety factor for a reliable design SRT, then size the tank from $X\,V=P_x\cdot SRT$.
Temperature-correct the nitrifier kinetics to $T=10\,{}^{\circ}\text{C}$.
$$\mu_{n,\max}(T)=\mu_{n,\max,20}\,\theta_\mu^{\,(T-20)}=0.75\times1.123^{-10}=0.235\ \text{d}^{-1},$$
$$k_{dn}(T)=k_{dn,20}\,\theta_{kdn}^{\,(T-20)}=0.08\times1.04^{-10}=0.0540\ \text{d}^{-1}.$$
Minimum SRT for nitrification. At (near) saturating ammonia and dissolved oxygen the net nitrifier specific growth rate approaches its maximum, $\mu_{n,\text{net}}\approx\mu_{n,\max}(T)-k_{dn}(T)$, and the minimum SRT that avoids nitrifier washout is its reciprocal:
$$\mu_{n,\text{net}}=0.235-0.0540=0.181\ \text{d}^{-1},\qquad SRT_{\min}=\frac{1}{\mu_{n,\text{net}}}=5.52\ \text{d}.$$
Apply a design safety factor. Reliable year-round nitrification (through diurnal ammonia peaks and temperature swings) is designed with $SF\approx1.5$–2.5 on the kinetic minimum; using $SF=2.0$:
$$SRT_{design}=SF\times SRT_{\min}=2.0\times5.52=11.0\ \text{d (rounded)}.$$
Aeration tank volume from the SRT mass balance. At steady state the total biomass held in the tank equals the net production rate ($P_x$, part I) times the SRT:
$$X\,V=P_x\cdot SRT\ \Longrightarrow\ V=\frac{P_x\cdot SRT}{X}=\frac{1560\ \tfrac{\text{kg}}{\text{d}}\times11.0\ \text{d}}{3.0\ \tfrac{\text{kg}}{\text{m}^3}}=\boxed{5{,}720\ \text{m}^3}.$$
Check: three design inputs are not stated in the question and are taken as typical published/engineering-judgment values (flagged above): the winter design temperature (10°C, a reasonable Canadian value), the safety factor (2.0, mid-range of the usual 1.5–2.5), and the design MLVSS (3,000 mg/L, a typical conventional activated-sludge value). The resulting hydraulic retention time is $V/Q=5{,}720/20{,}000=0.286\ \text{d}=6.9\ \text{h}$, squarely within the typical 4–8 h range for conventional nitrifying activated sludge (Metcalf & Eddy Table 8-9), which supports the assumption set.
