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18-Env-A4 Water and Wastewater Engineering · May 2015

Question 5 of 5: Activated Sludge WWTP — Process Flow Diagram

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Notes on this paper

National Exams / EGBC — May 2015 — 04-ENV-A4 Water and Wastewater Engineering. Three-hour exam; Question 1 is compulsory (25 marks) and any three of the remaining four questions are required (25 marks each); all five are solved below for completeness. Closed book, one double-sided aid sheet permitted, approved calculator permitted.

Reference texts: Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.) — nitrogen and solids characterization, BOD test theory, nitrification/alkalinity, disinfection chemistry, activated-sludge and sludge-processing design; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.) — the Streeter–Phelps oxygen sag, pH and coagulation–flocculation chemistry, jar testing; MWH's Water Treatment: Principles and Design (3rd ed.) — granular filtration (headloss, backwash) and chemical phosphorus removal.

Question 5: Activated Sludge WWTP — Process Flow Diagram (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. Average day flow $Q=20{,}000\ \text{m}^3/\text{d}$ with influent TSS 200, cBOD5 170, TKN 35, TP 4.5 mg/L. Effluent limits: cBOD5 10, TSS 10, total ammonia nitrogen (TAN) 3, TP 0.2 mg/L, and 200 CFU/100 mL E. coli.

ParameterInfluentEffluent limit
TSS200 mg/L10 mg/L
cBOD5170 mg/L10 mg/L
TKN / TAN35 mg/L (as TKN)3 mg/L (as TAN)
TP4.5 mg/L0.2 mg/L
E. coli—200 CFU/100 mL

Find. A process flow diagram covering every liquid and solids unit process, with approximate influent/effluent characteristics at each major stage, the sludge (digestion + dewatering) train with approximate stream volumes/%solids, and the chemicals used and their injection points.

Check: this is an open-ended conceptual-design sketch — the exam supplies only the influent loads and the four numeric effluent limits, not removal efficiencies, yields, or sludge concentrations. The process train below (conventional activated sludge with nitrification, tertiary filtration, chemical phosphorus removal and UV disinfection, with anaerobic digestion and mechanical dewatering of the combined sludge) is one defensible, EGBC-typical design that meets every stated limit; the unit-process removal fractions, observed sludge yield, VS destruction, and cake solids are stated design assumptions (typical literature ranges, Metcalf & Eddy) rather than exam-given values, and are called out explicitly at each step.

Approach. Convert the influent concentrations to mass loads, apply typical removal fractions at each treatment stage (primary clarification → nitrifying activated sludge → tertiary filtration with chemical P removal → UV disinfection) to confirm the train can meet every effluent limit, then size the parallel solids (sludge) train (primary sludge + thickened WAS → anaerobic digestion → dewatering) from the solids captured at each liquid-train step.

Screening &Grit RemovalPrimaryClarifierAeration Basin(Nitrification)SecondaryClarifierTertiaryFiltrationUVDisinfectionGravityThickenerDAFThickenerAnaerobicDigesterMechanicalDewateringInfluentQ=20,000 m3/dTSS200 BOD170TKN35 TP4.580 TSS/119 BOD531.5 TKN/4.05 TPRAS15 TSS/15 BOD52 TAN/3.0 TPFeCl3 dose22.4 mg/LEffluent8 TSS/8 BOD52 TAN/0.15 TP<200 CFU/100mLPrimary sludge60 m3/d @4%WAS (thickened)31.2 m3/d @4%Biosolids cake~10.6 t/d @22% solids
Figure 1 — Conventional activated-sludge WWTP: liquid train (top) is screening/grit → primary clarification → nitrifying aeration (with RAS return) → secondary clarification → tertiary filtration (FeCl3 chemical P removal) → UV disinfection; solids train (bottom) thickens primary sludge and WAS separately, combines them into anaerobic digestion, then mechanically dewaters the digested sludge to a land-applicable cake.
  1. Convert influent concentrations to mass loads. $\text{Load (kg/d)}=C(\text{mg/L})\times Q(\text{m}^3/\text{d})/1000$: TSS $=200(20{,}000)/1000=4000$ kg/d; BOD5 $=170(20{,}000)/1000=3400$ kg/d; TKN $=700$ kg/d; TP $=90$ kg/d.
  2. Primary clarifier. Assuming typical primary removal of 60% TSS, 30% BOD5, and 10% each of TKN/TP (removed with the settled solids): primary effluent is $200(1-0.60)=80$ mg/L TSS, $170(1-0.30)=119$ mg/L BOD5, $31.5$ mg/L TKN, $4.05$ mg/L TP. Primary sludge captured $=4000(0.60)=2400$ kg TSS/d; at an assumed 4% solids (40 kg/m³), $$V_\text{primary sludge}=\frac{2400}{40}=\boxed{60\ \text{m}^3/\text{d}}.$$
  3. Secondary treatment (nitrifying activated sludge). With adequate SRT for nitrification (Q2a/Q4d), the aeration basin plus secondary clarifier is assumed to reach a pre-tertiary effluent of 15 mg/L BOD5, 15 mg/L TSS, 2 mg/L TAN (safely under the 3 mg/L limit) and 3.0 mg/L TP (modest biological uptake only). BOD5 removed in this stage $=119-15=104$ mg/L, i.e. a load of $104(20{,}000)/1000=2080$ kg/d. Using a typical observed sludge yield $Y_\text{obs}\approx0.6$ kg WAS-TSS per kg BOD5 removed, $$\text{WAS}=0.6\times2080=\boxed{1248\ \text{kg TSS/d}},$$ which thickens (DAF) to 4% solids: $V=1248/40=\boxed{31.2\ \text{m}^3/\text{d}}.$
  4. Tertiary filtration + chemical P removal, and disinfection. Granular filtration polishes BOD5/TSS from 15 to about 8 mg/L each (comfortably under the 10 mg/L limits). To reach $TP=0.2$ mg/L, chemical precipitation with ferric chloride is dosed ahead of/at the filter, targeting a design residual of 0.15 mg/L (removing $3.0-0.15=2.85$ mg/L P) at a molar $Fe:P$ ratio of 1.5:1: $$\text{FeCl}_3\ \text{dose}=\frac{2.85}{31}\times1.5\times162.2=\boxed{22.4\ \text{mg/L}}\ (447\ \text{kg/d as 100\% product}).$$ UV disinfection (chosen over chlorination per Q2b, since ammonia is still present and the filtered water has low turbidity) reduces E. coli below the 200 CFU/100 mL limit with no added chemical.
  5. Solids train — digestion and dewatering. Combined solids to anaerobic digestion $=2400+1248=3648$ kg TSS/d ($\approx91$ m³/d at ~4% solids). Assuming 72% volatile solids (VS) and 50% VS destruction in the digester: VS in $=0.72(3648)=2627$ kg/d, VS destroyed $=1313$ kg/d, so digested solids out $=3648-1313=\boxed{2335\ \text{kg/d}}$. Mechanical dewatering (centrifuge/belt press) to a typical 22% cake solids gives $$\text{Cake}=\frac{2335}{0.22}=\boxed{10{,}612\ \text{kg/d}\approx10.6\ \text{t/d}}$$ of land-applicable biosolids.
StreamApproximate value
Primary effluent (TSS/BOD5/TKN/TP)80 / 119 / 31.5 / 4.05 mg/L
Secondary (pre-tertiary) effluent (TSS/BOD5/TAN/TP)15 / 15 / 2 / 3.0 mg/L
Final effluent (TSS/BOD5/TAN/TP)8 / 8 / 2 / 0.15 mg/L (all under limits)
Primary sludge2400 kg TSS/d, 60 m³/d @ 4% solids
Thickened WAS1248 kg TSS/d, 31.2 m³/d @ 4% solids
Digested solids2335 kg/d (50% VS destruction)
Dewatered cake≈ 10.6 t/d @ 22% solids
FeCl3 dose (chemical P removal)22.4 mg/L (≈447 kg/d), injected ahead of tertiary filtration
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