18-Env-A4 Water and Wastewater Engineering · Undated paper
Question 3 of 5: Secondary Clarifier Sizing, Solids Loading and SRT
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2019 — 18-Env-A4 / Water and Wastewater Engineering. 3 hours duration; closed book with one double-sided aid sheet; approved Casio/Sharp calculator permitted. Question 1 is compulsory; the paper instructs candidates to attempt any three of the remaining four (100 marks total); all five are solved below for completeness.
Reference texts. Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.) — BOD kinetics, activated-sludge/clarifier design, anaerobic digestion; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.) — alkalinity chemistry, disinfection, coagulation/flocculation; MWH’s Water Treatment: Principles and Design (3rd ed.) — ozonation, turbidity; Standard Methods for the Examination of Water and Wastewater — alkalinity titration (2320B).
Question 3: Secondary Clarifier Sizing, Solids Loading and SRT (25 marks: a 8, b 7, c 10)
Find. The diameter of each secondary clarifier, the solids loading rate at average flow, and the solids retention time (SRT) of the activated-sludge system.
Approach. Size total clarifier surface area from the plant flow and the given SOR, split it between the two clarifiers to get each diameter; compute solids loading rate as the MLSS mass carried by the combined (influent + RAS) flow entering the clarifiers, divided by that same total area; and close a solids mass balance on the aeration tank ($SRT=VX/Q_wX_w$, neglecting the small mass lost in the (assumed near-solids-free) clarified effluent) for part (c).
(a) Total and per-clarifier surface area. $$A_{tot}=\frac{Q}{SOR}=\frac{16{,}000}{16}=1{,}000\ \text{m}^2,\qquad A_{each}=\frac{A_{tot}}{n}=\frac{1{,}000}{2}=500\ \text{m}^2.$$
(a) Diameter. $$A_{each}=\frac{\pi D^2}{4}\ \Rightarrow\ D=\sqrt{\frac{4A_{each}}{\pi}}=\sqrt{\frac{4(500)}{\pi}}=\boxed{25.2\ \text{m}}\ \text{(each of the two clarifiers).}$$
(b) Combined flow entering the clarifiers. The full mixed-liquor stream (plant flow plus return activated sludge) crosses into the clarifiers carrying the MLSS concentration: $$Q+Q_r=16{,}000+10{,}000=26{,}000\ \text{m}^3/\text{d}.$$
(c) SRT from the sludge-wasting mass balance. With aeration-tank solids mass $VX$ divided by the daily solids mass removed via wasting $Q_wX_w$ (effluent solids neglected — a well-operated clarifier passes negligible TSS): $$SRT=\frac{V\,X}{Q_w\,X_w}=\frac{4{,}000\times3{,}000}{150\times8{,}000}=\frac{12{,}000{,}000}{1{,}200{,}000}=\boxed{10.0\ \text{days}}.$$
Check: part (c)'s SRT balance uses the aeration-tank MLSS of 3,000 mg/L given in part (b) as $X$ (the exam does not repeat it in part (c)'s own data list, but no other mixed-liquor concentration is given anywhere in Question 3, so it is the only value consistent with the stated aeration-tank volume). Effluent TSS is assumed negligible in the SRT balance, a standard simplification when no clarifier effluent solids concentration is supplied.
