18-Env-A4 Water and Wastewater Engineering · May 2017
Question 5 of 5: Primary Clarifier Design
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2017 — 04-Env-A4 / Water and Wastewater Engineering. 3 hours duration; closed book with one double-sided aid sheet; approved calculator permitted. Question 1 is compulsory; the paper instructs candidates to attempt any three of the remaining four questions — all five are solved below for completeness.
Reference texts. Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.) — oxygen sag/Streeter-Phelps, MLSS/MLVSS, population equivalent, primary clarifier design; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.) — turbidity, alkalinity chemistry; MWH’s Water Treatment: Principles and Design (3rd ed.) — coagulation-flocculation, softening, disinfection by-products, pH; Guidelines for Canadian Drinking Water Quality (Health Canada).
Find. The effluent TSS/VSS/BOD5 loads, the daily primary-sludge volume, the required clarifier surface area, and the HRT at average flow.
Approach. Apply the given 60% TSS removal to TSS directly; assume VSS is removed at the same fractional efficiency as TSS (settling does not discriminate strongly by volatility) and estimate BOD5 removal from the standard primary-clarifier TSS-vs-BOD5 removal correlation, then work the mass-balance, sludge-volume, overflow-rate and HRT calculations in sequence.
Check: BOD5 removal is not directly given. Metcalf & Eddy’s typical primary-clarifier performance curves put BOD5 removal roughly 25 percentage points below TSS removal over the normal design range (50–70% TSS / 25–40% BOD5); at 60% TSS removal this gives an assumed 35% BOD5 removal, used below. VSS removal is assumed equal to the stated TSS removal (60%), since a primary clarifier removes solids by settling velocity, not by volatile/fixed composition.
(II) Primary sludge volume. Mass of TSS removed, $M_{removed}=2{,}500-1{,}000=1{,}500$ kg/d (checks against $2{,}500\times0.60=1{,}500$). Sludge density $=SG\times1000=1{,}030$ kg/m³. $V_{sludge}=\dfrac{M_{removed}}{\rho_{sludge}\times P_s}=\dfrac{1{,}500}{1{,}030\times0.04}=\boxed{36.4\text{ m}^3\text{/d}}$.