Primary Clarifier Design: 18-Env-A4 Water and Wastewater Engineering, December 2015, Q5

Question 5: Primary Clarifier Design (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.

Primary clarifier mass balance: influent solids/BOD₅ split between the clarified effluent and the settled sludge withdrawn from the hopper.

Given.

Given data
Quantity	Symbol	Value
Average day flow	$Q$	20,000 m³/d
Influent TSS	$C_{TSS}$	200 mg/L
Influent VSS	$C_{VSS}$	170 mg/L
Influent BOD₅	$C_{BOD}$	160 mg/L
TSS removal efficiency	$R_{TSS}$	60%
Primary sludge solids concentration	$P_s$	4%
Primary sludge specific gravity	$SG$	1.03
Surface overflow rate (design, at peak flow)	$SOR$	80 m³/m²·d
Peaking factor	$PF$	2.25
Side water depth	$d$	3.0 m

Check: the source states a TSS removal efficiency but not a separate VSS or BOD₅ removal efficiency. VSS is treated as removed at the same 60% efficiency as TSS, since VSS is itself the volatile fraction of TSS and the two settle together as one particulate population. BOD₅ removal is read from the standard primary-clarifier TSS-vs-BOD₅ removal correlation (Metcalf & Eddy typical design ranges of 50–70% TSS removal against 25–40% BOD₅ removal, roughly $\%\text{BOD}\approx\%\text{TSS}-25$) — at 60% TSS removal this gives 35% BOD₅ removal, the "appropriate" figure the question asks the candidate to select.

I. TSS, VSS and BOD₅ Loads in Primary Effluent

Find. The mass loading (kg/d) of TSS, VSS and BOD₅ leaving the clarifier in the primary effluent.

Approach. Apply each substance's removal efficiency to its influent concentration to get the effluent concentration, then convert every concentration to a mass load using $\text{Load (kg/d)}=C(\text{mg/L})\times Q(\text{m}^3/\text{d})/1000$.

Effluent concentrations. $$C_{TSS,eff}=200(1-0.60)=80\ \text{mg/L},\quad C_{VSS,eff}=170(1-0.60)=68\ \text{mg/L},\quad C_{BOD,eff}=160(1-0.35)=104\ \text{mg/L}.$$
Convert to mass loads. $$\text{TSS load}=\frac{80\times20{,}000}{1000}=\boxed{1600\ \text{kg/d}},\qquad \text{VSS load}=\frac{68\times20{,}000}{1000}=\boxed{1360\ \text{kg/d}},\qquad \text{BOD}_5\ \text{load}=\frac{104\times20{,}000}{1000}=\boxed{2080\ \text{kg/d}}.$$
Cross-check via the influent side. Influent loads are 4000, 3400 and 3200 kg/d for TSS, VSS and BOD₅; removing 60%, 60% and 35% respectively leaves $4000-2400=1600$, $3400-2040=1360$ and $3200-1120=2080$ kg/d — matching step 2 exactly.

II. Volume of Primary Sludge Produced per Day

Find. The daily volume of primary sludge withdrawn from the clarifier.

Approach. The dry mass of solids captured equals the TSS removed (influent load minus effluent load); dividing that mass by the sludge's wet density and solids fraction gives the wet sludge volume.

Mass of TSS captured (removed) per day. $$m_{TSS,removed}=\frac{200\times20{,}000}{1000}\times0.60=4000\times0.60=2400\ \text{kg/d}.$$
Sludge (wet) density from its specific gravity. $$\rho_{sludge}=SG\times1000\ \text{kg/m}^3=1.03\times1000=1030\ \text{kg/m}^3.$$
Sludge volume from mass, density and solids fraction. $$V_{sludge}=\frac{m_{TSS,removed}}{\rho_{sludge}\times P_s}=\frac{2400}{1030\times0.04}=\frac{2400}{41.2}=\boxed{58.3\ \text{m}^3/\text{d}}.$$

III. Surface Area of the Primary Clarifier

Find. The clarifier surface area sized for the design SOR at peak day flow.

Approach. Scale the average day flow up to the peak day flow using the peaking factor, then size the area directly from $A=Q_{peak}/SOR$.

Peak day flow. $$Q_{peak}=Q\times PF=20{,}000\times2.25=\boxed{45{,}000\ \text{m}^3/\text{d}}.$$
Required surface area. $$A=\frac{Q_{peak}}{SOR}=\frac{45{,}000}{80}=\boxed{562.5\ \text{m}^2}.$$

IV. Hydraulic Retention Time at Average Day Flow

Find. The HRT of the clarifier basin sized in Part III, evaluated at the average day flow.

Approach. Multiply the Part III surface area by the given side water depth to get the basin volume, then divide by the average day flow.

Basin volume. $$V=A\times d=562.5\times3.0=1687.5\ \text{m}^3.$$
HRT at average day flow. $$\text{HRT}=\frac{V}{Q}=\frac{1687.5}{20{,}000}=0.0844\ \text{d}=0.0844\times24=\boxed{2.03\ \text{h}}.$$

Check: sizing the clarifier area on the PEAK flow (Part III) but then evaluating HRT at the AVERAGE flow (Part IV) is the standard design convention — the SOR criterion must hold even during the peak, while HRT is normally reported at average conditions since that is what the tank experiences most of the time; the two calculations therefore deliberately use different flows, not an inconsistency. A 2.03 h average-flow HRT sits at the low end of the typical 1.5–3 h range for primary clarifiers (Metcalf & Eddy) — a direct consequence of sizing area on this paper's relatively high SOR (80 m³/m²·d).

Quantity	Value
TSS load, primary effluent	1600 kg/d
VSS load, primary effluent	1360 kg/d
BOD₅ load, primary effluent	2080 kg/d
Primary sludge volume	58.3 m³/d
Peak day flow	45,000 m³/d
Clarifier surface area	562.5 m²
Basin volume	1687.5 m³
HRT at average day flow	2.03 h (0.0844 d)

Concept & Theory

Topic. Primary clarifier mass-balance sizing: solids/BOD₅ removal, sludge volume, surface-overflow-rate area, and hydraulic retention time.

Key relations. $\text{Load (kg/d)}=C(\text{mg/L})\,Q(\text{m}^3/\text{d})/1000$; $V_{sludge}=m_{solids}/(\rho_{sludge}P_s)$; $A=Q_{peak}/SOR$; $\text{HRT}=V/Q=Ad/Q$.

Why this works. A primary clarifier is sized by two independent physical criteria that must both be satisfied — particles must have time to settle out of the flow (governed by SOR $=Q/A$, a surface-loading criterion independent of depth) and the basin must hold the liquid long enough for that settling to actually occur (HRT $=V/Q$, which does depend on depth). Sizing area on SOR at the peak flow protects performance during the worst-case hydraulic loading, while reporting HRT at the average flow describes typical day-to-day operation; both numbers come from the same basin geometry, just evaluated at different flow conditions.

Common pitfall. Applying the TSS removal efficiency to BOD₅ (or vice versa) without adjustment, when the two substances are only correlated, not equal, in a primary clarifier; sizing area on the average flow instead of the peak flow (undersizes the clarifier for its worst-case condition); forgetting to convert the sludge's specific gravity into an actual density before dividing by the solids fraction.

Source. Metcalf & Eddy, Wastewater Engineering, ch. 5 (primary treatment, SOR/HRT design) & ch. 14 (sludge processing); Davis & Cornwell, Introduction to Environmental Engineering, primary sedimentation chapter.

Question 5 of 5: Primary Clarifier Design

Question 5: Primary Clarifier Design (25 marks)

I. TSS, VSS and BOD₅ Loads in Primary Effluent

II. Volume of Primary Sludge Produced per Day

III. Surface Area of the Primary Clarifier

IV. Hydraulic Retention Time at Average Day Flow