--- title: "Conventional ASP design, 0.104 m³/s" subject: "Sanitary Engineering" chapter: 7 key: "semester_6/sanitary_engineering:asp-0p104-conventional" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/asp-0p104-conventional site: BCT Engineering --- # Conventional ASP design, 0.104 m³/s Sanitary Engineering · Chapter 7: Wastewater Treatment - Asked in: 2081 Chaitra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/asp-0p104-conventional - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7 ## Question Design a conventional activated sludge plant to treat settled domestic sewage with diffused air aeration system for the given data: Sewage discharge = 0.104 m³/s; BOD5 of settled sewage = 220 mg/l; Effluent BOD allowed = 30 mg/l; F/M ratio = 0.2; MLSS = 3000 mg/l. ## Short answer Flow 8985.6 m³/d, aeration tank volume V=Q⋅S₀/(F/M⋅X)=3295 m³. Use liquid depth 4.0 m, provide two tanks each 40 m × 10.5 m × 4.5 m (depth includes 0.5 m freeboard). ## Answer #### Data and design basis Conventional ASP (plug flow, diffused air). The aeration tank volume follows from the food-to-micro-organism ratio: $$ \frac{F}{M}=\frac{Q\,S_0}{V\,X}\ \Rightarrow\ V=\frac{Q\,S_0}{(F/M)\,X} $$ Data: settled-sewage $BOD_5=S_0=220$ mg/l, effluent BOD $S_e=30$ mg/l, $F/M=0.2$ per day, MLSS $X=3000$ mg/l. The return sludge ratio is also checked (SVI assumed). #### Step 1: Design flow $$ Q = 0.104\times86400 = 8985.6\ \text{m}^3/\text{day} $$ #### Step 2: Volume of aeration tank $$ V = \frac{8985.6\times 220}{0.2\times 3000} = 3294.7\ \text{m}^3 $$ (units: $Q$ in m³/day and $S_0$, $X$ in mg/l = g/m³.) #### Step 3: Dimensions $$ \begin{aligned} \text{Adopt liquid depth } &= 4.0\ \text{m (usual range 3-4.5 m)}\\ \text{Plan area } A &= \frac{V}{4.0} = 823.7\ \text{m}^2 \end{aligned} $$ Provide 2 tanks (so one can be taken out of service), each with area $411.8$ m² and L : B about 4 : 1: $$ B=\sqrt{\frac{411.8}{4}} = 10.15\ \text{m (adopt 10.5 m)},\qquad L=\frac{411.8}{10.5} = 39.2\ \text{m (adopt 40 m)} $$ Adding 0.5 m free board, the tank depth is 4.5 m. The size of each tank is $40\ \text{m}\times 10.5\ \text{m}\times 4.5\ \text{m}$. #### Step 4: Checks $$ \begin{aligned} \text{Hydraulic retention time} &= \frac{V}{Q} = \frac{3294.7}{8985.6}\times 24 = 8.8\ \text{h}\ (\text{slightly above the usual 4-8 h; acceptable})\\ \text{Volumetric loading} &= \frac{Q\,S_0}{V\times 1000} = 0.60\ \text{kg BOD/m}^3\text{/day}\ (\text{usual 0.3-0.6}) \end{aligned} $$ BOD removal efficiency $=\dfrac{220-30}{220}\times100=86.4\%$, within the usual 85-95% of a conventional ASP. $$ \begin{aligned} \text{Return sludge ratio: } X_R &= \frac{10^6}{SVI} = \frac{10^6}{100} = 10000\ \text{mg/l}\ (SVI\ \text{assumed } 100\ \text{ml/g})\\ \frac{Q_R}{Q} &= \frac{X}{X_R-X} = \frac{3000}{10000-3000} = 0.429\ \ (\approx 43\%)\\ \text{Horizontal velocity} &= \frac{Q/N}{B\times H} = \frac{8986/(86400\times 2)}{10.5\times 4.0} = 0.0012\ \text{m/s}\ (= 0.07\ \text{m/min}) \end{aligned} $$ **Answer: aeration tank volume $\approx 3295$ m³ (detention 8.8 h, depth 4.0 m); provide 2 tanks, each $40\ \text{m}\times 10.5\ \text{m}$ in plan with liquid depth 4.0 m (tank depth 4.5 m).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is Q found from 0.104 m3/s? A. Q = 0.104 x 86400 = 8985.6 m3/day B. Q = 0.104 x 3600 = 374 m3/day C. Q = 0.104 x 1000 = 104 m3/day D. Q = 0.104 / 86400 Answer: A. Multiply the m3/s rate by 86400 s/day. (Key point: Design flow) 2. Which relation gives the aeration tank volume? A. V = Q (S0 - Se)/X B. V = Q S0 / ((F/M) X), from F/M = Q S0/(V X) C. V = Q x F/M x X D. V = Q / (F/M) Answer: B. F/M is BOD load per day over biomass in the tank, rearranged for V = 8985.6 x 220/(0.2 x 3000) = 3295 m3. (Key point: Volume formula) 3. How are the plan dimensions found? A. Adopt depth 10 m to reduce area, with a single tank B. Take L:B = 1:1 and no free board C. Adopt depth 4 m, area = V/4, split into 2 tanks of L:B about 4:1, adding 0.5 m free board D. Take depth from the HRT only Answer: C. Area = 823.7 m2, two tanks of 411.8 m2: 40 m x 10.5 m x 4.5 m including free board. (Key point: Tank dimensions) 4. HRT is 8.8 h versus the usual 4-8 h. How does the answer treat it? A. Rejects the design and halves the volume B. Accepts it, and also checks volumetric loading (0.60 kg/m3/day) and return ratio (about 0.43 for SVI 100) C. Ignores HRT as unimportant D. Raises MLSS to 8800 mg/l to compensate Answer: B. HRT = V/Q x 24 = 8.8 h; slightly above the range but accepted, with loading and return ratio within limits. 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