--- title: "Conventional ASP design, population 126000" subject: "Sanitary Engineering" chapter: 7 key: "semester_6/sanitary_engineering:asp-126000-checks" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/asp-126000-checks site: BCT Engineering --- # Conventional ASP design, population 126000 Sanitary Engineering · Chapter 7: Wastewater Treatment - Asked in: 2079 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-126000-checks - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7 ## Question Design a conventional activated sludge treatment plant in your Municipality to treat the domestic sewage with diffused air aeration with the following data. Population = 1,26,000; Per capita sewage flow = 160 lpcd; Settled sewage BOD5 = 200 mg/L; Effluent BOD5 required = 15 mg/L; Take F/M: 0.2 and MLSS = 3000 mg/L. Also check HRT, Volumetric Loading, Return Sludge ratio, Horizontal Velocity. ## Short answer Design flow 20160 m³/d, aeration tank volume 6720 m³ (detention 8.0 h). Provide three tanks each 47 m × 12.0 m × 4.5 m. Checks: HRT 8.0 h, volumetric loading 0.60 kg BOD/m³/d, return sludge ratio 43%, horizontal velocity 0.10 m/min. ## 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 $BOD_5=S_0=200$ mg/l; effluent $BOD_5=15$ mg/l (efficiency $\dfrac{200-15}{200}=92.5\%$); $F/M=0.2$/day; MLSS $=3000$ mg/l. Assumed: SVI $=100$ ml/g for the return sludge ratio. #### Step 1: Design flow $$ Q = 126000\times160/1000 = 20160.0\ \text{m}^3/\text{day} $$ #### Step 2: Volume of aeration tank $$ V = \frac{20160.0\times 200}{0.2\times 3000} = 6720.0\ \text{m}^3 $$ (units: $Q$ in m³/day and $S_0$, $X$ in mg/l = g/m³.) #### Step 3: Dimensions $$ \begin{aligned} \text{Adopt liquid depth } &= 4.0\ \text{m (usual range 3-4.5 m)}\\ \text{Plan area } A &= \frac{V}{4.0} = 1680.0\ \text{m}^2 \end{aligned} $$ Provide 3 tanks (so one can be taken out of service), each with area $560.0$ m² and L : B about 4 : 1: $$ B=\sqrt{\frac{560.0}{4}} = 11.83\ \text{m (adopt 12.0 m)},\qquad L=\frac{560.0}{12.0} = 46.7\ \text{m (adopt 47 m)} $$ Adding 0.5 m free board, the tank depth is 4.5 m. The size of each tank is $47\ \text{m}\times 12.0\ \text{m}\times 4.5\ \text{m}$. #### Step 4: Checks $$ \begin{aligned} \text{Hydraulic retention time} &= \frac{V}{Q} = \frac{6720.0}{20160.0}\times 24 = 8.0\ \text{h}\ (\text{usual 4-8 h; accepted})\\ \text{Volumetric loading} &= \frac{Q\,S_0}{V\times 1000} = 0.60\ \text{kg BOD/m}^3\text{/day}\ (\text{usual 0.3-0.6}) \end{aligned} $$ $$ \begin{aligned} \text{Return sludge ratio: } X_R &= \frac{10^6}{SVI} = \frac{10^6}{100} = 10000\ \text{mg/l}\ (SVI\ \text{assumed } 100\ \text{ml/g})\\ \frac{Q_R}{Q} &= \frac{X}{X_R-X} = \frac{3000}{10000-3000} = 0.429\ \ (\approx 43\%)\\ \text{Horizontal velocity} &= \frac{Q/N}{B\times H} = \frac{20160/(86400\times 3)}{12.0\times 4.0} = 0.0016\ \text{m/s}\ (= 0.10\ \text{m/min}) \end{aligned} $$ **Answer: aeration tank volume $\approx 6720$ m³ (detention 8.0 h, depth 4.0 m); provide 3 tanks, each $47\ \text{m}\times 12.0\ \text{m}$ 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. For 126 000 people at 160 lpcd, what is Q? A. 126 000 m3/day B. 20 160 m3/day C. 2016 m3/day D. 201 600 m3/day Answer: B. 126000 x 160 / 1000 = 20 160 m3/day. (Key point: Design flow) 2. What volume results with F/M = 0.2, MLSS = 3000 mg/l and S0 = 200 mg/l? A. 13 440 m3 B. 3360 m3 C. 6720 m3 D. 1008 m3 Answer: C. V = 20160 x 200 / (0.2 x 3000) = 6720 m3. (Key point: Aeration volume) 3. What are the HRT and volumetric loading? A. HRT = V/Q x 24 = 8.0 h; loading = Q S0/(V x 1000) = 0.60 kg BOD/m3/day B. HRT = 4.0 h; loading = 1.2 kg/m3/day C. HRT = 8.0 h; loading = 0.20 kg/m3/day D. HRT = 24 h; loading = 0.60 kg/m3/day Answer: A. 6720/20160 = 0.333 day = 8 h; 20160 x 200 / (6720 x 1000) = 0.60. (Key point: HRT and volumetric loading) 4. How is the return sludge ratio and horizontal velocity found? A. Qr/Q = Xr/X = 3.3; velocity = Q/(B x H) for the whole flow B. Qr/Q = X/Xr = 0.3; velocity = V/Q C. Qr/Q = SVI/100 = 1; velocity = Q/V D. Qr/Q = X/(Xr - X) with Xr = 10^6/SVI = 10 000 mg/l gives 0.43; velocity = (Q/N)/(B x H) for one tank Answer: D. Sludge balance gives 3000/(10000-3000) = 0.429; with 3 tanks, v = (20160/(86400 x 3))/(12 x 4) = 0.0016 m/s. 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