--- title: "Circular sedimentation tank, population 35675" subject: "Sanitary Engineering" chapter: 7 key: "semester_6/sanitary_engineering:sed-tank-35675" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sed-tank-35675 site: BCT Engineering --- # Circular sedimentation tank, population 35675 Sanitary Engineering · Chapter 7: Wastewater Treatment - Asked in: 2068 Magh (old course), 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sed-tank-35675 - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7 ## Question Design a circular sedimentation tank for a locality having population of 35675. The average water demand is 135 lpcd and 80% of the water consumed converts to sewage. Assume that maximum demand is 2.7 times the average demand. Check surface loading also. ## Short answer One circular tank of 19.5 m diameter, 3.0 m side water depth (total 4.0 m), detention 2.0 h, surface loading 34.8 m³/m²/day. ## Answer **Assumptions:** sewage $=0.8\times 135 = 108$ lpcd, so the average flow is $35675\times 108/1000 = 3852.9$ m³/day; the design flow is the maximum, $2.7\times$ average. Detention 2 h at maximum flow, depth 3 m. #### Step 1: Design flow $$ Q = 35675\times 135\times 0.8\times 2.7/1000 = 10402.8\ \text{m}^3/\text{day} $$ #### Step 2: Capacity Adopt detention period $t = 2.0$ h and side water depth $H = 3.0$ m (typical for plain sedimentation: 2-3 h, 2.5-4 m). $$ V = Q\,t = \frac{10402.8}{24}\times 2.0 = 866.9\ \text{m}^3,\qquad A = \frac{V}{H} = \frac{866.9}{3.0} = 289.0\ \text{m}^2 $$ #### Step 3: Diameter Provide one circular tank, so $A_1=A$. $$ D = \sqrt{\frac{4A_1}{\pi}} = \sqrt{\frac{4\times 289.0}{\pi}} = 19.18\ \text{m}\ \Rightarrow\ \text{adopt } D = 19.5\ \text{m} $$ Adopt a total depth of $H + 0.5\ (\text{free board}) + 0.5\ (\text{sludge zone}) = 4.0$ m, with a central inlet well and a peripheral weir. #### Step 4: Checks $$ \begin{aligned} \text{Surface loading} &= \frac{Q/N}{\pi D^2/4} = \frac{10402.8}{298.6} = 34.8\ \text{m}^3/\text{m}^2\text{/day}\ (\text{usual 25-40 for primary tanks})\\ \text{Weir loading} &= \frac{Q/N}{\pi D} = \frac{10402.8}{\pi\times 19.5} = 169.8\ \text{m}^3/\text{m/day}\ (\text{should be below about 300})\\ \text{Actual detention} &= \frac{(\pi D^2/4)H}{Q/N}\times 24 = 2.07\ \text{h} \end{aligned} $$ The surface loading lies within the usual range for primary settling (25-40 m³/m²/day), so the tank is satisfactory. The inlet well, scum board and sludge scraper are provided as shown. ```text Section (circular, radial flow) inlet well weir/launder -> |__|~~~~~~~~~~~~~~~~~~~~~|_ | \ / | H = 3.0 m | \______ ______/ | + sludge zone +----------\__/----------+ sludge hopper -> sludge pipe <---------- D = 19.5 m ----------> ``` **Answer: 1 circular sedimentation tank of diameter 19.5 m, side water depth 3.0 m (total depth about 4.0 m), detention 2.0 h, surface loading 34.8 m³/m²/day.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the design flow for 35675 people? A. 35675 x 135 x 0.8 / 1000 = 3853 m^3/day B. 35675 x 135 x 2.7 / 1000 C. 35675 x 135 / 0.8 x 2.7 D. 35675 x 135 x 0.8 x 2.7 / 1000 = 10402.8 m^3/day Answer: D. The tank is designed for the maximum flow, 2.7 times the average sewage flow. (Key point: Design flow) 2. How is the tank area found? A. A = V / H with V = Q t = 866.9 m^3, so A = 289 m^2 B. A = Q / H C. A = V x H D. A = Q x t / 24 x 3 Answer: A. Volume divided by side depth. (Key point: Area) 3. What is the surface loading and is it acceptable? A. 10402.8 m^3/m^2/day, too high B. 34.8 m^3/m^2/day, within 25-40 for primary tanks C. 2.07 m^3/m^2/day, too low D. 169.8 m^3/m^2/day, acceptable Answer: B. 10402.8 / 298.6 = 34.8, within the usual range. (Key point: Surface loading check) 4. Why is actual detention (2.07 h) slightly different from 2 h? A. The depth was changed B. The flow was reduced C. The diameter was rounded up to 19.5 m, giving a larger volume D. Free board was added Answer: C. A larger adopted D gives a larger actual volume for the same flow. 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