--- title: "Circular sedimentation tank, population 5500" subject: "Sanitary Engineering" chapter: 7 key: "semester_6/sanitary_engineering:sed-tank-industrial-5500" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sed-tank-industrial-5500 site: BCT Engineering --- # Circular sedimentation tank, population 5500 Sanitary Engineering · Chapter 7: Wastewater Treatment - Asked in: 2074 Bhadra, 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/sed-tank-industrial-5500 - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7 ## Question What will be the suitable dimensions of a circular sewage sedimentation tank for an industrial area having population of 5500? The average water demand is 180 lpcd. Assume that 75% water reaches the treatment plant and maximum demand is 2.4 times average demand. Dimension of the suspended silica particles available in influent water are larger than 0.14 mm. ## Short answer One circular tank of 8.0 m diameter, 3.0 m side water depth (total 4.0 m), detention 2.0 h, surface loading 35.5 m³/m²/day. ## Answer **Assumptions:** sewage reaching the plant $=0.75\times 180 = 135$ lpcd, designed for the maximum rate ($2.4\times$ average): $Q_{avg}=5500\times 135/1000 = 742.5$ m³/day. Detention 2 h and depth 3 m are adopted (usual design values). Silica grains of $d=0.14$ mm ($G=2.65$) at 20°C. #### Step 1: Design flow $$ Q = 5500\times 180\times 0.75\times 2.4/1000 = 1782\ \text{m}^3/\text{day} $$ #### 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). $$ V = Q\,t = \frac{1782}{24}\times 2.0 = 148.5\ \text{m}^3,\qquad A = \frac{V}{H} = \frac{148.5}{3.0} = 49.5\ \text{m}^2 $$ #### Step 3: Diameter Provide one circular tank, so $A_1=A$. $$ D = \sqrt{\frac{4A_1}{\pi}} = \sqrt{\frac{4\times 49.5}{\pi}} = 7.94\ \text{m}\ \Rightarrow\ \text{adopt } D = 8.0\ \text{m} $$ 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 $$ \begin{aligned} \text{Surface loading} &= \frac{Q/N}{\pi D^2/4} = \frac{1782}{50.3} = 35.5\ \text{m}^3/\text{m}^2\text{/day}\ (\text{usual 25-40 for primary tanks})\\ \text{Weir loading} &= \frac{Q/N}{\pi D} = \frac{1782}{\pi\times 8.0} = 70.9\ \text{m}^3/\text{m/day}\ (\text{should be below about 300})\\ \text{Actual detention} &= \frac{(\pi D^2/4)H}{Q/N}\times 24 = 2.03\ \text{h} \end{aligned} $$ **Check of removal of the 0.14 mm silica:** for $0.1 |__|~~~~~~~~~~~~~~~~~~~~~|_ | \ / | H = 3.0 m | \______ ______/ | + sludge zone +----------\__/----------+ sludge hopper -> sludge pipe <---------- D = 8.0 m ----------> ``` **Answer: 1 circular sedimentation tank of diameter 8.0 m, side water depth 3.0 m (total depth about 4.0 m), detention 2.0 h, surface loading 35.5 m³/m²/day.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the design flow computed? A. 5500 x 180 / 1000, average without factors B. 5500 x 135 / 1000 C. 5500 x 180 x 0.75 x 2.4 / 1000 = 1782 m^3/day D. 5500 x 180 x 2.4 / 0.75 Answer: C. Water demand x 75% reaching the plant x maximum factor 2.4. (Key point: Design flow) 2. With t = 2 h and H = 3 m, what is the area? A. A = Q x t = 3564 m^2 B. A = (1782/24 x 2)/3 = 49.5 m^2 C. A = 1782 / 3 = 594 m^2 D. A = 148.5 x 3 Answer: B. V = 148.5 m^3, then A = V/H = 49.5 m^2. (Key point: Volume and area) 3. What diameter results from A = 49.5 m^2? A. D = 49.5 / pi = 15.8 m B. D = 49.5 / 4 = 12.4 m C. D = sqrt(49.5) = 7.0 m D. D = sqrt(4 x 49.5 / pi) = 7.94 m, adopt 8.0 m Answer: D. From A = pi D^2/4. (Key point: Diameter) 4. Why does the check show the 0.14 mm silica is removed? A. Its Hazen settling velocity (about 1518 m/day) is far above the overflow rate of 35.5 m^3/m^2/day B. Its settling velocity is below the overflow rate C. Silica floats D. Settling velocity equals the overflow rate Answer: A. A particle whose settling velocity exceeds the overflow rate settles in the tank. (Key point: Silica settling check) ## Close variants set in other exams - [Circular sedimentation tank, population 80000](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-industrial-80000) — Sanitary Engineering, chapter 7, asked 2× ## Related in Sanitary Engineering - [Circular sedimentation tank, population 35675](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-35675) — Sanitary Engineering, chapter 7 - [Circular sedimentation tank for 58 MLD](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ased-tank-58mld-circular) — Sanitary Engineering, chapter 7 - [Degree of treatment, domestic + industrial](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Adegree-treatment-town-40000) — Sanitary Engineering, chapter 6 ## Related in other subjects - [Design pipes AB and BC, 75 lpcd, tank 6 m high](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-b81) — Water Supply Engineering, chapter 7 - [Rectangular sedimentation tank, 0.2 mm/s, 25000 population](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-bk79) — Water Supply Engineering, chapter 6, asked 2× - [Rectangular sedimentation tank, 9 million L peak demand](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-c75) — Water Supply Engineering, chapter 6 - [Rectangular sedimentation tank, 20000 population, 0.4 mm/s](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-s73) — Water Supply Engineering, chapter 6