--- title: "Continuous flow sedimentation tank particle size" subject: "Sanitary Engineering" chapter: 7 key: "semester_6/sanitary_engineering:sed-tank-continuous-4m-60m" source: https://www.bctengineer.com/q/semester_6/sanitary_engineering/sed-tank-continuous-4m-60m site: BCT Engineering --- # Continuous flow sedimentation tank particle size Sanitary Engineering · Chapter 7: Wastewater Treatment - Asked in: 2073 Magh, 8 marks - Video: [Settling tank problem, settling & flow velocity of particle, environmental engineering sedimentation](https://www.youtube.com/watch?v=3UHl93fc8wc) — Civil Engineering Exam - 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-continuous-4m-60m - Whole chapter: https://www.bctengineer.com/md/semester_6/sanitary_engineering/chapter-7 ## Question In a continuous flow sedimentation tank, 4 m deep, 60 m long, if flow velocity of sewage is observed as 1.20 cm/sec, what size of particles with specific gravity 2.65 may be effectively removed? Assume 25°C temperature and kinematic viscosity of water is 0.01 cm²/sec. If the particle size is half, determine the change in % of particles removed. ## Short answer The critical settling velocity is 0.080 cm/s, giving a particle size of 0.030 mm (Stokes' law). Halving the size reduces removal from 100% to 25%, a 75% decrease. ## Answer In an ideal continuous-flow (horizontal) sedimentation tank, a particle is removed if it settles the full depth $H$ before the water travels the length $L$. So the particle removed 100% has a settling velocity $v_s = v_0 = H/t_d$ with $t_d = L/v_h$. #### Step 1: Detention time and critical settling velocity $$ \begin{aligned} t_d &= \frac{L}{v_h} = \frac{60}{0.012} = 5000\ \text{s}\\ v_s &= \frac{H}{t_d} = \frac{4}{5000} = 0.00080\ \text{m/s} = 0.080\ \text{cm/s} \end{aligned} $$ #### Step 2: Size of the particle (Stokes' law) $$ v_s = \frac{g(G-1)d^2}{18\nu}\ \Rightarrow\ d=\sqrt{\frac{18\,\nu\,v_s}{g(G-1)}}=\sqrt{\frac{18\times 0.01\times 0.080}{981\times(2.65-1)}} = 0.00298\ \text{cm} = 0.030\ \text{mm} $$ Check: $Re = v_s d/\nu = 0.024 < 1$, so Stokes' law is valid. #### Step 3: Effect of halving the particle size Since $v_s\propto d^2$, a particle of half the size has $v_s' = v_s/4 = 0.0200$ cm/s. In an ideal tank the fraction of particles of a given size that is removed is $v_s'/v_0$: $$ \text{Removal} = \frac{v_s'}{v_0} = \frac{0.000200}{0.00080} = 25\% $$ The particles of the original size are removed 100%, so the removal of the half-size particles falls from 100% to 25%, i.e. a **decrease of 75%**. **Answer: the particle size effectively (100%) removed is about 0.030 mm; if the size is halved, only 25% are removed (a 75% reduction).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What condition must a particle meet to be 100% removed in the ideal tank? A. Its settling velocity is less than H/t_d B. It is carried the length L in less than t_d C. It settles the full depth H before the water travels the length L D. It has specific gravity of 1 Answer: C. So v_s >= v_0 = H/t_d with t_d = L/v_h. (Key point: Condition for removal) 2. What is the critical settling velocity here? A. v_s = L / t = 60 / 5000 B. v_s = H / t_d = 4 / 5000 = 0.0008 m/s C. v_s = v_h x H = 0.012 x 4 D. v_s = H x t_d Answer: B. Detention is 60/0.012 = 5000 s, so v_s = 4/5000 m/s = 0.08 cm/s. (Key point: Critical settling velocity) 3. How is the particle diameter found from v_s? A. d = sqrt(18 nu v_s / (g(G-1))), about 0.03 mm B. d = v_s x 18 C. d = g(G-1) / (18 nu v_s) D. d = v_s / G Answer: A. Rearranging Stokes' v_s = g(G-1)d^2/(18 nu) gives d about 0.030 mm. (Key point: Stokes law use) 4. Why does halving the particle size reduce removal to 25%? A. Because v_s is proportional to d, so it halves to 50% B. Because v_s is proportional to d^2, giving v_s/4, and removal = v_s'/v_0 C. Because removal does not depend on v_s D. Because v_s doubles Answer: B. v_s' = v_s/4, so the fraction removed is 1/4 = 25%, a 75% decrease. 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