--- title: "Settling basin 60 m³/s, 0.2 mm, w = 1.5 cm/s" subject: "Hydropower Engineering" chapter: 5 key: "semester_7/hydropower_engineering:settling-basin-60cumec-02mm-w15" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-60cumec-02mm-w15 site: BCT Engineering --- # Settling basin 60 m³/s, 0.2 mm, w = 1.5 cm/s Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants - Asked in: 2076 Chaitra, 4+4 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-60cumec-02mm-w15 - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5 ## Question Find out the dimension of a settling basin with turbulence flow for a high head hydropower plant, which utilizes a discharge of 60 m³/sec. The sediment particles coarser than 0.2 mm (fall velocity w = 1.5 cm/sec) have to be trapped in the basin. Draw plan and section showing major components and flushing arrangement, neat and proportionately. ## Short answer Three chambers, each 17.0 m wide, 270 m long, and 6.0 m deep. Design velocity is 0.197 m/s; length accounts for turbulence using effective fall velocity w’ = 0.0044 m/s to trap particles coarser than 0.2 mm with given fall velocity 1.5 cm/s. ## Answer **Given**: design discharge $Q=60.0$ m³/s; particles of size $d\ge 0.20$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (given). Turbulence is considered. (A fall velocity of 1.5 cm/s is lower than the usual value for 0.2 mm sand, so the basin comes out long; the given value is used.) #### Step 1: Horizontal velocity (Camp) For $0.1 \ / | chamber 1 | \ /===> to / guide \ +-----------------+ / weir \ headrace =====> \ vanes / | chamber 2 | \ / / \ +-----------------+ / \ | chamber 3 | +=================+ trash- | | flush gates rack v v --> flushing channel --> river ``` ```text LONGITUDINAL SECTION inlet settling zone (depth H) outlet ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ WL \ ---> ---> ---> ---> ---> ---> | \ ______________________________________ | weir \/ sediment storage / hopper slope 1:30 \__|__ to headrace flush gate -> flushing channel -> river ``` **Flushing arrangement**: the floor of each chamber slopes at about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. A flushing velocity of at least 3 m/s is kept in the channel. The chamber to be flushed is closed at the inlet, drained and flushed to the river, while the other chambers continue to supply the plant. **Answer**: 3 chambers, each about 17.0 m wide x 270 m long x 6.0 m deep (settling zone), with $v=0.197$ m/s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What horizontal velocity is used for d = 0.20 mm (Camp: v = a sqrt(d), a = 44, v in cm/s, d in mm)? A. v = 44 sqrt(0.20) = 19.68 cm/s = 0.1968 m/s B. v = 44 x 0.20 = 8.8 cm/s = 0.088 m/s C. v = 44 sqrt(0.20) = 19.68 m/s D. v = 44/sqrt(0.20) = 98 cm/s = 0.98 m/s Answer: A. For 0.1 < d < 1 mm Camp gives v = a sqrt(d); the result must be converted from cm/s to m/s and lies in the 0.1-0.4 m/s range so deposits are not resuspended. (Key point: Camp horizontal velocity) 2. With H = 6.0 m and flow 60.0 m^3/s, what is the total width B and why are chambers provided? A. B = Q/v = 305 m; chambers are provided to reduce the length B. B = Q/(vH) = 50.82 m; 3 chambers so that one can be flushed while the others work C. B = Q/(wH) = 667 m; chambers are provided to increase velocity D. B = Q/(vH) = 50.82 m; chambers are provided only to reduce construction cost Answer: B. Continuity Q = vBH gives the width, and splitting the basin into chambers allows flushing one chamber without stopping the plant. (Key point: Width of basin) 3. How does turbulence change the settling length? A. It increases the effective fall velocity, so the length decreases below 78.7 m B. It does not change the length; only the width is increased C. It reduces the effective fall velocity (w' = w - 0.132 v/sqrt(H)), so the length increases from 78.7 m to about 268 m D. It adds to the horizontal velocity, so the width is reduced to B/2 Answer: C. Turbulent eddies keep particles up, so effective fall velocity is w - alpha; since L = vH/w', a smaller w' gives a longer basin. (Key point: Effect of turbulence) 4. What are alpha, w' and the length L when turbulence is considered? A. alpha = 0.0106 m/s, w' = 0.0256 m/s, L = vH/w' = 46.1 m B. alpha = 0.0106 m/s, L = vH/alpha = 111 m C. No correction needed, L = vH/w = 78.7 m D. alpha = 0.0106 m/s, w' = 0.0044 m/s, L = vH/w' = 268.3 m (adopt about 270 m) Answer: D. alpha = 0.132 v/sqrt(H) is subtracted from w, giving w' and the longer length L = vH/w'. (Key point: Length with turbulence) ## Close variants set in other exams - [Settling basin 50 m³/s, 0.2 mm](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-50cumec-02mm-w15) — Hydropower Engineering, chapter 5, asked 4× - [Settling basin 25 m³/s, 0.2 mm](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-25cumec-02mm-w15) — Hydropower Engineering, chapter 5, asked 4× - [Settling basin 40 m³/s, 0.15 mm](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-40cumec-015mm-w15) — Hydropower Engineering, chapter 5, asked 4× ## Related in Hydropower Engineering - [Settling basin 60 m³/s, 0.15 mm, head 300 m](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-60cumec-300m-015mm-turbulence) — Hydropower Engineering, chapter 5, asked 2× - [Settling basin 60 m³/s, 0.15 mm, head 100 m](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-60cumec-head-100m-015mm) — Hydropower Engineering, chapter 5 - [Continuous flushing settling basin, 95% efficiency](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-continuous-60cumec-velikanov) — Hydropower Engineering, chapter 5 - [Periodic settling basin with and without turbulence](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-periodic-5cumec-with-without-turbulence) — Hydropower Engineering, chapter 5 - [Settling basin 30 m³/s with Velikanov check](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-30cumec-02mm-velikanov) — Hydropower Engineering, chapter 5 - [Settling basin 6 m³/s, 0.25 mm, depth 3 m](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-6cumec-025mm-depth3m) — Hydropower Engineering, chapter 5 ## Related in other subjects - [Settling basin design for 0.2 mm particles](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-design-0-2mm) — Hydro Power, chapter 5 - [Settling basin design for 0.15 mm particles](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-design-0-15mm) — Hydro Power, chapter 5 - [Settling basin design for 0.25 mm sediment](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asettling-basin-design-0-25mm) — Hydro Power, chapter 5 - [Components of a medium head high discharge plant](https://www.bctengineer.com/question.md?key=semester_8%2Fpower_plant_design%3Amedium-head-high-discharge-components) — Power Plant Design, chapter 3, asked 2×