--- title: "Settling basin 6 m³/s, 0.25 mm, depth 3 m" subject: "Hydropower Engineering" chapter: 5 key: "semester_7/hydropower_engineering:settling-basin-6cumec-025mm-depth3m" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-6cumec-025mm-depth3m site: BCT Engineering --- # Settling basin 6 m³/s, 0.25 mm, depth 3 m Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants - Asked in: 2080 Bhadra, 7 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-6cumec-025mm-depth3m - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5 ## Question Design a settling basin for a design discharge of 6 cumec. The basin is designed to remove particle size of 0.25 mm. If the depth of the basin is 3 m and settling velocity 2.5 cm/sec, find the dimensions of the basin considering turbulence. ## Short answer Camp velocity is 44 sqrt(0.25)=0.220 m/s. For Q=6 m3/s and H=3 m total width is 9.09 m; provide two chambers 5 m wide. Turbulence reduces fall velocity to 0.0082 m/s, requiring length 80.2 m, adopt 85 m. Final: two 5 m x 85 m x 3 m chambers. ## Answer **Given**: design discharge $Q=6.0$ m³/s; particles of size $d\ge 0.25$ mm to be trapped, fall velocity $w=2.50$ cm/s $=0.0250$ m/s (given). The depth of the basin is 3 m, and turbulence is considered. #### Step 1: Horizontal velocity (Camp) For $0.1 \ / | chamber 1 | \ /===> to / guide \ +-----------------+ / weir \ headrace =====> \ vanes / | chamber 2 | \ / / \ +=================+ / \ 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 maintained in the channel. The chamber to be flushed is closed at the inlet, drained and flushed to the river, while the other chamber continues to supply the plant. **Answer**: 2 chambers, each about 5.0 m wide x 85 m long x 3.0 m deep (settling zone), with $v=0.220$ 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 the 0.25 mm particle? A. 44 x 0.25 = 11 cm/s B. 44 sqrt(0.25) = 22 cm/s = 0.22 m/s C. 44 sqrt(0.25) = 22 m/s D. 2.5 cm/s, equal to the fall velocity Answer: B. v = a sqrt(d) = 44 x 0.5 = 22 cm/s = 0.22 m/s. (Key point: Camp velocity for 0.25 mm) 2. What total width is needed for Q = 6 m^3/s at H = 3 m? A. B = Q/(wH) = 80 m B. B = Q/(vH) = 6/(0.22 x 3) = 9.09 m, split into 2 chambers of about 5 m C. B = vH/w = 26.4 m D. B = Q/v = 27.3 m Answer: B. Continuity gives B = Q/(vH) = 9.09 m; with two chambers b = 4.55 m, adopted 5.0 m. (Key point: Basin width) 3. How is the turbulence-reduced fall velocity found? A. w' = w + 0.132 v/sqrt(H) = 0.0418 m/s B. w' = w - 0.132 v/sqrt(H) = 0.025 - 0.0168 = 0.0082 m/s C. w' = w/lambda = 0.0167 m/s D. w' = w - v = negative, so no settling Answer: B. alpha = 0.132 x 0.22 / sqrt(3) = 0.0168 m/s is subtracted from w, leaving w' = 0.0082 m/s. (Key point: Turbulence correction alpha) 4. What are the lengths without and with turbulence? A. 26.4 m without, 80.2 m with B. 80.2 m without, 26.4 m with C. 26.4 m without, 26.4 m with D. 9.1 m without, 26.4 m with Answer: A. L0 = vH/w = 0.22 x 3/0.025 = 26.4 m; with w' = 0.0082, L = 0.66/0.0082 = 80.2 m, so turbulence about triples the length (85 m adopted). (Key point: Length without and with turbulence) ## Related in Hydropower Engineering - [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 60 m³/s, 0.2 mm, w = 1.5 cm/s](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-60cumec-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 with Hazen coefficient and sediment depth](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-hazen-sediment-depth-5000ppm) — Hydropower Engineering, chapter 5 - [Settling basin 8 m³/s, 0.3 mm, 90% trap](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asettling-basin-8cumec-03mm-90-percent) — Hydropower Engineering, chapter 5, asked 2× - [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× ## Related in other subjects - [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 - [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 velocity of discrete particle 0.14 mm at 25°C](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Ased-j66) — Water Supply Engineering, chapter 6, asked 2× - [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