--- title: "Continuous settling basin (10 m³/s)" subject: "Hydropower Engineering" chapter: 5 key: "semester_7/hydropower_engineering:settling-basin-continuous-10cumec-015mm" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-continuous-10cumec-015mm site: BCT Engineering --- # Continuous settling basin (10 m³/s) Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants - Asked in: 2079 Bhadra, 8 marks - Video: [Design Of Settling Basin | Hazen's and Vetter's Equation |Hydropower Engineering numerical](https://www.youtube.com/watch?v=BKq6dZonTQU) — OPengineer - 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-continuous-10cumec-015mm - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5 ## Question Design a continuous type settling basin with neat sketches for a hydropower plant using the following data: settling velocity = 5 cm/sec; turbine discharge = 10 m³/sec; particle size to be removed = 0.15 mm. Assume other necessary data if necessary. ## Short answer Add 10% flushing, so design flow is 11 m3/s. Camp velocity is 0.1704 m/s; with H=3 m, provide two chambers 11 m wide. Turbulence gives effective fall velocity 0.037 m/s and length 13.8 m, adopt 15 m. Each chamber is 11 m x 15 m with hoppers. ## Answer **Given**: design discharge $Q=10.0$ m³/s; particles of size $d\ge 0.15$ mm to be trapped, fall velocity $w=5.00$ cm/s $=0.0500$ m/s (given). The basin is of the continuous flushing type, so a flushing discharge of 10% (1.0 m³/s) is added. Turbulence is included, and other data are assumed. (A fall velocity of 5 cm/s is high for 0.15 mm sand, so the basin is short; the given value is used.) Flushing water of 10% is added for continuous flushing, so the design flow of the basin is $Q'=10.0+1.0=11.0$ m³/s. #### Step 1: Horizontal velocity (Camp) For $0.1 \ / | chamber 1 | \ /===> to / guide \ +-----------------+ / weir \ headrace =====> \ vanes / | chamber 2 | \ / / \ +=================+ / \ trash- | | flushing conduits (0.5 m3/s each) rack v v --> flushing channel --> river ``` ```text LONGITUDINAL SECTION inlet settling zone (depth H) outlet ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ WL \ ---> ---> ---> ---> ---> ---> | \ ______________________________________ | weir \/ \/ \/ \/ \/ \/ hoppers with bottom outlets \__|__ to headrace ==== flushing conduit (continuous flow) ====> river ``` **Flushing arrangement**: the floor of each chamber is formed into hoppers with bottom openings leading to a flushing conduit. About 0.5 m³/s per chamber (1.0 m³/s in total) is drawn off continuously through these openings and carries the settled sediment back to the river, so the chambers do not have to be taken out of service for flushing. A gate on each conduit controls the flushing flow, and inlet gates allow one chamber to be isolated for inspection while the other keeps supplying the plant. **Answer**: 2 chambers, each about 11.0 m wide x 15 m long x 3.0 m deep (settling zone), with $v=0.170$ m/s and 1.0 m³/s of continuous flushing water. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the basin designed for 11.0 m^3/s instead of 10.0 m^3/s? A. A 10% flushing discharge is added because continuous flushing draws extra water B. Turbulence increases the discharge by 10% C. The turbine discharge is multiplied by a safety factor of 1.1 for the tailrace D. Evaporation losses of 10% are added Answer: A. In a continuous-flushing basin about 10% of the flow leaves with the sediment, so Q' = 10 + 1 = 11 m^3/s is used for sizing. (Key point: Design flow for continuous flushing) 2. What horizontal velocity is used and why is it acceptable? A. 0.0500 m/s, equal to the fall velocity B. 0.1704 m/s, within 0.1-0.4 m/s so deposits are not re-suspended C. 0.50 m/s, which sweeps sediment to the flush channel D. 1.70 m/s, to keep the flow turbulent Answer: B. v = 44 sqrt(0.15) = 17.04 cm/s = 0.1704 m/s, within the usual range that avoids scouring of settled sediment. (Key point: Camp velocity and its limit) 3. With H = 3.0 m, what total width results and how is it divided? A. B = Q'/(vH) = 21.52 m, divided in 2 chambers of about 11 m B. B = Q/(wH) = 66.7 m in one chamber C. B = Q'/(vH) = 21.52 m, kept as one chamber so flow is smooth D. B = vH/w = 10.2 m in two chambers of 5.1 m Answer: A. B = 11.0/(0.1704 x 3.0) = 21.52 m; two chambers give about 10.76 m each, adopted as 11.0 m. (Key point: Width of the basin) 4. How does turbulence change the length of the settling zone? A. It shortens the length, since turbulence helps particles reach the bed B. It has no effect on length, only on width C. It reduces the effective fall velocity (w' = w - alpha), so L rises from 10.2 m to 13.8 m D. It increases the fall velocity, so L drops to 6 m Answer: C. Turbulence lifts particles: w' = 0.050 - 0.013 = 0.037 m/s, so L = vH/w' = 13.8 m (adopted 15 m) instead of 10.2 m. 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