--- title: "Settling basin 60 m³/s, 0.15 mm, head 100 m" subject: "Hydropower Engineering" chapter: 5 key: "semester_7/hydropower_engineering:settling-basin-60cumec-head-100m-015mm" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-60cumec-head-100m-015mm site: BCT Engineering --- # Settling basin 60 m³/s, 0.15 mm, head 100 m Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants - Asked in: 2072 Kartik, 5+2 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-60cumec-head-100m-015mm - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5 ## Question Find the dimensions of a settling basin for a high head project of a Himalayan river which utilizes a discharge of 60 m³/s and a gross head of 100 m. The sediment size to be removed is up to 0.15 mm. Consider the turbulence effect also. Draw the plan and section. ## Short answer Use 3 parallel chambers, each 20.0 m wide, 180 m long, and 6.0 m deep settling zone. The design horizontal velocity is 0.170 m/s, determined by Camp’s formula and reduced effective fall velocity due to turbulence for trapping 0.15 mm particles. ## Answer **Given**: design discharge $Q=60.0$ m³/s and gross head 100 m; particles of size $d\ge 0.15$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (taken for 0.15 mm quartz sand). Turbulence is considered. #### 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 20.0 m wide x 180 m long x 6.0 m deep (settling zone), with $v=0.170$ 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 = 0.15 mm (Camp: v = a sqrt(d), a = 44, v in cm/s, d in mm)? A. v = 44 sqrt(0.15) = 17.04 cm/s = 0.1704 m/s B. v = 44 x 0.15 = 6.6 cm/s = 0.066 m/s C. v = 44 sqrt(0.15) = 17.04 m/s D. v = 44/sqrt(0.15) = 114 cm/s = 1.14 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 = 352 m; chambers are provided to reduce the length B. B = Q/(vH) = 58.68 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) = 58.68 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 68.2 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 68.2 m to about 176 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 = 0.0092 m/s, w' = 0.0242 m/s, L = vH/w' = 42.3 m B. alpha = 0.0092 m/s, L = vH/alpha = 111 m C. No correction needed, L = vH/w = 68.2 m D. alpha = 0.0092 m/s, w' = 0.0058 m/s, L = vH/w' = 176.3 m (adopt about 180 m) Answer: D. alpha = 0.132 v/sqrt(H) is subtracted from w, giving w' and the longer length L = vH/w'. 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