--- title: "Desilting basin 32 m³/s, 0.15 mm" subject: "Hydropower Engineering" chapter: 5 key: "semester_7/hydropower_engineering:desilting-basin-32cumec-lambda" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/desilting-basin-32cumec-lambda site: BCT Engineering --- # Desilting basin 32 m³/s, 0.15 mm Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants - Asked in: 2082 Baishakh, 5+2 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/desilting-basin-32cumec-lambda - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5 ## Question Determine the dimensions of a desilting basin for a high head hydropower plant with a discharge of 32 m³/s. The basin must trap sediment particles coarser than 0.15 mm. Take w = 0.025 m/s and λ = 1.5. Additionally, draw a plan and longitudinal section of the desilting basin, illustrating the major components and the flushing arrangement. ## Short answer Use Camp velocity v=44 sqrt(0.15)=0.1704 m/s. With two chambers and H=5 m, adopt 19 m width per chamber. Required length is 57.7 m by Velikanov, so adopt 60 m. Final: two chambers 19 m x 60 m, depth 5 m plus sediment storage. ## Answer **Given**: $Q=32$ m³/s, particles coarser than 0.15 mm, $w=0.025$ m/s, $\lambda=1.5$ (Velikanov's coefficient for turbulence and the required efficiency). For a high-head plant, two chambers are provided so that one can be flushed while the other works. #### Step 1: Horizontal velocity (Camp) $$ v=44\sqrt{0.15}=17.04\ \text{cm/s}=0.1704\ \text{m/s} $$ #### Step 2: Depth and width Assume a flow depth $H=5$ m: $$ B=\frac{Q}{vH}=\frac{32}{0.1704\times5}=37.56\ \text{m}\ (\text{two chambers of }18.78\ \text{m}) $$ Adopt a chamber width of 19.0 m. #### Step 3: Length by simple settling theory Without turbulence: $L_0=\dfrac{vH}{w}=\dfrac{0.1704\times5}{0.025}=34.1$ m. Using $\lambda$ as a factor on the surface area: $$ A=\frac{\lambda Q}{w}=\frac{1.5\times32}{0.025}=1920\ \text{m}^2,\qquad L=\frac{A}{B}=\frac{1920}{37.56}=51.1\ \text{m} $$ #### Step 4: Check by Velikanov's method $$ L_V=\frac{\lambda^2v^2(\sqrt H-0.2)^2}{7.51\,w^2}=\frac{1.5^2\times0.1704^2\times(\sqrt{5}-0.2)^2}{7.51\times0.025^2}=57.7\ \text{m} $$ The larger length governs. Adopt $L=60$ m, giving $L/b=60/19=3.2$. | Item | Value | |---|---| | Number of chambers | 2 | | Width of each chamber | 19.0 m | | Length | 60 m | | Flow depth | 5 m (+ 1.0 m sediment storage in the hopper) | | Velocity | 0.1704 m/s | ```text PLAN (flushing gates at the downstream end) inlet settling chambers outlet canal transition +=================+ transition =====> \ / | 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**: each chamber has a hopper floor sloping (about 1:30) to a central flushing channel, with a flushing gate at the downstream end. To flush a chamber, its inlet gate is closed and the deposit is flushed out while the other chamber carries the flow. The flushing water runs through the flushing tunnel to the river. **Answer**: 2 chambers, each 19.0 m wide x 60 m long x 5 m deep (plus storage). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the horizontal flow velocity in the basin found for the 0.15 mm particle? A. v = 44 sqrt(d) with d in mm, giving 17.04 cm/s B. v = Q/A using the surface area 1920 m^2 C. v = w x lambda = 0.0375 m/s D. v = 44 d with d in mm, giving 6.6 cm/s Answer: A. Camp's relation v = a sqrt(d) with a = 44 (v in cm/s, d in mm) gives 44 x sqrt(0.15) = 17.04 cm/s = 0.1704 m/s. (Key point: Camp horizontal velocity) 2. What is the required plan (surface) area of the basin and how is it obtained? A. A = Q/w = 1280 m^2, ignoring lambda B. A = lambda Q / w = 1.5 x 32 / 0.025 = 1920 m^2 C. A = Q/v = 32/0.1704 = 188 m^2 D. A = lambda w / Q = 0.00117 m^2 Answer: B. Settling theory gives A = Q/w; the turbulence/safety coefficient lambda multiplies it, so A = 1.5 x 32 / 0.025 = 1920 m^2. (Key point: Surface area with turbulence coefficient) 3. With assumed depth H = 5 m and v = 0.1704 m/s, how are the total width and length obtained? A. B = Q/(vH) = 37.56 m and L = A/B = 51.1 m B. B = A/H = 384 m and L = Q/B C. B = Q/(wH) = 256 m and L = vH/Q D. B = A/v and L = B/H Answer: A. Width follows from continuity B = Q/(vH) = 37.56 m, then length L = A/B = 51.1 m, which equals lambda v H / w. (Key point: Width and length from depth) 4. Why are two chambers provided and how is flushing done? A. Two chambers halve the velocity so the 0.15 mm particles settle faster B. Two chambers are needed to double the trap efficiency of each C. One chamber is flushed with its inlet closed through the flushing gate while the other carries the flow D. Both chambers are flushed together by opening the gates at the inlet Answer: C. The inlet of the chamber to be flushed is closed and its downstream flushing gate opened, while the other chamber keeps supplying water to the plant. 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