--- title: "Frequency of flushing of settling basin" subject: "Hydropower Engineering" chapter: 5 key: "semester_7/hydropower_engineering:sediment-flushing-frequency-10-percent-storage" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/sediment-flushing-frequency-10-percent-storage site: BCT Engineering --- # Frequency of flushing of settling basin Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants - Asked in: 2070 Ashad, 3 marks - Video: [Sediment in Settling Basins: Estimation of Flushing Planning](https://www.youtube.com/watch?v=5oByylOiALs) — Hydropower - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/sediment-flushing-frequency-10-percent-storage - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5 ## Question If you have allocated about 10% volume for sediment storage and the overall trapping efficiency of the settling basin is 40%, find out the frequency of flushing of the settling basin when the sediment concentration is 2000 ppm. ## Short answer Deposit volume per unit water: vs = (0.40×2.0)/1325 ≈ 6.04×10⁻⁴ m³ deposit/m³ water. Storage 0.1 Vb, water to fill = 165.6 Vb, so flush every 165.6 detention times (e.g., ~27.6 h for 10‑min detention). ## Answer **Given**: sediment storage $=10\%$ of the basin volume $V_b$; overall trap efficiency $\eta=40\%$; sediment concentration $C=2000$ ppm $=2.0$ kg/m³. Assumption: the settled sediment has a bulk density of $\rho_b=2650\times0.5=1325$ kg/m³ (particle density 2650 kg/m³, packing factor 0.5). #### Step 1: Volume of sediment trapped per m³ of water $$ v_s=\frac{\eta\,C}{\rho_b}=\frac{0.40\times2.0}{1325}=6.038 \times 10^{-4}\ \text{m}^3\ \text{of deposit per m}^3\ \text{of water} $$ #### Step 2: Storage volume $$ V_{st}=0.10\,V_b $$ #### Step 3: Volume of water that can pass before the storage is full $$ V_w=\frac{V_{st}}{v_s}=\frac{0.10\,V_b}{6.038 \times 10^{-4}}=165.6\,V_b $$ #### Step 4: Frequency of flushing The water passes through the basin at the rate $Q$, so the time between flushes is $$ T=\frac{V_w}{Q}=165.6\,\frac{V_b}{Q}=165.6\,t_d $$ where $t_d=V_b/Q$ is the detention time of the basin. So the basin should be flushed once after about **165.6 detention periods** (about every 166 times the time that water takes to pass through it). For example, if the detention time is 10 minutes, then $T=1656$ minutes $\approx27.6$ hours (about 0.9 flushes per day); if $t_d=5$ minutes, $T\approx13.8$ hours. **Answer**: flush every $\approx165.6\,V_b/Q$ (about 166 detention times; e.g. about 27.6 h for a 10-minute detention time). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the volume of deposit per m^3 of water found? A. eta C / rho_b = 0.4 x 2 / 1325 = 6.04e-4 m^3 B. C / eta = 5 m^3 C. eta rho_b / C = 265 m^3 D. C x rho_b = 2650 m^3 Answer: A. Trapped mass per m^3 of water is eta C = 0.8 kg; dividing by the bulk density 1325 kg/m^3 gives 6.04e-4 m^3. (Key point: Deposit volume per m^3 of water) 2. What bulk density of settled sediment is assumed and why? A. 2650 kg/m^3, the particle density B. 1325 kg/m^3, particle density times a packing factor of 0.5 C. 1000 kg/m^3, same as water D. 500 kg/m^3 Answer: B. Deposited sediment contains voids, so 2650 x 0.5 = 1325 kg/m^3 is used. (Key point: Bulk density assumption) 3. How much water can pass before the 10% storage is full? A. V_w = 0.10 V_b / 6.04e-4 = 165.6 V_b B. V_w = 0.10 V_b x 6.04e-4 C. V_w = 6.04e-4 V_b D. V_w = 0.4 V_b Answer: A. Storage volume divided by deposit per unit water volume gives 165.6 V_b. (Key point: Volume of water before storage fills) 4. What is the flushing interval? A. T = 165.6 V_b/Q, about 166 detention times B. T = Q/(165.6 V_b) C. T = 0.1 V_b/Q D. T = 1/(0.4 x 2000) Answer: A. Dividing the water volume by flow Q gives T = 165.6 t_d; for t_d = 10 min this is about 27.6 h. 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