--- title: "Single chamber settling basin design (15 m³/s)" subject: "Hydropower Engineering" chapter: 5 key: "semester_7/hydropower_engineering:settling-basin-single-chamber-15cumec" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-single-chamber-15cumec site: BCT Engineering --- # Single chamber settling basin design (15 m³/s) Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants - Asked in: 2082 Bhadra, 8 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-single-chamber-15cumec - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5 ## Question Design a single chamber settling basin using the following data: design discharge = 15 m³/sec; particle size to settle ≥ 0.2 mm; particle fall velocity = 0.02 m/sec; factor of safety for basin area = 1.5; length to breadth ratio = 8; sediment concentration = 1 kg/m³; density of sediment particles = 2,500 kg/m³; settling time of sediment particles = 8 hours; sediment packing factor = 0.5; additional discharge for flushing = 10%. ## Short answer Use design flow 16.5 m3/s. Basin area = 1.5 x 16.5 / 0.02 = 1237.5 m2; with L/B=8 adopt 13 m x 104 m. Depth 4.5 m gives velocity 0.282 m/s. Sediment per 8-h cycle is 380.2 m3, so provide 0.5 m storage. ## Answer **Given**: $Q=15$ m³/s, particles $\ge0.2$ mm, $w=0.02$ m/s, factor of safety for area $=1.5$, $L/B=8$, sediment concentration $C=1$ kg/m³, particle density 2500 kg/m³, flushing interval 8 h, packing factor 0.5, additional flushing discharge 10%. #### Step 1: Design flow $$ Q'=1.10\times15=16.5\ \text{m}^3/\text{s} $$ #### Step 2: Surface area of the basin For a particle to settle, the surface area must satisfy $A\ge Q/w$. With the factor of safety: $$ A=\frac{FS\times Q'}{w}=\frac{1.5\times16.5}{0.02}=1237.5\ \text{m}^2 $$ #### Step 3: Width and length ($L/B=8$) $$ B=\sqrt{\frac{A}{8}}=\sqrt{\frac{1237.5}{8}}=12.44\ \text{m},\qquad L=8B=99.5\ \text{m} $$ Adopt $B=13$ m and $L=104$ m ($A=1352$ m²). #### Step 4: Depth of flow Limit the horizontal velocity to about 0.3 m/s. Take the flow depth $H=4.5$ m: $$ v=\frac{Q'}{BH}=\frac{16.5}{13\times4.5}=0.282\ \text{m/s}<0.3\ \text{m/s}\quad\text{(OK)} $$ #### Step 5: Sediment storage (flushing every 8 h) $$ \text{Mass of sediment}=C\,Q'\,t=1\times16.5\times(8\times3600)=475200\ \text{kg} $$ $$ \text{Volume of deposit}=\frac{\text{mass}}{\rho_s\times p}=\frac{475200}{2500\times0.5}=380.2\ \text{m}^3 $$ $$ \text{Depth of deposit}=\frac{380.2}{1352}=0.28\ \text{m}\ \Rightarrow\ \text{provide }0.5\ \text{m storage} $$ (All the sediment entering is assumed to be trapped, which is conservative.) #### Step 6: Overall dimensions | Item | Value | |---|---| | Length of settling zone | 104 m | | Width | 13 m | | Flow depth | 4.5 m | | Sediment storage | 0.5 m (hopper with 1:30 slope to the flushing channel) | | Free board | 0.5 m | | Total depth at the upstream end | about 5.5 m | ```text PLAN (single chamber, flushing gate at the downstream end) inlet settling chamber outlet canal transition +========================+ transition =====> \ guide / | | \ weir /===> to / vanes \ | 13 m wide x 104 m long| / \ headrace +========================+ trash- | flush gate rack v --> flushing channel --> river ``` **Answer**: a single chamber of about 13 m x 104 m, with a flow depth of 4.5 m ($v=0.282$ m/s) and 0.5 m of sediment storage, flushed every 8 hours (deposit of 380.2 m³ per cycle). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What design flow is used for the single-chamber basin? A. Q' = 15 m^3/s only B. Q' = 1.10 x 15 = 16.5 m^3/s (10% extra for flushing) C. Q' = 15/1.10 = 13.6 m^3/s D. Q' = 15 + 1.5 x 15 = 37.5 m^3/s Answer: B. The additional flushing discharge of 10% is added to the plant discharge. (Key point: Design flow with flushing) 2. How is the surface area of the basin found? A. A = Q'/w = 825 m^2, ignoring the safety factor B. A = FS x Q'/v = 1.5 x 16.5/0.3 = 82.5 m^2 C. A = FS x Q'/w = 1.5 x 16.5/0.02 = 1237.5 m^2 D. A = FS x w/Q' = 0.0018 m^2 Answer: C. A particle settles if surface area is at least Q/w; the factor of safety increases it. (Key point: Surface area) 3. With L/B = 8 and A = 1237.5 m^2, what are B and L? A. B = A/8 = 154.7 m and L = 8B B. B = sqrt(A) x 8 = 281 m and L = B/8 C. B = 8L = 99.5 m and L = 12.44 m D. B = sqrt(A/8) = 12.44 m and L = 8B = 99.5 m Answer: D. Since A = B x L = 8B^2, B = sqrt(A/8) and L = 8B. (Key point: Width and length) 4. How much sediment deposits in the 8 h flushing interval (C = 1 kg/m^3, packing factor 0.5, density 2500 kg/m^3)? A. Mass = C Q' t = 475200 kg; volume = mass/(2500 x 0.5) = 380.2 m^3 B. Volume = 475200/2500 = 190.1 m^3, ignoring the packing factor C. Volume = 475200 x 0.5/2500 = 95.0 m^3 D. Volume = 475200 x 2500 x 0.5 m^3 Answer: A. Deposited sediment is less dense than solid grains, so the solid volume is divided by the packing factor to get the deposit volume. 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