--- title: "Settling basin with Hazen coefficient and sediment depth" subject: "Hydropower Engineering" chapter: 5 key: "semester_7/hydropower_engineering:settling-basin-hazen-sediment-depth-5000ppm" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-hazen-sediment-depth-5000ppm site: BCT Engineering --- # Settling basin with Hazen coefficient and sediment depth Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants - Asked in: 2081 Baishakh, 8+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/settling-basin-hazen-sediment-depth-5000ppm - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5 ## Question Design a settling basin for particle size = 6 mm, sp. gravity = 2.65, absolute viscosity = 1.34 gm/cm-s, temperature of water = 20°C, discharge = 12 m³/s. Calculate the depth of sediment assuming concentration is 5000 ppm. Assume 15% flushing discharge and performance coefficient of Hazen = 0.16. Draw a neat sketch of plan and section. ## Short answer Stokes fall velocity is 0.2416 m/s; with 15% flushing flow Q'=13.8 m³/s, Hazen minimum area is 159 m². Adopt 8 m wide by 32 m long by 6 m deep plus a 1.0 m hopper, giving v=0.29 m/s and trap efficiency 96.6%; sediment depth is about 0.71 m per hour. ## Answer **Given**: particle diameter $d=6$ mm $=0.6$ cm, $s=2.65$, $\mu=1.34$ g/cm·s, $T=20^\circ$C, $Q=12$ m³/s, concentration 5000 ppm (5 kg/m³), flushing discharge 15%, Hazen's performance coefficient $n=0.16$. Assumptions: required trap efficiency $\eta=90\%$; sediment bulk density 1325 kg/m³ (packing factor 0.5 of 2650 kg/m³); the basin is flushed every 1 hour; horizontal velocity not more than 0.3 m/s; $L/B=4$. #### Step 1: Fall velocity (Stokes, with the data given) $$ w=\frac{g\,d^2(\rho_s-\rho)}{18\mu}=\frac{981\times0.6^2\times(2.65-1)}{18\times1.34}=24.16\ \text{cm/s}=0.2416\ \text{m/s} $$ #### Step 2: Design flow with flushing $$ Q'=1.15\times12=13.8\ \text{m}^3/\text{s} $$ #### Step 3: Surface area (Hazen) Hazen's relation for the fraction removed is $\;1-\eta=\left[1+n\dfrac{w\,A}{Q'}\right]^{-1/n}$, so $$ A=\frac{Q'}{n\,w}\left[(1-\eta)^{-n}-1\right]=\frac{13.8}{0.16\times0.2416}\left[(0.1)^{-0.16}-1\right]=159.0\ \text{m}^2 $$ This is the minimum surface area. #### Step 4: Cross-section and plan size To stop the settled sediment from being scoured, the horizontal velocity is limited to 0.3 m/s: $$ A_c=\frac{Q'}{v}=\frac{13.8}{0.3}=46\ \text{m}^2 $$ With $L/B=4$, adopt $B=8$ m, flow depth $H=6$ m and $L=32$ m: $$ v=\frac{13.8}{8\times6}=0.29\ \text{m/s}\ (\text{OK}),\qquad A=8\times32=256\ \text{m}^2>159\ \text{m}^2\ (\text{OK}) $$ (A shallow 2 m deep basin of the Hazen area alone would give $v\approx1$ m/s, which is far too fast.) Trap efficiency of the adopted basin: $$ \eta=1-\left[1+0.16\times\frac{0.2416\times256}{13.8}\right]^{-1/0.16}=1-(1.717)^{-6.25}=96.6\% $$ #### Step 5: Depth of sediment Concentration $5000$ ppm $=5$ kg/m³. Sediment trapped per hour: $$ M=\eta\,C\,Q'\,t=0.966\times5\times13.8\times3600=239930\ \text{kg} $$ $$ V_s=\frac{M}{1325}=181.1\ \text{m}^3,\qquad h_s=\frac{V_s}{A}=\frac{181.1}{256}=0.71\ \text{m} $$ So the sediment builds up to about 0.71 m depth in one hour. A hopper depth of about 1.0 m is therefore provided, and the basin must be flushed at least once every hour (or continuously, using the 15% flushing discharge). ```text PLAN (flushing gate at the downstream end) inlet settling basin outlet canal transition +========================+ transition =====> \ guide / | | \ weir /===> to / vanes \ | 8 m wide x 32 m long | / \ headrace +========================+ trash- | flush gate rack v --> flushing channel --> river ``` ```text LONGITUDINAL SECTION inlet settling zone (H = 6 m) outlet ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ WL \ ---> ---> ---> ---> ---> ---> | \ ______________________________________ | weir \/ sediment hopper (about 1.0 m) \__|__ to headrace flush gate -> flushing channel -> river ``` **Answer**: basin about 8 m wide x 32 m long x 6 m flow depth (plus a 1.0 m hopper), $v=0.29$ m/s, trap efficiency about 96.6%; the sediment depth is about 0.71 m per hour of operation. Note: the 6 mm size and the viscosity of 1.34 g/cm·s are used exactly as given (water at 20°C actually has $\mu\approx0.01$ g/cm·s). Stokes' law is not strictly valid for such large particles, so the fall velocity is only indicative. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which expression gives the fall velocity of the particle? A. w = g d^2 (rho_s - rho) / (18 mu), giving about 24.2 cm/s B. w = 18 mu / (g d^2), giving 0.04 cm/s C. w = d sqrt(g), giving 4.3 cm/s D. w = Q/A Answer: A. Stokes' law with the given data gives w = 981 x 0.36 x 1.65 / (18 x 1.34) = 24.16 cm/s. (Key point: Stokes fall velocity) 2. What discharge is used in the Hazen area calculation? A. 12 m^3/s B. 13.8 m^3/s (12 + 15%) C. 10.2 m^3/s (12 - 15%) D. 15 m^3/s Answer: B. 15% flushing discharge is added: Q' = 1.15 x 12 = 13.8 m^3/s. (Key point: Design flow with flushing) 3. What does Hazen's performance coefficient n do in A = Q'/(n w) [(1 - eta)^(-n) - 1]? A. It accounts for non-ideal (turbulent, short-circuiting) settling, giving a larger area than the ideal Q/w B. It is the Manning roughness of the basin C. It is the fraction of flow used for flushing D. It is the sediment porosity Answer: A. n = 0.16 represents basin performance (poor flow conditions); as n goes to 0 the formula tends to the ideal result, and with n = 0.16 and eta = 90% A = 159 m^2. (Key point: Hazen relation) 4. How is the sediment depth in one hour found? A. Mass = eta C Q' t, volume = mass / bulk density, depth = volume / area = 1.06 m B. Depth = C x t / H C. Depth = Q' t / A = 312 m D. Depth = eta C / w = 0.19 m Answer: A. M = 0.9 x 5 x 13.8 x 3600 = 223,560 kg; V = M/1325 = 168.7 m^3; h = 168.7/159 = 1.06 m. 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