--- title: "Settling basin trap efficiency (Camp's diagram)" subject: "Hydropower Engineering" chapter: 5 key: "semester_7/hydropower_engineering:settling-basin-camp-diagram-80cumec" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-camp-diagram-80cumec site: BCT Engineering --- # Settling basin trap efficiency (Camp's diagram) Hydropower Engineering · Chapter 5: Headworks of Run-of- River (RoR) Plants - Asked in: 2071 Chaitra, 8 marks - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/settling-basin-camp-diagram-80cumec - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-5 ## Question Design the settling basin from the particle size and concentration approach and calculate the trap efficiency from the following data (refer to Figure 3 and 4: Camp's diagram; fall velocity of quartz spheres in water after Rouse): design discharge = 80 m³/s; number of basins = 2; installed capacity of the plant = 110 MW; water temperature = 12°C; particle size to be removed = 0.2 mm; Manning's constant (n) = 0.01; flushing discharge = 1 m³/s (if flushing system is continuous). Assume other necessary data if needed. If the flushing system is changed to intermittent with a single basin, what are the changes? Describe with suitable reason. ## Short answer Each of two basins is 42 m wide, 220 m long, and 5 m deep plus 1 m hopper, with v=0.1968 m/s and trap efficiency 90.2% for 0.2 mm particles. If changed to a single intermittent basin, width becomes about 82 m, length remains 220 m, sediment storage must be added, and generation stops during flushing unless a bypass is provided. ## Answer **Given**: $Q=80$ m³/s; 2 basins; installed capacity 110 MW (so a high-head Pelton plant); $T=12^\circ$C; particle size $d=0.2$ mm; continuous flushing of 1 m³/s. The head follows from $P=\rho g Q H\eta$ with $\eta=0.88$: $H=\dfrac{110\times10^6}{9810\times80\times0.88}=159$ m. At this head the particles above 0.2 mm must be removed. Assumptions: fall velocity of a 0.2 mm quartz sphere at 12°C (Rouse's curve) $w\approx2.2$ cm/s; flow depth $H_s=5$ m; design trap efficiency 90%; Manning's $n=0.01$ is only needed for the flushing channel. #### Step 1: Particle-size approach, design flow $$ Q'=80+1=81\ \text{m}^3/\text{s}\quad(40.5\ \text{m}^3/\text{s per basin}) $$ #### Step 2: Velocity and size Camp's velocity: $v=44\sqrt{0.2}=19.68$ cm/s $=0.1968$ m/s. $$ B_{total}=\frac{Q'}{vH_s}=\frac{81}{0.1968\times5}=82.3\ \text{m}\ \Rightarrow\ b=\frac{82.3}{2}=41.2\ \text{m per basin} $$ #### Step 3: Length (Camp-Vetter turbulence approach) Turbulence reduces the fall velocity: $\alpha=0.132v/\sqrt{H_s}=0.0116$ m/s, $w'=0.022-0.0116=0.0104$ m/s. $$ L_{ideal}=\frac{vH_s}{w'}=94.7\ \text{m} $$ For a trap efficiency $\eta$, Vetter's relation is $\eta=1-\exp\left(-\dfrac{w'L}{vH_s}\right)$. For $\eta=90\%$: $$ L=\ln(10)\times94.7=218.2\ \text{m}\ \Rightarrow\ \text{adopt }L=220\ \text{m} $$ #### Step 4: Trap efficiency of the adopted basin $$ \eta=1-\exp\left(-\frac{0.0104\times220}{0.1968\times5}\right)=90.2\% $$ Particles smaller than 0.2 mm have a lower fall velocity and are trapped with a lower efficiency, found in the same way with their own $w$. #### Step 5: Sediment and flushing The sediment trapped drops into the hoppers and leaves continuously with the 1 m³/s flushing water (0.5 m³/s per basin) through the flushing channel (slope about 1:50, $n=0.01$). Provide an extra depth of about 1.0 m for the hoppers. | Item | Value (each basin) | |---|---| | Width | 42 m | | Length | 220 m | | Flow depth | 5 m + 1 m hopper | | Velocity | 0.1968 m/s | | Trap efficiency (0.2 mm) | 90.2% | #### If the flushing is intermittent with a single basin - **Discharge**: the flushing discharge (1 m³/s) is no longer needed, so $Q'=80$ m³/s, but the whole flow passes through one basin: width $B=\dfrac{80}{0.1968\times5}=81.3$ m (about 82 m); the length is the same, so the structure is about twice as wide as one of the two basins. - **Storage**: the sediment is no longer removed continuously, so a sediment storage depth must be added (volume = sediment load x flushing interval / bulk density). - **Plant stoppage**: with a single basin, the plant (or the incoming water) must be stopped to flush it, or a bypass is needed; with two basins one can keep working. A single intermittent basin therefore interrupts generation during flushing unless a bypass or a storage pond is provided, and each flush needs a large discharge for a short time. - **Efficiency**: the trap efficiency during the flushing and just after is lower because of the disturbed flow and bed deposits. ```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 ``` **Answer**: 2 basins of about 42 m x 220 m x 5 m (+1 m hopper); $v=0.1968$ m/s; trap efficiency of the 0.2 mm particles about 90.2%. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What discharge is used for sizing each basin with continuous flushing? A. 80 m^3/s per basin B. 81 m^3/s total, 40.5 m^3/s per basin C. 79 m^3/s total, 39.5 per basin D. 40 m^3/s per basin, flushing ignored Answer: B. The 1 m^3/s flushing discharge is added to 80 giving 81 m^3/s, shared by the 2 basins. (Key point: Design flow per basin) 2. How is the length for 90% trap efficiency obtained? A. L = vH/w = 44 m with no correction B. Using w' = w - 0.132 v/sqrt(H), L_ideal = vH/w' = 94.7 m and L = ln(10) x 94.7 = 218 m C. L = 0.9 x 94.7 = 85 m D. L = B x 2 = 165 m Answer: B. Vetter's relation eta = 1 - exp(-w' L/(vH)) gives L = vH/w' x ln(1/(1 - eta)), i.e. 2.303 x 94.7 = 218 m, adopted 220 m. (Key point: Turbulence-corrected length) 3. What is the trap efficiency of 0.2 mm particles in the adopted 220 m basin, and how do smaller particles behave? A. About 90.2%; smaller particles have lower w and are trapped with lower efficiency B. 100%; smaller particles are trapped equally C. About 50%; smaller particles are trapped more D. About 90.2%; smaller particles are trapped better because they follow the flow Answer: A. eta = 1 - exp(-0.0104 x 220/(0.1968 x 5)) = 90.2%. Finer particles have smaller fall velocity so lower efficiency. (Key point: Trap efficiency of the adopted basin) 4. What changes if flushing is made intermittent with a single basin? A. Q' becomes 80 m^3/s but the width doubles to about 82 m, storage depth is added, and generation may stop during flushing unless bypassed B. Basin width halves and no storage is needed C. Length doubles and width is unchanged D. Nothing changes because flushing does not affect design Answer: A. No flushing discharge is added, but all flow passes one basin (B = 80/(0.1968 x 5) = 81 m); sediment storage is needed and flushing interrupts the plant without a bypass. 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