Settling basin trap efficiency (Camp's diagram)

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.

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.

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%.