Continuous flushing settling basin, 95% efficiency

Given: design discharge $Q=60.0$ m³/s and gross head 300 m; particles of size $d\ge 0.15$ mm to be trapped, fall velocity $w=1.50$ cm/s $=0.0150$ m/s (assumed for 0.15 mm sand, as in similar designs). The basin has continuous flushing, so 10% extra flow (6 m³/s) is added. The required trap efficiency is 95% for particles larger than 0.15 mm, and the effect of turbulence is included. The length is checked by Velikanov's relation with $\lambda=1.5$ (assumed) and silty water of density 1.105 t/m³.

Flushing water of 10% is added for continuous flushing, so the design flow of the basin is $Q'=60.0+6.0=66.0$ m³/s.

Step 1: Horizontal velocity (Camp)

For $0.1<d<1$ mm, $v=a\sqrt d$ with $a=44$ (v in cm/s, d in mm):

$$v=44\sqrt{0.15}=17.04\ \text{cm/s}=0.1704\ \text{m/s}$$

This is within the usual 0.1-0.4 m/s range, so deposited sediment is not re-suspended.

Step 2: Cross-section and width

Assume a flow depth $H=6.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q'}{vH}=\frac{66.0}{0.1704\times6.0}=64.55\ \text{m}$$

Provide $n=3$ chambers, so that one can be taken out for inspection while the others work: $b=B/n=21.52$ m each, adopt 22.0 m.

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.1704\times6.0}{0.0150}=68.2\ \text{m}$$

Step 4: Effect of turbulence

Turbulence reduces the effective fall velocity: $w'=w-\alpha$, with $\alpha=0.132\,v/\sqrt{H}$.

$$\begin{aligned} \alpha&=\frac{0.132\times0.1704}{\sqrt{6.0}}=0.0092\ \text{m/s}\\ w'&=0.0150-0.0092=0.0058\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1704\times6.0}{0.0058}=175.8\ \text{m} \end{aligned}$$

Step 5: Check by Velikanov's method

The silty water has a density of 1.105 t/m³ (sediment 2.65 t/m³). This reduces the effective fall velocity by the buoyancy ratio $\frac{2.65-1.105}{2.65-1}=0.936$, so $w_s=1.405$ cm/s.

$$L_V=\frac{\lambda^2v^2(\sqrt H-0.2)^2}{7.51\,w_s^2}=\frac{1.5^2\times0.1704^2\times(\sqrt{6.0}-0.2)^2}{7.51\times0.01405^2}=223.2\ \text{m}$$

The Velikanov length (223.2 m) is 27% greater than the turbulence length (175.8 m). The larger of the two is adopted and rounded up to 225 m.

Adopted dimensions

Total volume of water in the basin $\approx 89100$ m³.

PLAN (continuous flushing outlets along each chamber)
          inlet        settling chambers      outlet
 canal   transition  +=================+   transition
 =====>  \         / |  chamber 1      | \         /===> to
         /  guide  \ +-----------------+ /  weir   \    headrace
 =====>  \  vanes  / |  chamber 2      | \         /
         /         \ +-----------------+ /         \
                     |  chamber 3      |
                     +=================+
  trash-                  |  |  |  flushing conduits (2 m3/s each)
  rack                    v  v  v  --> flushing channel --> river

LONGITUDINAL SECTION
 inlet          settling zone (depth H)      outlet
 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  WL
 \    --->   --->   --->   --->   --->   --->  |
  \  ______________________________________   | weir
   \/ \/ \/ \/ \/ \/ hoppers with bottom outlets \__|__ to headrace
   ==== flushing conduit (continuous flow) ====> river

Flushing arrangement: the floor of each chamber is formed into longitudinal hoppers with a slot or a row of openings at the bottom leading to a flushing conduit below. About 2 m³/s per chamber (6 m³/s in total) is drawn off continuously through these openings and carries the settled sediment back to the river, so the chambers do not have to be taken out of service for flushing. A gate on each flushing conduit controls the flushing flow, and inlet gates allow one chamber to be isolated and emptied for inspection while the other two keep supplying the plant.

The adopted length is greater than the Velikanov length, so more than 95% of the particles coarser than 0.15 mm are expected to be trapped.

Answer: 3 chambers, each about 22.0 m wide x 225 m long x 6.0 m deep (settling zone), with $v=0.170$ m/s and 6 m³/s of continuous flushing water.