Settling basin 30 m³/s with Velikanov check

Given: design discharge $Q=30.0$ m³/s and gross head 100 m; particles of size $d\ge 0.20$ mm to be trapped, fall velocity $w=2.00$ cm/s $=0.0200$ m/s (given). The length is first found without turbulence, then with turbulence, and finally checked by Velikanov's method ($\lambda=1.5$).

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):

$$v=44\sqrt{0.20}=19.68\ \text{cm/s}=0.1968\ \text{m/s}$$

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 $H=5.0$ m in the settling zone (sediment storage is extra).

$$B=\frac{Q}{vH}=\frac{30.0}{0.1968\times5.0}=30.49\ \text{m}$$

Provide $n=2$ chambers so that one can be flushed while the other works: $b=B/n=15.25$ m each, adopt 15.5 m.

Step 3: Length without turbulence

$$L_0=\frac{vH}{w}=\frac{0.1968\times5.0}{0.0200}=49.2\ \text{m}$$

Step 4: Effect of turbulence

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

$$\begin{aligned} \alpha&=\frac{0.132\times0.1968}{\sqrt{5.0}}=0.0116\ \text{m/s}\\ w'&=0.0200-0.0116=0.0084\ \text{m/s}\\ L&=\frac{vH}{w'}=\frac{0.1968\times5.0}{0.0084}=117.4\ \text{m} \end{aligned}$$

Step 5: Check by Velikanov's method

$$L_V=\frac{\lambda^2v^2(\sqrt H-0.2)^2}{7.51\,w^2}=\frac{1.5^2\times0.1968^2\times(\sqrt{5.0}-0.2)^2}{7.51\times0.0200^2}=120.2\ \text{m}$$

The Velikanov length (120.2 m) is 2% greater than the turbulence length (117.4 m). The larger of the two is adopted and rounded up to 125 m.

Adopted dimensions

Total volume of water in the basin $\approx 19375$ m³.

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

LONGITUDINAL SECTION
 inlet          settling zone (depth H)      outlet
 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  WL
 \    --->   --->   --->   --->   --->   --->  |
  \  ______________________________________   | weir
   \/   sediment storage / hopper slope 1:30 \__|__ to headrace
          flush gate -> flushing channel -> river

Flushing arrangement: the floor of each chamber slopes at about 1:30-1:50 to a longitudinal flushing channel with a flushing gate at the downstream end. A flushing velocity of at least 3 m/s is maintained in the channel. The chamber to be flushed is closed at the inlet, drained and flushed to the river, while the other chamber continues to supply the plant.

Answer: 2 chambers, each about 15.5 m wide x 125 m long x 5.0 m deep (settling zone), with $v=0.197$ m/s.